chapter
stringlengths 1.97k
1.53M
| path
stringlengths 47
241
|
|---|---|
Learning Objectives
• To understand what information is obtained by each type of ionic equation
The chemical equations discussed in Chapter 7 showed the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little about what was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent.
Let’s consider the reaction of silver nitrate with potassium dichromate. As you learned in Example 9, when aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall chemical equationA chemical equation that shows all the reactants and products as undissociated, electrically neutral compounds. for the reaction shows each reactant and product as undissociated, electrically neutral compounds:
$2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq)\label{8.4.1}$
Although Equation $\ref{8.4.1}$ gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as AgNO3 and K2Cr2O7 are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag2Cr2O7 is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation, showing which ions and molecules are hydrated and which are present in other forms and phases:
$2Ag^+(aq) + 2NO_3^-(aq) + 2K^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^+(aq) + 2NO_3^-(aq)\label{8.4.2}$
Note that K+(aq) and NO3(aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectator ions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species that participate in the chemical reaction:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{8.4.3}$
Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation $\ref{8.4.3}$, the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag2Cr2O7 formula unit.
By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows:
$2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{8.4.4}$
The complete ionic equation for this reaction is as follows:
$2Ag^+(aq) + 2F^-(aq) + 2NH_4^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4^+(aq) + 2F^-(aq)\label{8.4.5}$
Because two NH4+(aq) and two F(aq) ions appear on both sides of Equation $\ref{8.4.5}$, they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation $\ref{8.4.6}$), which is identical to Equation $\ref{8.4.3}$:
$2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{8.4.6}$
If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction.
Example $1$
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate.
Given: reactants and products
Asked for: overall, complete ionic, and net ionic equations
Strategy:
Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation.
Solution:
From the information given, we can write the unbalanced chemical equation for the reaction:
$Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)$
Because the product is Ba3(PO4)2, which contains three Ba2+ ions and two PO43 ions per formula unit, we can balance the equation by inspection:
$3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq)$
This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form:
$3Ba^{2+}(aq) + 6NO_3^-(aq) + 6Na^+(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s) + 6Na^+(aq) + 6NO_3^-(aq)$
The six NO3(aq) ions and the six Na+(aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation:
$3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)$
Exercise $1$
Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride.
Answer
overall chemical equation:
$3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq)$
complete ionic equation:
$3Ag^+(aq) + 3F^-(aq) + 3Na^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^+(aq) + 3F^-(aq)$
net ionic equation:
$3Ag^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)$
So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that could occur and then assessing which is the most probable (or least improbable) outcome.
The most important step in analyzing an unknown reaction is to write down all the species—whether molecules or dissociated ions—that are actually present in the solution (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions introduced in Chapter 3 (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions.
Summary
The chemical equation for a reaction in solution can be written in three ways. The overall chemical equation shows all the substances present in their undissociated forms; the complete ionic equation shows all the substances present in the form in which they actually exist in solution; and the net ionic equation is derived from the complete ionic equation by omitting all spectator ions, ions that occur on both sides of the equation with the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction.
Key Takeaway
• A complete ionic equation consists of the net ionic equation and spectator ions.
Conceptual Problem
1. What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation? | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/09%3A_Introduction_to_Solutions_and_Aqueous_Reactions/9.06%3A_Representing_Aqueous_Reactions-_Molecular_Ionic_and_Complete_Ionic_Equations.txt |
Learning Objectives
• To know the characteristic properties of acids and bases.
Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases.
Definitions of Acids and Bases
We can define acids as substances that dissolve in water to produce H+ ions, whereas bases are defined as substances that dissolve in water to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of acid and base have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO3) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive.
The Arrhenius Definition of Acids and Bases
The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the Arrhenius definition, an acid is a substance like hydrochloric acid that dissolves in water to produce H+ ions (protons; Equation $\ref{4.3.1}$), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH) ions (Equation $\ref{4.3.2}$):
$\underset{an\: Arrhenius\: acid}{HCl_{(g)}} \xrightarrow {H_2 O_{(l)}} H^+_{(aq)} + Cl^-_{(aq)} \label{4.3.1}$
$\underset{an\: Arrhenius\: base}{NaOH_{(s)}} \xrightarrow {H_2O_{(l)}} Na^+_{(aq)} + OH^-_{(aq)} \label{4.3.2}$
According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H+ and OH ions, respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations:
1. First, because acids and bases were defined in terms of ions obtained from water, the Arrhenius concept applied only to substances in aqueous solution.
2. Second, and more important, the Arrhenius definition predicted that only substances that dissolve in water to produce $H^+$ and $OH^−$ ions should exhibit the properties of acids and bases, respectively. For example, according to the Arrhenius definition, the reaction of ammonia (a base) with gaseous HCl (an acid) to give ammonium chloride (Equation $\ref{4.3.3}$) is not an acid–base reaction because it does not involve $H^+$ and $OH^−$:
$NH_{3\;(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \label{4.3.3}$
The Brønsted–Lowry Definition of Acids and Bases
Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), who defined acid–base reactions in terms of the transfer of a proton (H+ ion) from one substance to another.
According to Brønsted and Lowry, an acid (A substance with at least one hydrogen atom that can dissociate to form an anion and an $H^+$ ion (a proton) in aqueous solution, thereby forming an acidic solution) is any substance that can donate a proton, and a base (a substance that produces one or more hydroxide ions ($OH^-$ and a cation when dissolved in aqueous solution, thereby forming a basic solution) is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form $NH_4^+$, so in Equation $\ref{4.3.3}$, $NH_3$ is a Brønsted–Lowry base and $HCl$ is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified.
Polyprotic Acids
Acids differ in the number of protons they can donate. For example, monoprotic acids (a compound that is capable of donating one proton per molecule) are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO3, and HNO2. All carboxylic acids that contain a single −CO2H group, such as acetic acid (CH3CO2H), are monoprotic acids, dissociating to form RCO2 and H+. A compound that can donate more than one proton per molecule is known as a polyprotic acid. For example, H2SO4 can donate two H+ ions in separate steps, so it is a diprotic acid (a compound that can donate two protons per molecule in separate steps) and H3PO4, which is capable of donating three protons in successive steps, is a triprotic acid (a compound that can donate three protons per molecule in separate steps), (Equation $\ref{4.3.4}$, Equation $\ref{4.3.5}$, and Equation $\ref{4.3.6}$):
$H_3 PO_4 (l) \overset{H_2 O(l)}{\rightleftharpoons} H ^+ ( a q ) + H_2 PO_4 ^- (aq) \label{4.3.4}$
$H_2 PO_4 ^- (aq) \rightleftharpoons H ^+ (aq) + HPO_4^{2-} (aq) \label{4.3.5}$
$HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq) \label{4.3.6}$
In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in equilibrium (the point at which the rates of the forward and reverse reactions become the same, so that the net composition of the system no longer changes with time).
Strengths of Acids and Bases
We will not discuss the strengths of acids and bases quantitatively until next semester. Qualitatively, however, we can state that strong acids react essentially completely with water to give $H^+$ and the corresponding anion. Similarly, strong bases dissociate essentially completely in water to give $OH^−$ and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids and weak bases react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form.
In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO3, HClO4, and H2SO4 (H3PO4 is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH)2. Common weak acids include HCN, H2S, HF, oxoacids such as HNO2 and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows:
$CH_3 CO_2 H(l) \overset{H_2 O(l)}{\rightleftharpoons} H^+ (aq) + CH_3 CO_2^- (aq) \label{4.3.7}$
Although acetic acid is very soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates). Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation $\ref{4.3.8}$) but a weak acid when it donates its second proton (Equation $\ref{4.3.9}$) as indicated by the single and double arrows, respectively:
$\underset{strong\: acid}{H_2 SO_4 (l)} \xrightarrow {H_2 O(l)} H ^+ (aq) + HSO_4 ^- (aq) \label{4.3.8}$
$\underset{weak\: acid}{HSO_4^- (aq)} \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq) \label{4.3.9}$
Consequently, an aqueous solution of sulfuric acid contains $H^+_{(aq)}$ ions and a mixture of $HSO^-_{4\;(aq)}$ and $SO^{2−}_{4\;(aq)}$ ions, but no $H_2SO_4$ molecules. All other polyprotic acids, such as H3PO4, are weak acids.
The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion:
$NH_3 (g) + H_2 O(l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \label{4.3.10}$
Most of the ammonia (>99%) is present in the form of NH3(g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S2−).
There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte.
Definition of Strong/Weak Acids & Bases: Definition of Strong/Weak Acids & Bases, YouTube (opens in new window) [Definition of Strong] [Definition of Strong] [youtu.be] (opens in new window)
Table $1$ lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that there is no correlation between solubility and whether a substance is a strong or a weak electrolyte! Many weak acids and bases are extremely soluble in water.
Table $1$: Common Strong Acids and Bases
Strong Acids Strong Bases
Hydrogen Halides Oxoacids Group 1 Hydroxides Hydroxides of Heavy Group 2 Elements
HCl HNO3 LiOH Ca(OH)2
HBr H2SO4 NaOH Sr(OH)2
HI HClO4 KOH Ba(OH)2
RbOH
CsOH
Example $1$: Acid Strength
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. CH3CH2CO2H
2. CH3OH
3. Sr(OH)2
4. CH3CH2NH2
5. HBrO4
Given: compound
Asked for: acid or base strength
Strategy:
A Determine whether the compound is organic or inorganic.
B If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H+ or OH ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H2SO4 are weak acids.
Solution:
1. A This compound is propionic acid, which is organic. B It contains a carboxylic acid group analogous to that in acetic acid, so it must be a weak acid.
2. A CH3OH is methanol, an organic compound that contains the −OH group. B As a covalent compound, it does not dissociate to form the OH ion. Because it does not contain a carboxylic acid (−CO2H) group, methanol also cannot dissociate to form H+(aq) ions. Thus we predict that in aqueous solution methanol is neither an acid nor a base.
3. A Sr(OH)2 is an inorganic compound that contains one Sr2+ and two OH ions per formula unit. B We therefore expect it to be a strong base, similar to Ca(OH)2.
4. A CH3CH2NH2 is an amine (ethylamine), an organic compound in which one hydrogen of ammonia has been replaced by an R group. B Consequently, we expect it to behave similarly to ammonia (Equation $\ref{4.3.7}$), reacting with water to produce small amounts of the OH ion. Ethylamine is therefore a weak base.
5. A HBrO4 is perbromic acid, an inorganic compound. B It is not listed in Table $1$ as one of the common strong acids, but that does not necessarily mean that it is a weak acid. If you examine the periodic table, you can see that Br lies directly below Cl in group 17. We might therefore expect that HBrO4 is chemically similar to HClO4, a strong acid—and, in fact, it is.
Exercise $1$: Acid Strength
Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these.
1. Ba(OH)2
2. HIO4
3. CH3CH2CH2CO2H
4. (CH3)2NH
5. CH2O
Answer a
strong base
Answer b
strong acid
Answer c
weak acid
Answer d
weak base
Answer e
none of these; formaldehyde is a neutral molecule | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/09%3A_Introduction_to_Solutions_and_Aqueous_Reactions/9.07%3A_Acid-Base.txt |
Learning Objectives
• To identify oxidation–reduction reactions in solution.
The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O2−). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short.
Any oxidation must ALWAYS be accompanied by a reduction and vice versa.
Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water vapor). The reaction is as follows:
$\ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1}$
Oxidation-reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants by a transfer of electrons, which follows the mnemonic "oxidation is loss, reduction is gain", or "oil rig". The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O2 or H2, are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is
$\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2}$
Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al2O3, electrons are transferred as follows (the small overset number emphasizes the oxidation state of the elements):
$4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} }\label{4.4.3}$
Equation $\ref{4.4.1}$ and Equation $\ref{4.4.2}$ are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In Equation $\ref{4.4.3}$, for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen:
\begin{align*} \text{electrons lost} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{lost} \over \ce{Al} \, \text{atom} } \[4pt] &= 12 \, e^- \, \text{lost} \label{4.4.4a} \end{align*}
\begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{O} \, \text{atom}} \[4pt] &= 12 \, e^- \, \text{gained} \label{4.4.4b}\end{align*}
The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with chlorine is illustrated in Figure $1$.
In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained.
Assigning Oxidation States
Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previously, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; Figure $1$), magnesium oxide (MgO), and calcium chloride (CaCl2). In covalent compounds, in contrast, atoms share electrons. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (that is, as if all the bonding electrons were transferred to the more attractive element). Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions.
A set of rules for assigning oxidation states to atoms in chemical compounds follows.
Rules for Assigning Oxidation States
1. The oxidation state of an atom in any pure element, whether monatomic, diatomic, or polyatomic, is zero.
2. The oxidation state of a monatomic ion is the same as its charge—for example, Na+ = +1, Cl = −1.
3. The oxidation state of fluorine in chemical compounds is always −1. Other halogens usually have oxidation states of −1 as well, except when combined with oxygen or other halogens.
4. Hydrogen is assigned an oxidation state of +1 in its compounds with nonmetals and −1 in its compounds with metals.
5. Oxygen is normally assigned an oxidation state of −2 in compounds, with two exceptions: in compounds that contain oxygen–fluorine or oxygen–oxygen bonds, the oxidation state of oxygen is determined by the oxidation states of the other elements present.
6. The sum of the oxidation states of all the atoms in a neutral molecule or ion must equal the charge on the molecule or ion.
Nonintegral (fractional) oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states.
In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound.
Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in Chapter 6. Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H+ and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF2 but −½ in KO2. Note that an oxidation state of −½ for O in KO2 is perfectly acceptable.
The reduction of copper(I) oxide shown in Equation $\ref{4.4.5}$ demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H2 and Cu. From rule 4, hydrogen in H2O has an oxidation state of +1, and from rule 5, oxygen in both Cu2O and H2O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu2O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows:
$\overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{\ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5}$
Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). Thus, this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge:
$\text{electrons lost} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{lost} \over H \, \text{atom} } = 2 \, e^- \, \text{lost} \label{4.4.6a}$
$\text{electrons gained} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gained} \over Cu \, \text{atom}} = 2 \, e^- \, \text{gained} \label{4.4.6b}$
Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo.
Example $1$: Oxidation States
Assign oxidation states to all atoms in each compound.
1. sulfur hexafluoride (SF6)
2. methanol (CH3OH)
3. ammonium sulfate [(NH4)2SO4]
4. magnetite (Fe3O4)
5. ethanoic (acetic) acid (CH3CO2H)
Given: molecular or empirical formula
Asked for: oxidation states
Strategy:
Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1.
Solution:
a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF6), the oxidation state of sulfur must be +6:
[(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0
b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2:
[(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0
c. Note that (NH4)2SO4 is an ionic compound that consists of both a polyatomic cation (NH4+) and a polyatomic anion (SO42) (see Table 2.4). We assign oxidation states to the atoms in each polyatomic ion separately. For NH4+, hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3:
[(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH4+ ion
For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6:
[(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion
d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron:
[(4 O atoms)(−2)]+[(3 Fe atoms)$\left (+{8 \over 3} \right )$]= 0
Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe3O4 can be viewed as having two Fe3+ ions and one Fe2+ ion per formula unit, giving a net positive charge of +8 per formula unit. Fe3O4 is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.”
e. Initially, we assign oxidation states to the components of CH3CO2H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of
[(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0
So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH3) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO2H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of
[(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3
To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus
$\underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \nonumber$
Thus the sum of the oxidation states of the two carbon atoms is indeed zero.
Exercise $1$: Oxidation States
Assign oxidation states to all atoms in each compound.
1. barium fluoride (BaF2)
2. formaldehyde (CH2O)
3. potassium dichromate (K2Cr2O7)
4. cesium oxide (CsO2)
5. ethanol (CH3CH2OH)
Answer a
Ba, +2; F, −1
Answer b
C, 0; H, +1; O, −2
Answer c
K, +1; Cr, +6; O, −2
Answer d
Cs, +1; O, −½
Answer e
C, −3; H, +1; C, −1; H, +1; O, −2; H, +1
Types of Redox Reactions
Many types of chemical reactions are classified as redox reactions, and it would be impossible to memorize all of them. However, there are a few important types of redox reactions that you are likely to encounter and should be familiar with. These include:
• Synthesis reactions: The formation of any compound directly from the elements is a redox reaction, for example, the formation of water from hydrogen and oxygen: $\ce{ 2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \nonumber$
• Decomposition reactions: Conversely, the decomposition of a compound to its elements is also a redox reaction, as in the electrolysis of water: $\ce{2H_2O(l) \rightarrow 2H_2(g) + O_2(g)} \nonumber$
• Combustion reactions: Many chemicals combust (burn) with oxygen. In particular, organic chemicals such as hydrocarbons burn in the presence of oxygen to produce carbon dioxide and water as the products: $\ce{ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)} \nonumber$
The following sections describe another important class of redox reactions: single-displacement reactions of metals in solution.
Redox Reactions of Solid Metals in Aqueous Solution
• A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure $2$). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas:
$\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81}$
In subsequent steps, $\ce{FeCl2}$ undergoes oxidation to form a reddish-brown precipitate of $\ce{Fe(OH)3}$.
Many metals dissolve through reactions of this type, which have the general form
$\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82}$
Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves:
$\ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83}$
Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul!
• Single-Displacement Reactions
Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation $\ref{4.4.84}$) and the reduction of silver salts by copper (Equation $\ref{4.4.85}$ and Figure $3$):
$\ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84}$
$\ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85}$
The reaction in Equation $\ref{4.4.84}$ is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object.
• The Activity Series
By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing $\ce{Zn2+}$. Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal:
$\ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10}$
$\ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11}$
Magnesium has a greater tendency to be oxidized than zinc does.
Pairwise reactions of this sort are the basis of the activity series (Figure $4$), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series).
When using the activity series to predict the outcome of a reaction, keep in mind that any element will reduce compounds of the elements below it in the series. Because magnesium is above zinc in Figure $4$, magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Only those metals that lie above hydrogen in the activity series dissolve in acids to produce H2. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example $2$ demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions.
Example $2$: Activity
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of aluminum foil is placed in an aqueous solution of silver nitrate.
2. A few drops of liquid mercury are added to an aqueous solution of lead(II) acetate.
3. Some sulfuric acid from a car battery is accidentally spilled on the lead cable terminals.
Given: reactants
Asked for: overall reaction and net ionic equation
Strategy:
1. Locate the reactants in the activity series in Figure $4$ and from their relative positions, predict whether a reaction will occur. If a reaction does occur, identify which metal is oxidized and which is reduced.
2. Write the net ionic equation for the redox reaction.
Solution:
1. A Aluminum is an active metal that lies above silver in the activity series, so we expect a reaction to occur. According to their relative positions, aluminum will be oxidized and dissolve, and silver ions will be reduced to silver metal. B The net ionic equation is as follows:
$\ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber$
Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction.
2. A Mercury lies below lead in the activity series, so no reaction will occur.
3. A Lead is above hydrogen in the activity series, so the lead terminals will be oxidized, and the acid will be reduced to form H2. B From our discussion of solubilities, recall that Pb2+ and SO42 form insoluble lead(II) sulfate. In this case, the sulfate ions are not spectator ions, and the reaction is as follows:
$\ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber$
Lead(II) sulfate is the white solid that forms on corroded battery terminals.
Corroded battery terminals. The white solid is lead(II) sulfate, formed from the reaction of solid lead with a solution of sulfuric acid.
Exercise $2$
Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation.
1. A strip of chromium metal is placed in an aqueous solution of aluminum chloride.
2. A strip of zinc is placed in an aqueous solution of chromium(III) nitrate.
3. A piece of aluminum foil is dropped into a glass that contains vinegar (the active ingredient is acetic acid).
Answer a
$no\: reaction$
Answer b
$3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)$
Answer c | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/09%3A_Introduction_to_Solutions_and_Aqueous_Reactions/9.09%3A_Oxidation-Reduction_Reactions.txt |
• 10.1: On Fire, But Not Consumed
• 10.2: The Nature of Energy - Key Definitions
All chemical changes are accompanied by the absorption or release of heat. In this unit we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics, which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes.
• 10.3: The First Law of Thermodynamics - There Is No Free Lunch
The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy.
• 10.4: Quantifying Heat and Work
Heat is the amount of energy that is transferred from one system to its surroundings because of a temperature difference. All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work.
• 10.5: Measuring ΔE for Chemical Reactions- Constant-Volume Calorimetry
A bomb calorimeter operates at constant volume and is particularly useful for measuring energies of combustion.
• 10.6: Enthalpy- The Heat Evolved in a Chemical Reaction at Constant Pressure
Enthalpy is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the change in enthalpy (ΔH) can be measured. A negative ΔH means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. Calorimetry measures enthalpy changes during chemical processes, where the magnitude of the temperature change depends on the amount of heat released or absorbed and on t
• 10.7: Measuring ΔH for Chemical Reactions- Constant-Pressure Calorimetry
a constant-pressure calorimeter, which gives ΔH values directly
• 10.8: Relationships Involving ΔHrxn
Hess's law argues that for a chemical reaction, the enthalpy of reaction (ΔHrxn) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of ΔHrxn. The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion (ΔHfus) and the enthalpy of vaporiz
• 10.9: Determining Enthalpies of Reaction from Bond Energies
• 10.10: Determining Enthalpies of Reaction from Standard Enthalpies of Formation
The standard state for measuring and reporting enthalpies of formation or reaction is 25 oC and 1 atm. The elemental form of each atom is that with the lowest enthalpy in the standard state. The standard state heat of formation for the elemental form of each atom is zero. The enthalpy of formation (ΔHf) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation (ΔHof) are determined under standard conditions: a pressure of 1 atm for ga
• 10.11: Lattice Energy
The Lattice energy, U, is the amount of energy requried to separate a mole of the solid (s) into a gas (g) of its ions.
10: Thermochemistry
; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of and molecules. "" usually refers to the energy that is stored in the of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy. In an chemical reaction, the electrons and nuclei within the reactants undergo rearrangement into products possessing lower energies, and the difference is released to the environment in the form of heat. chemical reaction heats surrounding , the kinetic energy gets dispersed into the molecular units in the environment. This "microscopic" form of kinetic energy, unlike that of a speeding bullet, is completely random in the kinds of motions it exhibits and in its direction. We refer to this as "thermalized" kinetic energy, or more commonly simply as . We observe the effects of this as a rise in the temperature of the . The temperature of a body is direct measure of the quantity of thermal energy is contains. , energy comes in different types. Energy can be converted from one form into another, but all of the energy present before a change occurs always exists in some form after the change is completed. This observation is expressed in the : (a type of potential energy) is stored in the molecules that compose gasoline. When gasoline is combusted within the cylinders of a car’s engine, the rapidly expanding gaseous products of this chemical reaction generate mechanical energy (a type of kinetic energy) when they move the cylinders’ pistons. The is also one version of the first law of thermodynamics, as you will learn later. in glucose to keep us warm and to move our muscles. In fact, life itself depends on the conversion of to other forms. So when you go uphill, your kinetic energy is transformed into potential energy, which gets changed back into kinetic energy as you coast down the other side. And where did the kinetic energy you expended in peddling uphill come from? By conversion of some of the chemical potential energy in your breakfast cereal. ). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed ( or first law of thermodynamics). has thermal energy due to the KE of its molecules and temperature that corresponds to the average KE of its molecules. Heat is energy that is transferred between objects at different temperatures; it flows from a high to a low temperature. Chemical and physical processes can absorb heat () or release heat (). The SI unit of energy, heat, and work is the . | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.02%3A_The_Nature_of_Energy_-_Key_Definitions.txt |
Learning Objectives
• To calculate changes in internal energy
To study the flow of energy during a chemical reaction, we need to distinguish between a system, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings, the rest of the universe, including the container in which the reaction is carried out (Figure $1$). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa.
Three kinds of systems are important in chemistry. An open system can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, the total energy of a system plus its surroundings is constant, which must be true if energy is conserved.
The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are path dependent. For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure $2$). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose.
Direction of Heat Flow
The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (Figure $3$). The balanced chemical equation for the reaction is as follows:
$\ce{ 2Al(s) + Fe_2O_3(s) -> 2Fe(s) + Al_2O_3(s)} \label{5.2.1}$
We can also write this chemical equation as
$\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s)} + \text{heat} \label{5.2.2}$
to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat (q) is transferred from a system to its surroundings is described as exothermic. By convention, $q < 0$ for an exothermic reaction.
When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:
$\ce{heat + H_2O(s) \rightarrow H_2O(l)} \label{5.2.3}$
When heat is transferred to a system from its surroundings, the process is endothermic. By convention, $q > 0$ for an endothermic reaction.
Heat is technically not a component in Chemical Reactions
Technically, it is poor form to have a $heat$ term in the chemical reaction like in Equations $\ref{5.2.2}$ and $\ref{5.2.3}$ since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists.
The First Law
The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics, which states that the energy of the universe is constant. We can express this law mathematically as follows:
$U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \label{5.2.4a}$
$\Delta{U_{sys}}=−ΔU_{surr} \label{5.2.4b}$
where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings.
The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy.
An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the $CO_2/H_2O$ product mixture is less than that of the isooctane/ $O_2$ reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w):
$ΔU_{sys} = q + w \label{5.2.5}$
Although $q$ and $w$ are not state functions on their own, their sum ($ΔU_{sys}$) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed.
Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings.
Although $q$ and $w$ are not state functions, their sum ($ΔU_{sys}$) is independent of the path taken and therefore is a state function.
Thus, because of the first law, we can determine $ΔU$ for any process if we can measure both $q$ and $w$. Heat, $q$, may be calculated by measuring a change in temperature of the surroundings. Work, $w$, may come in different forms, but it too can be measured. One important form of work for chemistry is pressure-volume work done by an expanding gas. At a constant external pressure (for example, atmospheric pressure)
$w = −PΔV \label{5.2.6}$
The negative sign associated with $PV$ work done indicates that the system loses energy when the volume increases. That is, an expanding gas does work on its surroundings, while a gas that is compressed has work done on it by the surroundings.
Example $1$
A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔU) of the gas in joules?
Given: initial volume, final volume, external pressure, and quantity of energy transferred as heat
Asked for: total change in internal energy
Strategy:
1. Determine the sign of $q$ to use in Equation $\ref{5.2.5}$.
2. From Equation $\ref{5.2.6}$ calculate $w$ from the values given. Substitute this value into Equation $\ref{5.2.5}$ to calculate $ΔU$.
Solution
A From Equation $\ref{5.2.5}$, we know that ΔU = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention.
B Because the gas is being compressed, we know that work is being done on the system, so $w$ must be positive. From Equation $\ref{5.2.5}$,
$w=-P_{\textrm{ext}}\Delta V=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)=284\textrm{ J} \nonumber$
Thus
\begin{align*} ΔU &= q + w \[4pt] &= −140 \,J + 284\, J \[4pt] &= 144\, J\end{align*} \nonumber
In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.
Exercise $1$
A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔU) of the gas in joules?
Answer
−216 J
By convention (to chemists), both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa.
Summary
In chemistry, the small part of the universe that we are studying is the system, and the rest of the universe is the surroundings. Open systems can exchange both matter and energy with their surroundings, closed systems can exchange energy but not matter with their surroundings, and isolated systems can exchange neither matter nor energy with their surroundings. A state function is a property of a system that depends on only its present state, not its history. A reaction or process in which heat is transferred from a system to its surroundings is exothermic. A reaction or process in which heat is transferred to a system from its surroundings is endothermic. The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.03%3A_The_First_Law_of_Thermodynamics_-_There_Is_No_Free_Lunch.txt |
Learning Objectives
• To calculate changes in internal energy
• Distinguish the related properties of heat, thermal energy, and temperature
• Define and distinguish specific heat and heat capacity, and describe the physical implications of both
• Perform calculations involving heat, specific heat, and temperature change
Thermal energy is kinetic energy associated with the random motion of atoms and molecules. Temperature is a quantitative measure of “hot” or “cold.” When the atoms and molecules in an object are moving or vibrating quickly, they have a higher average kinetic energy (KE), and we say that the object is “hot.” When the atoms and molecules are moving slowly, they have lower KE, and we say that the object is “cold” (Figure $1$). Assuming that no chemical reaction or phase change (such as melting or vaporizing) occurs, increasing the amount of thermal energy in a sample of matter will cause its temperature to increase. And, assuming that no chemical reaction or phase change (such as condensation or freezing) occurs, decreasing the amount of thermal energy in a sample of matter will cause its temperature to decrease.
Figure $1$: (a) The molecules in a sample of hot water move more rapidly than (b) those in a sample of cold water.
Heat (q) is the transfer of thermal energy between two bodies at different temperatures. Heat flow (a redundant term, but one commonly used) increases the thermal energy of one body and decreases the thermal energy of the other. Suppose we initially have a high temperature (and high thermal energy) substance (H) and a low temperature (and low thermal energy) substance (L). The atoms and molecules in H have a higher average KE than those in L. If we place substance H in contact with substance L, the thermal energy will flow spontaneously from substance H to substance L. The temperature of substance H will decrease, as will the average KE of its molecules; the temperature of substance L will increase, along with the average KE of its molecules. Heat flow will continue until the two substances are at the same temperature (Figure $2$).
Figure $2$: (a) Substances H and L are initially at different temperatures, and their atoms have different average kinetic energies. (b) When they are put into contact with each other, collisions between the molecules result in the transfer of kinetic (thermal) energy from the hotter to the cooler matter. (c) The two objects reach “thermal equilibrium” when both substances are at the same temperature, and their molecules have the same average kinetic energy.
Matter undergoing chemical reactions and physical changes can release or absorb heat. A change that releases heat is called an exothermic process. For example, the combustion reaction that occurs when using an oxyacetylene torch is an exothermic process—this process also releases energy in the form of light as evidenced by the torch’s flame (Figure $\PageIndex{3a}$). A reaction or change that absorbs heat is an endothermic process. A cold pack used to treat muscle strains provides an example of an endothermic process. When the substances in the cold pack (water and a salt like ammonium nitrate) are brought together, the resulting process absorbs heat, leading to the sensation of cold.
Figure $3$: (a) An oxyacetylene torch produces heat by the combustion of acetylene in oxygen. The energy released by this exothermic reaction heats and then melts the metal being cut. The sparks are tiny bits of the molten metal flying away. (b) A cold pack uses an endothermic process to create the sensation of cold. (credit a: modification of work by “Skatebiker”/Wikimedia commons).
Historically, energy was measured in units of calories (cal). A calorie is the amount of energy required to raise one gram of water by 1 degree C (1 kelvin). However, this quantity depends on the atmospheric pressure and the starting temperature of the water. The ease of measurement of energy changes in calories has meant that the calorie is still frequently used. The Calorie (with a capital C), or large calorie, commonly used in quantifying food energy content, is a kilocalorie. The SI unit of heat, work, and energy is the joule. A joule (J) is defined as the amount of energy used when a force of 1 newton moves an object 1 meter. It is named in honor of the English physicist James Prescott Joule. One joule is equivalent to 1 kg m2/s2, which is also called 1 newton–meter. A kilojoule (kJ) is 1000 joules. To standardize its definition, 1 calorie has been set to equal 4.184 joules.
$1\,cal \equiv 4.184\, J$
Application: Heat can do more than increase temperature
Most substances expand as their temperature increases and contract as their temperature decreases. This property can be used to measure temperature changes, as shown in Figure $4$. The operation of many thermometers depends on the expansion and contraction of substances in response to temperature changes.
Direction of Heat Flow: Endothermic vs. Exothermic Processes
The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel. The balanced chemical equation for the reaction is as follows:
$\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s) }\label{12.2.1}$
We can also write this chemical equation as
$\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s)} + \text{heat} \label{12.2.2}$
to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called thermochemical equations. In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat ($q$) is transferred from a system to its surroundings is described as exothermic. By convention, $q < 0$ for an exothermic reaction.
When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:
$\text{heat} + \ce{H_2O_{(s)} \rightarrow H_2O_{(l)}} \label{12.2.3}$
When heat is transferred to a system from its surroundings, the process is endothermic. By convention, $q > 0$ for an endothermic reaction.
By convention, $q < 0$ for an exothermic reaction and $q > 0$ for an endothermic reaction.
Exercise $1$
Decide whether the following are endothermic or exothermic processes
1. water evaporates off a shower door
2. an acid tablet being added to a pool and the surrounding water heats up
3. $\ce{NH_4Cl}$ is dissolved in water and the solution cools
4. the burning of a log in a campfire
Hint
During an endothermic process heat is absorbed from surroundings, causing them to cool, so in every case where there is cooling there is most likely an endothermic process taking place. For exothermic reactions energy is being released to the surroundings and so the surroundings feel like they have been heated by the process.
Answer a
endothermic
Answer b
exothermic
Answer c
endothermic
Answer d
exothermic
Heat is technically not a component in Chemical Reactions
Technically, it is poor form to have a $heat$ term in the chemical reaction like in Equations $\ref{12.2.2}$ and $\ref{12.2.3}$ since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists.
Contributors and Attributions
Learning Objectives
• To know the relationship between energy, work, and heat.
One definition of energy is the capacity to do work. There are many kinds of work, including mechanical work, electrical work, and work against a gravitational or a magnetic field. Here we will consider only mechanical work and focus on the work done during changes in the pressure or the volume of a gas.
Mechanical Work
The easiest form of work to visualize is mechanical work (Figure $5$), which is the energy required to move an object a distance $d$ when opposed by a force $F$, such as gravity:
$w=F\,d \label{12.3.1}$
with $w$ is work, $F$ is opposing force, and $d$ is distance.
Because the force ($F$) that opposes the action is equal to the mass ($m$) of the object times its acceleration ($a$), Equation $\ref{12.3.1}$ can be rewritten to:
$w = m\,a\,d \label{12.3.2}$
with $w$ is work, $m$ is mass, $a$ is acceleration, and $d$ is distance.
Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth. Hence for works against gravity (on Earth), $a$ can be set to $g=9.8\; m/s^2)$. Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the opposing force of gravity. The amount of work done (w) and thus the energy required depends on three things:
1. the height of the second floor (the distance $d$);
2. your mass, which must be raised that distance against the downward acceleration due to gravity; and
3. your path.
Pressure-Volume (PV) Work
To describe this pressure–volume work (PV work), we will use such imaginary oddities as frictionless pistons, which involve no component of resistance, and ideal gases, which have no attractive or repulsive interactions. Imagine, for example, an ideal gas, confined by a frictionless piston, with internal pressure $P_{int}$ and initial volume $V_i$ (Figure $6$). If $P_{ext} = P_{int}$, the system is at equilibrium; the piston does not move, and no work is done. If the external pressure on the piston ($P_{ext}$) is less than $P_{int}$, however, then the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings; that is, the final volume ($V_f$) will be greater than $V_i$. If $P_{ext} > P_{int}$, then the gas will be compressed, and the surroundings will perform work on the system.
Figure $6$: PV Work demonstrated with a frictionless piston. (a) if the external pressure is less than $P_{int}$, the ideal gas inside the piston will expand, forcing the piston to perform work on its surroundings. The final volume ($V_f$) will be greater than $V_i$. (b) Alternatively, if the external pressure is greater than $P_{int}$, the gas will be compressed, and the surroundings will perform work on the system.
If the piston has cross-sectional area $A$, the external pressure exerted by the piston is, by definition, the force per unit area:
$P_{ext} = \dfrac{F}{A} \label{eq5}$
The volume of any three-dimensional object with parallel sides (such as a cylinder) is the cross-sectional area times the height ($V = Ah$). Rearranging Equation \ref{eq5} to give
$F = P_{ext}A$
and defining the distance the piston moves ($d$) as $Δh$, we can calculate the magnitude of the work performed by the piston by substituting into Equation $\ref{12.3.1}$:
$w = F d = P_{ext}AΔh \label{12.3.3}$
The change in the volume of the cylinder ($ΔV$) as the piston moves a distance d is $ΔV = AΔh$, as shown in Figure $7$.
Figure $7$: Work Performed with a change in volume. The change in the volume ($ΔV$) of the cylinder housing a piston is $ΔV = AΔh$ as the piston moves. The work performed by the surroundings on the system as the piston moves inward is given by Equation \ref{12.3.4}.
The PV work performed is thus
$w = P_{ext}ΔV \label{12.3.4}$
The units of work obtained using this definition are correct for energy: pressure is force per unit area (newton/m2) and volume has units of cubic meters, so
$w=\left(\dfrac{F}{A}\right)_{\textrm{ext}}(\Delta V)=\dfrac{\textrm{newton}}{\textrm m^2}\times \textrm m^3=\mathrm{newton\cdot m}=\textrm{joule}$
If we use atmospheres for P and liters for V, we obtain units of L·atm for work. These units correspond to units of energy, as shown in the different values of the ideal gas constant R:
$R=\dfrac{0.08206\;\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}=\dfrac{8.314\textrm{ J}}{\mathrm{mol\cdot K}}$
Thus 0.08206 L·atm = 8.314 J and 1 L·atm = 101.3 J.
Exercise: Expansion (PV) work
1. How much work is done by a gas that expands from 2 liters to 5 liters against an external pressure of 750 mmHg?
2. How much work is done by 0.54 moles of a gas that has an initial volume of 8 liters and expands under the following conditions: 30 oC and 1.3 atm?
3. How much work is done by a gas (P=1.7 atm, V=1.56 L) that expands against an external pressure of 1.8 atm?
Solution a
$w = − PΔV$
$ΔV = V_{final} - V_{Initial} = 5 \,L - 2\, L = 3 L$
Convert 750 mmHg to atm:
$750 mmHg * 1/760 (atm/mmHg) = 0.9868 atm.$
$W = − pΔV = -(.9868\, atm)(3\,L) = -2.96 \,L\, atm.$
Solution b
First we must find the final volume using the ideal gas law:
$pV = nRT$
or
$V = \dfrac{nRT}{P} = \dfrac{ [(0.54 \,moles)(0.082057 (L\, atm)/ (mol\, K))(303\,K)] }{1.3\, atm} = 10.33\, L$
$ΔV = V_{final} - V_{initial} = 10.3\, L - 8\, L = 2.3\, L$
$w = − pΔV = - (1.3\, atm)(2.3 \,L) = -3\, L\, atm.$
Solution c
$w = - p * ΔV\) = - (1.8 \,atm )\, ΔV.$
Given $p_1$, $V_1$, and $p_2$, find $V_2$: $p_1V_1=p_2V_2$ (at constant $T$ and $n$)
$V_2= (V_1* P_1) / P_2$ = (1.56 L * 1.7 atm) / 1.8 atm = 1.47 L
Now,
$ΔV = V_2 - V_1=1.47\, L - 1.56\, L = -0.09$
$w = - (1.8 atm) * (-0.09 L) = 0.162 L atm.$
Whether work is defined as having a positive sign or a negative sign is a matter of convention. Heat flow is defined from a system to its surroundings as negative; using that same sign convention, we define work done by a system on its surroundings as having a negative sign because it results in a transfer of energy from a system to its surroundings. This is an arbitrary convention and one that is not universally used. Some engineering disciplines are more interested in the work done on the surroundings than in the work done by the system and therefore use the opposite convention. Because ΔV > 0 for an expansion, Equation $\ref{12.3.4}$ must be written with a negative sign to describe PV work done by the system as negative:
$w = −P_{ext}ΔV \label{12.3.5}$
The work done by a gas expanding against an external pressure is therefore negative, corresponding to work done by a system on its surroundings. Conversely, when a gas is compressed by an external pressure, ΔV < 0 and the work is positive because work is being done on a system by its surroundings.
A Matter of Convention
• Heat flow is defined from the system to its surroundings as negative
• Work is defined as by the system on its surroundings as negative
Suppose, for example, that the system under study is a mass of steam heated by the combustion of several hundred pounds of coal and enclosed within a cylinder housing a piston attached to the crankshaft of a large steam engine. The gas is not ideal, and the cylinder is not frictionless. Nonetheless, as steam enters the engine chamber and the expanding gas pushes against the piston, the piston moves, so useful work is performed. In fact, PV work launched the Industrial Revolution of the 19th century and powers the internal combustion engine on which most of us still rely for transportation.
In contrast to internal energy, work is not a state function. We can see this by examining Figure $8$, in which two different, two-step pathways take a gaseous system from an initial state to a final state with corresponding changes in temperature. In pathway A, the volume of a gas is initially increased while its pressure stays constant (step 1); then its pressure is decreased while the volume remains constant (step 2). In pathway B, the order of the steps is reversed. The temperatures, pressures, and volumes of the initial and final states are identical in both cases, but the amount of work done, indicated by the shaded areas in the figure, is substantially different. As we can see, the amount of work done depends on the pathway taken from ($V_1$, $P_1$) to ($V_2$, $P_2$), which means that work is not a state function.
Internal energy is a state function, whereas work is not.
Example $1$: Internal Combustion Engine
A small high-performance internal combustion engine has six cylinders with a total nominal displacement (volume) of 2.40 L and a 10:1 compression ratio (meaning that the volume of each cylinder decreases by a factor of 10 when the piston compresses the air–gas mixture inside the cylinder prior to ignition). How much work in joules is done when a gas in one cylinder of the engine expands at constant temperature against an opposing pressure of 40.0 atm during the engine cycle? Assume that the gas is ideal, the piston is frictionless, and no energy is lost as heat.
Given: final volume, compression ratio, and external pressure
Asked for: work done
Strategy:
Calculate the final volume of gas in a single cylinder. Then compute the initial volume of gas in a single cylinder from the compression ratio. Use Equation $\ref{12.3.5}$ to calculate the work done in liter-atmospheres. Convert from liter-atmospheres to joules.
Solution:
A To calculate the work done, we need to know the initial and final volumes. The final volume is the volume of one of the six cylinders with the piston all the way down: Vf = 2.40 L/6 = 0.400 L. With a 10:1 compression ratio, the volume of the same cylinder with the piston all the way up is Vi = 0.400 L/10 = 0.0400 L. Work is done by the system on its surroundings, so work is negative.
\begin{align} w &= −P_{ext}ΔV \nonumber \[4pt] &= −(40.0\, atm)(0.400\, L − 0.0400 \,L) \nonumber \[4pt] &= −14.4\, L·atm \nonumber \end{align} \nonumber
Converting from liter-atmospheres to joules,
\begin{align} w &=-(14.4\;\mathrm{L\cdot atm})[101.3\;\mathrm{J/(L\cdot atm)}] \nonumber \[4pt] &= -1.46\times10^3\textrm{ J} \nonumber \end{align} \nonumber
In the following exercise, you will see that the concept of work is not confined to engines and pistons. It is found in other applications as well.
Exercise $1$: Work to Breath
Breathing requires work, even if you are unaware of it. The lung volume of a 70 kg man at rest changed from 2200 mL to 2700 mL when he inhaled, while his lungs maintained a pressure of approximately 1.0 atm. How much work in liter-atmospheres and joules was required to take a single breath? During exercise, his lung volume changed from 2200 mL to 5200 mL on each in-breath. How much additional work in joules did he require to take a breath while exercising?
Answer
−0.500 L·atm, or −50.7 J; −304 J; if he takes a breath every three seconds, this corresponds to 1.4 Calories per minute (1.4 kcal).
Work and Chemical Reactions
We have stated that the change in energy ($ΔU$) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work, (or just PV work). Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
$Cu_{(s)} + 4HNO_{3(aq)} \rightarrow Cu(NO_3)_{2(aq)} + 2H_2O_{(l)} + 2NO_{2(g)} \label{12.3.5a}$
If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure $9$). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston (ΔV). At a constant external pressure (here, atmospheric pressure)
$w = −PΔV \label{12.3.6}$
The negative sign associated with $PV$ work done indicates that the system loses energy. If the volume increases at constant pressure (ΔV > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (ΔV < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.
The symbol $U$ represents the internal energy of a system, which is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy (H) (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy $U$ plus the product of its pressure $P$ and volume $V$:
$H =U + PV \label{12.3.7}$
Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function.
If a chemical change occurs at constant pressure (i.e., for a given P, ΔP = 0), the change in enthalpy (ΔH) is
$ΔH = Δ(U + PV) = ΔU + ΔPV = ΔU + PΔV \label{12.3.8}$
Substituting $q + w$ for $ΔU$ (Equation $\ref{12.3.8}$) and $−w$ for $PΔV$ (Equation $\ref{12.3.6}$), we obtain
$ΔH = ΔU + PΔV = q_p + w − w = q_p \label{12.3.9}$
The subscript $p$ is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation $\ref{12.3.9}$ we see that at constant pressure the change in enthalpy, $ΔH$ of the system, defined as $H_{final} − H_{initial}$, is equal to the heat gained or lost.
$ΔH = H_{final} − H_{initial} = q_p \label{12.3.10}$
Just as with $ΔU$, because enthalpy is a state function, the magnitude of $ΔH$ depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.
To find $ΔH$ for a reaction, measure $q_p$ under constant pressure.
Summary
All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work.
• Gasparro, Frances P. "Remembering the sign conventions for q and w in deltaU = q - w." J. Chem. Educ. 1976: 53, 389.
• Koubek, E. "PV work demonstration (TD)." J. Chem. Educ. 1980: 57, 374. ' | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.04%3A_Quantifying_Heat_and_Work.txt |
Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter(A device used to measure energy changes in chemical processes. shown schematically in Figure $1$). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated.
Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in internal energyE) rather than the enthalpy change (ΔH); ΔE is related to ΔH by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that ΔE < ΔH, the relationship between the measured temperature change and ΔHcomb is given in Equation $\ref{5.5.9}$, where Cbomb is the total heat capacity of the steel bomb and the water surrounding it:
$\Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{5.5.9}$
To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C6H5CO2H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its ΔHcomb = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in Equation $\ref{5.42}$ to determine Cbomb. The use of a bomb calorimeter to measure the ΔHcomb of a substance is illustrated in Example $4$.
Example $4$: Combustion of Glucose
The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the ΔHcomb of glucose?
Given: mass and ΔT for combustion of standard and sample
Asked for: ΔHcomb of glucose
Strategy:
1. Calculate the value of qrxn for benzoic acid by multiplying the mass of benzoic acid by its ΔHcomb. Then use Equation $ref{5.5.9}$ to determine the heat capacity of the calorimeter (Cbomb) from qcomb and ΔT.
2. Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the ΔHcomb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose.
Solution:
The first step is to use Equation $\ref{5.5.9}$ and the information obtained from the combustion of benzoic acid to calculate Cbomb. We are given ΔT, and we can calculate qcomb from the mass of benzoic acid:
$q{comb} = \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) = - 15.3 \; kJ$
From Equation $\ref{5.5.9}$,
$-C{bomb} = \dfrac{q_{comb}}{\Delta T} = \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} =- 7.34 \; kJ/^{o}C$
B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose:
$q_{comb}=-C_{bomb}\Delta T = \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right )=- 26.7 \; kJ$
Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the ΔHcomb of glucose is
$\Delta H_{comb}=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right )=-2780 \; kJ/mol =2.78 \times 10^{3} \; kJ/mol$
This result is in good agreement (< 1% error) with the value of ΔHcomb = −2803 kJ/mol that calculated using enthalpies of formation.
Exercise $4$: Combustion of Benzoic Acid
When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH3NHNH2) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the ΔHcomb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle.
Answer
−1.30 × 103 kJ/mol | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.05%3A_Measuring_E_for_Chemical_Reactions-_Constant-Volume_Calorimetry.txt |
Learning Objectives
• To understand how enthalpy pertains to chemical reactions
We have stated that the change in energy ($ΔU$) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work, (or just $PV$ work). Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
$\ce{Cu(s) + 4HNO3(aq) \rightarrow Cu(NO3)2(aq) + 2H_2O(l) + 2NO2(g)} \nonumber$
If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure $1$). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure). We find the amount of $PV$ work done by multiplying the external pressure $P$ by the change in volume caused by movement of the piston ($ΔV$). At a constant external pressure (here, atmospheric pressure),
$w = −PΔV \label{5.4.2}$
The negative sign associated with $PV$ work done indicates that the system loses energy when the volume increases. If the volume increases at constant pressure ($ΔV > 0$), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases ($ΔV < 0$), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy.
The internal energy $U$ of a system is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy ($H$) (from the Greek enthalpein, meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy $U$ plus the product of its pressure $P$ and volume $V$:
$H =U + PV \label{5.4.3}$
Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. So we can define a change in enthalpy ($\Delta H$) accordingly
$ΔH = H_{final} − H_{initial} \nonumber$
If a chemical change occurs at constant pressure (i.e., for a given $P$, $ΔP = 0$), the change in enthalpy ($ΔH$) is
\begin{align} ΔH &= Δ(U + PV) \[4pt] &= ΔU + ΔPV \[4pt] &= ΔU + PΔV \label{5.4.4} \end{align}
Substituting $q + w$ for $ΔU$ (First Law of Thermodynamics) and $−w$ for $PΔV$ (Equation $\ref{5.4.2}$) into Equation $\ref{5.4.4}$, we obtain
\begin{align} ΔH &= ΔU + PΔV \[4pt] &= q_p + \cancel{w} −\cancel{w} \[4pt] &= q_p \label{5.4.5} \end{align}
The subscript $p$ is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation $\ref{5.4.5}$ we see that at constant pressure the change in enthalpy, $ΔH$ of the system, is equal to the heat gained or lost.
\begin{align} ΔH &= H_{final} − H_{initial} \[4pt] &= q_p \label{5.4.6} \end{align}
Just as with $ΔU$, because enthalpy is a state function, the magnitude of $ΔH$ depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure.
To find $ΔH$ for a reaction, measure $q_p$.
When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction ($ΔH_{rxn}$), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so $ΔH_{rxn}$ is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so $ΔH_{rxn}$ is positive. Thus:
• $ΔH_{rxn} < 0$ for an exothermic reaction, and
• $ΔH_{rxn} > 0$ for an endothermic reaction.
In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table:
sign conventions for heat flow and enthalpy changes
Reaction Type q ΔHrxn
exothermic < 0 < 0 (heat flows from a system to its surroundings)
endothermic > 0 > 0 (heat flows from the surroundings to a system)
If ΔHrxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (Figure $2a$). Conversely, if ΔHrxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (Figure $\PageIndex{2b}$). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion.
Bond breaking ALWAYS requires an input of energy; bond making ALWAYS releases energy.y.
• Reversing a reaction or a process changes the sign of ΔH. Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed):
$\begin{matrix} heat+ H_{2}O(s) \rightarrow H_{2}O(l) & \Delta H > 0 \end{matrix} \label{5.4.7}$
$\begin{matrix} H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0 \end{matrix} \label{5.4.8}$
In both cases, the magnitude of the enthalpy change is the same; only the sign is different.
• Enthalpy is an extensive property (like mass). The magnitude of $ΔH$ for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation $\ref{5.4.9}$, which describes the reaction of aluminum with iron(III) oxide (Fe2O3) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al2O3, and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe2O3 consumed:
$\ce{2Al(s) + Fe2O3(s) -> 2Fe (s) + Al2O3 (s) } + 815.5 \; kJ \label{5.4.9}$
Thus ΔH = −851.5 kJ/mol of Fe2O3. We can also describe ΔH for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for ΔH, in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation $\ref{5.4.10}$, it is the value of ΔH corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:
$\ce{ 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3 (s)} \quad \Delta H_{rxn}= - 851.5 \; kJ \label{5.4.10}$
If 4 mol of Al and 2 mol of $\ce{Fe2O3}$ react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:
$- \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{5.4.6a}$
The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example $1$.
Example $1$: Melting Icebergs
Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If $ΔH$ is 6.01 kJ/mol for the reaction at 0°C and constant pressure:
$\ce{H2O(s) → H_2O(l)} \nonumber$
How much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 106 metric tons)? (A metric ton is 1000 kg.)
Given: energy per mole of ice and mass of iceberg
Asked for: energy required to melt iceberg
Strategy:
1. Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 106 metric tons) of ice.
2. Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice.
Solution:
A Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given ΔH for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by ΔH (+6.01 kJ/mol):
\begin{align*} moles \; H_{2}O & = 1.00\times 10^{6} \; \cancel{\text{metric ton }} \ce{H2O} \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{\text{metric ton}}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right ) \[4pt] & = 5.55\times 10^{10} \; mol \,\ce{H2O} \end{align*} \nonumber
B The energy needed to melt the iceberg is thus
$\left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \nonumber$
Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown below.
Possible sources of the approximately $3.34 \times 10^{11}\, kJ$ needed to melt a $1.00 \times 10^6$ metric ton iceberg
• Combustion of 3.8 × 103 ft3 of natural gas
• Combustion of 68,000 barrels of oil
• Combustion of 15,000 tons of coal
• $1.1 \times 10^8$ kilowatt-hours of electricity
Alternatively, we can rely on ambient temperatures to slowly melt the iceberg. The main issue with this idea is the cost of dragging the iceberg to the desired place.
Exercise $1$: Thermite Reaction
If 17.3 g of powdered aluminum are allowed to react with excess $\ce{Fe2O3}$, how much heat is produced?
Answer
273 kJ
Enthalpies of Reaction
One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, since enthalpy is a state function, all we have to know is the initial and final states of the reaction. This allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following:
• Enthalpy of combustion (ΔHcomb) The change in enthalpy that occurs during a combustion reaction. Enthalpy changes have been measured for the combustion of virtually any substance that will burn in oxygen; these values are usually reported as the enthalpy of combustion per mole of substance.
• Enthalpy of fusion (ΔHfus) The enthalpy change that acompanies the melting (fusion) of 1 mol of a substance. The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for almost all the elements and for most simple compounds.
• Enthalpy of vaporization (ΔHvap) The enthalpy change that accompanies the vaporization of 1 mol of a substance. The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for most volatile compounds.
• Enthalpy of solution (ΔHsoln) The change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent. The enthalpy change when a specified amount of solute dissolves in a given quantity of solvent.
Table $1$: Enthalpies of Vaporization and Fusion for Selected Substances at Their Boiling Points and Melting Points
Substance ΔHvap (kJ/mol) ΔHfus (kJ/mol)
argon (Ar) 6.3 1.3
methane (CH4) 9.2 0.84
ethanol (CH3CH2OH) 39.3 7.6
benzene (C6H6) 31.0 10.9
water (H2O) 40.7 6.0
mercury (Hg) 59.0 2.29
iron (Fe) 340 14
The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
Enthalpy of Reaction: Enthalpy of Reaction, YouTube(opens in new window) [youtu.be]
Summary
For a chemical reaction, the enthalpy of reaction ($ΔH_{rxn}$) is the difference in enthalpy between products and reactants; the units of $ΔH_{rxn}$ are kilojoules per mole. Reversing a chemical reaction reverses the sign of $ΔH_{rxn}$. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.06%3A_Enthalpy-_The_Heat_Evolved_in_a_Chemical_Reaction_at_Constant_Pressure.txt |
Because ΔH is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter (a device used to measure enthalpy changes in chemical processes at constant pressure) give ΔH values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure $1$), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10−6 °C). Because the heat released or absorbed at constant pressure is equal to ΔH, the relationship between heat and $ΔH_{rxn}$ is
$\Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mC_{s} \Delta T \label{5.5.8}$
The use of a constant-pressure calorimeter is illustrated in Example $3$.
Example $1$
When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm3. What is ΔHsoln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.
Given: mass of substance, volume of solvent, and initial and final temperatures
Asked for: ΔHsoln
Strategy:
1. Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution.
2. Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation 5.5.8.
3. Use the molar mass of KOH to calculate ΔHsoln.
Solution:
A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
$\left (100.0 \; mL\; H2O \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g$
The temperature change is (34.7°C − 23.0°C) = +11.7°C.
B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus
$q_{calorimater}=mC_{s} \Delta T =\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right )=5130 \; J =5.13 \; lJ$
The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation $\ref{5.5.8}$, we see that
ΔHrxn = −qcalorimeter = −5.13 kJ
This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.
C The last step is to use the molar mass of KOH to calculate ΔHsoln—the heat released when dissolving 1 mol of KOH:
$\Delta H_{soln}= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right )=-57.2 \; kJ/mol$
Exercise $1$
A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example $1$, find ΔHsoln for NH4Br (in kilojoules per mole).
16.6 kJ/mol | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.07%3A_Measuring_H_for_Chemical_Reactions-_Constant-Pressure_Calorimetry.txt |
Learning Objectives
• To use Hess’s law and thermochemical cycles to calculate enthalpy changes of chemical reactions.
Because enthalpy is a state function, the enthalpy change for a reaction depends on only two things: (1) the masses of the reacting substances and (2) the physical states of the reactants and products. It does not depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This principle is called Hess’s law, after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known ΔH values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps.
Hess's Law argues that ΔH for the net reaction is the sum of the ΔH values for the individual reactions. This is nothing more than arguing that ΔH is a state function.
We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation $\ref{5.6.1}$ can be viewed as occurring in three distinct steps with known ΔH values. As shown in Figure 5.6.1, the first reaction produces 1 mol of solid aluminum oxide (Al2O3) and 2 mol of liquid iron at its melting point of 1758°C (part (a) in Equation $\ref{5.6.1}$); the enthalpy change for this reaction is −732.5 kJ/mol of Fe2O3. The second reaction is the conversion of 2 mol of liquid iron at 1758°C to 2 mol of solid iron at 1758°C (part (b) in Equation 5.6.1); the enthalpy change for this reaction is −13.8 kJ/mol of Fe (−27.6 kJ per 2 mol Fe). In the third reaction, 2 mol of solid iron at 1758°C is converted to 2 mol of solid iron at 25°C (part (c) in Equation $\ref{5.6.1}$); the enthalpy change for this reaction is −45.5 kJ/mol of Fe (−91.0 kJ per 2 mol Fe). As you can see in Figure 5.6.1, the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding parts (a), (b), and (c) in Equation 5.6.1 gives the overall reaction, shown in part (d):
\small \newcommand{\Celsius}{^{\circ}\text{C}} \begin{align*} \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> 2 Fe (l, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius)} & \Delta H = - 732.5\,\text{kJ} && \text{(a)} \ \ce{2 Fe (l, 1758 \Celsius) &-> 2 Fe (s, 1758 \Celsius)} & \Delta H = -\phantom{0}27.6\,\text{kJ} && \text{(b)} \ \ce{2 Fe (s, 1758 \Celsius) + Al2O3 (s, 1758 \Celsius) &-> 2 Fe (s, 25 \Celsius) + Al2O3 (s, 25 \Celsius) } & \Delta H = -\phantom{0}91.0\,\text{kJ} && \text{(c)} \[2ex] \hline \ce{2 Al (s, 25 \Celsius) + Fe2O3 (s, 25 \Celsius) &-> Al2O3 (s, 25 \Celsius) + 2 Fe (s, 25 \Celsius) } & \Delta H = -851.1\,\text{kJ} && \text{(d)} \ \end{align*} \label{5.6.1} \tag{5.6.1}
The net reaction in part (d) in Equation $\ref{5.6.1}$ is identical to the equation for the thermite reaction that we saw in a previous section. By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure $1$.
Comparing parts (a) and (d) in Equation $\ref{5.6.1}$ also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in Equation $\ref{5.6.1}$), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation $\ref{5.6.1}$). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation.
When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure:
1. Identify the equation whose ΔH value is unknown and write individual reactions with known ΔH values that, when added together, will give the desired equation. We illustrate how to use this procedure in Example $1$.
2. Arrange the chemical equations so that the reaction of interest is the sum of the individual reactions.
3. If a reaction must be reversed, change the sign of ΔH for that reaction. Additionally, if a reaction must be multiplied by a factor to obtain the correct number of moles of a substance, multiply its ΔH value by that same factor.
4. Add together the individual reactions and their corresponding ΔH values to obtain the reaction of interest and the unknown ΔH.
Example $1$
When carbon is burned with limited amounts of oxygen gas (O2), carbon monoxide (CO) is the main product:
$\left ( 1 \right ) \; \ce{2C (s) + O2 (g) -> 2 CO (g)} \quad \Delta H=-221.0 \; \text{kJ} \nonumber$
When carbon is burned in excess O2, carbon dioxide (CO2) is produced:
$\left ( 2 \right ) \; \ce{C (s) + O2 (g) -> CO2 (g)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber$
Use this information to calculate the enthalpy change per mole of CO for the reaction of CO with O2 to give CO2.
Given: two balanced chemical equations and their ΔH values
Asked for: enthalpy change for a third reaction
Strategy:
1. After balancing the chemical equation for the overall reaction, write two equations whose ΔH values are known and that, when added together, give the equation for the overall reaction. (Reverse the direction of one or more of the equations as necessary, making sure to also reverse the sign of ΔH.)
2. Multiply the equations by appropriate factors to ensure that they give the desired overall chemical equation when added together. To obtain the enthalpy change per mole of CO, write the resulting equations as a sum, along with the enthalpy change for each.
Solution:
A We begin by writing the balanced chemical equation for the reaction of interest:
$\left ( 3 \right ) \; \ce{CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H_{rxn}=? \nonumber$
There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that CO, a reactant in Equation 3, is a product in Equation 1, we can reverse Equation (1) to give
$\ce{2 CO (g) -> 2 C (s) + O2 (g)} \quad \Delta H=+221.0 \; \text{kJ} \nonumber$
Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use Equation 2 as written because its product, CO2, is the product we want in Equation 3:
$\ce{C (s) + O2 (g) -> CO2 (s)} \quad \Delta H=-393.5 \; \text{kJ} \nonumber$
B Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant:
$\ce{2 C (s) + 2 O2 (g) -> 2 CO2 (s)} \quad \Delta H=-787.0 \; \text{kJ} \nonumber$
Writing the resulting equations as a sum, along with the enthalpy change for each, gives
\begin{align*} \ce{2 CO (g) &-> \cancel{2 C(s)} + \cancel{O_2 (g)} } & \Delta H & = -\Delta H_1 = +221.0 \; \text{kJ} \ \ce{\cancel{2 C (s)} + \cancel{2} O2 (g) &-> 2 CO2 (g)} & \Delta H & = -2\Delta H_2 =-787.0 \; \text{kJ} \[2ex] \hline \ce{2 CO (g) + O2 (g) &-> 2 CO2 (g)} & \Delta H &=-566.0 \; \text{kJ} \end{align*} \nonumber
Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of CO with O2, and the problem asks for the amount per mole of CO. Consequently, we must divide both sides of the final equation and the magnitude of ΔH by 2:
$\ce{ CO (g) + 1/2 O2 (g) -> CO2 (g)} \quad \Delta H = -283.0 \; \text{kJ} \nonumber$
An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms:
\begin{alignat*}{3} \text{(A)} \quad && \ce{ 2 C (s) + O2 (g) &-> \cancel{2 CO (g)}} \qquad & \Delta H_A &= \Delta H_1 &&= + 221.0 \; \text{kJ} \ \text{(B)} \quad && \ce{ \cancel{2 CO (g)} + O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H_B && &= ? \ \text{(C)} \quad && \ce{2 C (s) + 2 O2 (g) &-> 2 CO2 (g)} \qquad & \Delta H &= 2 \Delta H_2 &= 2 \times \left ( -393.5 \; \text{kJ} \right ) &= -787.0 \; \text{kJ} \end{alignat*}
The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO2. From Hess’s law, ΔHA + ΔHB = ΔHC, and we are given ΔH for reactions A and C. Substituting the appropriate values gives
$\begin{matrix} -221.0 \; kJ + \Delta H_{B} = -787.0 \; kJ \ \Delta H_{B} = -566.0 \end{matrix} \nonumber$
This is again the enthalpy change for the conversion of 2 mol of CO to CO2. The enthalpy change for the conversion of 1 mol of CO to CO2 is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem.
Exercise $1$
The reaction of acetylene (C2H2) with hydrogen (H2) can produce either ethylene (C2H4) or ethane (C2H6):
$\begin{matrix} C_{2}H_{2}\left ( g \right ) + H_{2}\left ( g \right ) \rightarrow C_{2}H_{4}\left ( g \right ) & \Delta H = -175.7 \; kJ/mol \; C_{2}H_{2} \ C_{2}H_{2}\left ( g \right ) + 2H_{2}\left ( g \right ) \rightarrow C_{2}H_{6}\left ( g \right ) & \Delta H = -312.0 \; kJ/mol \; C_{2}H_{2} \end{matrix} \nonumber$
What is ΔH for the reaction of C2H4 with H2 to form C2H6?
Answer
−136.3 kJ/mol of C2H4
Hess’s Law: Hess's Law, YouTube(opens in new window) [youtu.be]
Summary
Hess's law is arguing the overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions. For a chemical reaction, the enthalpy of reaction (ΔHrxn) is the difference in enthalpy between products and reactants; the units of ΔHrxn are kilojoules per mole. Reversing a chemical reaction reverses the sign of ΔHrxn. The magnitude of ΔHrxn also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the enthalpy of fusion (ΔHfus) and the enthalpy of vaporization (ΔHvap), respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is Hess’s law. The enthalpy of combustion (ΔHcomb) is the enthalpy change that occurs when a substance is burned in excess oxygen. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.08%3A_Relationships_Involving__Hrxn.txt |
The heat changes which accompany a chemical reaction are caused largely by changes in the electronic energy of the molecules. If we restrict our attention to gases, and hence to fairly simple molecules, we can go quite a long way toward predicting whether a reaction will be exothermic by considering the bonds which are broken and made in the course of the reaction. In order to do this we must first become familiar with the idea of a bond enthalpy.
In other sections we point out that when a chemical bond forms, negative charges move closer to positive charges than before, and so there is a lowering of the energy of the molecule relative to the atoms from which it was made. This means that energy is required to break a molecule into its constituent atoms. The bond enthalpy DX–Y of a diatomic molecule X—Y is the enthalpy change for the (usually hypothetical) process:
$\ce{XY(g) \rightarrow X(g) + Y(g)} \nonumber$ $\Delta H^{o} (298 K) = D_{x-y} \nonumber$
We have already used the term bond energy to describe this quantity, though strictly speaking the bond energy is a measure of ΔU rather than ΔH. As we have already seen, ΔU and ΔH are nearly equal, and so either term may be used.
As an example, let us consider the bond enthalpy for carbon monoxide. It is possible to establish the thermochemical equation
$\ce{CO(g) \rightarrow C(g) + O(g)} \nonumber$
$\Delta H^{o}(298 K) = 1073 kJ mol^{-1}\label{3}$
Accordingly we can write
$\ce{C_{C\equiv O}= 1073 kJ mol^{-1}} \nonumber$
even though the process to which Eq. $\ref{3}$ corresponds is hypothetical: Neither carbon nor oxygen exists as a monatomic gas at 298 K. For triatomic and polyatomic molecules, the bond enthalpy is usually defined as a mean. In the case of water, for instance, we have
$\ce{H_{2}O(g) \rightarrow 2H(g) + O(g)} \nonumber$
$\Delta H^{o} (298 K) = 927.2 \text{kJ mol}^{-1} \nonumber$
Since it requires 927.2 kJ to break open two O—H bonds, we take half this value as the mean bond enthalpy and write
$D_{O-H} = 463.6 \text{kJ mol}^{-1} \nonumber$
In methanol, CH3OH,however, a value of 427 kJ mol–1 for the O—H bond enthalpy fits the experimental data better. In other words the strength of the O—H varies somewhat from compound to compound. Because of this fact, we must expect to obtain only approximate results, accurate only to about ± 50 kJ mol–1, from the use of bond enthalpies. Bond enthalpies for both single and multiple bonds are given in Table $1$.
TABLE $1$ Average Bond Energies/kJ mol–1.
As an example of how a table of bond enthalpies can he used to predict the ΔH value for a reaction, let us take the simple case
$\ce{H_{2}(g) + F_{2}(g) \rightarrow 2HF(g)}\label{9}$
298 K, 1 atm
We can regard this reaction as occurring in two stages (Figure $1$ ). In the first stage all the reactant molecules are broken up into atoms:
$\ce{H_{2}(g) + F_{2}(g) \rightarrow 2H(g) + F(g)}\label{10}$ 298 K, 1 atm
For this stage
$\Delta H_I = H_{H-H}+ D_{F-F}\label{11}$
since 1 mol H2 and 1 mol F2 have been dissociated.
In the second stage the H and F atoms are reconstituted to form HF molecules:
$\ce{2H(g) + 2F(g) \rightarrow 2HF(g)} \nonumber$ 298 K, 1 atm
For which
$\Delta H_{II} = – 2D_{H-F} \nonumber$
where a negative sign is necessary since this stage corresponds to the reverse of dissociation.
Since Eq. $\ref{9}$ corresponds to the sum of Equations $\ref{10}$ and $\ref{11}$, Hess's law allows us to add ΔH values:
$\Delta H^{o}_{reaction} = \Delta H_{I} + \Delta H_{II} \nonumber$
$= D_{H-H} + D_{F-F} – 2D{H-F}$
$= (436 + 159 – 2 * 566) \text{kJ mol}^{-1}$
$= –539 \text{kJ mol}^{-1}$
We can work this same trick of subdividing a reaction into a bond-breaking stage followed by a bond-making stage for the general case of any gaseous reaction. In the first stage all the bonds joining the atoms in the reactant molecules are broken and a set of gaseous atoms results. For this stage
$\Delta H_{I} = \sum_{\text{bonds broken}} D \nonumber$
The enthalpy change is the sum of the bond enthalpies for all bonds broken. In the second stage these gaseous atoms are reconstituted into the product molecules. For this second stage therefore
$\Delta H_{II} = – \sum_{\text{bonds formed}} D \nonumber$
where the negative sign is necessary because the reverse of bond breaking is occurring in this stage. The total enthalpy change for the reaction at standard pressure is thus
$\Delta H^{o} = \Delta H_{I} + \Delta H_{II} \nonumber$
or
$\Delta H^{o} = \sum D \text{(bond broken)} – \sum D \text{(bond formed)} \nonumber$
The use of this equation is illustrated in the next example.
Example $1$: Enthalpy Change
Using Table $1$ calculate the value of ΔH°(298 K) for the reaction
$\ce{CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)} \nonumber$
Solution
It is best to sketch the molecules and their bonds in order to make sure that none are missed.
Thus $\Delta H^{o} = \sum D \text{(bond broken)} – \sum D \text{(bond formed)}$
$= 4 D_{C\bond{-}H} + 2 D_{D\bond{=}D} - 2 D_{C\bond{=}O} - 4 D_{O\bond{-}H}$
$= (4 * 416 + 2 * 498 – 2 * 803 – 4 * 467) \text{kJ mol}^{-1}$
$= – 814 \text{kJ mol}^{-1}$
The experimental value for this enthalpy change can be calculated from standard enthalpies of formation. It is –802.4 kJ mol–1. The discrepancy is due to the unavoidable use of mean bond enthalpies in the calculation. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.09%3A_Determining_Enthalpies_of_Reaction_from_Bond_Energies.txt |
Learning Objectives
• To understand Enthalpies of Formation and be able to use them to calculate Enthalpies of Reaction
One way to report the heat absorbed or released by chemical reactions would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, starting from the elemental forms of each atom at 25 oC and 1 atm pressure.
Enthalpy of formation ($ΔH_f$) is the enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The formation of any chemical can be as a reaction from the corresponding elements:
$\text{elements} \rightarrow \text{compound} \nonumber$
which in terms of the the Enthalpy of formation becomes
$\Delta H_{rxn} = \Delta H_{f} \label{7.8.1}$
For example, consider the combustion of carbon:
$\ce{ C(s) + O2 (g) -> CO2 (g)} \nonumber$
then
$\Delta H_{rxn} = \Delta H_{f}\left [CO_{2}\left ( g \right ) \right ] \nonumber$
The sign convention for ΔHf is the same as for any enthalpy change: $ΔH_f < 0$ if heat is released when elements combine to form a compound and $ΔH_f > 0$ if heat is absorbed.
The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system.
Standard Enthalpies of Formation
The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane.
The standard conditions for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state, which is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ($ΔH^o_f$) The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition.
The standard enthalpy of formation of any element in its standard state is zero by definition.
For example, although oxygen can exist as ozone (O3), atomic oxygen (O), and molecular oxygen (O2), O2 is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H2(g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure $1$). Therefore, $\ce{O2(g)}$, $\ce{H2(g)}$, and graphite have $ΔH^o_f$ values of zero.
The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction:
$6C\left (s, graphite \right ) + 6H_{2}\left (g \right ) + 3O_{2}\left (g \right ) \rightarrow C_{6}H_{12}O_{6}\left (s \right )\; \; \; \Delta H_{f}^{o} = - 1273.3 \; kJ \label{7.8.2}$
It is not possible to measure the value of $ΔH^oo_f$ for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, $\ce{O2}$, and $\ce{H2}$ and measuring the heat evolved as glucose is formed since the reaction shown in Equation $\ref{7.8.2}$ does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of $ΔH^oo_f$ are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of $ΔH^o_f$ for an extensive list of compounds are given in Table T1. Note that $ΔH^o_f$ values are always reported in kilojoules per mole of the substance of interest. Also notice in Table T1 that the standard enthalpy of formation of O2(g) is zero because it is the most stable form of oxygen in its standard state.
Example $1$: Enthalpy of Formation
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
1. $\ce{HCl(g)}$
2. $\ce{MgCO3(s)}$
3. $\ce{CH3(CH2)14CO2H(s)}$ (palmitic acid)
Given:
compound formula and phase.
Asked for:
balanced chemical equation for its formation from elements in standard states
Strategy:
Use Table T1 to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made.
Solution:
To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in Table T1: by a $ΔH^o_f$ value of 0 kJ/mol.
Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are $\ce{H2(g)}$ and $\ce{Cl2(g)}$, respectively, the unbalanced chemical equation is
$\ce{H2(g) + Cl2(g) \rightarrow HCl(g)} \nonumber$
Fractional coefficients are required in this case because ΔHof values are reported for 1 mol of the product, $\ce{HCl}$. Multiplying both $\ce{H2(g)}$ and $\ce{Cl2(g)}$ by 1/2 balances the equation:
$\ce{1/2 H_{2} (g) + 1/2 Cl_{2} (g) \rightarrow HCl (g)} \nonumber$
The standard states of the elements in this compound are $\ce{Mg(s)}$, $\ce{C(s, graphite)}$, and $\ce{O2(g)}$. The unbalanced chemical equation is thus
$\ce{Mg(s) + C (s, graphite) + O2 (g) \rightarrow MgCO3 (s)} \nonumber$
This equation can be balanced by inspection to give
$\ce{Mg (s) + C (s, graphite ) + 3/2 O2 (g)\rightarrow MgCO3 (s)} \nonumber$
Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows:
$\ce{C(s, graphite) + H2(g) + O2(g) \rightarrow CH3(CH2)14CO2H(s)} \nonumber$
There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is
$\ce{16C (s, graphite) + 16 H2(g) + O2(g) -> CH3(CH2)14CO2H(s) } \nonumber$
Exercise $1$
For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound.
1. $\ce{NaCl(s)}$
2. $\ce{H2SO4(l)}$
3. $\ce{CH3CO2H(l)}$ (acetic acid)
Answer a
$\ce{ Na (s) + 1/2 Cl2 (g) \rightarrow NaCl (s)} \nonumber$
Answer b
$\ce{H_{2} (g) + 1/8 S8 (s) + 2O2 ( g) \rightarrow H2 SO4( l) } \nonumber$
Answer c
$\ce{2C(s) + O2(g) + 2H2(g) -> CH3CO2H(l)} \nonumber$
Definition of Heat of Formation Reactions: https://youtu.be/A20k0CK4doI
Standard Enthalpies of Reaction
Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for any reaction involving substances whose $\Delta{H_f^o}$ values are known. The standard enthalpy of reaction $\Delta{H_{rxn}^o}$ is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction
$aA + bB \rightarrow cC + dD \label{7.8.3}$
where $A$, $B$, $C$, and $D$ are chemical substances and $a$, $b$, $c$, and $d$ are their stoichiometric coefficients. The magnitude of $ΔH^ο$ is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients:
$\Delta H_{rxn}^{o} = \underbrace{ \left [c\Delta H_{f}^{o}\left ( C \right ) + d\Delta H_{f}^{o}\left ( D \right ) \right ] }_{\text{products} } - \underbrace{ \left [a\Delta H_{f}^{o}\left ( A \right ) + b\Delta H_{f}^{o}\left ( B \right ) \right ]}_{\text{reactants }} \label{7.8.4}$
More generally, we can write
$\Delta H_{rxn}^{o} = \sum m\Delta H_{f}^{o}\left ( products \right ) - \sum n\Delta H_{f}^{o}\left ( reactants \right ) \label{7.8.5}$
where the symbol $\sum$ means “sum of” and $m$ and $n$ are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation $\ref{7.8.5}$ arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters.
"Products minus reactants" summations are typical of state functions.
To demonstrate the use of tabulated ΔHο values, we will use them to calculate $ΔH_{rxn}$ for the combustion of glucose, the reaction that provides energy for your brain:
$\ce{ C6H12O6 (s) + 6O2 (g) \rightarrow 6CO2 (g) + 6H2O (l)} \label{7.8.6}$
Using Equation $\ref{7.8.5}$, we write
$\Delta H_{f}^{o} =\left \{ 6\Delta H_{f}^{o}\left [ CO_{2}\left ( g \right ) \right ] + 6\Delta H_{f}^{o}\left [ H_{2}O\left ( g \right ) \right ] \right \} - \left \{ \Delta H_{f}^{o}\left [ C_{6}H_{12}O_{6}\left ( s \right ) \right ] + 6\Delta H_{f}^{o}\left [ O_{2}\left ( g \right ) \right ] \right \} \label{7.8.7}$
From Table T1, the relevant ΔHοf values are ΔHοf [CO2(g)] = -393.5 kJ/mol, ΔHοf [H2O(l)] = -285.8 kJ/mol, and ΔHοf [C6H12O6(s)] = -1273.3 kJ/mol. Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol. Inserting these values into Equation $\ref{7.8.7}$ and changing the subscript to indicate that this is a combustion reaction, we obtain
\begin{align} \Delta H_{comb}^{o} &= \left [ 6\left ( -393.5 \; kJ/mol \right ) + 6 \left ( -285.8 \; kJ/mol \right ) \right ] - \left [-1273.3 + 6\left ( 0 \; kJ\;mol \right ) \right ] \label{7.8.8} \[4pt] &= -2802.5 \; kJ/mol \end{align}
As illustrated in Figure $2$, we can use Equation $\ref{7.8.8}$ to calculate $ΔH^ο_f$ for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled $ΔH^ο_{comb}$. The alternative hypothetical pathway consists of four separate reactions that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the $ΔH^ο_f$ values of the reactants. Consequently, the enthalpy changes are
\begin{align} \Delta H_{1}^{o} &= \Delta H_{f}^{o} \left [ glucose \left ( s \right ) \right ] \nonumber \[4pt] &= -1 \; \cancel{mol \; glucose}\left ( \dfrac{1273.3 \; kJ}{1 \; \cancel{mol \; glucose}} \right ) \nonumber \[4pt] &= +1273.3 \; kJ \nonumber \[4pt] \Delta H_{2}^{o} &= 6 \Delta H_{f}^{o} \left [ O_{2} \left ( g \right ) \right ] \nonumber \[4pt] & =6 \; \cancel{mol \; O_{2}}\left ( \dfrac{0 \; kJ}{1 \; \cancel{mol \; O_{2}}} \right ) \nonumber \[4pt] &= 0 \; kJ \end{align} \label{7.8.9}
Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change (Equation \ref{7.8.4} since the products are now reactants and vice versa.
The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O2) to the elements is therefore +1273.3 kJ.
The reactions that convert the elements to final products (downward purple arrows in Figure $2$) are identical to those used to define the ΔHοf values of the products. Consequently, the enthalpy changes (from Table T1) are
$\begin{matrix} \Delta H_{3}^{o} = \Delta H_{f}^{o} \left [ CO_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; CO_{2}}\left ( \dfrac{393.5 \; kJ}{1 \; \cancel{mol \; CO_{2}}} \right ) = -2361.0 \; kJ \ \Delta H_{4}^{o} = 6 \Delta H_{f}^{o} \left [ H_{2}O \left ( l \right ) \right ] = 6 \; \cancel{mol \; H_{2}O}\left ( \dfrac{-285.8 \; kJ}{1 \; \cancel{mol \; H_{2}O}} \right ) = -1714.8 \; kJ \end{matrix}$
The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using any pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ):
$\Delta H_{comb}^{o} = +1273.3 \; kJ +\left ( -4075.8 \; kJ \right ) = -2802.5 \; kJ \label{7.8.10}$
This is the same result we obtained using the “products minus reactants” rule (Equation $\ref{7.8.5}$) and ΔHοf values. The two results must be the same because Equation $\ref{7.8.10}$ is just a more compact way of describing the thermochemical cycle shown in Figure $1$.
Example $2$: Heat of Combustion
Long-chain fatty acids such as palmitic acid ($\ce{CH3(CH2)14CO2H}$) are one of the two major sources of energy in our diet ($ΔH^o_f$ =−891.5 kJ/mol). Use the data in Table T1 to calculate ΔHοcomb for the combustion of palmitic acid. Based on the energy released in combustion per gram, which is the better fuel — glucose or palmitic acid?
Given: compound and $ΔH^ο_{f}$ values
Asked for: $ΔH^ο_{comb}$ per mole and per gram
Strategy:
1. After writing the balanced chemical equation for the reaction, use Equation $\ref{7.8.5}$ and the values from Table T1 to calculate $ΔH^ο_{comb}$ the energy released by the combustion of 1 mol of palmitic acid.
2. Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in Equation $\ref{7.8.8}$ for the combustion of glucose to determine which is the better fuel.
Solution:
A To determine the energy released by the combustion of palmitic acid, we need to calculate its $ΔH^ο_f$. As always, the first requirement is a balanced chemical equation:
$C_{16}H_{32}O_{2(s)} + 23O_{2(g)} \rightarrow 16CO_{2(g)} + 16H_2O_{(l)} \nonumber$
Using Equation $\ref{7.8.5}$ (“products minus reactants”) with ΔHοf values from Table T1 (and omitting the physical states of the reactants and products to save space) gives
\begin{align*} \Delta H_{comb}^{o} &= \sum m \Delta H^o_f\left( {products} \right) - \sum n \Delta H^o_f \left( {reactants} \right) \[4pt] &= \left [ 16\left ( -393.5 \; kJ/mol \; CO_{2} \right ) + 16\left ( -285.8 \; kJ/mol \; H_{2}O \; \right ) \right ] \[4pt] & - \left [ -891.5 \; kJ/mol \; C_{16}H_{32}O_{2} + 23\left ( 0 \; kJ/mol \; O_{2} \; \right ) \right ] \[4pt] &= -9977.3 \; kJ/mol \nonumber \end{align*}
This is the energy released by the combustion of 1 mol of palmitic acid.
B The energy released by the combustion of 1 g of palmitic acid is
$\Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{9977.3 \; kJ}{\cancel{1 \; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{256.42 \; g} \right )= -38.910 \; kJ/g \nonumber$
As calculated in Equation $\ref{7.8.8}$, $ΔH^o_f$ of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore
$\Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{-2802.5 \; kJ}{\cancel{1\; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{180.16\; g} \right ) = -15.556 \; kJ/g \nonumber$
The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight.
Exercise $2$: Water–gas shift reaction
Use Table T1 to calculate $ΔH^o_{rxn}$ for the water–gas shift reaction, which is used industrially on an enormous scale to obtain H2(g):
$\ce{ CO ( g ) + H2O (g ) -> CO2 (g) + H2 ( g )} \nonumber$
Answer
−41.2 kJ/mol
We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s $ΔH^ο_f$ which we cannot obtain otherwise. This procedure is illustrated in Example $3$.
Example $3$: Tetraethyllead
Beginning in 1923, tetraethyllead [$\ce{(C2H5)4Pb}$] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are $\ce{CO2(g)}$, $\ce{H2O(l)}$, and red $\ce{PbO(s)}$. What is the standard enthalpy of formation of tetraethyllead, given that $ΔH^ο_f$ is −19.29 kJ/g for the combustion of tetraethyllead and $ΔH^ο_f$ of red PbO(s) is −219.0 kJ/mol?
Given: reactant, products, and $ΔH^ο_{comb}$ values
Asked for: $ΔH^ο_f$ of the reactants
Strategy:
1. Write the balanced chemical equation for the combustion of tetraethyl lead. Then insert the appropriate quantities into Equation $\ref{7.8.5}$ to get the equation for ΔHοf of tetraethyl lead.
2. Convert $ΔH^ο_{comb}$ per gram given in the problem to $ΔH^ο_{comb}$ per mole by multiplying $ΔH^ο_{comb}$ per gram by the molar mass of tetraethyllead.
3. Use Table T1 to obtain values of $ΔH^ο_f$ for the other reactants and products. Insert these values into the equation for $ΔH^ο_f$ of tetraethyl lead and solve the equation.
Solution:
A The balanced chemical equation for the combustion reaction is as follows:
$\ce{2(C2H5)4Pb(l) + 27O2(g) → 2PbO(s) + 16CO2(g) + 20H2O(l)} \nonumber$
Using Equation $\ref{7.8.5}$ gives
$\Delta H_{comb}^{o} = \left [ 2 \Delta H_{f}^{o}\left ( PbO \right ) + 16 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 20 \Delta H_{f}^{o}\left ( H_{2}O \right )\right ] - \left [2 \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) + 27 \Delta H_{f}^{o}\left ( O_{2} \right ) \right ] \nonumber$
Solving for $ΔH^o_f [\ce{(C2H5)4Pb}]$ gives
$\Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) = \Delta H_{f}^{o}\left ( PbO \right ) + 8 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 10 \Delta H_{f}^{o}\left ( H_{2}O \right ) - \dfrac{27}{2} \Delta H_{f}^{o}\left ( O_{2} \right ) - \dfrac{\Delta H_{comb}^{o}}{2} \nonumber$
The values of all terms other than $ΔH^o_f [\ce{(C2H5)4Pb}]$ are given in Table T1.
B The magnitude of $ΔH^o_{comb}$ is given in the problem in kilojoules per gram of tetraethyl lead. We must therefore multiply this value by the molar mass of tetraethyl lead (323.44 g/mol) to get $ΔH^o_{comb}$ for 1 mol of tetraethyl lead:
\begin{align*} \Delta H_{comb}^{o} &= \left ( \dfrac{-19.29 \; kJ}{\cancel{g}} \right )\left ( \dfrac{323.44 \; \cancel{g}}{mol} \right ) \[4pt] &= -6329 \; kJ/mol \end{align*}
Because the balanced chemical equation contains 2 mol of tetraethyllead, $ΔH^o_{rxn}$ is
\begin{align*} \Delta H_{rxn}^{o} &= 2 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb} \left ( \dfrac{-6329 \; kJ}{1 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb }} \right ) \[4pt] &= -12,480 \; kJ \end{align*}
C Inserting the appropriate values into the equation for $ΔH^o_f [\ce{(C2H5)4Pb}]$ gives
\begin{align*} \Delta H_{f}^{o} \left [ \left (C_{2}H_{4} \right )_{4}Pb \right ] & = \left [1 \; mol \;PbO \;\times 219.0 \;kJ/mol \right ]+\left [8 \; mol \;CO_{2} \times \left (-393.5 \; kJ/mol \right )\right ] +\left [10 \; mol \; H_{2}O \times \left ( -285.8 \; kJ/mol \right )\right ] + \left [-27/2 \; mol \; O_{2}) \times 0 \; kJ/mol \; O_{2}\right ] \left [12,480.2 \; kJ/mol \; \left ( C_{2}H_{5} \right )_{4}Pb \right ]\[4pt] &= -219.0 \; kJ -3148 \; kJ - 2858 kJ - 0 kJ + 6240 \; kJ = 15 kJ/mol \end{align*}
Exercise $3$
Ammonium sulfate, $\ce{(NH4)2SO4}$, is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid:
$\ce{2NH3(g) + H2SO4(aq) \rightarrow (NH4)2SO4(s)} \nonumber$
The value of $ΔH^o_{rxn}$ is -179.4 kJ/mole $\ce{H2SO4}$. Use the data in Table T1 to calculate the standard enthalpy of formation of ammonium sulfate (in kilojoules per mole).
Answer
−1181 kJ/mol
Calculating DH° using DHf°: https://youtu.be/Y3aJJno9W2c
Summary
• The standard state for measuring and reporting enthalpies of formation or reaction is 25 oC and 1 atm.
• The elemental form of each atom is that with the lowest enthalpy in the standard state.
• The standard state heat of formation for the elemental form of each atom is zero.
The enthalpy of formation ($ΔH_{f}$) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation ($ΔH^o_{f}$) are determined under standard conditions: a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their standard states (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The standard enthalpy of reaction ($ΔH^o_{rxn}$) can be calculated from the sum of the standard enthalpies of formation of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The enthalpy of solution ($ΔH_{soln}$) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.10%3A_Determining_Enthalpies_of_Reaction_from_Standard_Enthalpies_of_Formation.txt |
Discussion Questions
• How is lattice energy estimated using Born-Haber cycle?
• How is lattice energy related to crystal structure?
The Lattice energy, $U$, is the amount of energy required to separate a mole of the solid (s) into a gas (g) of its ions.
$\ce{M_{a} L_{b} (s) \rightarrow a M^{b+} (g) + b X^{a-} (g) } \label{eq1}$
This quantity cannot be experimentally determined directly, but it can be estimated using a Hess Law approach in the form of Born-Haber cycle. It can also be calculated from the electrostatic consideration of its crystal structure. As defined in Equation \ref{eq1}, the lattice energy is positive, because energy is always required to separate the ions. For the reverse process of Equation \ref{eq1}:
$\ce{ a M^{b+} (g) + b X^{a-} (g) \rightarrow M_{a}L_{b}(s) }$
the energy released is called energy of crystallization ($E_{cryst}$). Therefore,
$U_{lattice} = - E_{cryst}$
Values of lattice energies for various solids have been given in literature, especially for some common solids. Some are given here.
Table $1$: Comparison of Lattice Energies (U in kJ/mol) of Some Salts
Solid U Solid U Solid U Solid U
LiF 1036 LiCl 853 LiBr 807 LiI 757
NaF 923 NaCl 786 NaBr 747 NaI 704
KF 821 KCl 715 KBr 682 KI 649
MgF2 2957 MgCl2 2526 MgBr2 2440 MgI2 2327
The following trends are obvious at a glance of the data in Table $1$:
• As the ionic radii of either the cation or anion increase, the lattice energies decrease.
• The solids consists of divalent ions have much larger lattice energies than solids with monovalent ions.
How is lattice energy estimated using Born-Haber cycle?
Estimating lattice energy using the Born-Haber cycle has been discussed in Ionic Solids. For a quick review, the following is an example that illustrate the estimate of the energy of crystallization of NaCl.
Hsub of Na = 108 kJ/mol (Heat of sublimation)
D of Cl2 = 244 (Bond dissociation energy)
IP of Na(g) = 496 (Ionization potential or energy)
EA of Cl(g) = -349 (Electron affinity of Cl)
Hf of NaCl = -411 (Enthalpy of formation)
The Born-Haber cycle to evaluate Elattice is shown below:
-----------Na+ + Cl(g)--------
|
| |-349
|496+244/2 ¯
| Na+(g) + Cl-(g)
| |
Na(g) + 0.5Cl2(g) |
|
|108 |
| |Ecryst= -788
Na(s) + 0.5Cl2(l) |
| |
|-411 |
¯ ¯
-------------- NaCl(s) --------------
Ecryst = -411-(108+496+244/2)-(-349) kJ/mol
= -788 kJ/mol.
Discussion
The value calculated for U depends on the data used. Data from various sources differ slightly, and so is the result. The lattice energies for NaCl most often quoted in other texts is about 765 kJ/mol.
Compare with the method shown below
Na(s) + 0.5 Cl2(l) ® NaCl(s) - 411 Hf
Na(g) ® Na(s) - 108 -Hsub
Na+(g) + e ® Na(g) - 496 -IP
Cl(g) ® 0.5 Cl2(g) - 0.5 * 244 -0.5*D
Cl-(g) ® Cl(g) + 2 e 349 -EA
Add all the above equations leading to
Na+(g) + Cl-(g) ® NaCl(s) -788 kJ/mol = Ecryst
Lattice Energy is Related to Crystal Structure
There are many other factors to be considered such as covalent character and electron-electron interactions in ionic solids. But for simplicity, let us consider the ionic solids as a collection of positive and negative ions. In this simple view, appropriate number of cations and anions come together to form a solid. The positive ions experience both attraction and repulson from ions of opposite charge and ions of the same charge.
As an example, let us consider the the NaCl crystal. In the following discussion, assume r be the distance between Na+ and Cl- ions. The nearest neighbors of Na+ are 6 Cl- ions at a distance 1r, 12 Na+ ions at a distance 2r, 8 Cl- at 3r, 6 Na+ at 4r, 24 Na+ at 5r, and so on. Thus, the energy due to one ion is
$E = \dfrac{Z^2e^2}{4\pi\epsilon_or} M \label{6.13.1}$
The Madelung constant, $M$, is a poorly converging series of interaction energies:
$M= \dfrac{6}{1} - \dfrac{12}{2} + \dfrac{8}{3} - \dfrac{6}{4} + \dfrac{24}{5} ... \label{6.13.2}$
with
• $Z$ is the number of charges of the ions, (e.g., 1 for NaCl),
• $e$ is the charge of an electron ($1.6022 \times 10^{-19}\; C$),
• $4\pi \epsilon_o$ is 1.11265x10-10 C2/(J m).
The above discussion is valid only for the sodium chloride (also called rock salt) structure type. This is a geometrical factor, depending on the arrangement of ions in the solid. The Madelung constant depends on the structure type, and its values for several structural types are given in Table 6.13.1.
A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.
Table $2$: Madelung Constants
Compound
Crystal Lattice
M
A : C Type
NaCl NaCl 1.74756 6 : 6 Rock salt
CsCl CsCl 1.76267 6 : 6 CsCl type
CaF2 Cubic 2.51939 8 : 4 Fluorite
CdCl2 Hexagonal 2.244
MgF2 Tetragonal 2.381
ZnS (wurtzite) Hexagonal 1.64132
TiO2 (rutile) Tetragonal 2.408 6 : 3 Rutile
bSiO2 Hexagonal 2.2197
Al2O3 Rhombohedral 4.1719 6 : 4 Corundum
A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.
Madelung constants for a few more types of crystal structures are available from the Handbook Menu. There are other factors to consider for the evaluation of energy of crystallization, and the treatment by M. Born led to the formula for the evaluation of crystallization energy $E_{cryst}$, for a mole of crystalline solid.
$E_{cryst} = \dfrac{N Z^2e^2}{4\pi \epsilon_o r} \left( 1 - \dfrac{1}{n} \right)\label{6.13.3a}$
where N is the Avogadro's number (6.022x10-23), and n is a number related to the electronic configurations of the ions involved. The n values and the electronic configurations (e.c.) of the corresponding inert gases are given below:
n = 5 7 9 10 12
e.c. He Ne Ar Kr Xe
The following values of n have been suggested for some common solids:
n = 5.9 8.0 8.7 9.1 9.5
e.c. LiF LiCl LiBr NaCl NaBr
Example $1$
Estimate the energy of crystallization for $\ce{NaCl}$.
Solution
Using the values giving in the discussion above, the estimation is given by Equation \ref{6.13.3a}:
\begin{align*} E_cryst &= \dfrac{(6.022 \times 10^{23} /mol (1.6022 \times 10 ^{-19})^2 (1.747558)}{ 4\pi \, (8.854 \times 10^{-12} C^2/m ) (282 \times 10^{-12}\; m} \left( 1 - \dfrac{1}{9.1} \right) \[4pt] &= - 766 kJ/mol \end{align*}
Discussion
Much more should be considered in order to evaluate the lattice energy accurately, but the above calculation leads you to a good start. When methods to evaluate the energy of crystallization or lattice energy lead to reliable values, these values can be used in the Born-Haber cycle to evaluate other chemical properties, for example the electron affinity, which is really difficult to determine directly by experiment.
Exercise $1$
Which one of the following has the largest lattice energy? LiF, NaF, CaF2, AlF3
Answer
Skill: Explain the trend of lattice energy.
Exercise $2$
Which one of the following has the largest lattice energy? LiCl, NaCl, CaCl2, Al2O3
Answer
Corrundum Al2O3 has some covalent character in the solid as well as the higher charge of the ions.
Exercise $3$
Lime, CaO, is know to have the same structure as NaCl and the edge length of the unit cell for CaO is 481 pm. Thus, Ca-O distance is 241 pm. Evaluate the energy of crystallization, Ecryst for CaO.
Answer
Energy of crystallization is -3527 kJ/mol
Skill: Evaluate the lattice energy and know what values are needed.
Exercise $4$
Assume the interionic distance for NaCl2 to be the same as those of NaCl (r = 282 pm), and assume the structure to be of the fluorite type (M = 2.512). Evaluate the energy of crystallization, Ecryst .
Answer
-515 kJ/mol
Discussion: This number has not been checked. If you get a different value, please let me know. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/10%3A_Thermochemistry/10.11%3A_Lattice_Energy.txt |
Learning Objectives
• Define the property of pressure
• Define and convert among the units of pressure measurements
• Describe the operation of common tools for measuring gas pressure
• Calculate pressure from manometer data
The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure $1$). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.
Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.
A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.
Pressure is defined as the force exerted on a given area:
$P=\dfrac{F}{A} \label{9.2.1}$
Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either increasing the amount of force or by decreasing the area over which it is applied. Correspondingly, pressure can be decreased by either decreasing the force or increasing the area.
Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure $2$.—the elephant or the figure skater?
A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in2), so the pressure exerted by each foot is about 14 lb/in2:
$\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2}$
The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in2, so the pressure exerted by each blade is about 30 lb/in2:
$\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3}$
Even though the elephant is more than one hundred times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall through thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted:
$\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4}$
The SI unit of pressure is the pascal (Pa), with 1 Pa = 1 N/m2, where N is the newton, a unit of force defined as 1 kg m/s2. One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch—pounds per square inch (psi)—for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table $1$ provides some information on these and a few other common units for pressure measurements
Table $1$: Pressure Units
Unit Name and Abbreviation Definition or Relation to Other Unit Comment
pascal (Pa) 1 Pa = 1 N/m2 recommended IUPAC unit
kilopascal (kPa) 1 kPa = 1000 Pa
pounds per square inch (psi) air pressure at sea level is ~14.7 psi
atmosphere (atm) 1 atm = 101,325 Pa air pressure at sea level is ~1 atm
bar (bar, or b) 1 bar = 100,000 Pa (exactly) commonly used in meteorology
millibar (mbar, or mb) 1000 mbar = 1 bar
inches of mercury (in. Hg) 1 in. Hg = 3386 Pa used by aviation industry, also some weather reports
torr $\mathrm{1\: torr=\dfrac{1}{760}\:atm}$ named after Evangelista Torricelli, inventor of the barometer
millimeters of mercury (mm Hg) 1 mm Hg ~1 torr
Example $1$: Conversion of Pressure Units
The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:
1. torr
2. atm
3. kPa
4. mbar
Solution
This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1.
1. $\mathrm{29.2\cancel{in\: Hg}×\dfrac{25.4\cancel{mm}}{1\cancel{in}} ×\dfrac{1\: torr}{1\cancel{mm\: Hg}} =742\: torr}$
2. $\mathrm{742\cancel{torr}×\dfrac{1\: atm}{760\cancel{torr}}=0.976\: atm}$
3. $\mathrm{742\cancel{torr}×\dfrac{101.325\: kPa}{760\cancel{torr}}=98.9\: kPa}$
4. $\mathrm{98.9\cancel{kPa}×\dfrac{1000\cancel{Pa}}{1\cancel{kPa}} \times \dfrac{1\cancel{bar}}{100,000\cancel{Pa}} \times\dfrac{1000\: mbar}{1\cancel{bar}}=989\: mbar}$
Exercise $1$
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?
Answer
0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar
We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure $3$). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere.
If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be $\dfrac{1}{13.6}$ as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, p:
$p=hρg \label{9.2.5}$
where
• $h$ is the height of the fluid,
• $ρ$ is the density of the fluid, and
• $g$ is acceleration due to gravity.
Example $2$: Calculation of Barometric Pressure
Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = $13.6 \,g/cm^3$.
Solution
The hydrostatic pressure is given by Equation \ref{9.2.5}, with $h = 760 \,mm$, $ρ = 13.6\, g/cm^3$, and $g = 9.81 \,m/s^2$. Plugging these values into the Equation \ref{9.2.5} and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:)
$\mathrm{101,325\:\mathit{N}/m^2=101,325\:\dfrac{kg·m/s^2}{m^2}=101,325\:\dfrac{kg}{m·s^2}} \nonumber$
\begin {align*} p&\mathrm{=\left(760\: mm×\dfrac{1\: m}{1000\: mm}\right)×\left(\dfrac{13.6\: g}{1\:cm^3}×\dfrac{1\: kg}{1000\: g}×\dfrac{( 100\: cm )^3}{( 1\: m )^3}\right)×\left(\dfrac{9.81\: m}{1\:s^2}\right)}\[4pt] &\mathrm{=(0.760\: m)(13,600\:kg/m^3)(9.81\:m/s^2)=1.01 \times 10^5\:kg/ms^2=1.01×10^5\mathit{N}/m^2} \[4pt] & \mathrm{=1.01×10^5\:Pa} \end {align*} \nonumber
Exercise $2$
Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm3.
Answer
10.3 m
A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube (h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure $3$) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.
Example $3$: Calculation of Pressure Using an Open-End Manometer
The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below. Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
Solution
The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)
1. In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg
2. $\mathrm{897\cancel{mm Hg}×\dfrac{1\: atm}{760\cancel{mm Hg}}=1.18\: atm}$
3. $\mathrm{1.18\cancel{atm}×\dfrac{101.325\: kPa}{1\cancel{atm}}=1.20×10^2\:kPa}$
Exercise $3$
The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown below Determine the pressure of the gas in:
1. mm Hg
2. atm
3. kPa
642 mm Hg
Answer b
0.845 atm
Answer c
85.6 kPa
Application: Measuring Blood Pressure
Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure $5$). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure—the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure—the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg).
Meteorology, Climatology, and Atmospheric Science
Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure $5$) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide.
In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events.
The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure $7$: the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease.
Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere.
Summary
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
Key Equations
• $P=\dfrac{F}{A}$
• p = hρg
Glossary
atmosphere (atm)
unit of pressure; 1 atm = 101,325 Pa
bar
(bar or b) unit of pressure; 1 bar = 100,000 Pa
barometer
device used to measure atmospheric pressure
hydrostatic pressure
pressure exerted by a fluid due to gravity
manometer
device used to measure the pressure of a gas trapped in a container
pascal (Pa)
SI unit of pressure; 1 Pa = 1 N/m2
pounds per square inch (psi)
unit of pressure common in the US
pressure
force exerted per unit area
torr
unit of pressure; $\mathrm{1\: torr=\dfrac{1}{760}\,atm}$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.02%3A_Pressure-_The_Result_of_Particle_Collisions.txt |
Learning Objectives
• To understand the relationships among pressure, temperature, volume, and the amount of a gas.
Early scientists explored the relationships among the pressure of a gas (P) and its temperature (T), volume (V), and amount (n) by holding two of the four variables constant (amount and temperature, for example), varying a third (such as pressure), and measuring the effect of the change on the fourth (in this case, volume). The history of their discoveries provides several excellent examples of the scientific method.
The Relationship between Pressure and Volume: Boyle's Law
As the pressure on a gas increases, the volume of the gas decreases because the gas particles are forced closer together. Conversely, as the pressure on a gas decreases, the gas volume increases because the gas particles can now move farther apart. Weather balloons get larger as they rise through the atmosphere to regions of lower pressure because the volume of the gas has increased; that is, the atmospheric gas exerts less pressure on the surface of the balloon, so the interior gas expands until the internal and external pressures are equal.
The Irish chemist Robert Boyle (1627–1691) carried out some of the earliest experiments that determined the quantitative relationship between the pressure and the volume of a gas. Boyle used a J-shaped tube partially filled with mercury, as shown in Figure $1$. In these experiments, a small amount of a gas or air is trapped above the mercury column, and its volume is measured at atmospheric pressure and constant temperature. More mercury is then poured into the open arm to increase the pressure on the gas sample. The pressure on the gas is atmospheric pressure plus the difference in the heights of the mercury columns, and the resulting volume is measured. This process is repeated until either there is no more room in the open arm or the volume of the gas is too small to be measured accurately. Data such as those from one of Boyle’s own experiments may be plotted in several ways (Figure $2$). A simple plot of $V$ versus $P$ gives a curve called a hyperbola and reveals an inverse relationship between pressure and volume: as the pressure is doubled, the volume decreases by a factor of two. This relationship between the two quantities is described as follows:
$PV = \rm constant \label{10.3.1}$
Dividing both sides by $P$ gives an equation illustrating the inverse relationship between $P$ and $V$:
$V=\dfrac{\rm const.}{P} = {\rm const.}\left(\dfrac{1}{P}\right) \label{10.3.2}$
or
$V \propto \dfrac{1}{P} \label{10.3.3}$
where the ∝ symbol is read “is proportional to.” A plot of V versus 1/P is thus a straight line whose slope is equal to the constant in Equations $\ref{10.3.1}$ and $\ref{10.3.3}$. Dividing both sides of Equation $\ref{10.3.1}$ by V instead of P gives a similar relationship between P and 1/V. The numerical value of the constant depends on the amount of gas used in the experiment and on the temperature at which the experiments are carried out. This relationship between pressure and volume is known as Boyle’s law, after its discoverer, and can be stated as follows: At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure. This law in practice is shown in Figure $2$.
At constant temperature, the volume of a fixed amount of a gas is inversely proportional to its pressure
The Relationship between Temperature and Volume: Charles's Law
Hot air rises, which is why hot-air balloons ascend through the atmosphere and why warm air collects near the ceiling and cooler air collects at ground level. Because of this behavior, heating registers are placed on or near the floor, and vents for air-conditioning are placed on or near the ceiling. The fundamental reason for this behavior is that gases expand when they are heated. Because the same amount of substance now occupies a greater volume, hot air is less dense than cold air. The substance with the lower density—in this case hot air—rises through the substance with the higher density, the cooler air.
The first experiments to quantify the relationship between the temperature and the volume of a gas were carried out in 1783 by an avid balloonist, the French chemist Jacques Alexandre César Charles (1746–1823). Charles’s initial experiments showed that a plot of the volume of a given sample of gas versus temperature (in degrees Celsius) at constant pressure is a straight line. Similar but more precise studies were carried out by another balloon enthusiast, the Frenchman Joseph-Louis Gay-Lussac (1778–1850), who showed that a plot of V versus T was a straight line that could be extrapolated to a point at zero volume, a theoretical condition now known to correspond to −273.15°C (Figure $3$).A sample of gas cannot really have a volume of zero because any sample of matter must have some volume. Furthermore, at 1 atm pressure all gases liquefy at temperatures well above −273.15°C. Note from part (a) in Figure $3$ that the slope of the plot of V versus T varies for the same gas at different pressures but that the intercept remains constant at −273.15°C. Similarly, as shown in part (b) in Figure $3$, plots of V versus T for different amounts of varied gases are straight lines with different slopes but the same intercept on the T axis.
The significance of the invariant T intercept in plots of V versus T was recognized in 1848 by the British physicist William Thomson (1824–1907), later named Lord Kelvin. He postulated that −273.15°C was the lowest possible temperature that could theoretically be achieved, for which he coined the term absolute zero (0 K).
We can state Charles’s and Gay-Lussac’s findings in simple terms: At constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature (in kelvins). This relationship, illustrated in part (b) in Figure $3$ is often referred to as Charles’s law and is stated mathematically as
$V ={\rm const.}\; T \label{10.3.4}$
or
$V \propto T \label{10.3.5}$
with temperature expressed in kelvins, not in degrees Celsius. Charles’s law is valid for virtually all gases at temperatures well above their boiling points.
The Relationship between Amount and Volume: Avogadro's Law
We can demonstrate the relationship between the volume and the amount of a gas by filling a balloon; as we add more gas, the balloon gets larger. The specific quantitative relationship was discovered by the Italian chemist Amedeo Avogadro, who recognized the importance of Gay-Lussac’s work on combining volumes of gases. In 1811, Avogadro postulated that, at the same temperature and pressure, equal volumes of gases contain the same number of gaseous particles (Figure $4$). This is the historic “Avogadro’s hypothesis.”
A logical corollary to Avogadro's hypothesis (sometimes called Avogadro’s law) describes the relationship between the volume and the amount of a gas: At constant temperature and pressure, the volume of a sample of gas is directly proportional to the number of moles of gas in the sample. Stated mathematically,
$V ={\rm const.} \; (n) \label{10.3.6}$
or
$V \propto.n \text{@ constant T and P} \label{10.3.7}$
This relationship is valid for most gases at relatively low pressures, but deviations from strict linearity are observed at elevated pressures.
For a sample of gas,
• V increases as P decreases (and vice versa)
• V increases as T increases (and vice versa)
• V increases as n increases (and vice versa)
The relationships among the volume of a gas and its pressure, temperature, and amount are summarized in Figure $5$. Volume increases with increasing temperature or amount, but decreases with increasing pressure.
Summary
The volume of a gas is inversely proportional to its pressure and directly proportional to its temperature and the amount of gas. Boyle showed that the volume of a sample of a gas is inversely proportional to its pressure (Boyle’s law), Charles and Gay-Lussac demonstrated that the volume of a gas is directly proportional to its temperature (in kelvins) at constant pressure (Charles’s law), and Avogadro postulated that the volume of a gas is directly proportional to the number of moles of gas present (Avogadro’s law). Plots of the volume of gases versus temperature extrapolate to zero volume at −273.15°C, which is absolute zero (0 K), the lowest temperature possible. Charles’s law implies that the volume of a gas is directly proportional to its absolute temperature. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.03%3A_The_Simple_Gas_Laws-_Boyles_Law_Charless_Law_and_Avogadros_Law.txt |
Learning Objectives
• Derive the ideal gas law from the constituent gas laws
• To use the ideal gas law to describe the behavior of a gas.
In this module, the relationship between pressure, temperature, volume, and amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas.
Deriving the Ideal Gas Law
Any set of relationships between a single quantity (such as $V$) and several other variables ($P$, $T$, and $n$) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions were derived previously:
• Boyle’s law
$V \propto \dfrac{1}{P} \;\; \text{@ constant n and T} \nonumber$
• Charles’s law
$V \propto T \;\; \text{@ constant n and P} \nonumber$
• Avogadro’s law
$V \propto n \;\; \text{@ constant T and P} \nonumber$
Combining these three expressions gives
$V \propto \dfrac{nT}{P} \label{10.4.1}$
which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as
$V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \label{10.4.2}$
By convention, the proportionality constant in Equation $\ref{10.4.1}$ is called the gas constant, which is represented by the letter $R$. Inserting R into Equation $\ref{10.4.2}$ gives
$V = \dfrac{RnT}{P} = \dfrac{nRT}{P} \label{10.4.3}$
Clearing the fractions by multiplying both sides of Equation $\ref{10.4.4}$ by $P$ gives
$PV = nRT \label{10.4.4}$
This equation is known as the ideal gas law.
An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed.
Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures.
Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then
$R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \label{10.4.5}$
Because the product PV has the units of energy, R can also have units of J/(K•mol):
$R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\label{10.4.6}$
Standard Conditions of Temperature and Pressure
Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and $\rm1\; bar = 100 \;kPa = 10^5\;Pa$ pressure, referred to as standard temperature and pressure (STP).
$\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa \nonumber$
Please note that STP was defined differently in the past. The old definition was based on a standard pressure of 1 atm.
We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation $\ref{10.4.4}$:
$V=\dfrac{nRT}{P}\label{10.4.7}$
Thus the volume of 1 mol of an ideal gas is 22.71 L at STP and 22.41 L at 0°C and 1 atm, approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed.
Table $1$: Molar Volumes of Selected Gases at 0°C and 1 atm
Gas Molar Volume (L)
He 22.434
Ar 22.397
H2 22.433
N2 22.402
O2 22.397
CO2 22.260
NH3 22.079
Applying the Ideal Gas Law
The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables (P, V, T, and n). It also allows us to predict the final state of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters (P, V, T, and n) are specified for an initial state. Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample.
Example $1$
The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft3), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon?
Given: volume, temperature, and pressure
Asked for: amount of gas
Strategy:
1. Solve the ideal gas law for the unknown quantity, in this case n.
2. Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed.
Solution:
A We are given values for P, T, and V and asked to calculate n. If we solve the ideal gas law (Equation $\ref{10.4.4}$) for $n$, we obtain
$\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm \nonumber$
B P and T are given in units that are not compatible with the units of the gas constant [R = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres:
$T=273+30=303{\rm K}\nonumber$
Substituting these values into the expression we derived for n, we obtain
\begin{align*} n &=\dfrac{PV}{RT} \[4pt] &=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K} \[4pt] &=1.23\times10^3\;mol \end{align*} \nonumber
Exercise $1$
Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO2. What is the pressure of the gas at 25°C?
Answer
1.5 atm
In Example $1$, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of changes in any of the specified conditions on any of the other parameters, as shown in Example $5$.
General Gas Equation
When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is:
$\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f) \ P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array} \nonumber$
Both equations can be rearranged to give:
$R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f} \nonumber$
The two equations are equal to each other since each is equal to the same constant $R$. Therefore, we have:
$\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\label{10.4.8}$
The equation is called the general gas equation. The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties.
Example $2$
Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example $1$?
Given: temperature, pressure, amount, and volume in August; temperature in January
Asked for: volume in January
Strategy:
1. Use the results from Example $1$ for August as the initial conditions and then calculate the change in volume due to the change in temperature from 30°C to −10°C. Begin by constructing a table showing the initial and final conditions.
2. Simplify the general gas equation by eliminating the quantities that are held constant between the initial and final conditions, in this case $P$ and $n$.
3. Solve for the unknown parameter.
Solution:
A To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions:
Solution to Example 10.4.2
Initial (August) Final (January)
$T_i=30\, °C = 303\, K$ $T_f=−10\,°C = 263\, K$
$P_i= 0.980 \, atm$ $P_f= 0.980\, atm$
$n_i=1.23 × 10^3\, mol$ $n_f= 1.23 × 10^3\, mol$
$V_i=31150\, L$ $V_f=?$
B Both $n$ and $P$ are the same in both cases ($n_i=n_f,P_i=P_f$). Therefore, Equation \ref{10.4.8} can be simplified to:
$\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f} \nonumber$
This is the relationship first noted by Charles.
C Solving the equation for $V_f$, we get:
\begin{align*} V_f &=V_i\times\dfrac{T_f}{T_i} \[4pt] &=\rm31150\;L\times\dfrac{263\;K}{303\;K} \[4pt] &=2.70\times10^4\;L \end{align*} \nonumber
It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change.
Exercise $2$
At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon?
Answer
0.52 L
Example $1$ illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle (PV = constant) and the relationship between volume and amount observed by Avogadro (V/n = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example $1$ can be applied in any such case, as we demonstrate in Example $2$ (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion).
Example $3$
Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise $1$ (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)?
Given: initial volume, amount, temperature, and pressure; final temperature
Asked for: final pressure
Strategy:
Follow the strategy outlined in Example $2$.
Solution:
Prepare a table to determine which parameters change and which are held constant:
Solution to Example 10.4.3
Initial Final
$V_i=0.406\;\rm L$ $V_f=0.406\;\rm L$
$n_i=0.025\;\rm mol$ $n_f=0.025\;\rm mol$
$T_i=\rm25\;^\circ C=298\;K$ $T_i=\rm750\;^\circ C=1023\;K$
$P_i=1.5\;\rm atm$ $P_f=?$
Both $V$ and $n$ are the same in both cases ($V_i=V_f,n_i=n_f$). Therefore, Equation can be simplified to:
$\dfrac{P_i}{T_i} = \dfrac{P_f}{T_f} \nonumber$
By solving the equation for $P_f$, we get:
\begin{align*} P_f &=P_i\times\dfrac{T_f}{T_i} \[4pt] &=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K} \[4pt] &=5.1\;atm \end{align*} \nonumber
This pressure is more than enough to rupture a thin sheet metal container and cause an explosion!
Exercise $3$
Suppose that a fire extinguisher, filled with CO2 to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher?
Answer
23.4 atm
In Examples $1$ and $2$, two of the four parameters (P, V, T, and n) were fixed while one was allowed to vary, and we were interested in the effect on the value of the fourth. In fact, we often encounter cases where two of the variables P, V, and T are allowed to vary for a given sample of gas (hence n is constant), and we are interested in the change in the value of the third under the new conditions.
Example $4$
We saw in Example $1$ that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 103 mol of H2 gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude?
Given: initial pressure, temperature, amount, and volume; final pressure and temperature
Asked for: final volume
Strategy:
Follow the strategy outlined in Example $3$.
Solution:
Begin by setting up a table of the two sets of conditions:
Solution to Example 10.4.4
Initial Final
$P_i=745\;\rm mmHg=0.980\;atm$ $P_f=312\;\rm mmHg=0.411\;atm$
$T_i=\rm30\;^\circ C=303\;K$ $T_f=\rm750-30\;^\circ C=243\;K$
$n_i=\rm1.2\times10^3\;mol$ $n_i=\rm1.2\times10^3\;mol$
$V_i=\rm31150\;L$ $V_f=?$
By eliminating the constant property ($n$) of the gas, Equation $\ref{10.4.8}$ is simplified to:
$\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f} \nonumber$
By solving the equation for $V_f$, we get:
\begin{align*} V_f &=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i} \[4pt] &=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K} \[4pt] &=5.96\times10^4\;L \end{align*} \nonumber
Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to increase the volume of the gas, while decreasing the temperature tends to decrease the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both T and 1/P, the variable that changes the most will have the greatest effect on V. In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation.
We could also have solved this problem by solving the ideal gas law for V and then substituting the relevant parameters for an altitude of 23,000 ft:
Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. There is often more than one “right” way to solve chemical problems.
Exercise $4$
A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.)
Answer
4.07 × 103
Using the Ideal Gas Law to Calculate Gas Densities and Molar Masses
The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain
$\dfrac{n}{V}=\dfrac{P}{RT}\label{10.4.9}$
The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass ($m$, in grams) divided by its molar mass ($M$, in grams per mole):
$n=\dfrac{m}{M}\label{10.4.10}$
Substituting this expression for $n$ into Equation $\ref{10.4.9}$ gives
$\dfrac{m}{MV}=\dfrac{P}{RT}\label{10.4.11}$
Because $m/V$ is the density $d$ of a substance, we can replace $m/V$ by $d$ and rearrange to give
$\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\label{10.4.12}$
The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL).
Example $5$
Calculate the density of butane at 25°C and a pressure of 750 mmHg.
Given: compound, temperature, and pressure
Asked for: density
Strategy:
1. Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant.
2. Substitute these values into Equation $\ref{10.4.12}$ to obtain the density.
Solution:
A The molar mass of butane (C4H10) is
$M=(4)(12.011) + (10)(1.0079) = 58.123 \rm g/mol \nonumber$
Using 0.08206 (L•atm)/(K•mol) for R means that we need to convert the temperature from degrees Celsius to kelvins (T = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres:
$P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \nonumber$
B Substituting these values into Equation $\ref{10.4.12}$ gives
$\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \nonumber$
Exercise $5$: Density of Radon
Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics.
Answer
radon, 9.23 g/L; N2, 1.17 g/L
A common use of Equation $\ref{10.4.12}$ is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example $6$.
Example $6$
The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound.
Given: pressure, temperature, mass, and volume
Asked for: molar mass and chemical formula
Strategy:
1. Solve Equation $\ref{10.4.12}$ for the molar mass of the gas and then calculate the density of the gas from the information given.
2. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass.
3. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas.
Solution:
A Solving Equation $\ref{10.4.12}$ for the molar mass gives
$M=\dfrac{mRT}{PV}=\dfrac{dRT}{P} \nonumber$
Density is the mass of the gas divided by its volume:
$\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.157\rm L}=1.84 \rm g/L\nonumber$
B We must convert the other quantities to the appropriate units before inserting them into the equation:
$T=18+273=291 K\nonumber$
$P=727 \, mmHg \times \dfrac{1\rm atm} {760\rm mmHg} =0.957\rm atm \nonumber$
The molar mass of the unknown gas is thus
$M=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol\nonumber$
C The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations:
$M({\rm NO})=14 + 16=30 \rm\; g/mol\nonumber$
$M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol\nonumber$
$M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol\nonumber$
The most likely choice is NO2 which is in agreement with the data. The red-brown color of smog also results from the presence of NO2 gas.
Exercise $6$
You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it.
Answer
44 g/mol; $CO_2$
Summary
The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known.
Ideal gas equation: $PV = nRT$,
where $R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}$
General gas equation: $\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}$
Density of a gas: $\rho=\dfrac{MP}{RT}$
The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an ideal gas, a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. Standard temperature and pressure (STP) is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the standard molar volume. All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity (P, V, T, or n) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of P, V, T, and n) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.04%3A_The_Ideal_Gas_Law.txt |
Learning Objectives
• Apply gas laws to solve stoichiometry problems.
• Apply principles of stoichiometry to calculate properties of gases.
The quantitative relationship of reactants and products is called stoichiometry. Stoichiometric problems require you to calculate the amounts of reactants required for certain amounts of products, or amounts of products produced from certain amounts of reactants. If, in a chemical reaction, one or more reactants or products are gases, gas laws must be considered for the calculation. Usually, the applications of the ideal gas law give results within 5% precision. Below, we review several important concepts that are helpful for solving Stoichiometry Problems Involving Gases.
The Mole Concept and Molar Volume
The mole concept is the key to both stoichiometry and gas laws. A mole is a definite amount of substance. Mole is a unit based on the number of identities (i.e. atoms, molecules, ions, or particles). A mole of anything has the same number of identities as the number of atoms in exactly 12 grams of carbon-12, the most abundant isotope of carbon.
Molar volume is defined as the volume occupied by one mole of a gas. Using the ideal gas law and assuming standard pressure and temperature (STP), the volume of one mole of gas can be calculated:
$PV=nRT$
$V = \dfrac{nRT}{P}$
${V = \rm \dfrac{1.00 mol\ \cdot 0.08206 \dfrac{L atm}{mol K}\cdot 273 K}{1.00 atm }}$
$V = 22.4 \rm L$
In other words, 1 mole of a gas will occupy 22.4 L at STP, assuming ideal gas behavior.
At STP, the volume of a gas is only dependent on number of moles of that gas and is independent of molar mass. With this information we can calculate the density ($\rho$) of a gas using only its molar mass. First, starting with the definition of density
$\rho =\dfrac{m}{V}\label {where D = density, m=mass and V= volume}$
we rearrange for volume:
$V=\dfrac{m}{ \rho }$
We then substitute $V$ into the ideal gas equation and rearrange for density:
$PV=nRT$
$P\dfrac{m}{ \rho }=nRT$
$\rho=\dfrac{mP}{nRT}$
Finally, we remember that molar mass is equal to mass divided by number of moles:
$MM =\dfrac{m}{n}$
and substitute this into our expression for density to give:
$\rho = \dfrac{MM\cdot P}{RT}$
This equation can further be simplified if we assume STP:
$\rho = \dfrac{MM\cdot 1 \rm atm}{(0.08206 \dfrac{\rm L atm}{\rm mol K}) \rm 273 K}$
$\rho = \dfrac{MM}{\rm 22.4 \dfrac{L}{mol}}$
Using this information, we can calculate the density of a gas using the gas's molar mass.
Example 1
Calculate the density of N2 gas at STP.
What we know: Pressure (1 atm),temperature (273 K), the identity of the gas (N2).
Asked for: Density of N2
Strategy:
1. Calculate the molar mass of N2
2. Solve for the density of using the equation relating density and molar mass at STP
Solution:
A The molar mass of N2:
$MM_{N2}=\rm2\cdot 14.0 g/mol = 28.0 g/mol$
B Calculate the density of N2
$\rho = \dfrac{MM}{\rm 22.4 \dfrac{L}{mol}}$
$\rho = \dfrac{28.0 \rm g/mol}{\rm 22.4 \dfrac{L}{mol}}$
$\rho = 1.25 \rm g/L$
Example 2
Calculate the density of Ne gas at 143 ºC and 4.3 atm.
What we know: Pressure (4.3 atm), temperature (143 ºC ), the identity of the gas (Ne), the molar mass of Ne from the periodic table (20.2 g/mol).
Asked for: Density of Ne
Strategy:
1. The temperature is given in degrees Celsius. This must be converted to Kelvin
2. Solve for the density
Solution:
A Calculate temperature in Kelvin:
$T = \rm 143 C + 273 = 416 K$
B Calculate the density of Ne:
$\rho = \dfrac{MM\cdot P}{RT}$
$\rho = \dfrac{\rm 20.2 g/mol \cdot 4.3 \rm atm}{(0.08206 \dfrac{\rm L atm}{\rm mol K}) \rm 416 K}$
$\rho = \rm 2.54 g/L$
Molar Mass of a Gas
The equations for calculating the density of a gas can rearranged to calculate the molar mass of a gas:
$MM = \dfrac{ \rho RT}{P}\label {where MM= molar mass and D = density}$
this can be further simplified if we work at STP:
$MM = \rho \cdot 22.4 L/mol$
We can use these equations to identify an unknown gas, as shown below:
Example 3
A unknown gas has density of 1.78 g/L at STP. What is the identify of this gas?
What we know: Pressure (1.00 atm), temperature (273 K ), density of the gas (1.783 g/L)
Asked for: Identity of the unknown gas
Strategy:
1. First calculate the molar mass of the unknown gas
2. Determine the identity of the gas by comparing the calculated molar mass to molar masses of known gases.
Solution:
A Since we are at STP, we can use the following equation to calculate molar mass:
$MM = \rho \cdot 22.4 L/mol$
$MM = \rm 1.783 g/L \cdot 22.4 L/mol$
$MM = \rm 39.9 g/mol$
B The calculated molar mass is 33.9 g/mol. Examination of the periodic table reveals that Argon has a mass of 39.948 g/mol. Therefore, the unknown gas is most likely argon.
Stoichiometry and Gas Laws
Stoichiometry is the theme of the previous block of modules, and the ideal gas law is the theme of this block of modules. These subjects are related. Be prepared to solve problems requiring concepts or principles of stoichiometry and gases. For example, we can calculate the number of moles from a certain volume, temperature and pressure of a $\ce{HCl}$ gas. When n moles are dissolved in V L solution, its concentration is n/V M.
Three examples are given to illustrate some calculations of stoichiometry involving gas laws and more are given in question form for you to practice.
Example 1
If 500 mL of $\ce{HCl}$ gas at 300 K and 100 kPa dissolve in 100 mL of pure water, what is the concentration? Data required: R value 8.314 kPa L / (K mol).
Solution
\begin{align} n_{\textrm{HCl}} &= \mathrm{\dfrac{0.50\: L \times 100\: kPa}{8.314\: \dfrac{kPa\: L}{K\: mol} \times 300\: K}}\ &= \mathrm{0.02\: mol} \end{align}
Concentration of $\ce{HCl}$, $\ce{[HCl]}$
$\mathrm{[HCl] = \dfrac{0.02\: mol}{0.1\: L} = 0.2\: mol/L}$
Discussion
Note that R = 0.08205 L atm /(K mol) will not be suitable in this case. If you have difficulty, review Solutions.
Example 2
If 500 mL of $\ce{HCl}$ gas at 300 K and 100 kPa dissolved in pure water requires 12.50 mL of the $\ce{NaOH}$ solution to neutralize in a titration experiment, what is the concentration of the $\ce{NaOH}$ solution?
Solution
Solution in Example 1 showed nHCl = 0.02 mol. From the titration experiment, we can conclude that there were 0.02 moles of $\ce{NaOH}$ in 12.50 mL. Thus,
$\mathrm{[NaOH] = \dfrac{0.02\: mol}{0.0125\: L} = 1.60\: mol/L}$
Discussion
Think in terms of reaction,
\begin{align} \mathrm{HCl + NaOH \rightarrow NaCl + H_2O} &\Leftarrow \mathrm{Reaction}\ \mathrm{0.02\: mol \hspace{8px} 0.02\: mol \hspace{103px}} &\Leftarrow \mathrm{Quantities\: reacted} \end{align}
Note that 0.02 mol of $\ce{NaOH}$ is in 0.0125 mL solution.
Example 3
A 5.0-L air sample containing $\ce{H2S}$ at STP is treated with a catalyst to promote the reaction
$\mathrm{H_2S + O_2 \rightarrow H_2O + S_{(solid)}}$.
If 3.2 g of solid $\ce{S}$ was collected, calculate the volume percentage of $\ce{H2S}$ in the original sample.
Solution
$\mathrm{3.2\: g\: S\times\dfrac{1\: mol\: H_2S}{32\: g\: S}= 0.10\: mol\: H_2S}$
\begin{align} \mathrm{V_{H_2S}} &= \mathrm{0.10\: mol \times 22.4\: L/mol}\ &= \mathrm{2.24\: L} \end{align}
\begin{align} \mathrm{Volume\: \%} &= \mathrm{\dfrac{2.25\: L}{5.0\: L}}\ &= 0.45\ &= 45 \% \end{align}
Discussion
Data required: Atomic mass: $\mathrm{H = 1}$; $\mathrm{O = 16}$; $\mathrm{S = 32}$. R = 0.08205 L atm /(K mol) is now suitable R values or molar volume at STP (22.4 L/mol)
The volume percentage is also the mole percentage, but not the weight percentage.
Example 4
Hydrogen sulfide reacts with sulfur dioxide to give $\ce{H2O}$ and $\ce{S}$,
$\mathrm{H_2S + SO_2 \rightarrow H_2O + S_{(solid)}}$, unbalanced.
If 6.0 L of $\ce{H2S}$ gas at 750 torr produced 3.2 g of sulfur, calculate the temperature in C.
Solution
Balanced reaction:
$\mathrm{2 H_2S + SO_2 \rightarrow 2 H_2O + {3 S_{(solid)}}}\ \mathrm{2\: mol \hspace{130px} 3\times32 = 96\: g}$
$\mathrm{3.2\: g\: S\times\dfrac{2\: mol\: H_2S}{96\: g\: S}= 0.067\: mol\: H_2S}$
$\mathrm{P = \dfrac{750}{760} = 0.987\: atm}$
\begin{align} T =\dfrac{PV}{n R}&=\mathrm{\dfrac{0.987\: atm \times 6\: L}{0.067\: mol \times 0.08205\: \dfrac{atm\: L}{mol\: K}}}\ &= \mathrm{1085\: K}\ &= \mathrm{812^\circ C} \end{align}
Discussion
Atomic mass: $\mathrm{H = 1.0}$; $\mathrm{O = 16.0}$; $\mathrm{S = 32.0}$. R = 0.08205 L atm /(K mol) is OK but watch units used for pressure.
Example 5
When 50.0 mL of $\ce{AgNO3}$ solution is treated with an excess amount of $\ce{HI}$ gas to give 2.35 g of $\ce{AgI}$, what is the concentration of the $\ce{AgNO3}$ solution?
Solution
$\mathrm{2.35\: g\: AgI \times\dfrac{1\: mol\: Ag^+}{234.8\: g\: AgI}\times\dfrac{1\: mol\: AgNO_3}{1\: mol\: Ag^+}= 0.010\: mol\: AgNO_3}$
\begin{align} \mathrm{[AgNO_3]} &= \mathrm{\dfrac{0.01\: mol\: AgNO_3}{0.050\: L}}\ &= \mathrm{0.20\: M\: AgNO_3} \end{align}
Discussion
A gas is involved, but there is no need to consider the gas law. At. mass: $\mathrm{Ag = 107.9}$; $\mathrm{N = 14.0}$; $\mathrm{O = 16.0}$; $\mathrm{I = 126.9}$
Example 6
What volume (L) will 0.20 mol $\ce{HI}$ occupy at 300 K and 100.0 kPa? $\mathrm{R = 8.314\: \dfrac{kPa\: L}{K\: mol} = 0.08205\: \dfrac{atm\: L}{mol\: K}}$
Solution
\begin{align} V &=\dfrac{n RT}{P}\ &= \mathrm{\dfrac{0.20\: mol \times 8.314\, \dfrac{kPa\:L}{mol\: K} \times 300\: K}{100\: kPa}}\ &= \mathrm{\mathrm{5\: L} }\end{align}
Example 7
A 3.66-g sample containing $\ce{Zn}$ (at.wt. 65.4) and $\ce{Mg}$ (24.3) reacted with a dilute acid to produce 2.470 L $\ce{H2}$ gas at 101.0 kPa and 300 K. Calculate the percentage of $\ce{Zn}$ in the sample.
Solution
The number of moles of gas produced is the number of moles of metals in the sample. Once you know the number of moles, set up an equation to give the number of moles of metal in the sample.
\begin{align} n &= \mathrm{\dfrac{101\: kPa \times 2.470\: L}{8.3145\: \dfrac{kPa\: L}{mol\: K} \times 300\: K}}\ &= \mathrm{0.100\: mol} \end{align}
Let x be the mass of $\ce{Zn}$, then the mass of $\ce{Mg}$ is 3.66 - x g. Thus, we have
$\dfrac{x}{65.4}+\dfrac{3.66 - x}{24.3}= \mathrm{0.100\: mole}$
Solving for x gives x = 1.96 g $\ce{Zn}$,
and the $\mathrm{weight\: percent = 100 \times \dfrac{1.96}{3.66} = 53.6 \%}$
Discussion
Find the mole percent of $\ce{Zn}$ in the sample.
$\mathrm{\#\: mol\: of\: Zn = \dfrac{1.96}{65.4} = 0.03\: mol}$
$\mathrm{\#\: mol\: of\: Mg = \dfrac{1.70}{24.3} = 0.07\: mol}$
$\mathrm{mole\: percent = 100 \times \dfrac{0.03}{0.03 + 0.07} = 30 \%}$
Example 8
When a 2.00 g mixture of $\ce{Na}$ and $\ce{Ca}$ reacted with water, 1.164 L hydrogen was produced at 300.0 K and 100.0 kPa. What is the percentage of $\ce{Na}$ in the sample?
Solution
$\ce{2 Na + 2 H2O \rightarrow 2 Na(OH) + H2_{\large{(g)}}}$
$\ce{Ca + H2O \rightarrow Ca(OH) + H2_{\large{(g)}}}$
Let x be the mass of $\ce{Na}$, then (2.00-x) is the mass of $\ce{Ca}$.
We have the following relationship
$\mathrm{\dfrac{x\: g}{23.0\: g/mol}\times\dfrac{1\: mol\: H_2}{2\: mol\: Na}+\dfrac{(2.0 - x)\: g\: Ca}{40.1\: g\: Ca/mol}\times\dfrac{1\: mol\: H_2}{1\: mol\: Ca}=\dfrac{1.164\: L\: H_2 \times 100.0\: kPa}{8.3145\: kPa\: L\: mol^{-1}\: K^{-1}\: 300.0\: K}}$
Simplify to give
$\mathrm{\dfrac{x}{46.0}+\dfrac{2}{40.1}-\dfrac{x}{40.1}= 0.0467\:all\: in\: mol}$
Multiply all terms by (40.1 * 46.0)
$\mathrm{40.1\, x + 2 \times 46.0 - 46.0\, x = 86.1}$
Simplify
$\mathrm{-5.9\, x = 86.1 - 92.0 = -5.91}$
Thus,
$\mathrm{Mass\: of\: Na = x = 1.0\: g}$
$\mathrm{Mass\: of\: Ca = 2.0 - x = 1.0\: g}$
$\mathrm{Mass\: Percentage\: of\: Na = 100\times \dfrac{1}{2.0} = 50\%}$
Discussion
$\mathrm{Mole\: of\: Na = \dfrac{1}{23} = 0.0435\: mol}$
$\mathrm{Mole\: percentage = \dfrac{\dfrac{1}{23}}{\dfrac{1}{23} + \dfrac{1}{40.1}} = 0.635 = 63.5\%}$
Compare this example with gravimetric analyses using the reaction
$\mathrm{Ag^+_{\large{(aq)}} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl} \rightarrow AgCl_{\large{(s)}}}$
where $\mathrm{\sideset{ }{_{\large{(aq)}}^{-}}{Cl}}$ comes from the disolution of two salts such as $\ce{NaCl}$ and $\ce{MgCl2}$.
Also compare with analyses making use of the reaction
$\mathrm{Ba^{2+}_{\large{(aq)}} + SO^{2-}_{4\large{(aq)}} \rightarrow BaSO_{4\large{(s)}}}$
where the anion $\mathrm{SO^{2-}_{4\large{(aq)}}}$ comes from the dissolution of two sulfate salts.
This example is very similar to Example 7. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.05%3A_Applications_of_the_Ideal_Gas_Law-_Molar_Volume_Density_and_Molar_Mass_of_a_Gas.txt |
In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture. The Learning Objective of this module is to determine the contribution of each component gas to the total pressure of a mixture of gases.
Partial Pressures
The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:
$P=n \bigg(\dfrac{RT}{V}\bigg) = n \times \rm const. \tag{10.6.1}$
Nothing in the equation depends on the nature of the gas—only the amount.
With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure, the pressure the gas would exert if it were the only one present (at the same temperature and volume).
To summarize,the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressures. We can write it mathematically as
$P_{tot}= P_1+P_2+P_3+P_4 \; ... = \sum_{i=1}^n{P_i} \tag{10.6.2}$
where $P_{tot}$ is the total pressure and the other terms are the partial pressures of the individual gases (up to $n$ component gases).
Figure Dalton’s Law. The total pressure of a mixture of gases is the sum of the partial pressures of the individual gases.
For a mixture of two ideal gases, $A$ and $B$, we can write an expression for the total pressure:
$P_{tot}=P_A+P_B=n_A\bigg(\dfrac{RT}{V}\bigg) + n_B\bigg(\dfrac{RT}{V}\bigg)=(n_A+n_B)\bigg(\dfrac{RT}{V}\bigg) \tag{10.6.3}$
More generally, for a mixture of $n$ component gases, the total pressure is given by
$P_{tot}=(P_1+P_2+P_3+ \; \cdots +P_n)\bigg(\dfrac{RT}{V}\bigg)\tag{10.6.2a}$
$P_{tot}=\sum_{i=1}^n{n_i}\bigg(\dfrac{RT}{V}\bigg)\tag{10.6.2b}$
Equation 10.6.4 restates Equation 10.6.3 in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation 10.6.4 to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in the example below:
Example
Deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of $O_2$ and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?
Given: masses of components, total volume, and temperature
Asked for: partial pressures and total pressure
Strategy:
1. Solution:
$n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol$
The number of moles of $O_2$ is
$n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol$
B We can now use the ideal gas law to calculate the partial pressure of each:
$P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm$
$P_{\rm O_2}=\dfrac{n_{\rm O_2}RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm$
The total pressure is the sum of the two partial pressures:
$P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm$
Exercise 10.6.1
A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.
Answer: $P_{CH_4}=137 \; atm$; $P_{C_2H_6}=13.4\; atm$; $P_{tot}=151\; atm$
Mole Fractions of Gas Mixtures
The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction ($X$) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture ($n_{tot}$):
$x_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\tag{10.6.5}$
The mole fraction is a dimensionless quantity between 0 and 1. If $x_A = 1.0$, then the sample is pure $A$, not a mixture. If $x_A = 0$, then no $A$ is present in the mixture. The sum of the mole fractions of all the components present must equal 1.
To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas $A$ to the total pressure of a gas mixture that contains $A$. We can use the ideal gas law to describe the pressures of both gas $A$ and the mixture: $P_A = n_ART/V$ and $P_{tot} = n_tRT/V$. The ratio of the two is thus
$\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=x_A \tag{10.6.6}$
Rearranging this equation gives
$P_A = x_AP_{tot} \tag{10.6.7}$
That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. By volume, Earth’s atmosphere is about 78% $N_2$, 21% $O_2$, and 0.9% $Ar$, with trace amounts of gases such as $CO_2$, $H_2O$, and others. This means that 78% of the particles present in the atmosphere are $N_2$; hence the mole fraction of $N_2$ is 78%/100% = 0.78. Similarly, the mole fractions of $O_2$ and $Ar$ are 0.21 and 0.009, respectively. Using Equation 10.6.7, we therefore know that the partial pressure of N2 is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of $O_2$ and $Ar$ are 0.21 and 0.009 atm, respectively.
Example
We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we exhale yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air.
Inhaled Air / mmHg Exhaled Air / mmHg
$P_{\rm N_2}$ 597 568
$P_{\rm O_2}$ 158 116
$P_{\rm H_2O}$ 0.3 28
$P_{\rm CO_2}$ 5 48
$P_{\rm Ar}$ 8 8
$P_{tot}$ 767 767
Given: pressures of gases in inhaled and exhaled air
Asked for: mole fractions of gases in exhaled air
Strategy:
Calculate the mole fraction of each gas using Equation 10.6.7.
Solution:
The mole fraction of any gas $A$ is given by
$x_A=\dfrac{P_A}{P_{tot}}$
where $P_A$ is the partial pressure of $A$ and $P_{tot}$ is the total pressure. For example, the mole fraction of $CO_2$ is given as:
$x_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063$
The following table gives the values of $x_A$ for the gases in the exhaled air.
Gas Mole Fraction
${\rm N_2}$ 0.741
${\rm O_2}$ 0.151
${\rm H_2O}$ 0.037
${\rm CO_2}$ 0.063
${\rm Ar}$ 0.010
Exercise 10.6.2
Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO2 and 3% N2, with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO2 and N2.
Answer
$P_{\rm CO_2}=\rm86\; atm$, $P_{\rm N_2}=\rm2.7\;atm$
Summary
• The pressure exerted by each gas in a gas mixture (its partial pressure) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components (Dalton’s law of partial pressures). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The mole fraction of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.06%3A_Mixtures_of_Gases_and_Partial_Pressures.txt |
The Learning Objective of this Module is to understand the significance of the kinetic molecular theory of gases.
The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles.
A Molecular Description
The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates:
1. Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In Section 10.8, we explain how this theory must be modified to account for the behavior of real gases.
The Relationships among Pressure, Volume, and Temperature
We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously.
• $P_{1}V_{1}=P_{2}V_{2}$
• $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$
• $P_{total}=P_{a}+P_{b}+P_{c} +...$
11.08: Temperature and Molecular Velocities
Boltzmann Distributions
At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? . This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure 10.7.1. Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the value of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure 10.7.1 were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases.
Figure 10.7.2 The Distributions of Molecular Speeds for a Sample of Nitrogen Gas at Various Temperatures. Increasing the temperature increases both the most probable speed (given at the peak of the curve) and the width of the curve. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.07%3A_A_Particulate_Model_for_Gases-_Kinetic_Molecular_Theory.txt |
Diffusion and Effusion
As you have learned, the molecules of a gas are not stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment.
Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space.
The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship is referred to as Graham’s law, after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses:
$\dfrac{\text{rate of effusion A}}{\text{rate of effusion B}}=\sqrt{\dfrac{M_B}{M_A}}.$
Heavy molecules effuse through a porous material more slowly than light molecules, as illustrated schematically in Figure 10.8.1 for ethylene oxide and helium. Helium (M = 4.00 g/mol) effuses much more rapidly than ethylene oxide (M = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon $\sqrt{\dfrac{29}{4.00}}=2.7$ times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days.
Figure 10.8.1 The Relative Rates of Effusion of Two Gases with Different Masses. The lighter He atoms (M = 4.00 g/mol) effuse through the small hole more rapidly than the heavier ethylene oxide (C2H4O) molecules (M = 44.0 g/mol), as predicted by Graham’s law.
Note
At a given temperature, heavier molecules move more slowly than lighter molecules.
Example 10.8.1
During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of 235U. Naturally occurring uranium is only 0.720% 235U, whereas most of the rest (99.275%) is 238U, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound UF6 (boiling point = 56°C).
1. Given: isotopic content of naturally occurring uranium and atomic masses of 235U and 238U
Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure 235UF6
1. Solution:
1. The first step is to calculate the molar mass of UF6 containing 235U and 238U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of 235UF6 is 234.04 + (6)(18.998) = 349.03 g/mol 238.05 + (6)(18.998) = 352.04 g/mol
The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law.
$\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043$rate U235F6rate U238F6=352.04=1.0043
Figure A Portion of a Plant for Separating Uranium Isotopes by Effusion of UF6. The large cylindrical objects (note the human for scale) are so-called diffuser (actually effuser) units, in which gaseous UF6 is pumped through a porous barrier to partially separate the isotopes. The UF6 must be passed through multiple units to become substantially enriched in 235U.
Rates of Diffusion or Effusion
Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses:
$KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\tag{10.8.1}$
Multiplying both sides by 2 and rearranging give
11.10: Gases in Chemical Reactions- Stoichiometry Revisited
Earlier in the course we performed stoichiometric calculations with chemical reactions using quantities of moles and mass (typically in grams). These same principles can be applied to chemical reactions involving gases except that we first have to convert volumes of gases into moles.
Example
Hydrogen gas reacts with oxygen gas to produce water vapor via the following balanced chemical equation:
2H2(g) + O2(g) -> 2H2O(g)
If the temperature is 320 K and pressure is 1.34 atm, what volume of oxygen is required to produce 65.0 g of water?
Strategy: Since we are given the temperature and pressure, to find the volume of oxygen using the ideal gas law we need to first calculate the moles of oxygen. To find the moles of oxygen required, we can first calculate the moles of water in 65.0g.
$n_{water}=\frac{m_{water}}{mm_{water}}=\frac{\text{65.0 g}}{\text{18.0g/mol}}=\text{3.61 mol of water}$
From the balanced chemical equation, we can see that 1 equivalent of oxygen produces 2 equivalents of water. We can therefore write the following ratio:
1 mol O2 : 2 mol H2O
We can now solve for the amount of oxygen:
$(\text{3.61 mol } H_{2}O)\times(\frac{ \text{1 mol } O_{2}}{\text{2 mol } H_{2}O })= \text{1.81 mol } O_{2}$
Finally, now that we know how many moles of oxygen are required, we can calculate the volume of the oxygen using the ideal gas law and the temperature&pressure provided in question:
$PV=nRT$
$V=\frac{nRT}{P}$
${V = \rm \frac{1.81 mol\ \cdot 0.08206 \frac{L atm}{mol K}\cdot 320K}{1.34 atm }}$
$V = 35.5 \rm L$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.09%3A_Mean_Free_Path_Diffusion_and_Effusion_of_Gases.txt |
Learning Objectives
• To recognize the differences between the behavior of an ideal gas and a real gas
• To understand how molecular volumes and intermolecular attractions cause the properties of real gases to deviate from those predicted by the ideal gas law.
The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases.
Pressure, Volume, and Temperature Relationships in Real Gases
For an ideal gas, a plot of $PV/nRT$ versus $P$ gives a horizontal line with an intercept of 1 on the $PV/nRT$ axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (Figure $\PageIndex{1a}$). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (Figure $\PageIndex{1b}$).
Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure $2$ for $N_2$. Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid.
Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure $3$). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.
Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is less than expected (Figure $4$). Thus as shown in Figure $2$, at low temperatures, the ratio of $PV/nRT$ is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the $PV/nRT$ versus $P$ plot for many gases.
Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted.
At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid).
The van der Waals Equation
The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation,
$\underbrace{ \left(P + \dfrac{an^2}{V^2}\right)}_{\text{Pressure Term}} \overbrace{(V − nb)}^{\text{Pressure Term}} =nRT \label{10.9.1}$
a and b are empirical constants that are different for each gas. The values of $a$ and $b$ are listed in Table $1$ for several common gases.
Table $1$:: van der Waals Constants for Some Common Gases (see Table A8 for more complete list)
Gas a ((L2·atm)/mol2) b (L/mol)
He 0.03410 0.0238
Ne 0.205 0.0167
Ar 1.337 0.032
H2 0.2420 0.0265
N2 1.352 0.0387
O2 1.364 0.0319
Cl2 6.260 0.0542
NH3 4.170 0.0371
CH4 2.273 0.0430
CO2 3.610 0.0429
The pressure term in Equation $\ref{10.9.1}$ corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, $n^2/V^2$ represents the concentration of the gas ($n/V$) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure $4$. The volume term corrects for the volume occupied by the gaseous molecules.
The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on V and P, respectively. Because nonzero molecular volumes produce a measured volume that is larger than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is less than that expected based on the ideal gas law, so the $an^2/V^2$ term must be added to the measured pressure to correct for these effects.
Example $1$
You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)?
Given: volume of cylinder, mass of compound, pressure, and temperature
Asked for: safety
Strategy:
A Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law.
B Obtain a and b values for Cl2 from Table $1$. Use the van der Waals equation ($\ref{10.9.1}$) to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture.
Solution:
A We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol):
\begin{align} n &=\dfrac{m}{M} \[4pt] &= \rm\dfrac{500\;g}{70.906\;g/mol} \[4pt] &=7.052\;mol\nonumber \end{align} \nonumber
Using the ideal gas law and the temperature in kelvin (298 K), we calculate the pressure:
\begin{align} P &=\dfrac{nRT}{V} \[4pt] &=\rm\dfrac{7.052\;mol\times 0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L} \[4pt] &= 43.1\;atm \end{align} \nonumber
If chlorine behaves like an ideal gas, you have a real problem!
B Now let’s use the van der Waals equation with the a and b values for Cl2 from Table $1$. Solving for $P$ gives
\begin{align}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\&=\rm28.2\;atm\end{align} \nonumber
This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation.
Exercise $1$
A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the
1. ideal gas law.
2. van der Waals equation.
Answer a
77 atm
Answer b
67 atm
Liquefaction of Gases
Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure).
Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals $a$ coefficients are relatively easy to liquefy because large a coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small a coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O2, N2, Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points.
A large value of a in the van der Waals equation indicates the presence of relatively strong intermolecular attractive interactions.
The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek kryo, meaning “cold,” and genes, meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called cryosurgery, which selectively destroys tissues with a minimal loss of blood by the use of extreme cold.
Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure $5$). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles.
Summary
No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of PV/nRT versus P at a given temperature; for an ideal gas, PV/nRT versus P = 1 under all conditions. At high pressures, most real gases exhibit larger PV/nRT values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit PV/nRT values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the van der Waals equation, which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, liquefaction occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold cryogenic liquids.
11.E: Exercises
11.5: Applications of the Ideal Gas Law: Molar Volume, Density and Molar Mass of a Gas
1. If 100 mL of $\ce{HCl}$ gas at 300 K and 100 kPa dissolves in 20 mL of pure water, what is the concentration?
Hint: 0.4 mol/L
Skill:
Calculate n using ideal gas law.
2. If 100 mL of $\ce{HCl}$ gas at 300 K and 100 kPa dissolved in pure water requires 12.50 mL of the $\ce{NaOH}$ solution to neutralize in a titration experiment, what is the concentration of the $\ce{NaOH}$ solution?
Hint: 0.32 mol/L
Skill:
Apply ideal gas law to solve stoichiometry problems.
3. If 100 mL of $\ce{HCl}$ gas at 300 K and 200 kPa dissolved in pure water requires 12.50 mL of the $\ce{NaOH}$ solution to neutralize in a titration experiment, what is the concentration of the $\ce{NaOH}$ solution?
Hint: 0.64 mol/L
Skill:
Solve stoichiometric problem.
4. Hydrogen sulfide reacts with sulfur dioxide to give $\ce{H2O}$ and $\ce{S}$,
$\ce{H2S + SO2 \rightarrow H2O + S_{(solid)}}$, unbalanced
If 3.0 L of $\ce{H2S}$ gas at 760 torr produced 4.8 g of sulfur, calculate the temperature in C.
Hint: 93 degrees C
Skill:
Apply ideal gas law to solve stoichiometry problems.
5. When 10.0 mL of $\ce{AgNO3}$ solution is treated with excess amount of $\ce{HI}$ gas to give 0.235 g of $\ce{AgI}$, what is the concentration of the $\ce{AgNO3}$ solution?
Hint: 0.10 M
6. When an $\ce{AgNO3}$ solution is treated with 50.0 mL of $\ce{HI}$ gas to give 0.235 g of $\ce{AgI}$, what is the concentration of the $\ce{HI}$ gas?
Hint: 0.020 mol/L
7. When an $\ce{AgNO3}$ solution is treated with 50.0 mL of $\ce{HI}$ gas at 300 K to give 0.235 g of $\ce{AgI}$, what is the pressure of the $\ce{HI}$ gas?
Hint: 0.49 atm
Discussion:
Depending on the numerical values you use, you may get the pressure in other units.
8. When an $\ce{AgNO3}$ solution is treated with 50.0 mL of $\ce{HI}$ gas at 374 torr to give 0.235 g of $\ce{AgI}$, what is the temperature of the $\ce{HI}$ gas?
Hint: 300 K
Discussion:
Note the relationship of this problem with the previous one. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.11%3A_Real_Gases-_The_Effects_of_Size_and_Intermolecular_Forces.txt |
• 12.1: Structure Determines Properties
If you go far enough out in space, for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth. Just how might water act in a place of zero gravity?
• 12.2: Solids, Liquids, and Gases- A Molecular Comparison
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance and the intermolecular forces try to draw the particles together.
• 12.3: Intermolecular Forces- The Forces that Hold Condensed Phases Together
Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold molecules and polyatomic ions together. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds.
• 12.4: Intermolecular Forces in Action- Surface Tension, Viscosity, and Capillary Action
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid. Surfactants are molecules that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. The viscosity of a liquid is its resistance to flow.
• 12.5: Vaporization and Vapor Pressure
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state or dynamic equilibrium is reached.
• 12.6: Sublimation and Fusion
The heat energy which a solid absorbs when it melts is called the enthalpy of fusion or heat of fusion and is usually quoted on a molar basis. (The word fusion means the same thing as “melting.”)
• 12.7: Heating Curve for Water
Freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve.
• 12.8: Water - An Extraordinary Substance
Water is an unusual compound with unique physical properties. As a result, its the compound of life. Yet, its the most abundant compound in the biosphere of Earth. These properties are related to its electronic structure, bonding, and chemistry. However, due to its affinity for a variety of substances, ordinary water contains other substances. Few of us has used, seen or tested pure water, based on which we discuss its chemistry.
12: Liquids Solids and Intermolecular Forces
Here on Earth, we all live in a state of gravity. Not only us, but everything around us, including water, is being pulled towards the center of the planet by gravity. True, it is nice that our dogs don't float off into space, but when a child drops their ice cream (which is full of water, by the way) they don't have to know about gravity to be upset
Floating Water in Zero Gravity: Once again, astronauts on the International Space Station dissolved an effervescent tablet in a floating ball of water, and captured images using a camera capable of recording four times the resolution of normal high-definition cameras. The higher resolution images and higher frame rate videos can reveal more information when used on science investigations, giving researchers a valuable new tool aboard the space station. This footage is one of the first of its kind. The cameras are being evaluated for capturing science data and vehicle operations by engineers at NASA's Marshall Space Flight Center in Huntsville, Alabama.
If you go far enough out in space, for instance, onto the International Space Station, gravity becomes negligible, and the laws of physics act differently than here on Earth. Just how might water act in a place of zero gravity? This video above from NASA gives you a good idea of how different water behaves when the effects if gravity are counteracted.
Actually, on the International Space Station, there is plenty of gravity—according to NASA scientists, the pull of Earth's gravity on the space station and its occupants is substantial: about 90 percent of the force at the Earth's surface. But since the space station is continuously falling around our planet, the astronauts and objects on board are in a kind of free-fall, too, and feel nearly weightless. Water on the space station behaves as if in a zero-gravity environment.
This unique picture shows not only a water drop but also an air bubble inside of the water drop. Notice they both behave the same....according to the laws of physics in space. They both form spheres. This makes sense, as without gravity to tug downward, the forces governing the objects are all the same. So, the water drop (and air bubble) form themselves so they occupy a shape having the least amount of surface area, which is a sphere. On Earth, gravity distorts the shape, but not in space.
Consider what would happen on Earth: The air bubble, lighter than water, would race upward to burst through the surface of the droplet. In space, the air bubble doesn't rise because it is no lighter than the water around it—there's no buoyancy. The droplet doesn't fall from the leaf because there's no force to pull it off. It's stuck there by molecular adhesion.
Sticky water. No buoyancy. These are some of the factors space-farers must take into account when they plan their space gardens. If water is sprayed onto the base of the plant will it trickle down to the roots? More likely it will stick to the stem or adhere to the material in which the plant grows. As humans spend more time and go farther out in space in the future, the physics of "space water" will need to be well understood. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.01%3A_Structure_Determines_Properties.txt |
Learning Objectives
• To be familiar with the kinetic molecular description of liquids.
The physical properties of a substance depends upon its physical state. Water vapor, liquid water and ice all have the same chemical properties, but their physical properties are considerably different. In general covalent bonds determine: molecular shape, bond energies, chemical properties, while intermolecular forces (non-covalent bonds) influence the physical properties of liquids and solids. The kinetic molecular theory of gases gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces.
The state of a substance depends on the balance between the kinetic energy of the individual particles (molecules or atoms) and the intermolecular forces. The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. The intermolecular forces are attractive forces that try to draw the particles together (Figure \(2\)). A discussed previously, gasses are very sensitive to temperatures and pressure. However, these also affect liquids and solids too. Heating and cooling can change the kinetic energy of the particles in a substance, and so, we can change the physical state of a substance by heating or cooling it. Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.
Below is an overview of the general properties of the three different phases of matter.
Properties of Gases
• A collection of widely separated molecules
• The kinetic energy of the molecules is greater than any attractive forces between the molecules
• The lack of any significant attractive force between molecules allows a gas to expand to fill its container
• If attractive forces become large enough, then the gases exhibit non-ideal behavior
Properties of Liquids
• The intermolecular attractive forces are strong enough to hold molecules close together
• Liquids are more dense and less compressible than gasses
• Liquids have a definite volume, independent of the size and shape of their container
• The attractive forces are not strong enough, however, to keep neighboring molecules in a fixed position and molecules are free to move past or slide over one another
Thus, liquids can be poured and assume the shape of their containers.
Properties of Solids
• The intermolecular forces between neighboring molecules are strong enough to keep them locked in position
• Solids (like liquids) are not very compressible due to the lack of space between molecules
• If the molecules in a solid adopt a highly ordered packing arrangement, the structures are said to be crystalline
Due to the strong intermolecular forces between neighboring molecules, solids are rigid.
• Cooling a gas may change the state to a liquid
• Cooling a liquid may change the state to a solid
• Increasing the pressure on a gas may change the state to a liquid
• Increasing the pressure on a liquid may change the state to a solid
Physical Properties of Liquids
In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them. As in gases, however, the molecules in liquids are in constant motion, and their kinetic energy (and hence their speed) depends on their temperature. We begin our discussion by examining some of the characteristic properties of liquids to see how each is consistent with a modified kinetic molecular description.
The properties of liquids can be explained using a modified version of the kinetic molecular theory of gases described previously. This model explains the higher density, greater order, and lower compressibility of liquids versus gases; the thermal expansion of liquids; why they diffuse; and why they adopt the shape (but not the volume) of their containers. A kinetic molecular description of liquids must take into account both the nonzero volumes of particles and the presence of strong intermolecular attractive forces. Solids and liquids have particles that are fairly close to one another, and are thus called "condensed phases" to distinguish them from gases
• Density: The molecules of a liquid are packed relatively close together. Consequently, liquids are much denser than gases. The density of a liquid is typically about the same as the density of the solid state of the substance. Densities of liquids are therefore more commonly measured in units of grams per cubic centimeter (g/cm3) or grams per milliliter (g/mL) than in grams per liter (g/L), the unit commonly used for gases.
• Molecular Order: Liquids exhibit short-range order because strong intermolecular attractive forces cause the molecules to pack together rather tightly. Because of their higher kinetic energy compared to the molecules in a solid, however, the molecules in a liquid move rapidly with respect to one another. Thus unlike the ions in the ionic solids, the molecules in liquids are not arranged in a repeating three-dimensional array. Unlike the molecules in gases, however, the arrangement of the molecules in a liquid is not completely random.
• Compressibility: Liquids have so little empty space between their component molecules that they cannot be readily compressed. Compression would force the atoms on adjacent molecules to occupy the same region of space.
• Thermal Expansion: The intermolecular forces in liquids are strong enough to keep them from expanding significantly when heated (typically only a few percent over a 100°C temperature range). Thus the volumes of liquids are somewhat fixed. Notice from Table S1 (with a shorten version in Table \(1\)) that the density of water, for example, changes by only about 3% over a 90-degree temperature range.
Table \(1\): The Density of Water at Various Temperatures
T (°C) Density (g/cm3)
0 0.99984
30 0.99565
60 0.98320
90 0.96535
• Diffusion: Molecules in liquids diffuse because they are in constant motion. A molecule in a liquid cannot move far before colliding with another molecule, however, so the mean free path in liquids is very short, and the rate of diffusion is much slower than in gases.
• Fluidity: Liquids can flow, adjusting to the shape of their containers, because their molecules are free to move. This freedom of motion and their close spacing allow the molecules in a liquid to move rapidly into the openings left by other molecules, in turn generating more openings, and so forth (Figure \(3\)). | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.02%3A_Solids_Liquids_and_Gases-_A_Molecular_Comparison.txt |
Learning Objectives
• To describe the intermolecular forces in liquids.
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids.
Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures.
In this section, we explicitly consider three kinds of intermolecular interactions. There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding, and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\).
These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions, whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions. Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(2\). On average, however, the attractive interactions dominate.
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/r, where r is the distance between the ions. Doubling the distance (r → 2r) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/r3, so doubling the distance between the dipoles decreases the strength of the interaction by 23, or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure. Conversely, \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(1\).
Table \(1\): Relationships Between the Dipole Moment and the Boiling Point for Organic Compounds of Similar Molar Mass
Compound Molar Mass (g/mol) Dipole Moment (D) Boiling Point (K)
C3H6 (cyclopropane) 42 0 240
CH3OCH3 (dimethyl ether) 46 1.30 248
CH3CN (acetonitrile) 41 3.9 355
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Video Discussing Dipole Intermolecular Forces. Source: Dipole Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(1\)
Arrange ethyl methyl ether (CH3OCH2CH3), 2-methylpropane [isobutane, (CH3)2CHCH3], and acetone (CH3COCH3) in order of increasing boiling points. Their structures are as follows:
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds.
The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.
Ethyl methyl ether has a structure similar to H2O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point.
Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point.
Thus we predict the following order of boiling points:
2-methylpropane < ethyl methyl ether < acetone
This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Exercise \(1\)
Arrange carbon tetrafluoride (CF4), ethyl methyl sulfide (CH3SC2H5), dimethyl sulfoxide [(CH3)2S=O], and 2-methylbutane [isopentane, (CH3)2CHCH2CH3] in order of decreasing boiling points.
Answer
dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far, we have considered only interactions between polar molecules. Other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature; why others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(2\)).
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
Table \(2\): Normal Melting and Boiling Points of Some Elements and Nonpolar Compounds
Substance Molar Mass (g/mol) Melting Point (°C) Boiling Point (°C)
Ar 40 −189.4 −185.9
Xe 131 −111.8 −108.1
N2 28 −210 −195.8
O2 32 −218.8 −183.0
F2 38 −219.7 −188.1
I2 254 113.7 184.4
CH4 16 −182.5 −161.5
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(3\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/r6. Doubling the distance therefore decreases the attractive energy by 26, or 64-fold.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H2 molecules in part (b) in Figure \(3\), tends to become more pronounced as atomic and molecular masses increase (Table \(2\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus, London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(4\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(4\) shows 2,2-dimethylpropane (neopentane) and n-pentane, both of which have the empirical formula C5H12. Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n-pentane has an extended conformation that enables it to come into close contact with other n-pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n-pentane (36.1°C).
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Video Discussing London/Dispersion Intermolecular Forces. Source: Dispersion Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(2\)
Arrange n-butane, propane, 2-methylpropane [isobutene, (CH3)2CHCH3], and n-pentane in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Determine the intermolecular forces in the compounds, and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n-pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n-butane has the more extended shape. Consequently, we expect intermolecular interactions for n-butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n-butane (−0.5°C) < n-pentane (36.1°C).
Exercise \(2\)
Arrange GeH4, SiCl4, SiH4, CH4, and GeCl4 in order of decreasing boiling points.
Answer
GeCl4 (87°C) > SiCl4 (57.6°C) > GeH4 (−88.5°C) > SiH4 (−111.8°C) > CH4 (−161°C)
Hydrogen Bonds
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent, Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(5\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(6\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water, rather than sinks.
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cage like structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Video Discussing Hydrogen Bonding Intermolecular Forces. Source: Hydrogen Bonding Intermolecular Force, YouTube(opens in new window) [youtu.be]
Example \(3\)
Considering CH3OH, C2H6, Xe, and (CH3)3N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
1. Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
2. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A. Of the species listed, xenon (Xe), ethane (C2H6), and trimethylamine [(CH3)3N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B. The one compound that can act as a hydrogen bond donor, methanol (CH3OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Exercise \(3\)
Considering CH3CO2H, (CH3)3N, NH3, and CH3F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Answer
CH3CO2H and NH3;
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH3. Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Example \(4\): Buckyballs
Arrange C60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N2O in order of increasing boiling points.
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N2O have very similar molar masses (40 and 44 g/mol, respectively), but N2O is polar while Ar is not. Consequently, N2O should have a higher boiling point. A C60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N2O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses:
He (−269°C) < Ar (−185.7°C) < N2O (−88.5°C) < C60 (>280°C) < NaCl (1465°C).
Exercise \(4\)
Arrange 2,4-dimethylheptane, Ne, CS2, Cl2, and KBr in order of decreasing boiling points.
Answer
KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS2 (46.6°C) > Cl2 (−34.6°C) > Ne (−246°C)
Example \(5\)
Identify the most significant intermolecular force in each substance.
1. C3H8
2. CH3OH
3. H2S
Solution
1. Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
2. This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
3. Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Exercise \(6\)
Identify the most significant intermolecular force in each substance.
1. HF
2. HCl
Answer a
hydrogen bonding
Answer b
dipole-dipole interactions
Summary
Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid, or the liquid and gas phases, are due to changes in intermolecular interactions, but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/r3, where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules; their energy falls off as 1/r6. Larger atoms tend to be more polarizable than smaller ones, because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cage like structure that is less dense than liquid water. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.03%3A_Intermolecular_Forces-_The_Forces_that_Hold_Condensed_Phases_Together.txt |
Learning Objectives
• To describe the unique properties of liquids.
Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to three unique properties of liquids that intimately depend on the nature of intermolecular interactions:
• surface tension,
• capillary action, and
• viscosity.
Surface Tension
If liquids tend to adopt the shapes of their containers, then why do small amounts of water on a freshly waxed car form raised droplets instead of a thin, continuous film? The answer lies in a property called surface tension, which depends on intermolecular forces. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, e.g., water with hydrogen bonds has a surface tension of 7.29 x 10-2 J/m2 (at 20°C), while mercury with metallic bonds has as surface tension that is 15 times higher: 4.86 x 10-1 J/m2 (at 20°C).
Figure \(1\) presents a microscopic view of a liquid droplet. A typical molecule in the interior of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no net force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the surface experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads. A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads.
The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends above the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water. Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking. This is even observable in the zero gravity conditions of space as shown in Figure \(2\) (and more so in the video link) where water wrung from a wet towel continues to float along the towel's surface!
Such phenomena are manifestations of surface tension, which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m2) or dyne per centimeter (dyn/cm), where 1 dyn = 1 × 10−5 N. The values of the surface tension of some representative liquids are listed in Table \(1\). Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding.
Table \(1\): Surface Tension, Viscosity, Vapor Pressure (at 25°C Unless Otherwise Indicated), and Normal Boiling Points of Common Liquids
Substance Surface Tension (× 10−3 J/m2) Viscosity (mPa•s) Vapor Pressure (mmHg) Normal Boiling Point (°C)
Organic Compounds
diethyl ether 17 0.22 531 34.6
n-hexane 18 0.30 149 68.7
acetone 23 0.31 227 56.5
ethanol 22 1.07 59 78.3
ethylene glycol 48 16.1 ~0.08 198.9
Liquid Elements
bromine 41 0.94 218 58.8
mercury 486 1.53 0.0020 357
Water
0°C 75.6 1.79 4.6
20°C 72.8 1.00 17.5
60°C 66.2 0.47 149
100°C 58.9 0.28 760
Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactants. In the 1960s, US Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids.
Capillary Action
Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a capillary), as shown in Figure \(3\). When a glass capillary is is placed in liquid water, water rises up into the capillary. The height to which the water rises depends on the diameter of the tube and the temperature of the water but not on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises.
• Cohesive forces bind molecules of the same type together
• Adhesive forces bind a substance to a surface
Capillary action is the net result of two opposing sets of forces: cohesive forces, which are the intermolecular forces that hold a liquid together, and adhesive forces, which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar SiOH groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (Figure \(4\)). The upper surface of a liquid in a tube is called the meniscus, and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (Figure \(4\)).
Polar substances are drawn up a glass capillary and generally have a concave meniscus.
Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body.
Viscosity
Viscosity (η) is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a low viscosity. Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a high viscosity. The two most common methods for evaluating the viscosity of a liquid are (1) to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and (2) to measure the time it takes steel balls to fall through a given volume of the liquid. The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa•s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in Table 11.3.1 and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH2CH2OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces.
There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous.
Viscosity increases as intermolecular interactions or molecular size increases.
Video Discussing Surface Tension and Viscosity. Video Link: Surface Tension, Viscosity, & Melting Point, YouTube(opens in new window) [youtu.be]
Application: Motor Oils
Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing. As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures.
The viscosity of motor oils is described by an SAE (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity (Figure \(5\)). So-called single-grade oils can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are multigrade, with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils.
Example \(1\)
Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.)
Given: substance and composition of the glass surface
Asked for: behavior of oil and the shape of meniscus
Strategy:
1. Identify the cohesive forces in the motor oil.
2. Determine whether the forces interact with the surface of glass. From the strength of this interaction, predict the behavior of the oil and the shape of the meniscus.
Solution
A Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains.
B Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury.
Exercise \(1\)
Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)?
Answer
Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave.
Summary
Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. Surface tension is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. Surfactants are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. Capillary action is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when cohesive forces, the intermolecular forces in the liquid, are weaker than adhesive forces, the attraction between a liquid and the surface of the capillary. The shape of the meniscus, the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The viscosity of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.04%3A_Intermolecular_Forces_in_Action-_Surface_Tension_Viscosity_and_Capillary_Action.txt |
Learning Objectives
• To know how and why the vapor pressure of a liquid varies with temperature.
• To understand that the equilibrium vapor pressure of a liquid depends on the temperature and the intermolecular forces present.
• To understand that the relationship between pressure, enthalpy of vaporization, and temperature is given by the Clausius-Clapeyron equation.
Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid.
Evaporation and Condensation
Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy (KE) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure $1$), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ($E_0$) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than $E_0$. The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than $E_0$ has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure.
To understand the causes of vapor pressure, consider the apparatus shown in Figure $2$. When a liquid is introduced into an evacuated chamber (part (a) in Figure $2$), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with $KE > E_0$ will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase.
As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure $2$). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a steady state will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure $3$.
Equilibrium Vapor Pressure
Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no net change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid.
If a liquid is in an open container, however, most of the molecules that escape into the vapor phase will not collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure $4$) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile.
The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does not depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure $4$. Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is much steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release.
Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions.
A Video Discussing Vapor Pressure and Boiling Points. Video Source: Vapor Pressure & Boiling Point(opens in new window) [youtu.be]
The exponential rise in vapor pressure with increasing temperature in Figure $4$ allows us to use natural logarithms to express the nonlinear relationship as a linear one.
$\boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1}$
where
• $\ln P$ is the natural logarithm of the vapor pressure,
• $ΔH_{vap}$ is the enthalpy of vaporization,
• $R$ is the universal gas constant [8.314 J/(mol•K)],
• $T$ is the temperature in kelvins, and
• $C$ is the y-intercept, which is a constant for any given line.
Plotting $\ln P$ versus the inverse of the absolute temperature ($1/T$) is a straight line with a slope of −ΔHvap/R. Equation $\ref{Eq1}$, called the Clausius–Clapeyron Equation, can be used to calculate the $ΔH_{vap}$ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine $ΔH_{vap}$ is to measure the vapor pressure of a liquid at two temperatures and insert the values of $P$ and $T$ for these points into Equation $\ref{Eq2}$, which is derived from the Clausius–Clapeyron equation:
$\ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2}$
Conversely, if we know ΔHvap and the vapor pressure $P_1$ at any temperature $T_1$, we can use Equation $\ref{Eq2}$ to calculate the vapor pressure $P_2$ at any other temperature $T_2$, as shown in Example $1$.
A Video Discussing the Clausius-Clapeyron Equation. Video Link: The Clausius-Clapeyron Equation(opens in new window) [youtu.be]
Example $1$: Vapor Pressure of Mercury
The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table:
experimentally measured vapor pressures of liquid Hg at four temperatures
T (°C) 80.0 100 120 140
P (torr) 0.0888 0.2729 0.7457 1.845
From these data, calculate the enthalpy of vaporization (ΔHvap) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.)
Given: vapor pressures at four temperatures
Asked for: ΔHvap of mercury and vapor pressure at 160°C
Strategy:
1. Use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values in the table, making sure to convert all values to the appropriate units.
2. Substitute the calculated value of ΔHvap into Equation $\ref{Eq2}$ to obtain the unknown pressure (P2).
Solution:
A The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation $\ref{Eq1}$ and find the value of ΔHvap from the slope of the line, an alternative approach is to use Equation $\ref{Eq2}$ to obtain ΔHvap directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C (T1) and 120.0°C (T2) into Equation $\ref{Eq2}$ gives
\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber
B We can now use this value of ΔHvap to calculate the vapor pressure of the liquid (P2) at 160.0°C (T2):
$\ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber$
Using the relationship $e^{\ln x} = x$, we have
\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber
At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect.
Exercise $1$: Vapor Pressure of Nickel
The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr?
Answer
1896°C
Boiling Points
As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure $4$ are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the actual boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C.
Table $1$: The Boiling Points of Water at Various Locations on Earth
Place Altitude above Sea Level (ft) Atmospheric Pressure (mmHg) Boiling Point of Water (°C)
Mt. Everest, Nepal/Tibet 29,028 240 70
Bogota, Colombia 11,490 495 88
Denver, Colorado 5280 633 95
Washington, DC 25 759 100
Dead Sea, Israel/Jordan −1312 799 101.4
Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table $1$ lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked.
As pressure increases, the boiling point of a liquid increases and vice versa.
Example $2$: Boiling Mercury
Use Figure $4$ to estimate the following.
1. the boiling point of water in a pressure cooker operating at 1000 mmHg
2. the pressure required for mercury to boil at 250°C
Given: Data in Figure $4$, pressure, and boiling point
Asked for: corresponding boiling point and pressure
Strategy:
1. To estimate the boiling point of water at 1000 mmHg, refer to Figure $4$ and find the point where the vapor pressure curve of water intersects the line corresponding to a pressure of 1000 mmHg.
2. To estimate the pressure required for mercury to boil at 250°C, find the point where the vapor pressure curve of mercury intersects the line corresponding to a temperature of 250°C.
Solution:
1. A The vapor pressure curve of water intersects the P = 1000 mmHg line at about 110°C; this is therefore the boiling point of water at 1000 mmHg.
2. B The vertical line corresponding to 250°C intersects the vapor pressure curve of mercury at P ≈ 75 mmHg. Hence this is the pressure required for mercury to boil at 250°C.
Exercise $2$: Boiling Ethlyene Glycol
Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure $4$ to estimate the following.
1. the normal boiling point of ethylene glycol
2. the pressure required for diethyl ether to boil at 20°C.
Answer a
200°C
Answer b
450 mmHg
Summary
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via condensation. Eventually, a steady state is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of dynamic equilibrium. Under these conditions, a liquid exhibits a characteristic equilibrium vapor pressure that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the Clausius–Clapeyron equation. This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. Volatile liquids are liquids with high vapor pressures, which tend to evaporate readily from an open container; nonvolatile liquids have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its normal boiling point. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.05%3A_Vaporization_and_Vapor_Pressure.txt |
Sublimation
Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid CO2) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure $6$). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.
Figure $6$: Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)
Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:
$\ce{CO2}(s)⟶\ce{CO2}(g)\hspace{20px}ΔH_\ce{sub}=\mathrm{26.1\: kJ/mol}$
Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:
$\ce{CO2}(g)⟶\ce{CO2}(s)\hspace{20px}ΔH_\ce{dep}=−ΔH_\ce{sub}=\mathrm{−26.1\:kJ/mol}$
Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law.
$\mathrm{solid⟶liquid}\hspace{20px}ΔH_\ce{fus}\\underline{\mathrm{liquid⟶gas}\hspace{20px}ΔH_\ce{vap}}\\mathrm{solid⟶gas}\hspace{20px}ΔH_\ce{sub}=ΔH_\ce{fus}+ΔH_\ce{vap}$
Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure $7$. For example:
Fusion
When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure $5$.
Figure $5$: (a) This beaker of ice has a temperature of −12.0 °C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 °C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 °C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 °C. (credit: modification of work by Mark Ott).
If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal process of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).
The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.
The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, $ΔH_{fus}$ of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process:
$\ce{H2O}_{(s)} \rightarrow \ce{H2O}_{(l)} \;\; ΔH_\ce{fus}=\mathrm{6.01\; kJ/mol} \label{10.4.9}$
The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C:
$\ce{H_2O}_{(l)} \rightarrow \ce{H_2O}_{(s)}\;\; ΔH_\ce{frz}=−ΔH_\ce{fus}=−6.01\;\mathrm{kJ/mol} \label{10.4.10}$
Selected molar enthalpies of fusion are tabulated in Table $1$. Solids like ice which have strong intermolecular forces have much higher values than those like CH4 with weak ones. Note that the enthalpies of fusion and vaporization change with temperature.
Table $1$: Molar Enthalpies of Fusion and Vaporization of Selected Substances.
Substance Formula ΔH(fusion)
/ kJ mol1
Melting Point / K ΔH(vaporization) / kJ mol-1 Boiling Point / K (ΔHv/Tb)
/ JK-1 mol-1
Neon Ne 0.33 24 1.80 27 67
Oxygen O2 0.44 54 6.82 90.2 76
Methane CH4 0.94 90.7 8.18 112 73
Ethane C2H6 2.85 90.0 14.72 184 80
Chlorine Cl2 6.40 172.2 20.41 239 85
Carbon tetrachloride CCl4 2.67 250.0 30.00 350 86
Water* H2O 6.00678 at 0°C, 101kPa
6.354 at 81.6 °C, 2.50 MPa
273.1 40.657 at 100 °C,
45.051 at 0 °C,
46.567 at -33 °C
373.1 109
n-Nonane C9H20 19.3 353 40.5 491 82
Mercury Hg 2.30 234 58.6 630 91
Sodium Na 2.60 371 98 1158 85
Aluminum Al 10.9 933 284 2600 109
Lead Pb 4.77 601 178 2022 88
*www1.lsbu.ac.uk/water/data.html | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.06%3A_Sublimation_and_Fusion.txt |
Freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve.
Heating Curves
Figure \(3\) shows a heating curve, a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat (Cs) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does not increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is greater than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a faster rate than seen in the other phases because the heat capacity of steam is less than that of ice or water.
Thus the temperature of a system does not change during a phase change. In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does not affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils.
The temperature of a sample does not change during a phase change.
If heat is added at a constant rate, as in Figure \(3\), then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure \(3\), the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion.
A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form.
Cooling Curves
The cooling curve, a plot of temperature versus cooling time, in Figure \(4\) plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in Figure \(3\), the cooling curve is not an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system.
Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a nucleus) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during seeding (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO2 (dry ice) into the cloud from an airplane. Solid CO2 sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO2 sublimes, it absorbs heat from the cloud, often with the desired results.
Example \(1\): Cooling Hot Tea
If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol.
Given: mass, volume, initial temperature, density, specific heats, and \(ΔH_{fus}\)
Asked for: final temperature
Strategy:
Substitute the values given into the general equation relating heat gained to heat lost (Equation 5.39) to obtain the final temperature of the mixture.
Solution:
When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by
\[q=mC_sΔT\]
where q is heat, m is mass, Cs is the specific heat, and ΔT is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C.
Exercise \(1\): Death by Freezing
Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example \(1\)
Answer
200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.07%3A_Heating_Curve_for_Water.txt |
Learning Objectives
• Identify three special properties of water that make it unusual for a molecule of its size, and explain how these result from hydrogen bonding.
• Explain what is meant by hydrogen bonding and the molecular structural features that bring it about.
• Describe the "structure", such as it is, of liquid water.
• Sketch out structural examples of hydrogen bonding in three small molecules other than H2O.
• Describe the roles of hydrogen bonding in proteins and in DNA.
Most students of chemistry quickly learn to relate the structure of a molecule to its general properties. Thus we generally expect small molecules to form gases or liquids, and large ones to exist as solids under ordinary conditions. And then we come to H2O, and are shocked to find that many of the predictions are way off, and that water (and by implication, life itself) should not even exist on our planet! In this section we will learn why this tiny combination of three nuclei and ten electrons possesses special properties that make it unique among the more than 15 million chemical species we presently know.
In water, each hydrogen nucleus is covalently bound to the central oxygen atom by a pair of electrons that are shared between them. In H2O, only two of the six outer-shell electrons of oxygen are used for this purpose, leaving four electrons which are organized into two non-bonding pairs. The four electron pairs surrounding the oxygen tend to arrange themselves as far from each other as possible in order to minimize repulsions between these clouds of negative charge. This would ordinarily result in a tetrahedral geometry in which the angle between electron pairs (and therefore the H-O-H bond angle) is 109.5°. However, because the two non-bonding pairs remain closer to the oxygen atom, these exert a stronger repulsion against the two covalent bonding pairs, effectively pushing the two hydrogen atoms closer together. The result is a distorted tetrahedral arrangement in which the H—O—H angle is 104.5°.
Water's large dipole moment leads to hydrogen bonding
The H2O molecule is electrically neutral, but the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an electric dipole, represented by the arrow at the bottom; you can think of this dipole as the electrical "image" of a water molecule.
Opposite charges attract, so it is not surprising that the negative end of one water molecule will tend to orient itself so as to be close to the positive end of another molecule that happens to be nearby. The strength of this dipole-dipole attraction is less than that of a normal chemical bond, and so it is completely overwhelmed by ordinary thermal motions in the gas phase. However, when the H2O molecules are crowded together in the liquid, these attractive forces exert a very noticeable effect, which we call (somewhat misleadingly) hydrogen bonding. And at temperatures low enough to turn off the disruptive effects of thermal motions, water freezes into ice in which the hydrogen bonds form a rigid and stable network.
Notice that the hydrogen bond (shown by the dashed green line) is somewhat longer than the covalent O—H bond. It is also much weaker, about 23 kJ mol–1 compared to the O–H covalent bond strength of 492 kJ mol–1.
Water has long been known to exhibit many physical properties that distinguish it from other small molecules of comparable mass. Although chemists refer to these as the "anomalous" properties of water, they are by no means mysterious; all are entirely predictable consequences of the way the size and nuclear charge of the oxygen atom conspire to distort the electronic charge clouds of the atoms of other elements when these are chemically bonded to the oxygen.
The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(6\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor) and the atom that has the lone pair of electrons (the hydrogen bond acceptor). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks.
Figure \(6\): The Hydrogen-Bonded Structure of Ice.
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Boiling Point
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(5\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H2Te and H2Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles.
Ice Floats on Water
The most energetically favorable configuration of H2O molecules is one in which each molecule is hydrogen-bonded to four neighboring molecules. Owing to the thermal motions described above, this ideal is never achieved in the liquid, but when water freezes to ice, the molecules settle into exactly this kind of an arrangement in the ice crystal. This arrangement requires that the molecules be somewhat farther apart then would otherwise be the case; as a consequence, ice, in which hydrogen bonding is at its maximum, has a more open structure, and thus a lower density than water.
Here are three-dimensional views of a typical local structure of water (left) and ice (right.) Notice the greater openness of the ice structure which is necessary to ensure the strongest degree of hydrogen bonding in a uniform, extended crystal lattice. The more crowded and jumbled arrangement in liquid water can be sustained only by the greater amount of thermal energy available above the freezing point.
When ice melts, the more vigorous thermal motion disrupts much of the hydrogen-bonded structure, allowing the molecules to pack more closely. Water is thus one of the very few substances whose solid form has a lower density than the liquid at the freezing point. Localized clusters of hydrogen bonds still remain, however; these are continually breaking and reforming as the thermal motions jiggle and shove the individual molecules. As the temperature of the water is raised above freezing, the extent and lifetimes of these clusters diminish, so the density of the water increases.
At higher temperatures, another effect, common to all substances, begins to dominate: as the temperature increases, so does the amplitude of thermal motions. This more vigorous jostling causes the average distance between the molecules to increase, reducing the density of the liquid; this is ordinary thermal expansion. Because the two competing effects (hydrogen bonding at low temperatures and thermal expansion at higher temperatures) both lead to a decrease in density, it follows that there must be some temperature at which the density of water passes through a maximum. This temperature is 4° C; this is the temperature of the water you will find at the bottom of an ice-covered lake in which this most dense of all water has displaced the colder water and pushed it nearer to the surface.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.08%3A_Water_-_An_Extraordinary_Substance.txt |
• 13.1: Sliding Glaciers
• 13.2: Phase Diagrams
The states of matter exhibited by a substance under different temperatures and pressures can be summarized graphically in a phase diagram, which is a plot of pressure versus temperature. Phase diagrams contain discrete regions corresponding to the solid, liquid, and gas phases. The solid and liquid regions are separated by the melting curve of the substance, and the liquid and gas regions are separated by its vapor pressure curve, which ends at the critical point.
• 13.3: Crystalline Solids- Determining Their Structure by X-Ray Crystallography
Since X-ray photons are very energetic, they have relatively short wavelengths. Thus, typical X-ray photons act like rays when they encounter macroscopic objects, like teeth, and produce sharp shadows. However, since atoms are on the order of 0.1 nm in size, X-rays can be used to detect the location, shape, and size of atoms and molecules. The process is called X-ray diffraction, and it involves the interference of X-rays to produce patterns.
• 13.4: Crystalline Solids- Unit Cells and Basic Structures
When substances form solids, they tend to pack together to form ordered arrays of atoms, ions, or molecules that we call crystals. Why does this order arise, and what kinds of arrangements are possible? We will limit our discussion to cubic crystals, which form the simplest and most symmetric of all the lattice types. Cubic lattices are also very common — they are formed by many metallic crystals, and also by most of the alkali halides, several of which we will study as examples.
• 13.5: Crystalline Solids- The Fundamental Types
Some substances form crystalline solids consisting of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions bet
• 13.6: The Structure of Ionic Solids
In this section we deal mainly with a very small but imporant class of solids that are commonly regarded as composed of ions. We will see how the relative sizes of the ions determine the energetics of such compounds. And finally, we will point out that not all solids that are formally derived from ions can really be considered "ionic" at all.
• 13.7: Network Covalent Atomic Solids- Carbon and Silicates
Covalent solids are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule.
13: Phase Diagrams and Crystalline Solids
Learning Objectives
• To understand the basics of a one-component phase diagram as a function of temperature and pressure in a closed system.
• To be able to identify the triple point, the critical point, and four regions: solid, liquid, gas, and a supercritical fluid.
The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system.
Introduction
A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (Figure $1$). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of Figure $1$) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid.
The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure.
The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In Figure $1$, the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in Section 11.5. It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the solid phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point; it is the only combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point.
Remember that a phase diagram, such as the one in Figure $1$, is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error.
The Phase Diagram of Water
Figure $2$ shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee).
The phase diagram for water illustrated in Figure $\PageIndex{2b}$ shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in Figure $1$; that is, the melting point of ice decreases with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is less dense than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice.
In Figure $\PageIndex{2b}$ point A is located at P = 1 atm and T = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines.
Ice Skating: An Incorrect Hypothesis of Phase Transitions
Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation.
Recall that pressure (P) is the force (F) applied per unit area (A):
$P=\dfrac{F}{A} \nonumber$
To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is
$F = mg \nonumber$
where m is the mass and g is the acceleration due to Earth’s gravity (9.81 m/s2). Thus the force is
$F = (75.0\; kg)(9.81\; m/s^2) = 736\; (kg•m)/s^2 = 736 N \nonumber$
If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is
$A = (2.0 \times 10^{−3}\; m)(25 \times 10^{−2}\; m) = 5.0 \times 10^{−4} m^2 \nonumber$
If the skater is gliding on one foot, the pressure exerted on the ice is
$P= \dfrac{736\;N}{5.0 \times 10^{-4} \; m^2} = 1.5 \times 10^6 \; N/m^2 = 1.5 \times 10^6\; Pa =15 \; atm \nonumber$
The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases.
Example $1$: Water
Referring to the phase diagram of water in Figure $2$:
1. predict the physical form of a sample of water at 400°C and 150 atm.
2. describe the changes that occur as the sample in part (a) is slowly allowed to cool to −50°C at a constant pressure of 150 atm.
Given: phase diagram, temperature, and pressure
Asked for: physical form and physical changes
Strategy:
1. Identify the region of the phase diagram corresponding to the initial conditions and identify the phase that exists in this region.
2. Draw a line corresponding to the given pressure. Move along that line in the appropriate direction (in this case cooling) and describe the phase changes.
Solution:
1. A Locate the starting point on the phase diagram in part (a) in Figure $2$. The initial conditions correspond to point A, which lies in the region of the phase diagram representing water vapor. Thus water at T = 400°C and P = 150 atm is a gas.
2. B Cooling the sample at constant pressure corresponds to moving left along the horizontal line in part (a) in Figure $2$. At about 340°C (point B), we cross the vapor pressure curve, at which point water vapor will begin to condense and the sample will consist of a mixture of vapor and liquid. When all of the vapor has condensed, the temperature drops further, and we enter the region corresponding to liquid water (indicated by point C). Further cooling brings us to the melting curve, the line that separates the liquid and solid phases at a little below 0°C (point D), at which point the sample will consist of a mixture of liquid and solid water (ice). When all of the water has frozen, cooling the sample to −50°C takes us along the horizontal line to point E, which lies within the region corresponding to solid water. At P = 150 atm and T = −50°C, therefore, the sample is solid ice.
Exercise $2$
Referring to the phase diagram of water in Figure $2$, predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm.
Answer
The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid.
The Phase Diagram of Carbon Dioxide
In contrast to the phase diagram of water, the phase diagram of CO2 (Figure $3$) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO2 cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO2 sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO2 is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed.
Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps.
The Critical Point
As the phase diagrams above demonstrate, a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature (Tc), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure (Pc), the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical point. The critical temperatures and pressures of several common substances are listed in Figure $1$.
Figure $1$: Critical Temperatures and Pressures of Some Simple Substances
Substance Tc (°C) Pc (atm)
NH3 132.4 113.5
CO2 31.0 73.8
CH3CH2OH (ethanol) 240.9 61.4
He −267.96 2.27
Hg 1477 1587
CH4 −82.6 46.0
N2 −146.9 33.9
H2O 374.0 217.7
High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa.
Supercritical Fluids
A Video Discussing Phase Diagrams. Video Source: Phase Diagrams(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/13%3A_Phase_Diagrams_and_Crystalline_Solids/13.02%3A_Phase_Diagrams.txt |
Learning Objectives
By the end of this section, you will be able to:
• Describe interference and diffraction effects exhibited by X-rays in interaction with atomic-scale structures
Since X-ray photons are very energetic, they have relatively short wavelengths, on the order of $10^{-8}$ m to $10^{-12}$ m. Thus, typical X-ray photons act like rays when they encounter macroscopic objects, like teeth, and produce sharp shadows. However, since atoms are on the order of 0.1 nm in size, X-rays can be used to detect the location, shape, and size of atoms and molecules. The process is called X-ray diffraction, and it involves the interference of X-rays to produce patterns that can be analyzed for information about the structures that scattered the X-rays.
Perhaps the most famous example of X-ray diffraction is the discovery of the double-helical structure of DNA in 1953 by an international team of scientists working at England’s Cavendish Laboratory—American James Watson, Englishman Francis Crick, and New Zealand-born Maurice Wilkins. Using X-ray diffraction data produced by Rosalind Franklin, they were the first to model the double-helix structure of DNA that is so crucial to life. For this work, Watson, Crick, and Wilkins were awarded the 1962 Nobel Prize in Physiology or Medicine. (There is some debate and controversy over the issue that Rosalind Franklin was not included in the prize, although she died in 1958, before the prize was awarded.)
Figure $1$ shows a diffraction pattern produced by the scattering of X-rays from a crystal. This process is known as X-ray crystallography because of the information it can yield about crystal structure, and it was the type of data Rosalind Franklin supplied to Watson and Crick for DNA. Not only do X-rays confirm the size and shape of atoms, they give information about the atomic arrangements in materials. For example, more recent research in high-temperature superconductors involves complex materials whose lattice arrangements are crucial to obtaining a superconducting material. These can be studied using X-ray crystallography.
Figure $1$: X-ray diffraction from the crystal of a protein (hen egg lysozyme) produced this interference pattern. Analysis of the pattern yields information about the structure of the protein. (credit: “Del45”/Wikimedia Commons)
Historically, the scattering of X-rays from crystals was used to prove that X-rays are energetic electromagnetic (EM) waves. This was suspected from the time of the discovery of X-rays in 1895, but it was not until 1912 that the German Max von Laue (1879–1960) convinced two of his colleagues to scatter X-rays from crystals. If a diffraction pattern is obtained, he reasoned, then the X-rays must be waves, and their wavelength could be determined. (The spacing of atoms in various crystals was reasonably well known at the time, based on good values for Avogadro’s number.) The experiments were convincing, and the 1914 Nobel Prize in Physics was given to von Laue for his suggestion leading to the proof that X-rays are EM waves. In 1915, the unique father-and-son team of Sir William Henry Bragg and his son Sir William Lawrence Bragg were awarded a joint Nobel Prize for inventing the X-ray spectrometer and the then-new science of X-ray analysis.
In ways reminiscent of thin-film interference, we consider two plane waves at X-ray wavelengths, each one reflecting off a different plane of atoms within a crystal’s lattice, as shown in Figure $2$. From the geometry, the difference in path lengths is $2d \, \sin \, \theta$. Constructive interference results when this distance is an integer multiple of the wavelength. This condition is captured by the Bragg equation,
$m\lambda = 2d \, \sin \, \theta,\label{Bragg}$
for $m = 1,2,3, ...$.
where $m$ is a positive integer and $d$ is the spacing between the planes. Following the Law of Reflection, both the incident and reflected waves are described by the same angle, $θ$, but unlike the general practice in geometric optics, $θ$ is measured with respect to the surface itself, rather than the normal.
Figure $2$: X-ray diffraction with a crystal. Two incident waves reflect off two planes of a crystal. The difference in path lengths is indicated by the dashed line.
Example $1$: X-Ray Diffraction with Salt Crystals
Common table salt is composed mainly of $\ce{NaCl}$ crystals. In a $\ce{NaCl}$ crystal, there is a family of planes 0.252 nm apart. If the first-order maximum is observed at an incidence angle of 18.1°, what is the wavelength of the X-ray scattering from this crystal?
Strategy:
Use the Bragg equation, Equation \ref{Bragg}, to solve for $θ$.
Solution
For first-order, $m = 1$, and the plane spacing $d$ is known. Solving the Bragg equation for wavelength yields
\begin{align*} \lambda &= \dfrac{2d \, \sin \, \theta}{m} \[4pt] &= \dfrac{2(0.252 \times 10^{-9} m) \, \sin \, (18.1^o)}{1} \[4pt] &= 1.57 \times 10^{-10} m, \, or \, 0.157 \, nm \end{align*} \nonumber
Significance
The determined wavelength fits within the X-ray region of the electromagnetic spectrum. Once again, the wave nature of light makes itself prominent when the wavelength ($\lambda = 0.157 \, nm$) is comparable to the size of the physical structures ($d = 0.252 \, nm$) it interacts with.
Exercise $1$
For the experiment described in Example $1$, what are the two other angles where interference maxima may be observed? What limits the number of maxima?
Answer
$38.4^o$ and $68.8^o$; Between $\theta = 0^o \rightarrow 90^o$, orders 1, 2, and 3, are all that exist.
Although Figure $2$ depicts a crystal as a two-dimensional array of scattering centers for simplicity, real crystals are structures in three dimensions. Scattering can occur simultaneously from different families of planes at different orientations and spacing patterns known as called Bragg planes, as shown in Figure $3$. The resulting interference pattern can be quite complex.
Figure $3$: Because of the regularity that makes a crystal structure, one crystal can have many families of planes within its geometry, each one giving rise to X-ray diffraction. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/13%3A_Phase_Diagrams_and_Crystalline_Solids/13.03%3A_Crystalline_Solids-_Determining_Their_Structure_by_X-Ray_Crystallography.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• The difference between square and hexagonal packing in two dimensions.
• The definition and significance of the unit cell.
• Sketch the three Bravais lattices of the cubic system, and calculate the number of atoms contained in each of these unit cells.
• Show how alternative ways of stacking three close-packed layers can lead to the hexagonal or cubic close packed structures.
• Explain the origin and significance of octahedral and tetrahedral holes in stacked close-packed layers, and show how they can arise.
Close-Packing of Identical Spheres
Crystals are of course three-dimensional objects, but we will begin by exploring the properties of arrays in two-dimensional space. This will make it easier to develop some of the basic ideas without the added complication of getting you to visualize in 3-D — something that often requires a bit of practice. Suppose you have a dozen or so marbles. How can you arrange them in a single compact layer on a table top? Obviously, they must be in contact with each other in order to minimize the area they cover. It turns out that there are two efficient ways of achieving this:
The essential difference here is that any marble within the interior of the square-packed array is in contact with four other marbles, while this number rises to six in the hexagonal-packed arrangement. It should also be apparent that the latter scheme covers a smaller area (contains less empty space) and is therefore a more efficient packing arrangement. If you are good at geometry, you can show that square packing covers 78 percent of the area, while hexagonal packing yields 91 percent coverage.
If we go from the world of marbles to that of atoms, which kind of packing would the atoms of a given element prefer?
If the atoms are identical and are bound together mainly by dispersion forces which are completely non-directional, they will favor a structure in which as many atoms can be in direct contact as possible. This will, of course, be the hexagonal arrangement.
Directed chemical bonds between atoms have a major effect on the packing. The version of hexagonal packing shown at the right occurs in the form of carbon known as graphite which forms 2-dimensional sheets. Each carbon atom within a sheet is bonded to three other carbon atoms. The result is just the basic hexagonal structure with some atoms missing.
The coordination number of 3 reflects the sp2-hybridization of carbon in graphite, resulting in plane-trigonal bonding and thus the sheet structure. Adjacent sheets are bound by weak dispersion forces, allowing the sheets to slip over one another and giving rise to the lubricating and flaking properties of graphite.
Lattices
The underlying order of a crystalline solid can be represented by an array of regularly spaced points that indicate the locations of the crystal's basic structural units. This array is called a crystal lattice. Crystal lattices can be thought of as being built up from repeating units containing just a few atoms. These repeating units act much as a rubber stamp: press it on the paper, move ("translate") it by an amount equal to the lattice spacing, and stamp the paper again.
The gray circles represent a square array of lattice points.
The orange square is the simplest unit cell that can be used to define the 2-dimensional lattice.
Building out the lattice by moving ("translating") the unit cell in a series of steps,
Although real crystals do not actually grow in this manner, this process is conceptually important because it allows us to classify a lattice type in terms of the simple repeating unit that is used to "build" it. We call this shape the unit cell. Any number of primitive shapes can be used to define the unit cell of a given crystal lattice. The one that is actually used is largely a matter of convenience, and it may contain a lattice point in its center, as you see in two of the unit cells shown here. In general, the best unit cell is the simplest one that is capable of building out the lattice.
Shown above are unit cells for the close-packed square and hexagonal lattices we discussed near the start of this lesson. Although we could use a hexagon for the second of these lattices, the rhombus is preferred because it is simpler.
Notice that in both of these lattices, the corners of the unit cells are centered on a lattice point. This means that an atom or molecule located on this point in a real crystal lattice is shared with its neighboring cells. As is shown more clearly here for a two-dimensional square-packed lattice, a single unit cell can claim "ownership" of only one-quarter of each molecule, and thus "contains" 4 × ¼ = 1 molecule.
The unit cell of the graphite form of carbon is also a rhombus, in keeping with the hexagonal symmetry of this arrangement. Notice that to generate this structure from the unit cell, we need to shift the cell in both the x- and y- directions in order to leave empty spaces at the correct spots. We could alternatively use regular hexagons as the unit cells, but the x+y shifts would still be required, so the simpler rhombus is usually preferred. As you will see in the next section, the empty spaces within these unit cells play an important role when we move from two- to three-dimensional lattices.
Cubic crystals
In order to keep this lesson within reasonable bounds, we are limiting it mostly to crystals belonging to the so-called cubic system. In doing so, we can develop the major concepts that are useful for understanding more complicated structures (as if there are not enough complications in cubics alone!) But in addition, it happens that cubic crystals are very commonly encountered; most metallic elements have cubic structures, and so does ordinary salt, sodium chloride.
We usually think of a cubic shape in terms of the equality of its edge lengths and the 90° angles between its sides, but there is another way of classifying shapes that chemists find very useful. This is to look at what geometric transformations (such as rotations around an axis) we can perform that leave the appearance unchanged. For example, you can rotate a cube 90° around an axis perpendicular to any pair of its six faces without making any apparent change to it. We say that the cube possesses three mutually perpendicular four-fold rotational axes, abbreviated C4 axes. But if you think about it, a cube can also be rotated around the axes that extend between opposite corners; in this case, it takes three 120° rotations to go through a complete circle, so these axes (also four in number) are three-fold or C3 axes.
Cubic crystals belong to one of the seven crystal systems whose lattice points can be extended indefinitely to fill three-dimensional space and which can be constructed by successive translations (movements) of a primitive unit cell in three dimensions. As we will see below, the cubic system, as well as some of the others, can have variants in which additional lattice points can be placed at the center of the unit or at the center of each face.
The three types of cubic lattices
The three Bravais lattices which form the cubic crystal system are shown here.
Structural examples of all three are known, with body- and face-centered (BCC and FCC) being much more common; most metallic elements crystallize in one of these latter forms. But although the simple cubic structure is uncommon by itself, it turns out that many BCC and FCC structures composed of ions can be regarded as interpenetrating combinations of two simple cubic lattices, one made up of positive ions and the other of negative ions. Notice that only the FCC structure, which we will describe below, is a close-packed lattice within the cubic system.
Close-packed lattices in three dimensions
Close-packed lattices allow the maximum amount of interaction between atoms. If these interactions are mainly attractive, then close-packing usually leads to more energetically stable structures. These lattice geometries are widely seen in metallic, atomic, and simple ionic crystals.
As we pointed out above, hexagonal packing of a single layer is more efficient than square-packing, so this is where we begin. Imagine that we start with the single layer of green atoms shown below. We will call this the A layer. If we place a second layer of atoms (orange) on top of the A-layer, we would expect the atoms of the new layer to nestle in the hollows in the first layer. But if all the atoms are identical, only some of these void spaces will be accessible.
In the diagram on the left, notice that there are two classes of void spaces between the A atoms; one set (colored blue) has a vertex pointing up, while the other set (not colored) has down-pointing vertices. Each void space constitutes a depression in which atoms of a second layer (the B-layer) can nest. The two sets of void spaces are completely equivalent, but only one of these sets can be occupied by a second layer of atoms whose size is similar to those in the bottom layer. In the illustration on the right above we have arbitrarily placed the B-layer atoms in the blue voids, but could just as well have selected the white ones.
Two choices for the third layer lead to two different close-packed lattice types
Now consider what happens when we lay down a third layer of atoms. These will fit into the void spaces within the B-layer. As before, there are two sets of these positions, but unlike the case described above, they are not equivalent.
The atoms in the third layer are represented by open blue circles in order to avoid obscuring the layers underneath. In the illustration on the left, this third layer is placed on the B-layer at locations that are directly above the atoms of the A-layer, so our third layer is just a another A layer. If we add still more layers, the vertical sequence A-B-A-B-A-B-A... repeats indefinitely.
In the diagram on the right above, the blue atoms have been placed above the white (unoccupied) void spaces in layer A. Because this third layer is displaced horizontally (in our view) from layer A, we will call it layer C. As we add more layers of atoms, the sequence of layers is A-B-C-A-B-C-A-B-C..., so we call it ABC packing.
For the purposes of clarity, only three atoms of the A and C layers are shown in these diagrams. But in reality, each layer consists of an extended hexagonal array; the two layers are simply displaced from one another.
These two diagrams that show exploded views of the vertical stacking further illustrate the rather small fundamental difference between these two arrangements— but, as you will see below, they have widely divergent structural consequences. Note the opposite orientations of the A and C layers
The Hexagonal closed-packed structure
The HCP stacking shown on the left just above takes us out of the cubic crystal system into the hexagonal system, so we will not say much more about it here except to point out each atom has 12 nearest neighbors: six in its own layer, and three in each layer above and below it.
The cubic close-packed structure
Below we reproduce the FCC structure that was shown above.
You will notice that the B-layer atoms form a hexagon, but this is a cubic structure. How can this be? The answer is that the FCC stack is inclined with respect to the faces of the cube, and is in fact coincident with one of the three-fold axes that passes through opposite corners. It requires a bit of study to see the relationship, and we have provided two views to help you. The one on the left shows the cube in the normal isometric projection; the one on the right looks down upon the top of the cube at a slightly inclined angle.
Both the CCP and HCP structures fill 74 percent of the available space when the atoms have the same size. You should see that the two shaded planes cutting along diagonals within the interior of the cube contain atoms of different colors, meaning that they belong to different layers of the CCP stack. Each plane contains three atoms from the B layer and three from the C layer, thus reducing the symmetry to C3, which a cubic lattice must have.
The FCC unit cell
The figure below shows the the face-centered cubic unit cell of a cubic-close packed lattice.
How many atoms are contained in a unit cell? Each corner atom is shared with eight adjacent unit cells and so a single unit cell can claim only 1/8 of each of the eight corner atoms. Similarly, each of the six atoms centered on a face is only half-owned by the cell. The grand total is then (8 × 1/8) + (6 × ½) = 4 atoms per unit cell.
Interstitial Void Spaces
The atoms in each layer in these close-packing stacks sit in a depression in the layer below it. As we explained above, these void spaces are not completely filled. (It is geometrically impossible for more than two identical spheres to be in contact at a single point.) We will see later that these interstitial void spaces can sometimes accommodate additional (but generally smaller) atoms or ions.
If we look down on top of two layers of close-packed spheres, we can pick out two classes of void spaces which we call tetrahedral and octahedral holes.
Tetrahedral holes
If we direct our attention to a region in the above diagram where a single atom is in contact with the three atoms in the layers directly below it, the void space is known as a tetrahedral hole. A similar space will be be found between this single atom and the three atoms (not shown) that would lie on top of it in an extended lattice. Any interstitial atom that might occupy this site will interact with the four atoms surrounding it, so this is also called a four-coordinate interstitial space.
Don't be misled by this name; the boundaries of the void space are spherical sections, not tetrahedra. The tetrahedron is just an imaginary construction whose four corners point to the centers of the four atoms that are in contact.
Octahedral holes
Similarly, when two sets of three trigonally-oriented spheres are in close-packed contact, they will be oriented 60° apart and the centers of the spheres will define the six corners of an imaginary octahedron centered in the void space between the two layers, so we call these octahedral holes or six-coordinate interstitial sites. Octahedral sites are larger than tetrahedral sites.
An octahedron has six corners and eight sides. We usually draw octahedra as a double square pyramid standing on one corner (left), but in order to visualize the octahedral shape in a close-packed lattice, it is better to think of the octahedron as lying on one of its faces (right).
Each sphere in a close-packed lattice is associated with one octahedral site, whereas there are only half as many tetrahedral sites. This can be seen in this diagram that shows the central atom in the B layer in alignment with the hollows in the C and A layers above and below.
The face-centered cubic unit cell contains a single octahedral hole within itself, but octahedral holes shared with adjacent cells exist at the centers of each edge. Each of these twelve edge-located sites is shared with four adjacent cells, and thus contributes (12 × ¼) = 3 atoms to the cell. Added to the single hole contained in the middle of the cell, this makes a total of 4 octahedral sites per unit cell. This is the same as the number we calculated above for the number of atoms in the cell.
Common cubic close-packed structures
It can be shown from elementary trigonometry that an atom will fit exactly into an octahedral site if its radius is 0.414 as great as that of the host atoms. The corresponding figure for the smaller tetrahedral holes is 0.225.
Many pure metals and compounds form face-centered cubic (cubic close- packed) structures. The existence of tetrahedral and octahedral holes in these lattices presents an opportunity for "foreign" atoms to occupy some or all of these interstitial sites. In order to retain close-packing, the interstitial atoms must be small enough to fit into these holes without disrupting the host CCP lattice. When these atoms are too large, which is commonly the case in ionic compounds, the atoms in the interstitial sites will push the host atoms apart so that the face-centered cubic lattice is somewhat opened up and loses its close-packing character.
The rock-salt structure
Alkali halides that crystallize with the "rock-salt" structure exemplified by sodium chloride can be regarded either as a FCC structure of one kind of ion in which the octahedral holes are occupied by ions of opposite charge, or as two interpenetrating FCC lattices made up of the two kinds of ions. The two shaded octahedra illustrate the identical coordination of the two kinds of ions; each atom or ion of a given kind is surrounded by six of the opposite kind, resulting in a coordination expressed as (6:6).
How many NaCl units are contained in the unit cell? If we ignore the atoms that were placed outside the cell in order to construct the octahedra, you should be able to count fourteen "orange" atoms and thirteen "blue" ones. But many of these are shared with adjacent unit cells.
An atom at the corner of the cube is shared by eight adjacent cubes, and thus makes a 1/8 contribution to any one cell. Similarly, the center of an edge is common to four other cells, and an atom centered in a face is shared with two cells. Taking all this into consideration, you should be able to confirm the following tally showing that there are four AB units in a unit cell of this kind.
Orange Blue
8 at corners: 8 x 1/8 = 1 12 at edge centers: 12 x ¼ = 3
6 at face centers: 6 x ½ = 3 1 at body center = 1
total: 4 total: 4
If we take into consideration the actual sizes of the ions (Na+ = 116 pm, Cl = 167 pm), it is apparent that neither ion will fit into the octahedral holes with a CCP lattice composed of the other ion, so the actual structure of NaCl is somewhat expanded beyond the close-packed model.
The space-filling model on the right depicts a face-centered cubic unit cell of chloride ions (purple), with the sodium ions (green) occupying the octahedral sites.
The zinc-blende structure: using some tetrahedral holes
Since there are two tetrahedral sites for every atom in a close-packed lattice, we can have binary compounds of 1:1 or 1:2 stoichiometry depending on whether half or all of the tetrahedral holes are occupied. Zinc-blende is the mineralogical name for zinc sulfide, ZnS. An impure form known as sphalerite is the major ore from which zinc is obtained.
This structure consists essentially of a FCC (CCP) lattice of sulfur atoms (orange) (equivalent to the lattice of chloride ions in NaCl) in which zinc ions (green) occupy half of the tetrahedral sites. As with any FCC lattice, there are four atoms of sulfur per unit cell, and the the four zinc atoms are totally contained in the unit cell. Each atom in this structure has four nearest neighbors, and is thus tetrahedrally coordinated.
It is interesting to note that if all the atoms are replaced with carbon, this would correspond to the diamond structure.
The fluorite structure: all tetrahedral sites occupied
Fluorite, CaF2, having twice as many ions of fluoride as of calcium, makes use of all eight tetrahedral holes in the CPP lattice of calcium ions (orange) depicted here. To help you understand this structure, we have shown some of the octahedral sites in the next cell on the right; you can see that the calcium ion at A is surrounded by eight fluoride ions, and this is of course the case for all of the calcium sites. Since each fluoride ion has four nearest-neighbor calcium ions, the coordination in this structure is described as (8:4).
Although the radii of the two ions (F= 117 pm, Ca2+ = 126 pm does not allow true close packing, they are similar enough that one could just as well describe the structure as a FCC lattice of fluoride ions with calcium ions in the octahedral holes.
Simple- and body-centered cubic structures
In Section 4 we saw that the only cubic lattice that can allow close packing is the face-centered cubic structure. The simplest of the three cubic lattice types, the simple cubic lattice, lacks the hexagonally-arranged layers that are required for close packing. But as shown in this exploded view, the void space between the two square-packed layers of this cell constitutes an octahedral hole that can accommodate another atom, yielding a packing arrangement that in favorable cases can approximate true close-packing. Each second-layer B atom (blue) resides within the unit cell defined the A layers above and below it.
The A and B atoms can be of the same kind or they can be different. If they are the same, we have a body-centered cubic lattice. If they are different, and especially if they are oppositely-charged ions (as in the CsCl structure), there are size restrictions: if the B atom is too large to fit into the interstitial space, or if it is so small that the A layers (which all carry the same electric charge) come into contact without sufficient A-B coulombic attractions, this structural arrangement may not be stable.
The cesium chloride structure
CsCl is the common model for the BCC structure. As with so many other structures involving two different atoms or ions, we can regard the same basic structure in different ways. Thus if we look beyond a single unit cell, we see that CsCl can be represented as two interpenetrating simple cubic lattices in which each atom occupies an octahedral hole within the cubes of the other lattice. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/13%3A_Phase_Diagrams_and_Crystalline_Solids/13.04%3A_Crystalline_Solids-_Unit_Cells_and_Basic_Structures.txt |
Learning Objectives
• Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids
• Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids
• Explain the ways in which crystal defects can occur in a solid
When most liquids are cooled, they eventually freeze and form crystalline solids, solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (Figure \(1\)).
Metals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as boron oxide (Figure \(2\)), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions.
Crystalline solids are generally classified according the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular.
Ionic Solids
Ionic solids, such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (Figure \(3\)). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions—in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic.
Metallic Solids
Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms Figure \(4\). The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below 200 °C. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above 1000 °C. These differences reflect differences in strengths of metallic bonding among the metals.
Covalent Network Solids
Covalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in Figure \(5\). To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above 3500 °C.
Molecular Solids
Molecular solids, such as ice, sucrose (table sugar), and iodine, as shown in Figure \(6\), are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as H2, N2, O2, and F2, have weak attractive forces and form molecular solids with very low melting points (below −200 °C). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar molecules) melt at still higher temperatures. Examples include ice (melting point, 0 °C) and table sugar (melting point, 185 °C).
Properties of Solids
A crystalline solid, like those listed in Table \(1\) has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures.
Table \(1\): Types of Crystalline Solids and Their Properties
Type of Solid Type of Particles Type of Attractions Properties Examples
ionic ions ionic bonds hard, brittle, conducts electricity as a liquid but not as a solid, high to very high melting points NaCl, Al2O3
metallic atoms of electropositive elements metallic bonds shiny, malleable, ductile, conducts heat and electricity well, variable hardness and melting temperature Cu, Fe, Ti, Pb, U
covalent network atoms of electronegative elements covalent bonds very hard, not conductive, very high melting points C (diamond), SiO2, SiC
molecular molecules (or atoms) IMFs variable hardness, variable brittleness, not conductive, low melting points H2O, CO2, I2, C12H22O11
Graphene: Material of the Future
Carbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in Figure \(7\). You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes.
You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in Figure \(8\), is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene.
Crystal Defects
In a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in Figure \(9\). Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites, located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips.
Summary
Some substances form crystalline solids consisting of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips.
Glossary
amorphous solid
(also, noncrystalline solid) solid in which the particles lack an ordered internal structure
covalent network solid
solid whose particles are held together by covalent bonds
crystalline solid
solid in which the particles are arranged in a definite repeating pattern
interstitial sites
spaces between the regular particle positions in any array of atoms or ions
ionic solid
solid composed of positive and negative ions held together by strong electrostatic attractions
metallic solid
solid composed of metal atoms
molecular solid
solid composed of neutral molecules held together by intermolecular forces of attraction
vacancy
defect that occurs when a position that should contain an atom or ion is vacant | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/13%3A_Phase_Diagrams_and_Crystalline_Solids/13.05%3A_Crystalline_Solids-_The_Fundamental_Types.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above.
• What is an ionic solid, what are its typical physical properties, and what kinds of elements does it contain?
• Define the lattice energy of an ionic solid in terms of the energetic properties of its component elements.
• Make a rough sketch that describes the structure of solid sodium chloride.
• Describe the role that the relative ionic radii play in contributing to the stability of an ionic solids.
• Give examples of some solids that can form when ionic solutions are evaporated, but which do not fall into the category of "ionic" solids.
In this section we deal mainly with a very small but important class of solids that are commonly regarded as composed of ions. We will see how the relative sizes of the ions determine the energetics of such compounds. And finally, we will point out that not all solids that are formally derived from ions can really be considered "ionic" at all.
Ionic Solids
The idealized ionic solid consists of two interpenetrating lattices of oppositely-charged point charges that are held in place by a balance of coulombic forces. But because real ions occupy space, no such "perfect" ionic solid exists in nature. Nevertheless, this model serves as a useful starting point for understanding the structure and properties of a small group of compounds between elements having large differences in electronegativity.
Chemists usually apply the term "ionic solid" to binary compounds of the metallic elements of Groups 1-2 with one of the halogen elements or oxygen. As can be seen from the diagram, the differences in electronegativity between the elements of Groups 1-2 and those of Group 17 (as well as oxygen in Group 16) are sufficiently great that the binding in these solids is usually dominated by Coulombic forces and the crystals can be regarded as built up by aggregation of oppositely-charged ions.
Sodium Chloride (rock-salt) Structure
The most well known ionic solid is sodium chloride, also known by its geological names as rock-salt or halite. We can look at this compound in both structural and energetic terms.
Structurally, each ion in sodium chloride is surrounded and held in tension by six neighboring ions of opposite charge. The resulting crystal lattice is of a type known as simple cubic, meaning that the lattice points are equally spaced in all three dimensions and all cell angles are 90°.
In Figure $2$, we have drawn two imaginary octahedra centered on ions of different kinds and extending partially into regions outside of the diagram. (We could equally well have drawn them at any of the lattice points, but show only two in order to reduce clutter.) Our object in doing this is to show that each ion is surrounded by six other ions of opposite charge; this is known as (6,6) coordination. Another way of stating this is that each ion resides in an octahedral hole within the cubic lattice.
How can one sodium ion surrounded by six chloride ions (or vice versa) be consistent with the simplest formula NaCl? The answer is that each of those six chloride ions also sits at the center of its own octahedron defined by another six sodium ions. You might think that this corresponds to Na6Cl6, but note that the central sodium ion shown in the diagram can claim only a one-sixth share of each of its chloride ion neighbors, so the formula NaCl is not just the simplest formula, but correctly reflects the 1:1 stoichiometry of the compound. But of course, as in all ionic structures, there are no distinguishable "molecular" units that correspond to the NaCl simplest formula. Bear in mind that large amount of empty space in diagrams depicting a crystal lattice structure can be misleading, and that the ions are really in direct contact with each other to the extent that this is geometrically possible.
Sodium Chloride Energetics
Sodium chloride, like virtually all salts, is a more energetically favored configuration of sodium and chlorine than are these elements themselves; in other words, the reaction
$Na_{(s)} + ½Cl_{2(g)} \rightarrow NaCl_{(s)}$
is accompanied by a release of energy in the form of heat. How much heat, and why? To help us understand, we can imagine the formation of one mole of sodium chloride from its elements proceeding in these hypothetical steps in which we show the energies explicitly:
Step 1: Atomization of sodium (breaking one mole of metallic sodium into isolated sodium atoms)
$\ce{ Na(s) + 108 kJ → Na(g)} \label{Step1}$
Step 2: Same thing with chlorine. This requires more energy because it involves breaking a covalent bond.
$\ce{ ½Cl2(g) + 127\, kJ → Cl(g)} \label{Step2}$
Step 3: We strip an electron from one mole of sodium atoms (this costs a lot of energy!)
$\ce{ Na(g) + 496\, kJ → Na^{+}(g) + e^{–}} \label{Step3}$
Step 4: Feeding these electrons to the chlorine atoms gives most of this energy back.
$\ce{ Cl(g) + e^{–} → Cl^{–}(g) + 348\, kJ}\label{Step4}$
Step 5: Finally, we bring one mole of the ions together to make the crystal lattice — with a huge release of energy.
$\ce{ Na^{+}(g) + Cl^{–}(g) → NaCl(s) + 787\, kJ} \label{Step5}$
If we add all of these equations together, we get
$\ce{Na(s) + 1/2Cl2(g) → NaCl(s)} + 404\; kJ$
In other words, the formation of solid sodium chloride from its elements is highly exothermic. As this energy is released in the form of heat, it spreads out into the environment and will remain unavailable to push the reaction in reverse. We express this by saying that "sodium chloride is more stable than its elements".
Looking at the equations above, you can see that Equation \ref{Step5} constitutes the big payoff in energy. The 787 kj/mol noted there is known as the NaCl lattice energy. Its large magnitude should be no surprise, given the strength of the coulombic force between ions of opposite charge.
It turns out that it is the lattice energy that renders the gift of stability to all ionic solids.Note that this lattice energy, while due principally to coulombic attraction between each ion and its eight nearest neighbors, is really the sum of all the interactions with the crystal. Lattice energies cannot be measured directly, but they can be estimated fairly well from the energies of the other processes described in the table immediately above.
How Geometry and Periodic Properties Interact
The most energetically stable arrangement of solids made up of identical molecular units (as in the noble gas elements and pure metals) are generally those in which there is a minimum of empty space; these are known as close-packed structures, and there are several kinds. In the case of ionic solids of even the simplest 1:1 stoichiometry, the positive and negative ions usually differ so much in size that packing is often much less efficient. This may cause the solid to assume lattice geometries that differ from the one illustrated above for sodium chloride.
By way of illustration, consider the structure of cesium chloride (the spelling cæsium is also used), CsCl. The radius of the Cs+ ion is 168 pm compared to 98 pm for Na+ and cannot possibly fit into the octahedral hole of a simple cubic lattice of chloride ions. The CsCl lattice therefore assumes a different arrangement.
Figure $3$ focuses on two of these cubic lattice elements whose tops and bottoms are shaded for clarity. It should be easy to see that each cesium ion now has eight nearest-neighbor chloride ions. Each chloride ion is also surrounded by eight cesium ions, so all the lattice points are still geometrically equivalent. We therefore describe this structure as having (8,8) coordination.
The two kinds of lattice arrangements exemplified by NaCl ("rock salt") and CsCl are found in a large number of other 1:1 ionic solids, and these names are used generically to describe the structures of these other compounds. There are of course many other fundamental lattice arrangements (not all of them cubic), but the two we have described here are sufficient to illustrate the point that the radius ratio (the ratio of the radii of the positive to the negative ion) plays an important role in the structures of simple ionic solids.
The Alkali Halides
The interaction of the many atomic properties that influence ionic binding are nicely illustrated by looking at a series of alkali halides, especially those involving extreme differences in atomic radii. The latter are all drawn to the same scale. On the energetic plots at the right, the lattice energies are shown in green. We will start with the one you already know very well.
Sodium chloride - NaCl ("rock-salt") mp/bp 801/1413 °C; coordination (6,6)
Lithium Fluoride - LiF - mp/bp 846/1676 °C, rock-salt lattice structure (6,6).
Tiny-tiny makes strong-strong! This is the most "ionic" of the alkali halides, with the largest lattice energy and highest melting and boiling points. The small size of these ions (and consequent high charge densities) together with the large electronegativity difference between the two elements places a lot of electronic charge between the atoms.
Even in this highly ionic solid, the electron that is "lost" by the lithium atom turns out to be closer to the Li nucleus than when it resides in the 2s shell of the neutral atom.
Cesium Fluoride, CsF - mp/bp 703/1231 °C, (8,8) coordination.
With five shells of electrons shielding its nucleus, the Cs+ ion with its low charge density resembles a big puff-ball which can be distorted by the highly polarizing fluoride ion. The resulting ion-induced dipoles (blue arrows) account for much of the lattice energy here.
The reverse of this would be a tiny metal ion trying to hold onto four relatively huge iodide ions like Lithium iodide.
Lithium iodide, LiI - mp/bp 745/1410 °C.
Negative ions can make even bigger puff-balls. The tiny lithium ion can't get very close to any of the iodides to generate a very strong coulombic binding, but does polarize them to create an ion-induced dipole component. It does not help that the negative ions are in contact with each other. The structural geometry is the same (6,6) coordination as NaCl.
Cesium iodide, CsI - mp/bp 626/1280 °C.
Even with the (8,8) coordination afforded by the CsCl structure, this is a pretty sorry combination owing to the low charge densities. The weakness of coulombic- compared to van der Waals interactions makes this the least-"ionic" of all the alkali halide solids.
Conclusion: Many of the alkali halide solids are not all that "ionic" in the sense that coulombic forces are the predominant actors; in many, such as the CsI illustrated above, ion-induced dipole forces are more important.
Some Properties of Ionic Solids
As noted above, ionic solids are generally hard and brittle. Both of these properties reflect the strength of the coulombic force. Hardness measures resistance to deformation. Because the ions are tightly bound to their oppositely-charged neighbors and, a mechanical force exerted on one part of the solid is resisted by the electrostatic forces operating over an extended volume of the crystal.
But by applying sufficient force, one layer of ions can be made to slip over another; this is the origin of brittleness. This slippage quickly propagates along a plane of the crystal (more readily in some directions than in others), weakening their attraction and leading to physical cleavage. Because the "ions" in ionic solids lack mobility, the solids themselves are electrical insulators.
Not all ion-derived solids are "ionic".
Even within the alkali halides, the role of coulombic attraction diminishes as the ions become larger and more polarizable or differ greatly in radii. This is especially true of the anions, which tend to be larger and whose electron clouds are more easily distorted. In solids composed of polyatomic ions such as (NH4)2SO4, SrClO4, NH4CO3, ion-dipole and ion-induced dipole forces may actually be stronger than the coulombic force. Higher ionic charges help, especially if the ions are relatively small. This is especially evident in the extremely high melting points of the Group 2 and higher oxides:
MgO (magnesia) CaO (lime) SrO (strontia) Al2O3 (alumina) ZrO2 (zirconia)
2830 °C 2610 °C 2430 °C 2050 °C 2715 °C
These substances are known as refractories, meaning that they retain their essential properties at high temperatures. Magnesia, for example, is used to insulate electric heating elements and, in the form of fire bricks, to line high-temperature furnaces. No boiling points have been observed for these compounds; on further heating, they simply dissociate into their elements. Their crystal structures can be very complex, and some (notably Al2O3) can have several solid forms. Even in the most highly ionic solids there is some electron sharing, so the idea of a “pure” ionic bond is an abstraction.
Many solids that are formally derived from ions cannot really be said to form "ionic" solids at all. For example, anhydrous copper(II) chloride consists of layers of copper atoms surrounded by four chlorine atoms in a square arrangement. Neighboring chains are offset so as to create an octahedral coordination of each copper atom. Similar structures are commonly encountered for other salts of transition metals. Similarly, most oxides and sulfides of metals beyond Group 2 tend to have structures dominated by other than ion-ion attractions.
Aluminum Halides
The trihalides of aluminum offer another example of the dangers of assuming ionic character of solids that are formally derived from ions. Aqueous solutions of what we assume to be AlF3, AlCl3, AlBr3, and AlI3 all exhibit the normal properties ionic solutions (they are electrically conductive, for example), but the solids are quite different: the melting point of AlF3 is 1290°C, suggesting that it is indeed ionic. But AlCl3 melts at 192°C — hardly consistent with ionic bonding, and the other two halides are also rather low-melting. Structural studies show that when AlCl3 vaporizes or dissolves in a non-polar solvent it forms a dimer Al2Cl6. The two other halides exist only as dimers in all states.
The structural formula of the Al2Cl6 molecule shows that the aluminum atoms are bonded to four chlorines, two of which are shared between the two metal atoms. The arrows represent coordinate covalent bonds in which the bonding electrons both come from the same atom (chlorine in this case.)
As shown at the right above, the aluminum atoms can be considered to be located at the centers of two tetrahedra that possess one edge in common. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/13%3A_Phase_Diagrams_and_Crystalline_Solids/13.06%3A_The_Structure_of_Ionic_Solids.txt |
Covalent Network Solids
Covalent solids are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule. For example, the structure of diamond, shown in part (a) in Figure $1$, consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. The carbon atoms form six-membered rings.
The unit cell of diamond can be described as an fcc array of carbon atoms with four additional carbon atoms inserted into four of the tetrahedral holes. It thus has the zinc blende structure described in Section 12.3, except that in zinc blende the atoms that compose the fcc array are sulfur and the atoms in the tetrahedral holes are zinc. Elemental silicon has the same structure, as does silicon carbide (SiC), which has alternating C and Si atoms. The structure of crystalline quartz (SiO2), shown in Section 12.1, can be viewed as being derived from the structure of silicon by inserting an oxygen atom between each pair of silicon atoms.
All compounds with the diamond and related structures are hard, high-melting-point solids that are not easily deformed. Instead, they tend to shatter when subjected to large stresses, and they usually do not conduct electricity very well. In fact, diamond (melting point = 3500°C at 63.5 atm) is one of the hardest substances known, and silicon carbide (melting point = 2986°C) is used commercially as an abrasive in sandpaper and grinding wheels. It is difficult to deform or melt these and related compounds because strong covalent (C–C or Si–Si) or polar covalent (Si–C or Si–O) bonds must be broken, which requires a large input of energy.
Other covalent solids have very different structures. For example, graphite, the other common allotrope of carbon, has the structure shown in part (b) in Figure $1$. It contains planar networks of six-membered rings of sp2 hybridized carbon atoms in which each carbon is bonded to three others. This leaves a single electron in an unhybridized 2pz orbital that can be used to form C=C double bonds, resulting in a ring with alternating double and single bonds. Because of its resonance structures, the bonding in graphite is best viewed as consisting of a network of C–C single bonds with one-third of a π bond holding the carbons together, similar to the bonding in benzene.
To completely describe the bonding in graphite, we need a molecular orbital approach similar to the one used for benzene in Chapter 9. In fact, the C–C distance in graphite (141.5 pm) is slightly longer than the distance in benzene (139.5 pm), consistent with a net carbon–carbon bond order of 1.33. In graphite, the two-dimensional planes of carbon atoms are stacked to form a three-dimensional solid; only London dispersion forces hold the layers together. As a result, graphite exhibits properties typical of both covalent and molecular solids. Due to strong covalent bonding within the layers, graphite has a very high melting point, as expected for a covalent solid (it actually sublimes at about 3915°C). It is also very soft; the layers can easily slide past one another because of the weak interlayer interactions. Consequently, graphite is used as a lubricant and as the “lead” in pencils; the friction between graphite and a piece of paper is sufficient to leave a thin layer of carbon on the paper. Graphite is unusual among covalent solids in that its electrical conductivity is very high parallel to the planes of carbon atoms because of delocalized C–C π bonding. Finally, graphite is black because it contains an immense number of alternating double bonds, which results in a very small energy difference between the individual molecular orbitals. Thus light of virtually all wavelengths is absorbed. Diamond, on the other hand, is colorless when pure because it has no delocalized electrons. Table $2$ compares the strengths of the intermolecular and intramolecular interactions for three covalent solids, showing the comparative weakness of the interlayer interactions.
Table $2$: A Comparison of Intermolecular (ΔHsub) and Intramolecular Interactions
Substance ΔHsub (kJ/mol) Average Bond Energy (kJ/mol)
phosphorus (s) 58.98 201
sulfur (s) 64.22 226
iodine (s) 62.42 149
Carbon: An example of an Covalent Network Solid
In network solids, conventional chemical bonds hold the chemical subunits together. The bonding between chemical subunits, however, is identical to that within the subunits, resulting in a continuous network of chemical bonds. One common examples of network solids are diamond (a form of pure carbon) Carbon exists as a pure element at room temperature in three different forms: graphite (the most stable form), diamond, and fullerene.
Diamonds
The structure of diamond is shown at the right in a "ball-and-stick" format. The balls represent the carbon atoms and the sticks represent a covalent bond. Be aware that in the "ball-and-stick" representation the size of the balls do not accurately represent the size of carbon atoms. In addition, a single stick is drawn to represent a covalent bond irrespective of whether the bond is a single, double, or triple bond or requires resonance structures to represent. In the diamond structure, all bonds are single covalent bonds ($\sigma$ bonds). The "space-filling" format is an alternate representation that displays atoms as spheres with a radius equal to the van der Waals radius, thus providing a better sense of the size of the atoms.
Notice that diamond is a network solid. The entire solid is an "endless" repetition of carbon atoms bonded to each other by covalent bonds. (In the display at the right, the structure is truncated to fit in the display area.)
Questions to consider:
• What is the bonding geometry around each carbon?
• What is the hybridization of carbon in diamond?
• The diamond structure consists of a repeating series of rings. How many carbon atoms are in a ring?
• Diamond are renowned for its hardness. Explain why this property is expected on the basis of the structure of diamond.
Graphite
The most stable form of carbon is graphite. Graphite consists of sheets of carbon atoms covalently bonded together. These sheets are then stacked to form graphite. Figure $3$ shows a ball-and-stick representation of graphite with sheets that extended "indefinitely" in the xy plane, but the structure has been truncated for display purposed. Graphite may also be regarded as a network solid, even though there is no bonding in the z direction. Each layer, however, is an "endless" bonded network of carbon atoms.
Questions to consider:
• What is the bonding geometry around each carbon?
• What is the hybridization of carbon in graphite?
• The a layer of the graphite structure consists of a repeating series of rings. How many carbon atoms are in a ring?
• What force holds the carbon sheets together in graphite?
• Graphite is very slippery and is often used in lubricants. Explain why this property is expected on the basis of the structure of graphite.
• The slipperiness of graphite is enhanced by the introduction of impurities. Where would such impurities be located and why would they make graphite a better lubricant?
Fullerenes
Until the mid 1980's, pure carbon was thought to exist in two forms: graphite and diamond. The discovery of C60 molecules in interstellar dust in 1985 added a third form to this list. The existence of C60, which resembles a soccer ball, had been hypothesized by theoretians for many years. In the late 1980's synthetic methods were developed for the synthesis of C60, and the ready availability of this form of carbon led to extensive research into its properties.
The C60 molecule (Figure $4$; left), is called buckminsterfullerene, though the shorter name fullerene is often used. The name is a tribute to the American architect R. Buckminster Fuller, who is famous for designing and constructing geodesic domes which bear a close similarity to the structure of C60. As is evident from the display, C60 is a sphere composed of six-member and five-member carbon rings. These balls are sometimes fondly referred to as "Bucky balls". It should be noted that fullerenes are an entire class of pure carbon compounds rather than a single compound. A distorted sphere containing more than 60 carbon atoms have also been found, and it is also possible to create long tubes (Figure $4$; right). All of these substances are pure carbon.
to Consider
• What is the bonding geometry around each carbon? (Note that this geometry is distorted in $C_{60}$.)
• What is the hybridization of carbon in fullerene?
• A single crystal of C60 falls into which class of crystalline solids?
• It has been hypothesized that C60 would make a good lubricant. Why might C60 make a good lubricant? | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/13%3A_Phase_Diagrams_and_Crystalline_Solids/13.07%3A_Network_Covalent_Atomic_Solids-_Carbon_and_Silicates.txt |
• 14.2: Types of Solutions and Solubility
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible.
• 14.3: Energetics of Solution Formation
• 14.4: Solution Equilibrium and Factors Affecting Solubility
The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas.
• 14.5: Expressing Solution Concentration
Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways.
• 14.6: Colligative Properties- Freezing Point Depression, Boiling Point Elevation, and Osmosis
Colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law.
• 14.7: The Colligative Properties of Strong Electrolyte Solutions
Ion-pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH all affect solubility. There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) ion pair formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes
14: Solutions
Learning Objectives
• To understand how enthalpy and entropy changes affect solution formation.
• To use the magnitude of the changes in both enthalpy and entropy to predict whether a given solute–solvent combination will spontaneously form a solution.
In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table $1$ lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each.
Table $1$: Types of Solutions
Solution Solute Solvent Examples
gas gas gas air, natural gas
liquid gas liquid seltzer water ($CO_2$ gas in water)
liquid liquid liquid alcoholic beverage (ethanol in water), gasoline
liquid solid liquid tea, salt water
solid gas solid $H_2$ in Pd (used for $H_2$ storage)
solid solid liquid mercury in silver or gold (amalgam often used in dentistry)
solid solid solid alloys and other "solid solutions"
Forming a Solution
The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate:
$\ce{Zn(NO3)2(s) + H2O(l) \rightarrow Zn^{2+}(aq) + 2NO^{-}3(aq)} \label{13.1.1}$
Because $Zn(NO_3)_2$ can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas:
$\ce{ Zn(s) + 2H^{+}(aq) + 2Cl^{-}(aq) \rightarrow Zn^{2+}(aq) + 2Cl^{-}(aq) + H2(g)} \label{13.1.2}$
When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation (that it is a physical change).
Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation.
Substances that form a single homogeneous phase in all proportions are said to be completely miscible in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are immiscible. Examples of gaseous solutions that we have already discussed include Earth’s atmosphere.
The Role of Enthalpy in Solution Formation
Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called solvation (or hydration when the solvent is water). Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process.
Because enthalpy is a state function, we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by $ΔH_1$, $ΔH_2$, and $ΔH_3$ in Figure $2$. The overall enthalpy change in the formation of the solution ($\Delta H_{soln}$) is the sum of the enthalpy changes in the three steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.1.3}$
When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent ($\Delta H_1$) and the solute ($\Delta H_2$). Because $ΔH$ is positive for both steps 1 and 2, the solute–solvent interactions ($\Delta H_3$) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic ($\Delta H_{soln} < 0$). When the solute is an ionic solid, $ΔH_2$ corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as $MgO$ (−3791 kJ/mol), are generally insoluble in all solvents.
A positive value for $ΔH_{soln}$ does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high $ΔH_{soln}$ is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive $ΔH_{soln}$ of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure $3$).
Entropy and Solution Formation
The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detail elsewhere, but for now we can state that entropy ($S$) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, has an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds.
The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, $ΔH_{soln}$ should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative $ΔH_{soln}$ but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with $ΔH \ge 0$ are characterized by an increase in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form.
All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy.
Table $2$ summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily.
Table $2$: Relative Changes in Enthalpies for Different Solute–Solvent Combinations*
$ΔH_1$ (separation of solvent molecules) $ΔH_2$ (separation of solute particles) $ΔH_3$ (solute–solvent interactions) $ΔH_{soln}$ ($ΔH_1$ + $ΔH_2$ +$ΔH_3$) Result of Mixing Solute and Solvent†
large; positive large; positive large; negative small; positive or negative solution will usually form
small; positive large; positive small; negative large; positive solution will not form
large; positive small; positive small; negative large; positive solution will not form
small; positive small; positive small; negative small; positive or negative solution will usually form
*$ΔH_1$, $ΔH_2$, and $ΔH_3$ refer to the processes indicated in the thermochemical cycle shown in Figure $2$.
In all four cases, entropy increases.
In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process ($ΔH_{soln} \approx 0$), and the entropic factor due to the increase in disorder is dominant (Figure $4$). Consequently, all gases dissolve readily in one another in all proportions to form solutions.
Example $1$
Considering $\ce{LiCl}$, benzoic acid ($\ce{C6H5CO2H}$), and naphthalene, which will be most soluble and which will be least soluble in water?
Given: three compounds
Asked for: relative solubilities in water
Strategy: Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure $2$. Then use Table $2$ to predict the solubility of each compound in water and arrange them in order of decreasing solubility.
Solution:
The first substance, $\ce{LiCl}$, is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH2 is far greater than zero in Equation $\ref{13.1.1}$). Because water is a polar substance, the interactions between both Li+ and Cl ions and water should be favorable and strong. Thus we expect $ΔH_3$ to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect $ΔH_2$ to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak $ΔH_3 \approx 0$. Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH2) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than $\ce{LiCl}$. We thus predict $\ce{LiCl}$ to be the most soluble in water and naphthalene to be the least soluble.
Exercise $1$
Considering ammonium chloride, cyclohexane, and ethylene glycol ($HOCH_2CH_2OH$), which will be most soluble and which will be least soluble in benzene?
Answer
The most soluble is cyclohexane; the least soluble is ammonium chloride.
Summary
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, $ΔH_{soln}$, is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic ($ΔH_{soln} < 0$) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/14%3A_Solutions/14.02%3A_Types_of_Solutions_and_Solubility.txt |
Learning Objectives
• Understand entropy drives solution formation but necessary enthalpy changes can prevent solution formation.
• Define heat of solute and heat of solvent, know these are always endothermic. ("the price you pay to pull substances apart")
• Define heat of mixing, know this is always exothermic. ("energy that splashes out when you drop atoms into each others energy well")
• Define heat of solvation, identify when it will be exothermic and when it will be endothermic.
• Define heat of hydration, know this is the sum of heat of mixing and heat of solvent when the solvent is water.
• Relate lattice energy to heat of solute.
• Given a substances lattice energy and heat of hydration, determine heat of solvation for that substance in water.
• Use the magnitude of changes in enthalpy to predict whether a given solute–solvent combination can form a solution.
In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table $1$ lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each.
The Role of Enthalpy in Solution Formation
Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called solvation (or hydration when the solvent is water). Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process.
Because enthalpy is a state function, we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by $ΔH_1$, $ΔH_2$, and $ΔH_3$ in Figure $2$. The overall enthalpy change in the formation of the solution ($\Delta H_{soln}$) is the sum of the enthalpy changes in the three steps:
$\Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.1.3}$
When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent ($\Delta H_1$) and the solute ($\Delta H_2$). Because $ΔH$ is positive for both steps 1 and 2, the solute–solvent interactions ($\Delta H_3$) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic ($\Delta H_{soln} < 0$). When the solute is an ionic solid, $ΔH_2$ corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as $MgO$ (−3791 kJ/mol), are generally insoluble in all solvents.
A positive value for $ΔH_{soln}$ does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high $ΔH_{soln}$ is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive $ΔH_{soln}$ of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure $3$).
Entropy and Solution Formation
The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detail elsewhere, but for now we can state that entropy ($S$) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, has an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds.
The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, $ΔH_{soln}$ should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative $ΔH_{soln}$ but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with $ΔH \ge 0$ are characterized by an increase in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form.
All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy.
Table $2$ summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily.
Table $2$: Relative Changes in Enthalpies for Different Solute–Solvent Combinations*
$ΔH_1$ (separation of solvent molecules) $ΔH_2$ (separation of solute particles) $ΔH_3$ (solute–solvent interactions) $ΔH_{soln}$ ($ΔH_1$ + $ΔH_2$ +$ΔH_3$) Result of Mixing Solute and Solvent†
large; positive large; positive large; negative small; positive or negative solution will usually form
small; positive large; positive small; negative large; positive solution will not form
large; positive small; positive small; negative large; positive solution will not form
small; positive small; positive small; negative small; positive or negative solution will usually form
*$ΔH_1$, $ΔH_2$, and $ΔH_3$ refer to the processes indicated in the thermochemical cycle shown in Figure $2$.
In all four cases, entropy increases.
In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process ($ΔH_{soln} \approx 0$), and the entropic factor due to the increase in disorder is dominant (Figure $4$). Consequently, all gases dissolve readily in one another in all proportions to form solutions.
Example $1$
Considering $\ce{LiCl}$, benzoic acid ($\ce{C6H5CO2H}$), and naphthalene, which will be most soluble and which will be least soluble in water?
Given: three compounds
Asked for: relative solubilities in water
Strategy: Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure $2$. Then use Table $2$ to predict the solubility of each compound in water and arrange them in order of decreasing solubility.
Solution:
The first substance, $\ce{LiCl}$, is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH2 is far greater than zero in Equation $\ref{13.1.1}$). Because water is a polar substance, the interactions between both Li+ and Cl ions and water should be favorable and strong. Thus we expect $ΔH_3$ to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect $ΔH_2$ to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak $ΔH_3 \approx 0$. Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH2) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than $\ce{LiCl}$. We thus predict $\ce{LiCl}$ to be the most soluble in water and naphthalene to be the least soluble.
Exercise $1$
Considering ammonium chloride, cyclohexane, and ethylene glycol ($HOCH_2CH_2OH$), which will be most soluble and which will be least soluble in benzene?
Answer
The most soluble is cyclohexane; the least soluble is ammonium chloride.
Summary
Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, $ΔH_{soln}$, is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic ($ΔH_{soln} < 0$) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/14%3A_Solutions/14.03%3A_Energetics_of_Solution_Formation.txt |
Learning Objectives
• To understand the relationship among temperature, pressure, and solubility.
• The understand that the solubility of a solid may increase or decrease with increasing temperature,
• To understand that the solubility of a gas decreases with an increase in temperature and a decrease in pressure.
Experimentally it is found that the solubility of most compounds depends strongly on temperature and, if a gas, on pressure as well. As we shall see, the ability to manipulate the solubility by changing the temperature and pressure has several important consequences.
Effect of Temperature on the Solubility of Solids
Figure $1$ shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and $\ce{CH_3CO_2Na}$) exhibit a dramatic increase in solubility with increasing temperature. Others (such as $\ce{NaCl}$ and $\ce{K_2SO_4}$) exhibit little variation, and still others (such as $\ce{Li_2SO_4}$) become less soluble with increasing temperature.
Notice in particular the curves for $\ce{NH4NO3}$ and $\ce{CaCl2}$. The dissolution of ammonium nitrate in water is endothermic ($ΔH_{soln} = +25.7\; kJ/mol$), whereas the dissolution of calcium chloride is exothermic ($ΔH_{soln} = −68.2 \;kJ/mol$), yet Figure $1$ shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature.
The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by fractional crystallization, the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate ($\ce{CH_3CO_2Na}$) and 50 g of $\ce{KBr}$, we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure $1$, both compounds dissolve in water at 80°C, and all 50 g of $\ce{KBr}$ remains in solution at 0°C. Only about 36 g of $\ce{CH3CO2Na}$ are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of $\ce{CH_3CO_2Na}$ crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original $\ce{CH_3CO_2Na}$ in essentially pure form in only one step.
Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure $1$ and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was $\ce{KBr}$ in this example) and preferably present in relatively small amounts.
Effect of Temperature on the Solubility of Gases
The solubility of gases in liquids decreases with increasing temperature, as shown in Figure $2$. Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process ($ΔH_{soln} < 0$). Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas. The phenomenon is similar to that involved in the increase in the vapor pressure of a pure liquid with increasing temperature. In the case of vapor pressure, however, it is attractive forces between solvent molecules that are being overcome by the added thermal energy when the temperature is increased.
The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (Figure $3$). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex and will be described elsewhere, but the driving force is the loss of dissolved $\ce{CO2}$ from solution. Hard water contains dissolved $\ce{Ca^{2+}}$ and $\ce{HCO3^{-}}$ (bicarbonate) ions. Calcium bicarbonate ($\ce{Ca(HCO3)2}$ is rather soluble in water, but calcium carbonate ($\ce{CaCO3}$) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water:
$\ce{2HCO3^{-}(aq) -> CO3^{2-}(aq) + H2O(l) + CO2(aq)} \label{13.9}$
Heating the solution decreases the solubility of $\ce{CO2}$, which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale.
Thermal Pollution
In thermal pollution, lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of $\ce{O2}$ at higher temperatures (Figure $2$), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water.
A similar effect is seen in the rising temperatures of bodies of water such as the k0oi89Chesapeake Bay, the largest estuary in North America, where \lobal warming has been implicated as the cause. For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality.
Effect of Pressure on the Solubility of Gases: Henry’s Law
External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure $4$, which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures.
The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836):
$C = kP \label{13.3.1}$
where
• $C$ is the concentration of dissolved gas at equilibrium,
• $P$ is the partial pressure of the gas, and
• $k$ is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature.
Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table $1$.
As the data in Table $1$ demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the Group 18 elements, the Henry’s law constants increase smoothly from $\ce{He}$ to $\ce{Ne}$ to $\ce{Ar}$.
Table $1$: Henry’s Law Constants for Selected Gases in Water at 20°C
Gas Henry’s Law Constant [mol/(L·atm)] × 10−4
$\ce{He}$ 3.9
$\ce{Ne}$ 4.7
$\ce{Ar}$ 15
$\ce{H_2}$ 8.1
$\ce{N_2}$ 7.1
$\ce{O_2}$ 14
$\ce{CO_2}$ 392
Oxygen is Especially Soluble
Nitrogen and oxygen are the two most prominent gases in the Earth’s atmosphere and they share many similar physical properties. However, as Table $1$ shows, $\ce{O2}$ is twice as soluble in water as $\ce{N2}$. Many factors contribute to solubility including the nature of the intermolecular forces at play. For a details discussion, see "The O2/N2 Ratio Gas Solubility Mystery" by Rubin Battino and Paul G. Seybold (J. Chem. Eng. Data 2011, 56, 5036–5044),
Gases that react chemically with water, such as $\ce{HCl}$ and the other hydrogen halides, $\ce{H2S}$, and $\ce{NH3}$, do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, $\ce{HCl}$ reacts with water to give $\ce{H^{+}(aq)}$ and $\ce{Cl^{-}(aq)}$, not dissolved $\ce{HCl}$ molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule.
Gases that react with water do not obey Henry’s law.
Henry’s law has important applications. For example, bubbles of $\ce{CO2}$ form as soon as a carbonated beverage is opened because the drink was bottled under $\ce{CO2}$ at a pressure greater than 1 atm. When the bottle is opened, the pressure of $\ce{CO2}$ above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of $\ce{N2}$ to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases.
Due to the low Henry’s law constant for $\ce{O2}$ in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the $\ce{O2}$ concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind $\ce{O2}$ reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds $\ce{O2}$ and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four $\ce{O2}$ molecules. Although the concentration of dissolved $\ce{O2}$ in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved $\ce{O2}$ concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for $\ce{O2}$ result in dissolved oxygen concentrations comparable to those in normal blood.
A Video Discussing Henry's Law. Video Link: Henry's Law (The Solubility of Gases in Solvents), YouTube(opens in new window) [youtu.be]
Example $1$
The Henry’s law constant for $\ce{O2}$ in water at 25°C is $1.27 \times 10^{-3} M/atm$, and the mole fraction of $\ce{O2}$ in the atmosphere is 0.21. Calculate the solubility of $\ce{O2}$ in water at 25°C at an atmospheric pressure of 1.00 atm.
Given: Henry’s law constant, mole fraction of $\ce{O2}$, and pressure
Asked for: solubility
Strategy:
1. Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen. (For more information about Dalton’s law of partial pressures)
2. Use Henry’s law to calculate the solubility, expressed as the concentration of dissolved gas.
Solution:
A According to Dalton’s law, the partial pressure of $\ce{O2}$ is proportional to the mole fraction of $\ce{O2}$:
\begin{align*} P_A &= \chi_A P_t \[4pt] &= (0.21)(1.00\; atm) \[4pt] &= 0.21\; atm \end{align*} \nonumber
B From Henry’s law, the concentration of dissolved oxygen under these conditions is
\begin{align*} [\ce{CO2}] &= k P_{\ce{O2}} \[4pt] &=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) \[4pt] &=2.7 \times 10^{-4}\; M \end{align*} \nonumber
Exercise $1$
To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved $\ce{CO2}$ in a soft drink:
1. bottled under a pressure of 5.0 atm of $\ce{CO2}$
2. in equilibrium with the normal partial pressure of $\ce{CO_2}$ in the atmosphere (approximately $3 \times 10^{-4} \;atm$). The Henry’s law constant for $\ce{CO2}$ in water at 25°C is $3.4 \times 10^{-2}\; M/atm$.
Answer a
$0.17 M$
Answer b
$1 \times 10^{-5} M$
Summary
The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/14%3A_Solutions/14.04%3A_Solution_Equilibrium_and_Factors_Affecting_Solubility.txt |
Learning Objectives
• To describe the concentration of a solution in the way that is most appropriate for a particular problem or application.
• To be familiar with the different units used to express the concentrations of a solution.
There are several different ways to quantitatively describe the concentration of a solution. For example, molarity is a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example $1$ reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.
Example $1$: Molarity and Mole Fraction
Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Given: mass of substance and mass and density of solution
Asked for: molarity and mole fraction
Strategy:
1. Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.
2. Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.
Solution:
A: The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass.
\begin{align*} \text{moles } \ce{CH_3CO_2H} &=\dfrac{3.78\; \cancel{\ce{g}}\; \ce{CH_3CO_2H}}{60.05\; \cancel{\ce{g}}/\ce{mol}} \[4pt] &=0.0629 \; \ce{mol} \end{align*} \nonumber
The volume of the solution equals its mass divided by its density.
\begin{align*} \text{volume} &=\dfrac{\text{mass}}{\text{density}} \[4pt] &=\dfrac{100.0\; \cancel{\ce{g}}\; \text{solution}}{1.00\; \cancel{\ce{g}}/\ce{mL}}=100\; mL\nonumber \end{align*} \nonumber
Then calculate the molarity directly.
\begin{align*} \text{molarity of } \ce{CH_3CO_2H} &=\dfrac{\text{moles } \ce{CH3CO2H} }{\text{liter solution}} \[4pt] &=\dfrac{0.0629\; mol\; \ce{CH_3CO_2H}}{(100\; \cancel{\ce{mL}})(1\; L/1000\; \cancel{\ce{mL}})}=0.629\; M \; \ce{CH_3CO_2H} \end{align*} \nonumber
This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than $\ce{ 1/2}$ mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.
B: To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have
$moles\; \ce{H_2O}=\dfrac{96.2\; \cancel{\ce{g}}\; \ce{H_2O}}{18.02\; \cancel{\ce{g}}/mol}=5.34\; mol\; \ce{H_2O}\nonumber$
The mole fraction $\chi$ of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:
\begin{align*} \chi_{\ce{CH3CO2H}} &=\dfrac{moles\; \ce{CH_3CO_2H}}{moles \; \ce{CH_3CO_2H} + moles\; \ce{H_2O}} \[4pt] &=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol} \[4pt] &=0.0116=1.16 \times 10^{−2} \end{align*} \nonumber
This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.
Exercise $1$: Molarity and Mole Fraction
A solution of $\ce{HCl}$ gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of $\ce{HCl}$ per 100.0 g of solution, and its density is 1.10 g/mL.
1. What is its molarity?
2. What is its mole fraction?
Answer a
6.10 M HCl
Answer b
$\chi_{HCl} = 0.111$
The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent:
$\text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \label{Eq1}$
Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of $H_2O$ under these conditions is very close to 1.0 L, and a 0.50 M solution of $KBr$ in water, for example, has approximately the same concentration as a 0.50 m solution.
Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb):
\begin{align} \text{mass percentage}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \label{Eq2} \[4pt] \text{parts per million (ppm)} &=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \label{Eq3} \[4pt] \text{parts per billion (ppb)}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \label{Eq4} \end{align}
In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.
The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of $\ce{H_2SO_4}$ per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).
Example $2$: Molarity and Mass
Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.
1. What is the molarity of the solution?
2. What is the mass of benzene in the sample?
Given: volume of sample, solute concentration, and density of solution
Asked for: molarity of solute and mass of solute in 250 mL
Strategy:
1. Use the concentration of the solute in parts per million to calculate the molarity.
2. Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.
Solution:
a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore
\begin{align*} \text{molarity}&=\dfrac{\text{moles}}{\text{liter solution}} \[4pt] &=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L} \[4pt] &=1.63 \times 10^{-4} M\end{align*} \nonumber
b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is
\begin{align*} \text{mass of benzene} &=\dfrac{(12.7\; mg\; \text{benzene})(250\; \cancel{mL})}{1000\; \cancel{mL}} \[4pt] &=3.18\; mg \[4pt] &=3.18 \times 10^{-3}\; g\; \text{benzene} \end{align*} \nonumber
Exercise $2$: Molarity of Lead Solution
The maximum allowable concentration of lead in drinking water is 9.0 ppb.
1. What is the molarity of $\ce{Pb^{2+}}$ in a 9.0 ppb aqueous solution?
2. Use your calculated concentration to determine how many grams of $\ce{Pb^{2+}}$ are in an 8 oz glass of water.
Answer a
4.3 × 10−8 M
Answer b
2 × 10−6 g
How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature.
Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as discussed previously. Mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies.
Table $1$ summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example $3$.
Table $1$: Different Units for Expressing the Concentrations of Solutions*
Unit Definition Application
*The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature.
molarity (M) moles of solute/liter of solution (mol/L) Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known.
mole fraction ($\chi$) moles of solute/total moles present (mol/mol) Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known.
molality (m) moles of solute/kg of solvent (mol/kg) Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known.
mass percentage (%) [mass of solute (g)/mass of solution (g)] × 100 Useful when masses are known but molecular masses are unknown.
parts per thousand (ppt) [mass of solute/mass of solution] × 103 (g solute/kg solution) Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000.
parts per million (ppm) [mass of solute/mass of solution] × 106 (mg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
parts per billion (ppb) [mass of solute/mass of solution] × 109 (µg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
Example $3$: Vodka
Vodka is essentially a solution of ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka.
1. the mass percentage
2. the mole fraction
3. the molarity
4. the molality
Given: volume percent and density
Asked for: mass percentage, mole fraction, molarity, and molality
Strategy:
1. Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.
2. Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.
3. Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.
4. Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.
Solution:
The key to this problem is to use the density of pure ethanol to determine the mass of ethanol ($CH_3CH_2OH$), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density:
$mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\nonumber$
If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution.
B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage:
\begin{align*} \%EtOH &=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100) \[4pt] &=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) \[4pt]&= 34.5\%\end{align*} \nonumber
C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL:
\begin{align*} moles\; \ce{CH_3CH_2OH}&=(31.6\; \cancel{g\; \ce{CH_3CH_2OH}}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; \ce{CH_3CH_2OH}}}\right) \[4pt] &=0.686 \;mol\; \ce{CH_3CH_2OH} \end{align*} \nonumber
Similarly, the number of moles of water is
$moles \;\ce{H_2O}=(60.0\; \cancel{g \; \ce{H_2O}}) \left(\dfrac{1 \;mol\; \ce{H_2O}}{18.02\; \cancel{g\; \ce{H_2O}}}\right)=3.33\; mol\; \ce{H_2O}\nonumber$
The mole fraction of ethanol is thus
$\chi_{\ce{CH_3CH_2OH}}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\nonumber$
D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is
$M_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol}{100\; \cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=6.86 \;M\nonumber$
The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward:
$m_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m\nonumber$
Exercise $3$: Toluene/Benzene Solution
A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.
1. mass percentage
2. mole fraction
3. molarity
4. molality
Answer a
mass percentage toluene = 24.8%
Answer b
$\chi_{toluene} = 0.219$
Answer c
2.35 M toluene
Answer d
3.59 m toluene
A Video Discussing Different Measures of Concentration. Video Link: Measures of Concentration, YouTube (opens in new window) [youtu.be]
A Video Discussing how to Convert Measures of Concentration. Video Link: Converting Units of Concentration, YouTube(opens in new window) [youtu.be]
Summary
Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/14%3A_Solutions/14.05%3A_Expressing_Solution_Concentration.txt |
Learning Objectives
• To describe the relationship between solute concentration and the physical properties of a solution.
• To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.
Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode.
Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties.
Counting concentrations
When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, $\ce{NaCl}$, and $\ce{CaCl_2}$. Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both $\ce{NaCl}$ and $\ce{CaCl_2}$ are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of $\ce{NaCl}$ contains 0.01 M $\ce{Na^{+}}$ ions and 0.01 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.02 M. Similarly, the $\ce{CaCl_2}$ solution contains 0.01 M $\ce{Ca^{2+}}$ ions and 0.02 M $\ce{Cl^{−}}$ ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete.
At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as $\ce{Mg^{2+}}$ or $\ce{Al^{3+}}$), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind.
Vapor Pressure of Solutions and Raoult’s Law
Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure $1$, which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent.
Figure $2$ shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant.
If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore
$P_A=\chi_AP^0_A \label{13.5.1}$
where $P_A$ is the vapor pressure of component A of the solution (in this case the solvent), $\chi_A$ is the mole fraction of $A$ in solution, and $P^0_A$ is the vapor pressure of pure $A$. Equation $\ref{13.5.1}$ is known as Raoult’s law, after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then $\chi_A + \chi_B = 1$, and we can substitute $\chi_A = 1 − \chi_B$ to obtain
\begin{align} P_A &=(1−\chi_B)P^0_A \[4pt] &=P^0_A−\chi_BP^0_A \label{13.5.2} \end{align}
Rearranging and defining $ΔP_A=P^0_A−P_A$, we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute:
\begin{align} P^0_A−P_A &=ΔP_A \[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align}
We can solve vapor pressure problems in either of two ways: by using Equation $\ref{13.5.1}$ to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation $\ref{13.5.3}$ to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute.
Example $1$: Anti-Freeze
Ethylene glycol ($\ce{HOCH_2CH_2OH}$), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter.
Given: identity of solute, percentage by mass, and vapor pressure of pure solvent
Asked for: vapor pressure of solution
Strategy:
1. Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water.
2. Use Raoult’s law to calculate the vapor pressure of the solution.
Solution:
A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present:
$moles \;EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber$
$moles \; H_2O=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber$
The mole fraction of water is thus
$\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber$
B From Raoult’s law (Equation $\ref{13.5.1}$), the vapor pressure of the solution is
\begin{align*} P_{H_2O} &=(\chi_{H2_O})(P^0_{H_2O}) \[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*} \nonumber
Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation $\ref{13.5.3}$ to calculate the resulting decrease in vapor pressure:
$\chi_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber$
$ΔP_{H2_O}=(\chi_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber$
$P_{H_2O}=P^0_{H_2O}−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber$
The same result is obtained using either method.
Exercise $1$
Seawater is an approximately 3.0% aqueous solution of $\ce{NaCl}$ by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of $\ce{NaCl}$, remembering that 1 mol of $\ce{NaCl}$ produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg.
Answer
0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling.
Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution ($P_{tot}$) is the sum of the vapor pressures of the components:
$P_{tot}=P_A+P_B=\chi_AP^0_A+\chi_BP^0_B \label{13.5.4}$
Because $\chi_B = 1 − \chi_A$ for a two-component system,
$P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \label{13.5.5}$
Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is
$P_{C_6H_6}=\chi_{C_6H_6}P^0_{C_6H_6} \label{13.5.6}$
and the vapor pressure of toluene in the solution is
$P{C_6H_5CH_3}=\chi_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.5.7}$
Equations $\ref{13.5.6}$ and $\ref{13.5.7}$ are both in the form of the equation for a straight line: $y = mx + b$, where $b = 0$. Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure $3$. Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component.
A solution of two volatile components that behaves like the solution in Figure $3$, which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero ($ΔH_{soln} ≈ 0$), which is one of the defining characteristics of an ideal solution.
Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions.
Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit negative deviations from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of $\ce{CCl_4}$ and methanol, for example, the nonpolar $\ce{CCl_4}$ molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier $\ce{CCl_4}$ molecules. Consequently, solutions of $\ce{CCl_4}$ and methanol exhibit positive deviations from Raoult’s law.
Example $2$
For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation).
1. cyclohexane and ethanol
2. methanol and acetone
3. n-hexane and isooctane (2,2,4-trimethylpentane)
Given: identity of pure liquids
Asked for: predicted deviation from Raoult’s law (Equation \ref{13.5.1})
Strategy:
Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution.
Solution:
1. Liquid ethanol contains an extensive hydrogen bonding network, and cyclohexane is nonpolar. Because the cyclohexane molecules cannot interact favorably with the polar ethanol molecules, they will disrupt the hydrogen bonding. Hence the A–B interactions will be weaker than the A–A and B–B interactions, leading to a higher vapor pressure than predicted by Raoult’s law (a positive deviation).
2. Methanol contains an extensive hydrogen bonding network, but in this case the polar acetone molecules create A–B interactions that are stronger than the A–A or B–B interactions, leading to a negative enthalpy of solution and a lower vapor pressure than predicted by Raoult’s law (a negative deviation).
3. Hexane and isooctane are both nonpolar molecules (isooctane actually has a very small dipole moment, but it is so small that it can be ignored). Hence the predominant intermolecular forces in both liquids are London dispersion forces. We expect the A–B interactions to be comparable in strength to the A–A and B–B interactions, leading to a vapor pressure in good agreement with that predicted by Raoult’s law (an ideal solution).
Exercise $2$
For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation):
1. benzene and n-hexane
2. ethylene glycol and $\ce{CCl_4}$
3. acetic acid and n-propanol
Answer a
approximately equal
Answer b
positive deviation (vapor pressure greater than predicted)
Answer c
negative deviation (vapor pressure less than predicted)
A Video Discussing Roult's Law. Video Link: Introduction to the Vapor Pressure of a Solution (Raoult's Law), YouTube(opens in new window) [youtu.be]
A Video Discussing How to find the Vapor Pressure of a Solution. Video Link: Finding the Vapor Pressure of a Solution (Nonionic-Nonvolatile Solute), YouTube(opens in new window) [youtu.be]
Boiling Point Elevation
Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure $4$). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water.
Figure $4$: Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute. The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution.
The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent.
The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure $5$).
We can define the boiling point elevation ($ΔT_b$) as the difference between the boiling points of the solution and the pure solvent:
$ΔT_b=T_b−T^0_b \label{13.5.8}$
where $T_b$ is the boiling point of the solution and $T^0_b$ is the boiling point of the pure solvent. We can express the relationship between $ΔT_b$ and concentration as follows
$ΔT_b = mK_b \label{13.5.9}$
where m is the concentration of the solute expressed in molality, and $K_b$ is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table $1$ lists characteristic $K_b$ values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration.
Table $1$: Boiling Point Elevation Constants (Kb) and Freezing Point Depression Constants (Kf) for Some Solvents
Solvent Boiling Point (°C) Kb (°C/m) Freezing Point (°C) Kf (°C/m)
acetic acid 117.90 3.22 16.64 3.63
benzene 80.09 2.64 5.49 5.07
d-(+)-camphor 207.4 4.91 178.8 37.8
carbon disulfide 46.2 2.42 −112.1 3.74
carbon tetrachloride 76.8 5.26 −22.62 31.4
chloroform 61.17 3.80 −63.41 4.60
nitrobenzene 210.8 5.24 5.70 6.87
water 100.00 0.51 0.00 1.86
The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros.
According to Table $1$, the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous $\ce{NaCl}$ solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved ions. Hence a 1.00 m $\ce{NaCl}$ solution will have a boiling point of about 101.02°C.
Example $3$
In Example $1$, we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution.
Given: composition of solution
Asked for: boiling point
Strategy:
Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation $\ref{13.5.9}$ to calculate the increase in boiling point.
Solution:
From Example $1$, we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus
$\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98\, m \nonumber$
From Equation $\ref{13.5.9}$, the increase in boiling point is therefore
$ΔT_b=m K_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C \nonumber$
The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation $\ref{13.5.9}$ may not be valid.
Exercise $3$
Assume that a tablespoon (5.00 g) of $\ce{NaCl}$ is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil?
Answer
100.04°C, or 100°C to three significant figures. (Recall that 1 mol of $\ce{NaCl}$ produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.)
A Video Discussing Boiling Point Elevation and Freezing Point Depression. Video Link: Boiling Point Elevation and Freezing Point Depression, YouTube(opens in new window) [youtu.be] (opens in new window)
Freezing Point Depression
The phase diagram in Figure $4$ shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to $P = 1\,atm$ at a lower temperature than the curve for pure water.
We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice.
By analogy to our treatment of boiling point elevation,the freezing point depression ($ΔT_f$) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
$ΔT_f=T^0_f−T_f \label{13.5.10}$
where $T^0_f$ is the freezing point of the pure solvent and $T_f$ is the freezing point of the solution.
The order of the terms is reversed compared with Equation $\ref{13.5.8}$ to express the freezing point depression as a positive number. The relationship between $ΔT_f$ and the solute concentration is given by an equation analogous to Equation $\ref{13.5.9}$:
$ΔT_f = mK_f \label{13.5.11}$
where $m$ is the molality of the solution and $K_f$ is the molal freezing point depression constant for the solvent (in units of °C/m).
Like $K_b$, each solvent has a characteristic value of $K_f$ (see Table $1$). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous $\ce{NaCl}$ solution has twice as large a freezing point depression as a glucose solution of the same molality.
People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing.
The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles.
Example $4$: Salting the Roads
In colder regions of the United States, $\ce{NaCl}$ or $\ce{CaCl_2}$ is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of $\ce{NaCl}$ and one of $\ce{CaCl_2}$, and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice.
Given: solubilities of two compounds
Asked for: concentrations and freezing points
Strategy:
1. Estimate the solubility of each salt in 100 g of water from the figure. Determine the number of moles of each in 100 g and calculate the molalities.
2. Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation $\ref{13.5.11}$ to calculate the freezing point depressions of the solutions.
Solution:
A From Figure above, we can estimate the solubilities of $\ce{NaCl}$ and $\ce{CaCl_2}$ to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are
$m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber$
$m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber$
The lower formula mass of $\ce{NaCl}$ more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than $CaCl_2$.
B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of $\ce{NaCl}$ and $CaCl_2$, respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for $\ce{NaCl}$ and 3 × 5.4 m = 16 m for $CaCl_2$. The resulting freezing point depressions can be calculated using Equation $\ref{13.5.11}$:
$\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber$
$\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber$
Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that $CaCl_2$ is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, $CaCl_2$ is the salt usually sold for home use, and it is also often used on highways.
Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer
Exercise $4$
Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples $5$ and $5$.
Answer
−13.0°C
Example $5$
Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m $KCl$, 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl.
Given: molalities of six solutions
Asked for: relative freezing points
Strategy:
1. Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced.
2. Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression.
Solution:
A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. $\ce{KCl}$, $\ce{SrCl_2}$, and $\ce{HCl}$ are strong electrolytes, producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule).
B The molalities of the solutions in terms of the total particles of solute are: $\ce{KCl}$ and $\ce{HCl}$, 0.2 m; $SrCl_2$, 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > $\ce{HCl}$ = $\ce{KCl}$ > $\ce{SrCl_2}$.
Exercise $5$
Arrange these aqueous solutions in order of increasing freezing points: 0.2 m $\ce{NaCl}$, 0.3 m acetic acid, 0.1 m $\ce{CaCl_2}$, and 0.2 m sucrose.
Answer
0.2 m $\ce{NaCl}$ (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m $\ce{CaCl_2}$ < 0.2 m sucrose (highest freezing point)
Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing $K_b$ and $K_f$ values in Table $1$, we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of $K_f$ (37.8°C/m), d-(+)-camphor (Table $1$) is often used to determine the molar mass of organic compounds by this method.
Example $6$: Sulfur
A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of $CS_2$ to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved $\ce{S_n}$ molecules (i.e., what is the value of $n$?).
Given: masses of solute and solvent and freezing point
Asked for: molar mass and number of $\ce{S}$ atoms per molecule
Strategy:
1. Use Equation $\ref{13.5.10}$, the measured freezing point of the solution, and the freezing point of $CS_2$ from Table $1$ to calculate the freezing point depression. Then use Equation $\ref{13.5.11}$ and the value of $K_f$ from Table $1$ to calculate the molality of the solution.
2. From the calculated molality, determine the number of moles of solute present.
3. Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain $n$, the number of sulfur atoms per mole of dissolved sulfur.
Solution:
A The first step is to calculate the freezing point depression using Equation $\ref{13.5.10}$:
$ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber$
Then Equation $\ref{13.5.11}$ gives
$m=\dfrac{ΔT_f}{K_f}=\dfrac{1.4° \cancel{C}}{3.74° \cancel{C}/m}=0.37\;m \nonumber$
B The total number of moles of solute present in the solution is
$\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber$
C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus
$\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber$
The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of $\ce{S_8}$.
Exercise $6$
One of the byproducts formed during the synthesis of $C_{60}$ is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of $CCl_4$ has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance?
Answer
847 g/mol; $\ce{C_{70}}$
A Video Discussing how to find the Molecular Weight of an Unknown using Colligative Properties. Video Link: Finding the Molecular Weight of an Unknown using Colligative Properties, YouTube(opens in new window) [youtu.be]
Osmotic Pressure
Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration.
Osmosis can be demonstrated using a U-tube like the one shown in Figure $6$, which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure ($\Pi$) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution.
Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation:
$\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12}$
where $M$ is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), $R$ is the ideal gas constant, and $T$ is the absolute temperature.
As shown in Example $7$, osmotic pressures tend to be quite high, even for rather dilute solutions.
Example $7$
When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% $\ce{NaCl}$ by mass; the solution density is 1.02 g/mL at 25°C.
1. Calculate the osmotic pressure of a 4.0% aqueous $\ce{NaCl}$ solution at 25°C.
2. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C?
Given: concentration, density, and temperature of $\ce{NaCl}$ solution; internal osmotic pressure of cell
Asked for: osmotic pressure of $\ce{NaCl}$ solution and concentration of glycerol needed
Strategy:
1. Calculate the molarity of the $\ce{NaCl}$ solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles.
2. Use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution.
3. Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation \ref{13.5.12} to calculate the molarity of glycerol needed to create this osmotic pressure.
Solution:
A The solution contains 4.0 g of $\ce{NaCl}$ per 100 g of solution. Using the formula mass of $\ce{NaCl}$ (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity:
\begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*} \nonumber
Because 1 mol of $\ce{NaCl}$ produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M.
B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution:
\begin{align*} \Pi &=MRT \[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\[4pt] &=34 \;atm\end{align*} \nonumber
C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure:
\begin{align*} M&=\dfrac{\Pi}{RT}\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*} \nonumber
In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the $\ce{NaCl}$ solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol.
Exercise $7$
Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them.
Answer
24 atm
Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins.
The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure $7$, which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells.
In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves.
Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure $8$, applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats.
A Video Discussing Osmotic Pressure. Video Link: Osmotic Pressure, YouTube(opens in new window) [youtu.be] (opens in new window)
Colligative Properties of Electrolyte Solutions
Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of $\ce{NaCl}$ in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of $\ce{NaCl}$ and $\ce{KCl}$ are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution.
The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor ($i$) and is defined as follows:
$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13}$
Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.
As the solute concentration increases, the van’t Hoff factor decreases.
The van’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van’t Hoff factor, the greater the deviation. As the data in Table $2$ show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution.
Table $2$: van’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C
Compound i (measured) i (ideal)
glucose 1.0 1.0
sucrose 1.0 1.0
$NaCl$ 1.9 2.0
$HCl$ 1.9 2.0
$MgCl_2$ 2.7 3.0
$FeCl_3$ 3.4 4.0
$Ca(NO_3)_2$ 2.5 3.0
$AlCl_3$ 3.2 4.0
$MgSO_4$ 1.4 2.0
Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure $9$). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as $Mg^{2+}$, $Al^{3+}$, $SO_4^{2−}$, and $PO_4^{3−}$ have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations.
Example $8$: Iron Chloride in Water
A 0.0500 M aqueous solution of $FeCl_3$ has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor $i$ for the solution.
Given: solute concentration, osmotic pressure, and temperature
Asked for: van’t Hoff factor
Strategy:
1. Use Equation $\ref{13.5.12}$ to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent.
2. Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation $\ref{13.5.13}$ to calculate the van’t Hoff factor.
Solution:
A If $\ce{FeCl_3}$ dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be
\begin{align*} \Pi &=MRT \[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*} \nonumber
B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of $\ce{FeCl_3}$ dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:
$4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber$
or after rearranging
$M = 0.170 mol \nonumber$
The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of $\ce{FeCl_3}$ dissolved. From Equation $\ref{13.5.13}$, the van’t Hoff factor for the solution is
$i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber$
Exercise $8$: Magnesium Chloride in Water
Calculate the van’t Hoff factor for a 0.050 m aqueous solution of $MgCl_2$ that has a measured freezing point of −0.25°C.
Answer
2.7 (versus an ideal value of 3).
A Video Discussing the Colligative Properties in Solutions. Video Link: Colligative Properties in Solutions, YouTube(opens in new window) [youtu.be]
Summary
The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation ($ΔT_b$) and freezing point depression ($ΔT_f$) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt.
• Henry’s law: $C = kP \nonumber$
• Raoult’s law: $P_A=\chi_AP^0_A \nonumber$
• vapor pressure lowering: $P^0_A−P_A=ΔP_A=\chi_BP^0_A \nonumber$
• vapor pressure of a system containing two volatile components: $P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \nonumber$
• boiling point elevation: $ΔT_b = mK_b \nonumber$
• freezing point depression: $ΔT_f = mK_f \nonumber$
• osmotic pressure: $\Pi=nRTV=MRT \nonumber$
• van ’t Hoff factor: $i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/14%3A_Solutions/14.06%3A_Colligative_Properties-_Freezing_Point_Depression_Boiling_Point_Elevation_and_Osmosis.txt |
Learning Objectives
• To understand the factors that determine the solubility of ionic compounds.
The solubility product of an ionic compound describes the concentrations of ions in equilibrium with a solid, but what happens if some of the cations become associated with anions rather than being completely surrounded by solvent? Then predictions of the total solubility of the compound based on the assumption that the solute exists solely as discrete ions would differ substantially from the actual solubility, as would predictions of ionic concentrations. In general, four situations explain why the solubility of a compound may be other than expected: ion pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH.
Ion-Pair Formation
An ion pair consists of a cation and an anion that are in intimate contact in solution, rather than separated by solvent (Figure $1$). The ions in an ion pair are held together by the same attractive electrostatic force in ionic solids. As a result, the ions in an ion pair migrate as a single unit, whose net charge is the sum of the charges on the ions. In many ways, we can view an ion pair as a species intermediate between the ionic solid (in which each ion participates in many cation–anion interactions that hold the ions in a rigid array) and the completely dissociated ions in solution (where each is fully surrounded by water molecules and free to migrate independently).
As illustrated for calcium sulfate in the following equation, a second equilibrium must be included to describe the solubility of salts that form ion pairs:
$\mathrm{CaSO_4(s)}\rightleftharpoons \underbrace{\mathrm{Ca^{2+}}\cdot \mathrm{SO_4^{2-}(aq)}}_ {\textrm{ion pair}} \rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{SO_4^{2-}(aq)} \label{17.5.3.1}$
The ion pair is represented by the symbols of the individual ions separated by a dot, which indicates that they are associated in solution. The formation of an ion pair is a dynamic process, just like any other equilibrium, so a particular ion pair may exist only briefly before dissociating into the free ions, each of which may later associate briefly with other ions.
Ion-pair formation can have a major effect on the measured solubility of a salt. For example, the measured Ksp for calcium sulfate is 4.93 × 10−5 at 25°C. The solubility of CaSO4 should be 7.02 × 10−3 M if the only equilibrium involved were as follows:
$\ce{ CaSO4(s) <=> Ca^{2+}(aq) + SO^{2−}4(aq)} \label{17.5.2}$
In fact, the experimentally measured solubility of calcium sulfate at 25°C is 1.6 × 10−2 M, almost twice the value predicted from its Ksp. The reason for the discrepancy is that the concentration of ion pairs in a saturated $\ce{CaSO4}$ solution is almost as high as the concentration of the hydrated ions. Recall that the magnitude of attractive electrostatic interactions is greatest for small, highly charged ions. Hence ion pair formation is most important for salts that contain M2+ and M3+ ions, such as Ca2+ and La3+, and is relatively unimportant for salts that contain monopositive cations, except for the smallest, Li+. We therefore expect a saturated solution of $\ce{CaSO4}$ to contain a high concentration of ion pairs and its solubility to be greater than predicted from its Ksp.
The formation of ion pairs increases the solubility of a salt.
Incomplete Dissociation
A molecular solute may also be more soluble than predicted by the measured concentrations of ions in solution due to incomplete dissociation. This is particularly common with weak organic acids. Although strong acids (HA) dissociate completely into their constituent ions (H+ and A) in water, weak acids such as carboxylic acids do not (Ka = 1.5 × 10−5). However, the molecular (undissociated) form of a weak acid (HA) is often quite soluble in water; for example, acetic acid (CH3CO2H) is completely miscible with water. Many carboxylic acids, however, have only limited solubility in water, such as benzoic acid (C6H5CO2H), with Ka = 6.25 × 10−5. Just as with calcium sulfate, we need to include an additional equilibrium to describe the solubility of benzoic acid:
$\ce{ C6H5CO2H(s) \rightleftharpoons C6H5CO2H(aq) \rightleftharpoons C6H5CO^{-}2(aq) + H^{+}(aq)} \nonumber$
In a case like this, measuring only the concentration of the ions grossly underestimates the total concentration of the organic acid in solution. In the case of benzoic acid, for example, the pH of a saturated solution at 25°C is 2.85, corresponding to [H+] = [C6H5CO2] = 1.4 × 10−3 M. The total concentration of benzoic acid in the solution, however, is 2.8 × 10−2 M. Thus approximately 95% of the benzoic acid in solution is in the form of hydrated neutral molecules—$C_6H_5CO_2H_{(aq)}$—and only about 5% is present as the dissociated ions (Figure $2$).
Although ion pairs, such as Ca2+·SO42, and undissociated electrolytes, such as C6H5CO2H, are both electrically neutral, there is a major difference in the forces responsible for their formation. Simple electrostatic attractive forces between the cation and the anion hold the ion pair together, whereas a polar covalent O−H bond holds together the undissociated electrolyte.
Incomplete dissociation of a molecular solute that is miscible with water can increase the solubility of the solute.
Complex Ion Formation
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion.
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $1$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
The Formation Constant
The replacement of water molecules from $\ce{[Cu(H2O)6]^{2+}}$ by ammonia occurs in sequential steps. Omitting the water molecules bound to $\ce{Cu^{2+}}$ for simplicity, we can write the equilibrium reactions as follows:
\begin{align*}\mathrm{Cu^{2+}(aq)}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)]^{2+}_{(aq)}}\quad \quad K_1 \[4pt] \mathrm{[Cu(NH_3)]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}} \quad \quad K_2 \[4pt] \mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}} \quad \quad K_3 \[4pt] \mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons \mathrm{[Cu(NH_3)_4]^{2+}_{(aq)}} \quad \quad K_4 \end{align*} \nonumber
The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated $\ce{Cu^{2+}}$ ion contains six H2O ligands, but the complex ion that is produced contains only four $NH_3$ ligands, not six.
$\ce{Cu^{2+}(aq) + 4NH3(aq) <=> [Cu(NH3)4]^{2+}(aq)} \label{17.3.2}$
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant ($K_f$). The equilibrium constant expression for $K_f$ has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
$K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{17.3.3}$
The formation constant (Kf) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table $1$.
Table $1$: Formation Constants for Selected Complex Ions in Aqueous Solution*
Complex Ion Equilibrium Equation Kf
*Reported values are overall formation constants. Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
Ammonia Complexes [Ag(NH3)2]+ Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ 1.1 × 107
[Cu(NH3)4]2+ Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ 2.1 × 1013
[Ni(NH3)6]2+ Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ 5.5 × 108
Cyanide Complexes [Ag(CN)2] Ag+ + 2CN ⇌ [Ag(CN)2] 1.1 × 1018
[Ni(CN)4]2− Ni2+ + 4CN ⇌ [Ni(CN)4]2− 2.2 × 1031
[Fe(CN)6]3− Fe3+ + 6CN ⇌ [Fe(CN)6]3− 1 × 1042
Hydroxide Complexes [Zn(OH)4]2− Zn2+ + 4OH ⇌ [Zn(OH)4]2− 4.6 × 1017
[Cr(OH)4] Cr3+ + 4OH ⇌ [Cr(OH)4] 8.0 × 1029
Halide Complexes [HgCl4]2− Hg2+ + 4Cl ⇌ [HgCl4]2− 1.2 × 1015
[CdI4]2− Cd2+ + 4I ⇌ [CdI4]2− 2.6 × 105
[AlF6]3− Al3+ + 6F ⇌ [AlF6]3− 6.9 × 1019
Other Complexes [Ag(S2O3)2]3− Ag+ + 2S2O32 ⇌ [Ag(S2O3)2]3− 2.9 × 1013
[Fe(C2O4)3]3− Fe3+ + 3C2O42 ⇌ [Fe(C2O4)3]3− 2.0 × 1020
Example $1$
If 12.5 g of Cu(NO3)2•6H2O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
1. Calculate the initial concentration of Cu2+ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation $\ref{17.3.2}$ to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
2. Substitute the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) to calculate the equilibrium concentration of Cu2+(aq).
Solution
Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation $\ref{17.3.2}$. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows:
$12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M} \nonumber$
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
$\ce{Cu^{2+} + 4NH3 <=> [Cu(NH3)4]^{2+}} \nonumber$
Solutions to Example 17.5.1
[Cu2+] [NH3] [[Cu(NH3)4]2+]
initial 0.0846 1.00 0
after complete reaction 0 0.66 0.0846
change +x +4x x
final x 0.66 + 4x 0.0846 − x
B Substituting the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) and assuming that $x \ll 0.0846$, which allows us to remove x from the sum and difference,
\begin{align*}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \x&=2.1\times 10^{-14}\end{align*} \nonumber
The value of $x$ indicates that our assumption was justified. The equilibrium concentration of $\ce{Cu^{2+}(aq)}$ in a 1.00 M ammonia solution is therefore $2.1 \times 10^{−14} M$.
Exercise $1$
The ferrocyanide ion $\ce{[Fe(CN)6]^{4−}}$ is very stable, with a Kf = 1 × 1035. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of $\ce{K4[Fe(CN)6]}$.
Answer
2 × 10−6 M
The Effect of the Formation of Complex Ions on Solubility
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted $\ce{AgBr}$ on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
$\ce{AgBr(s) <=> Ag^{+}(aq) + Br^{-}(aq)} \nonumber$
with $K_{sp} = 5.35 \times 10^{−13}$ at 25°C.
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
$\ce{Ag^{+}(aq) + 2S2O^{2-}3(aq) <=> [Ag(S2O3)2]^{3-}(aq)} \nonumber$
with $K_f = 2.9 \times 10^{13}$.
The magnitude of the equilibrium constant indicates that almost all $\ce{Ag^{+}}$ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together:
\begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13} \ \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}\hspace{3mm}K_\textrm f&=2.9\times10^{13} \ \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=15\end{align} \label{17.3.6}
Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Example $2$
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2). Calculate the solubility of AgCl in each situation:
1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2 is the only Ag+ complex that forms in significant concentrations
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2.
Given: Ksp of AgCl, Kf of AgCl2, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
1. Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl in water.
2. Calculate the concentration of Ag+ in the KCl solution.
3. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2 complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
4. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution
1. A If we let x equal the solubility of AgCl, then at equilibrium [Ag+] = [Cl] = x M. Substituting this value into the solubility product expression,
Ksp = [Ag+][Cl] = (x)(x) = x2 =1.77×10−10
x = 1.33×10−5
Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M.
1. B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,
Ksp = [Ag+][Cl] = (x)(1.0 + x) ≈ x(1.0) = 1.77×10−10 = x
If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
1. C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf:
\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10} \ \mathrm{Ag^+(aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5} \ \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align}
D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2] = x and [Cl] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0-x}\approx1.9\times10^{-5}=x$
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2 complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise $2$
Calculate the solubility of mercury(II) iodide ($\ce{HgI2}$) in each situation:
1. pure water
2. a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer
1. 1.9 × 10−10 M
2. 1.4 M
Solubility of Complex Ions: Solubility of Complex Ions(opens in new window) [youtu.be]
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34, or P2O74) or triphosphate (P3O105) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
$\ce{Ca^{2+}(aq) + O_3POPO^{4−}4(aq) <=> [Ca(O_3POPO_3)]^{2−}(aq)} \label{17.3.7a}$
with $K_f = 4\times 10^4$.
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $4$). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5 (diethylene triamine pentaacetic acid), whose fully protonated form is shown here.
Summary
Ion-pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH all affect solubility. There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) ion pair formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes in pH. An ion pair is held together by electrostatic attractive forces between the cation and the anion, whereas incomplete dissociation results from intramolecular forces, such as polar covalent O–H bonds.
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/14%3A_Solutions/14.07%3A_The_Colligative_Properties_of_Strong_Electrolyte_Solutions.txt |
• 15.1: Catching Lizards
• 15.2: Rates of Reaction and the Particulate Nature of Matter
Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time.
• 15.3: Defining and Measuring the Rate of a Chemical Reaction
• 15.4: The Rate Law- The Effect of Concentration on Reaction Rate
The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time. The rate constant and reaction order are extracted directly from the rate law.
• 15.5: The Integrated Rate Law- The Dependence of Concentration on Time
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
• 15.6: The Effect of Temperature on Reaction Rate
A minimum energy (activation energy,Ea) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the Ea, the slower the reaction.
• 15.7: Reaction Mechanisms
A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity. The slowest step in a reaction mechanism is the rate-determining step.
• 15.8: Catalysis
Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts.
15: Chemical Kinetics
Learning Objectives
• To determine the reaction rate of a reaction.
Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time.
The progress of a simple reaction (A → B) is shown in Figure $1$; the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in Figure $1$. The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time.
$\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1}$
Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation $\ref{Eq1}$ is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first.
Reaction rates generally decrease with time as reactant concentrations decrease.
A Video Discussing Average Reaction Rates. Video Link: Introduction to Chemical Reaction Kinetics(opens in new window) [youtu.be] (opens in new window)
Determining the Reaction Rate of Hydrolysis of Aspirin
We can use Equation $\ref{Eq1}$ to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure $2$.
Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table $1$ and are shown in the graph in Figure $3$.
Table $1$: Data for Aspirin Hydrolysis in Aqueous Solution at pH 7.0 and 37°C*
Time (h) [Aspirin] (M) [Salicylic Acid] (M)
*The reaction at pH 7.0 is very slow. It is much faster under acidic conditions, such as those found in the stomach.
0 5.55 × 10−3 0
2.0 5.51 × 10−3 0.040 × 10−3
5.0 5.45 × 10−3 0.10 × 10−3
10 5.35 × 10−3 0.20 × 10−3
20 5.15 × 10−3 0.40 × 10−3
30 4.96 × 10−3 0.59 × 10−3
40 4.78 × 10−3 0.77 × 10−3
50 4.61 × 10−3 0.94 × 10−3
100 3.83 × 10−3 1.72 × 10−3
200 2.64 × 10−3 2.91 × 10−3
300 1.82 × 10−3 3.73 × 10−3
The data in Table $1$ were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid).
The average reaction rate for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t0) and at the end of the interval (t1). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=\frac{[\textrm{salicyclic acid}]_2-[\textrm{salicyclic acid}]_0}{\textrm{2.0 h}-\textrm{0 h}} \&=\frac{0.040\times10^{-3}\textrm{ M}-0\textrm{ M}}{\textrm{2.0 h}}=2.0\times10^{-5}\textrm{ M/h} \end{align*} \nonumber
The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases:
\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=-\dfrac{[\textrm{aspirin}]_2-[\textrm{aspirin}]_0}{\mathrm{2.0\,h-0\,h}} \&=-\dfrac{(5.51\times10^{-3}\textrm{ M})-(5.55\times10^{-3}\textrm{ M})}{\textrm{2.0 h}} \&=2\times10^{-5}\textrm{ M/h}\end{align*} \nonumber
If the reaction rate is calculated during the last interval given in Table $1$(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h):
\begin{align*}\textrm{rate}_{(t=200-300\textrm{h})}&=\dfrac{[\textrm{salicyclic acid}]_{300}-[\textrm{salicyclic acid}]_{200}}{\mathrm{300\,h-200\,h}} \&=-\dfrac{(3.73\times10^{-3}\textrm{ M})-(2.91\times10^{-3}\textrm{ M})}{\textrm{100 h}} \&=8.2\times10^{-6}\textrm{ M/h}\end{align*} \nonumber
Calculating the Reaction Rate of Fermentation of Sucrose
In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide:
$\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2}$
The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as explained in Chapter 10, the volume of CO2 gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed:
$\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3}$
The concentration of the reactant—in this case sucrose—decreases with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation $\ref{Eq3}$ so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with time, so its rate of change is automatically expressed as a positive value.
Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation $\ref{Eq2}$) corresponds to sucrose, so the reaction rate is generally defined as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4}$
Example $1$: Decomposition Reaction I
Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation:
$\mathrm{2N_2O_5(g)}\xrightarrow{\,\Delta\,}\mathrm{4NO_2(g)}+\mathrm{O_2(g)} \nonumber$
Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time.
Given: balanced chemical equation
Asked for: reaction rate expressions
Strategy:
1. Choose the species in the equation that has the smallest coefficient. Then write an expression for the rate of change of that species with time.
2. For the remaining species in the equation, use molar ratios to obtain equivalent expressions for the reaction rate.
Solution
A Because O2 has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O2 and write that expression.
B The balanced chemical equation shows that 2 mol of N2O5 must decompose for each 1 mol of O2 produced and that 4 mol of NO2 are produced for every 1 mol of O2 produced. The molar ratios of O2 to N2O5 and to NO2 are thus 1:2 and 1:4, respectively. This means that the rate of change of [N2O5] and [NO2] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO2 is produced at four times the rate of O2, the rate of production of NO2 is divided by 4. The reaction rate expressions are as follows:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}$
Exercise $1$: Contact Process I
The contact process is used in the manufacture of sulfuric acid. A key step in this process is the reaction of $SO_2$ with $O_2$ to produce $SO_3$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
Write expressions for the reaction rate in terms of the rate of change of the concentration of each species.
Answer
$\textrm{rate}=-\dfrac{\Delta[\mathrm O_2]}{\Delta t}=-\dfrac{\Delta[\mathrm{SO_2}]}{2\Delta t}=\dfrac{\Delta[\mathrm{SO_3}]}{2\Delta t}$
Instantaneous Rates of Reaction
The instantaneous rate of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time.
The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.
Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0.
Example $2$: Decomposition Reaction II
Using the reaction shown in Example $1$, calculate the reaction rate from the following data taken at 56°C:
$2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber$
calculate the reaction rate from the following data taken at 56°C:
Time (s) [N2O5] (M) [NO2] (M) [O2] (M)
240 0.0388 0.0314 0.00792
600 0.0197 0.0699 0.0175
Given: balanced chemical equation and concentrations at specific times
Asked for: reaction rate
Strategy:
1. Using the equations in Example $1$, subtract the initial concentration of a species from its final concentration and substitute that value into the equation for that species.
2. Substitute the value for the time interval into the equation. Make sure your units are consistent.
Solution
A Calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example $1$, the reaction rate can be evaluated using any of three expressions:
$\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t} \nonumber$
Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5,
$\textrm{rate}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}=-\dfrac{[\mathrm{N_2O_5}]_{600}-[\mathrm{N_2O_5}]_{240}}{2(600\textrm{ s}-240\textrm{ s})} \nonumber$
B Substituting actual values into the expression,
$\textrm{rate}=-\dfrac{\mathrm{\mathrm{0.0197\;M-0.0388\;M}}}{2(360\textrm{ s})}=2.65\times10^{-5} \textrm{ M/s}$
Similarly, NO2 can be used to calculate the reaction rate:
$\textrm{rate}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=\dfrac{[\mathrm{NO_2}]_{600}-[\mathrm{NO_2}]_{240}}{4(\mathrm{600\;s-240\;s})}=\dfrac{\mathrm{0.0699\;M-0.0314\;M}}{4(\mathrm{360\;s})}=\mathrm{2.67\times10^{-5}\;M/s} \nonumber$
Allowing for experimental error, this is the same rate obtained using the data for N2O5. The data for O2 can also be used:
$\textrm{rate}=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=\dfrac{[\mathrm{O_2}]_{600}-[\mathrm{O_2}]_{240}}{\mathrm{600\;s-240\;s}}=\dfrac{\mathrm{0.0175\;M-0.00792\;M}}{\mathrm{360\;s}}=\mathrm{2.66\times10^{-5}\;M/s} \nonumber$
Again, this is the same value obtained from the N2O5 and NO2 data. Thus, the reaction rate does not depend on which reactant or product is used to measure it.
Exercise $2$: Contact Process II
Using the data in the following table, calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber$
calculate the reaction rate of $SO_2(g)$ with $O_2(g)$ to give $SO_3(g)$.
Time (s) [SO2] (M) [O2] (M) [SO3] (M)
300 0.0270 0.0500 0.0072
720 0.0194 0.0462 0.0148
Answer:
9.0 × 10−6 M/s
Summary
In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time.
• General definition of rate for A → B: $\textrm{rate}=\frac{\Delta [\textrm B]}{\Delta t}=-\frac{\Delta [\textrm A]}{\Delta t} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/15%3A_Chemical_Kinetics/15.02%3A_Rates_of_Reaction_and_the_Particulate_Nature_of_Matter.txt |
Learning Objectives
• To understand the meaning of the rate law.
The factors that affect the reaction rate of a chemical reaction, which may determine whether a desired product is formed. In this section, we will show you how to quantitatively determine the reaction rate.
Rate Laws
Typically, reaction rates decrease with time because reactant concentrations decrease as reactants are converted to products. Reaction rates generally increase when reactant concentrations are increased. This section examines mathematical expressions called rate laws, which describe the relationships between reactant rates and reactant concentrations. Rate laws are mathematical descriptions of experimentally verifiable data.
Rate laws may be written from either of two different but related perspectives. A differential rate law expresses the reaction rate in terms of changes in the concentration of one or more reactants (Δ[R]) over a specific time interval (Δt). In contrast, an integrated rate law describes the reaction rate in terms of the initial concentration ([R]0) and the measured concentration of one or more reactants ([R]) after a given amount of time (t); integrated rate laws are discussed in more detail later. The integrated rate law is derived by using calculus to integrate the differential rate law. Whether using a differential rate law or integrated rate law, always make sure that the rate law gives the proper units for the reaction rate, usually moles per liter per second (M/s).
Reaction Orders
For a reaction with the general equation:
$aA + bB \rightarrow cC + dD \label{14.3.1}$
the experimentally determined rate law usually has the following form:
$\text{rate} = k[A]^m[B]^n \label{14.3.2}$
The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent; varying the temperature or the solvent usually changes the value of the rate constant. The numerical value of k, however, does not change as the reaction progresses under a given set of conditions.
Under a given set of conditions, the value of the rate constant does not change as the reaction progresses.
The reaction rate thus depends on the rate constant for the given set of reaction conditions and the concentration of A and B raised to the powers m and n, respectively. The values of m and n are derived from experimental measurements of the changes in reactant concentrations over time and indicate the reaction order, the degree to which the reaction rate depends on the concentration of each reactant; m and n need not be integers. For example, Equation $\ref{14.3.2}$ tells us that Equation $\ref{14.3.1}$ is mth order in reactant A and nth order in reactant B. It is important to remember that n and m are not related to the stoichiometric coefficients a and b in the balanced chemical equation and must be determined experimentally. The overall reaction order is the sum of all the exponents in the rate law: m + n.
The orders of the reactions (e.g. n and m) are not related to the stoichiometric coefficients in the balanced chemical (e.g., a and b).
To illustrate how chemists interpret a differential rate law, consider the experimentally derived rate law for the hydrolysis of t-butyl bromide in 70% aqueous acetone.
This reaction produces t-butanol according to the following equation:
$(CH_3)_3CBr_{(soln)} + H_2O_{(soln)} \rightarrow (CH_3)_3COH_{(soln)} + HBr_{(soln)} \label{14.3.3}$
Combining the rate expression in Equation $\ref{14.3.2}$ with the definition of average reaction rate
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t} \nonumber$
gives a general expression for the differential rate law:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^m[\textrm B]^n \label{14.3.4}$
Inserting the identities of the reactants into Equation $\ref{14.3.4}$ gives the following expression for the differential rate law for the reaction:
$\textrm{rate}=-\dfrac{\Delta[\mathrm{(CH_3)_3CBr}]}{\Delta t}=k[\mathrm{(CH_3)_3CBr}]^m[\mathrm{H_2O}]^n \label{14.3.5}$
Experiments to determine the rate law for the hydrolysis of t-butyl bromide show that the reaction rate is directly proportional to the concentration of (CH3)3CBr but is independent of the concentration of water. Therefore, m and n in Equation $\ref{14.3.4}$ are 1 and 0, respectively, and,
$\text{rate} = k[(CH_3)_3CBr]^1[H_2O]^0 = k[(CH_3)_3CBr] \label{14.3.6}$
Because the exponent for the reactant is 1, the reaction is first order in (CH3)3CBr. It is zeroth order in water because the exponent for [H2O] is 0. (Recall that anything raised to the zeroth power equals 1.) Thus, the overall reaction order is 1 + 0 = 1. The reaction orders state in practical terms that doubling the concentration of (CH3)3CBr doubles the reaction rate of the hydrolysis reaction, halving the concentration of (CH3)3CBr halves the reaction rate, and so on. Conversely, increasing or decreasing the concentration of water has no effect on the reaction rate. (Again, when working with rate laws, there is no simple correlation between the stoichiometry of the reaction and the rate law. The values of k, m, and n in the rate law must be determined experimentally.) Experimental data show that k has the value 5.15 × 10−4 s−1 at 25°C. The rate constant has units of reciprocal seconds (s−1) because the reaction rate is defined in units of concentration per unit time (M/s). The units of a rate constant depend on the rate law for a particular reaction.
Under conditions identical to those for the t-butyl bromide reaction, the experimentally derived differential rate law for the hydrolysis of methyl bromide (CH3Br) is as follows:
$\textrm{rate}=-\dfrac{\Delta[\mathrm{CH_3Br}]}{\Delta t}=k'[\mathrm{CH_3Br}] \label{14.3.7}$
This reaction also has an overall reaction order of 1, but the rate constant in Equation $\ref{14.3.7}$ is approximately 106 times smaller than that for t-butyl bromide. Thus, methyl bromide hydrolyzes about 1 million times more slowly than t-butyl bromide, and this information tells chemists how the reactions differ on a molecular level.
Frequently, changes in reaction conditions also produce changes in a rate law. In fact, chemists often alter reaction conditions to study the mechanics of a reaction. For example, when t-butyl bromide is hydrolyzed in an aqueous acetone solution containing OH ions rather than in aqueous acetone alone, the differential rate law for the hydrolysis reaction does not change. In contrast, for methyl bromide, the differential rate law becomes
$\text{rate} =k″[CH_3Br][OH^−] \nonumber$
with an overall reaction order of 2. Although the two reactions proceed similarly in neutral solution, they proceed very differently in the presence of a base, providing clues as to how the reactions differ on a molecular level.
Example $1$: Writing Rate Laws from Reaction Orders
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
$\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \nonumber$
is second order in NO2 and zero order in CO at 100 °C. What is the rate law for the reaction?
Solution
The reaction will have the form:
$\ce{rate}=k[\ce{NO2}]^m[\ce{CO}]^n \nonumber$
The reaction is second order in NO2; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:
$\ce{rate}=k[\ce{NO2}]^2[\ce{CO}]^0=k[\ce{NO2}]^2 \nonumber$
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO2. When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.
Exercise $\PageIndex{1A}$
The rate law for the reaction:
$\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g) \nonumber$
has been experimentally determined to be $rate = k[NO]^2[H_2]$. What are the orders with respect to each reactant, and what is the overall order of the reaction?
Answer
• order in NO = 2
• order in H2 = 1
• overall order = 3
Exercise $\PageIndex{1B}$
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH3OH) and ethyl acetate (CH3CH2OCOCH3) as a sample reaction before studying the chemical reactions that produce biodiesel:
$\ce{CH3OH + CH3CH2OCOCH3 ⟶ CH3OCOCH3 + CH3CH2OH} \nonumber$
The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, experimentally determined to be:
$\ce{rate}=k[\ce{CH3OH}] \nonumber$
What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?
Answer
• order in CH3OH = 1
• order in CH3CH2OCOCH3 = 0
• overall order = 1
Example $2$: Differential Rate Laws
Below are three reactions and their experimentally determined differential rate laws. For each reaction, give the units of the rate constant, give the reaction order with respect to each reactant, give the overall reaction order, and predict what happens to the reaction rate when the concentration of the first species in each chemical equation is doubled.
1. $\mathrm{2HI(g)}\xrightarrow{\textrm{Pt}}\mathrm{H_2(g)}+\mathrm{I_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{HI}]}{\Delta t} \right )=k[\textrm{HI}]^2 \nonumber$
2. $\mathrm{2N_2O(g)}\xrightarrow{\Delta}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \ \textrm{rate}=-\frac{1}{2}\left (\frac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=k \nonumber$
3. $\mathrm{cyclopropane(g)}\rightarrow\mathrm{propane(g)} \ \textrm{rate}=-\frac{\Delta[\mathrm{cyclopropane}]}{\Delta t}=k[\mathrm{cyclopropane}] \nonumber$
Given: balanced chemical equations and differential rate laws
Asked for: units of rate constant, reaction orders, and effect of doubling reactant concentration
Strategy:
1. Express the reaction rate as moles per liter per second [mol/(L·s), or M/s]. Then determine the units of each chemical species in the rate law. Divide the units for the reaction rate by the units for all species in the rate law to obtain the units for the rate constant.
2. Identify the exponent of each species in the rate law to determine the reaction order with respect to that species. Add all exponents to obtain the overall reaction order.
3. Use the mathematical relationships as expressed in the rate law to determine the effect of doubling the concentration of a single species on the reaction rate.
Solution
1. A [HI]2 will give units of (moles per liter)2. For the reaction rate to have units of moles per liter per second, the rate constant must have reciprocal units [1/(M·s)]:
$k\textrm M^2=\dfrac{\textrm M}{\textrm s}k=\dfrac{\textrm{M/s}}{\textrm M^2}=\dfrac{1}{\mathrm{M\cdot s}}=\mathrm{M^{-1}\cdot s^{-1}} \nonumber$
B The exponent in the rate law is 2, so the reaction is second order in HI. Because HI is the only reactant and the only species that appears in the rate law, the reaction is also second order overall.
C If the concentration of HI is doubled, the reaction rate will increase from k[HI]02 to k(2[HI])02 = 4k[HI]02. The reaction rate will therefore quadruple.
1. A Because no concentration term appears in the rate law, the rate constant must have M/s units for the reaction rate to have M/s units.
B The rate law tells us that the reaction rate is constant and independent of the N2O concentration. That is, the reaction is zeroth order in N2O and zeroth order overall.
C Because the reaction rate is independent of the N2O concentration, doubling the concentration will have no effect on the reaction rate.
1. A The rate law contains only one concentration term raised to the first power. Hence the rate constant must have units of reciprocal seconds (s−1) to have units of moles per liter per second for the reaction rate: M·s−1 = M/s.
B The only concentration in the rate law is that of cyclopropane, and its exponent is 1. This means that the reaction is first order in cyclopropane. Cyclopropane is the only species that appears in the rate law, so the reaction is also first order overall.
C Doubling the initial cyclopropane concentration will increase the reaction rate from k[cyclopropane]0 to 2k[cyclopropane]0. This doubles the reaction rate.
Exercise $2$
Given the following two reactions and their experimentally determined differential rate laws: determine the units of the rate constant if time is in seconds, determine the reaction order with respect to each reactant, give the overall reaction order, and predict what will happen to the reaction rate when the concentration of the first species in each equation is doubled.
1. \begin{align}\textrm{CH}_3\textrm N\textrm{=NCH}_3\textrm{(g)}\rightarrow\mathrm{C_2H_6(g)}+\mathrm{N_2(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\textrm{CH}_3\textrm N\textrm{=NCH}_3]}{\Delta t} \&=k[\textrm{CH}_3\textrm N\textrm{=NCH}_3]\end{align}
2. \begin{align}\mathrm{2NO_2(g)}+\mathrm{F_2(g)}\rightarrow\mathrm{2NO_2F(g)}\hspace{5mm}&\textrm{rate}=-\frac{\Delta[\mathrm{F_2}]}{\Delta t}=-\frac{1}{2}\left ( \frac{\Delta[\mathrm{NO_2}]}{\Delta t} \right ) \&=k[\mathrm{NO_2}][\mathrm{F_2}]\end{align}
Answer a
s−1; first order in CH3N=NCH3; first order overall; doubling [CH3N=NCH3] will double the reaction rate.
Answer b
M−1·s−1; first order in NO2, first order in F2; second order overall; doubling [NO2] will double the reaction rate.
Determining the Rate Law of a Reaction
The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanisms can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction $A + B \rightarrow products$, for example, we need to determine k and the exponents m and n in the following equation:
$\text{rate} = k[A]^m[B]^n \label{14.4.11}$
To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments. Rate data for a hypothetical reaction of the type $A + B \rightarrow products$ are given in Table $1$.
Table $1$: Rate Data for a Hypothetical Reaction of the Form $A + B \rightarrow products$
Experiment [A] (M) [B] (M) Initial Rate (M/min)
1 0.50 0.50 8.5 × 10−3
2 0.75 0.50 19 × 10−3
3 1.00 0.50 34 × 10−3
4 0.50 0.75 8.5 × 10−3
5 0.50 1.00 8.5 × 10−3
The general rate law for the reaction is given in Equation $\ref{14.4.11}$. We can obtain m or n directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in Table $3$.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_3}}=\dfrac{k[\textrm A_1]^m[\textrm B_1]^n}{k[\textrm A_3]^m[\textrm B_3]^n} \nonumber$
Inserting the appropriate values from Table $3$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{34\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{1.00 M}]^m[\textrm{0.50 M}]^n} \nonumber$
Because 1.00 to any power is 1, [1.00 M]m = 1.00 M. We can cancel like terms to give 0.25 = [0.50]m, which can also be written as 1/4 = [1/2]m. Thus we can conclude that m = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for m.
Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for n.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_5}}=\dfrac{k[\mathrm{A_1}]^m[\mathrm{B_1}]^n}{k[\mathrm{A_5}]^m[\mathrm{B_5}]^n}$
Substituting the appropriate values from Table $3$,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{8.5\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{0.50 M}]^m[\textrm{1.00 M}]^n} \nonumber$
Canceling leaves 1.0 = [0.50]n, which gives $n = 0$; that is, the reaction is zeroth order in $B$. The experimentally determined rate law is therefore
rate = k[A]2[B]0 = k[A]2
We can now calculate the rate constant by inserting the data from any row of Table $3$ into the experimentally determined rate law and solving for $k$. Using Experiment 2, we obtain
19 × 10−3 M/min = k(0.75 M)2
3.4 × 10−2 M−1·min−1 = k
You should verify that using data from any other row of Table $1$ gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same.
Example $3$
Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with $O_2$ to give $NO_2$, which then reacts rapidly with water to give nitrous acid and nitric acid:
These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O2 at 25°C:
$2NO(g) + O_2(g) \rightarrow 2NO_2(g) \nonumber$
Determine the rate law for the reaction and calculate the rate constant.
rate law for the reaction and calculate the rate constant.
Experiment [NO]0 (M) [O2]0 (M) Initial Rate (M/s)
1 0.0235 0.0125 7.98 × 10−3
2 0.0235 0.0250 15.9 × 10−3
3 0.0470 0.0125 32.0 × 10−3
4 0.0470 0.0250 63.5 × 10−3
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: rate law and rate constant
Strategy:
1. Compare the changes in initial concentrations with the corresponding changes in rates of reaction to determine the reaction order for each species. Write the rate law for the reaction.
2. Using data from any experiment, substitute appropriate values into the rate law. Solve the rate equation for k.
Solution
A Comparing Experiments 1 and 2 shows that as [O2] is doubled at a constant value of [NO2], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O2]1, so the reaction is first order in O2. Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O2] is held constant. That is, the reaction rate is proportional to [NO]2, which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction:
rate = k[NO]2[O2]
B The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{7.98\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0125\textrm{ M})}=1.16\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
Alternatively, using Experiment 2 gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{15.9\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0250\textrm{ M})}=1.15\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
The difference is minor and associated with significant digits and likely experimental error in making the table.
The overall reaction order $(m + n) = 3$, so this is a third-order reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases.
Exercise $3$
The peroxydisulfate ion (S2O82) is a potent oxidizing agent that reacts rapidly with iodide ion in water:
$S_2O^{2−}_{8(aq)} + 3I^−_{(aq)} \rightarrow 2SO^{2−}_{4(aq)} + I^−_{3(aq)} \nonumber$
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.
kinetics data for this reaction at 25°C.
Experiment [S2O82]0 (M) [I]0 (M) Initial Rate (M/s)
1 0.27 0.38 2.05
2 0.40 0.38 3.06
3 0.40 0.22 1.76
Answer:
rate = k[S2O82][I]; k = 20 M−1·s−1
A Video Discussing Initial Rates and Rate Law Expressions. Video Link: Initial Rates and Rate Law Expressions(opens in new window) [youtu.be]
Summary
The rate law for a reaction is a mathematical relationship between the reaction rate and the concentrations of species in solution. Rate laws can be expressed either as a differential rate law, describing the change in reactant or product concentrations as a function of time, or as an integrated rate law, describing the actual concentrations of reactants or products as a function of time. The rate constant (k) of a rate law is a constant of proportionality between the reaction rate and the reactant concentration. The exponent to which a concentration is raised in a rate law indicates the reaction order, the degree to which the reaction rate depends on the concentration of a particular reactant. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/15%3A_Chemical_Kinetics/15.04%3A_The_Rate_Law-_The_Effect_of_Concentration_on_Reaction_Rate.txt |
Learning Objectives
• To apply rate laws to zeroth, first and second order reactions.
Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. The learning objective of this Module is to know how to determine the reaction order from experimental data.
Zeroth-Order Reactions
A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is
$\text{rate} = k. \nonumber$
We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1}$
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of $−k$. The value of $k$ is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of $k$, a positive value.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
$[A] = [A]_0 − kt \label{14.4.2}$
where $[A]_0$ is the initial concentration of reactant $A$. Equation \ref{14.4.2} has the form of the algebraic equation for a straight line,
$y = mx + b, \nonumber$
with $y = [A]$, $mx = −kt$, and $b = [A]_0$.)
Units
In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:
$\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3}$
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate. At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:
$\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4}$
Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in Figure $2$, the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzyme alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
where \ce{NAD^{+}}\) (nicotinamide adenine dinucleotide) and $\ce{NADH}$ (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (Figure $\PageIndex{3a}$). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure $3$).
These examples illustrate two important points:
1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.
First-Order Reactions
In a first-order reaction, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5}$
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$[A] = [A]_0e^{−kt} \label{14.4.6}$
where $[A]_0$ is the initial concentration of reactant $A$ at $t = 0$; $k$ is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation $\ref{14.4.6}$ predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation $\ref{14.4.6}$ and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of $A$ and $t$:
$\ln[A] = \ln[A]_0 − kt \label{14.4.7}$
Because Equation $\ref{14.4.7}$ has the form of the algebraic equation for a straight line,
$y = mx + b, \nonumber$
with $y = \ln[A]$ and $b = \ln[A]_0$, a plot of $\ln[A]$ versus $t$ for a first-order reaction should give a straight line with a slope of $−k$ and an intercept of $\ln[A]_0$. Either the differential rate law (Equation $\ref{14.4.5}$) or the integrated rate law (Equation $\ref{14.4.7}$) can be used to determine whether a particular reaction is first order.
First-order reactions are very common. One reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure $5$ is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure $4$ have been studied extensively to find ways of maximizing the concentration of the active species.
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table $1$. The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin.
Table $1$: Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C
Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5
Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table $1$ shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table $1$. For example, substituting the values for Experiment 3 into Equation $\ref{14.4.5}$,
3.6 × 10−5 M/min = k(0.024 M)
1.5 × 10−3 min−1 = k
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
Example $1$
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
$\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber$
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
data for the reaction at 650°C
Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
1. Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
2. Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].
C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10−8 M/s = k(0.010 M)
1.6 × 10−6 s−1 = k
Exercise $1$
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
$SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber$
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Data for the reaction at 320°C
Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7
Answer
first order; k = 2.2 × 10−5 s−1
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Figure $\PageIndex{6a}$ shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C.
The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure $6$. The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure $6$ for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}} \[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}} \[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*} \nonumber
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure $6$ are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
Video Example Using the First-Order Integrated Rate Law Equation:
Example $2$
If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M (k = 1.6 × 10−6 s−1) ?
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
1. Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.
2. Given a concentration [A], solve the integrated rate law for time t.
Solution
The exponential form of the integrated rate law for a first-order reaction (Equation $\ref{14.4.6}$) is [A] = [A]0ekt.
A Having been given the initial concentration of ethyl chloride ([A]0) and having the rate constant of k = 1.6 × 10−6 s−1, we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,
\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt} \[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}}) \[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber
We could also have used the logarithmic form of the integrated rate law (Equation $\ref{14.4.7}$):
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt \[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})] \[4pt] &=-3.912-0.0576=-3.970 \nonumber \[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber \[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for $t$. Equation $\ref{14.4.7}$ gives the following:
\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt \[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right) \[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*} \nonumber
Exercise $2$
In the exercise in Example $1$, you found that the decomposition of sulfuryl chloride ($\ce{SO2Cl2}$) is first order, and you calculated the rate constant at 320°C.
1. Use the form(s) of the integrated rate law to find the amount of $\ce{SO2Cl2}$ that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C.
2. How long would it take for 90% of the SO2Cl2 to decompose?
Answer a
0.0252 M
Answer b
29 h
Second-Order Reactions
The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form
$\ce{2A → products.}\nonumber$
A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8}$
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
$\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9}$
Because Equation $\ref{14.4.9}$ has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.
Second-order reactions generally have the form 2A → products or A + B → products.
Video Discussing the Second-Order Integrated Rate Law Equation: Second-Order Integrated Rate Law Equation(opens in new window) [youtu.be]
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
Figure $7$
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation $\ref{14.4.8}$) or the integrated rate law (Equation $\ref{14.4.9}$).
Table $2$: Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table $2$. From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
$\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7 \nonumber$
Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
rate ∝ [monomer]2
This means that the reaction is second order in the monomer. Using Equation $\ref{14.4.8}$ and the data from any row in Table $2$, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
\begin{align}\textrm{rate}&=k[\textrm A]^2 \8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align} \nonumber
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in Figure $\PageIndex{8a}$. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in Figure $\PageIndex{8b}$. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders.
Example $3$
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experimental data for the reaction at 300°C and four initial concentrations of NO2
Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
1. From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
2. Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).
Solution
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.
B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2 \5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2 \0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*} \nonumber
Exercise $3$
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
$2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber$
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Some initial rate data at 25°C
Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6
Determine the reaction order and the rate constant.
Answer
second order in HO2; k = 1.4 × 109 M−1·s−1
If a plot of reactant concentration versus time is not linear, but a plot of 1/(reactant concentration) versus time is linear, then the reaction is second order.
Example $4$
If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation \ref{14.4.9}) and the rate constant calculated above.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
1. Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
2. Setting [A] equal to 1/10 of [A]0, use the same equation to solve for $t$.
Solution
A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation \ref{14.4.9},
\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*} \nonumber
Thus [NO2]3600 = 5.1 × 10−4 M.
B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation $\ref{14.4.9}$ for t, using the concentrations given.
\begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*} \nonumber
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise $4$
In the previous exercise, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation $\ref{14.4.9}$) and the rate constant calculated in the exercise in Example $3$.
Answer
2.0 × 10−13 M; 6.4 × 10−6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form $A + B \rightarrow products$, in which the reaction is first order in $A$ and first order in $B$. The differential rate law for this reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \nonumber$
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.
Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
• zeroth-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k \nonumber$ $[A] = [A]_0 − kt \nonumber$
• first-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \nonumber$ $[A] = [A]_0e^{−kt} \nonumber$ $\ln[A] = \ln[A]_0 − kt \nonumber$
• second-order reaction: $\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2 \nonumber$ $\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/15%3A_Chemical_Kinetics/15.05%3A_The_Integrated_Rate_Law-_The_Dependence_of_Concentration_on_Time.txt |
Learning Objectives
• To understand why and how chemical reactions occur.
It is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. The collision model explains why chemical reactions often occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates. Before delving into the relationship between temperature and reaction rate, we must discuss three microscopic factors that influence the observed macroscopic reaction rates.
Microscopic Factor 1: Collisional Frequency
Central to collision model is that a chemical reaction can occur only when the reactant molecules, atoms, or ions collide. Hence, the observed rate is influence by the frequency of collisions between the reactants. The collisional frequency is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system. While deriving the collisional frequency ($Z_{AB}$) between two species in a gas is straightforward, it is beyond the scope of this text and the equation for collisional frequency of $A$ and $B$ is the following:
$Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{ \dfrac{8\pi k_{B}T}{\mu_{AB}}} \label{freq}$
with
• $N_A$ and $N_B$ are the numbers of $A$ and $B$ molecules in the system, respectively
• $r_a$ and $r_b$ are the radii of molecule $A$ and $B$, respectively
• $k_B$ is the Boltzmann constant $k_B$ =1.380 x 10-23 Joules Kelvin
• $T$ is the temperature in Kelvin
• $\mu_{AB}$ is calculated via $\mu_{AB} = \frac{m_Am_B}{m_A + m_B}$
The specifics of Equation \ref{freq} are not important for this conversation, but it is important to identify that $Z_{AB}$ increases with increasing density (i.e., increasing $N_A$ and $N_B$), with increasing reactant size ($r_a$ and $r_b$), with increasing velocities (predicted via Kinetic Molecular Theory), and with increasing temperature (although weakly because of the square root function).
A Video Discussing Collision Theory of Kinetics: Collusion Theory of Kinetics (opens in new window) [youtu.be]
Microscopic Factor 2: Activation Energy
Previously, we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time.
The collision model of chemical kinetics explains this behavior by introducing the concept of activation energy ($E_a$). We will define this concept using the reaction of $\ce{NO}$ with ozone, which plays an important role in the depletion of ozone in the ozone layer:
$\ce{NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)} \nonumber$
Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate.
Experimental rate law for this reaction is
$\text{rate} = k [\ce{NO}][\ce{O3}] \nonumber$
and is used to identify how the reaction rate (not the rate constant) vares with concentration. The rate constant, however, does vary with temperature. Figure $1$ shows a plot of the rate constant of the reaction of $\ce{NO}$ with $\ce{O3}$ at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (e.g, the Clausius-Claperyon equation). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier.
In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the activation energy, was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the activated complex or the transition state of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily.
Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence.
We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. Figure $2$ shows a plot for the NO–O3 system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction ($ΔE$) is negative, which means that the reaction releases energy. (In this case, $ΔE$ is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction ($E_a$ is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur.
Figure $\PageIndex{3a}$ illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, Figure $\PageIndex{3b}$ illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and $ΔE > 0$. Although the energy changes that result from a reaction can be positive, negative, or even zero, in most cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is almost always positive; there is a class of reactions called barrierless reactions, but those are discussed elsewhere.
For similar reactions under comparable conditions, the one with the smallest Ea will occur most rapidly.
Whereas $ΔE$ is related to the tendency of a reaction to occur spontaneously, $E_a$ gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. For two similar reactions under comparable conditions, the reaction with the smallest $E_a$ will occur more rapidly.
Figure $4$ shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than Ea; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than Ea. Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier.
Video Discussing Transition State Theory: Transition State Theory(opens in new window) [youtu.be]
Microscopic Factor 3: Sterics
Even when the energy of collisions between two reactant species is greater than $E_a$, most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For $\ce{NO}$ and $\ce{O3}$ to produce $\ce{NO2}$ and $\ce{O2}$, a terminal oxygen atom of $\ce{O3}$ must collide with the nitrogen atom of $\ce{NO}$ at an angle that allows $\ce{O3}$ to transfer an oxygen atom to $\ce{NO}$ to produce $\ce{NO2}$ (Figure $4$). All other collisions produce no reaction. Because fewer than 1% of all possible orientations of $\ce{NO}$ and $\ce{O3}$ result in a reaction at kinetic energies greater than $E_a$, most collisions of $\ce{NO}$ and $\ce{O3}$ are unproductive. The fraction of orientations that result in a reaction is called the steric factor ($\rho$) and its value can range from $\rho=0$ (no orientations of molecules result in reaction) to $\rho=1$ (all orientations result in reaction).
Macroscopic Behavior: The Arrhenius Equation
The collision model explains why most collisions between molecules do not result in a chemical reaction. For example, nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 1030 times per second. If every collision produced two molecules of $\ce{NO}$, the atmosphere would have been converted to $\ce{NO}$ and then $\ce{NO2}$ a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide.
For an $A + B$ elementary reaction, all three microscopic factors discussed above that affect the reaction rate can be summarized in a single relationship:
$\text{rate} = (\text{collision frequency}) \times (\text{steric factor}) \times (\text{fraction of collisions with } E > E_a ) \nonumber$
where
$\text{rate} = k[A][B] \label{14.5.2}$
Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, $A$, called the frequency factor:
$k=Ae^{-E_{\Large a}/RT} \label{14.5.3}$
The frequency factor is used to convert concentrations to collisions per second (scaled by the steric factor). Because the frequency of collisions depends on the temperature, $A$ is actually not constant (Equation \ref{freq}). Instead, $A$ increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time.
Equation $\ref{14.5.3}$ is known as the Arrhenius equation and summarizes the collision model of chemical kinetics, where $T$ is the absolute temperature (in K) and R is the ideal gas constant [8.314 J/(K·mol)]. $E_a$ indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large $E_a$ increases rapidly with increasing temperature, whereas the reaction rate with a smaller $E_a$ increases much more slowly with increasing temperature.
If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of Equation $\ref{14.5.3}$,
\begin{align} \ln k &=\ln A+\left(-\dfrac{E_{\textrm a}}{RT}\right) \[4pt] &=\ln A+\left[\left(-\dfrac{E_{\textrm a}}{R}\right)\left(\dfrac{1}{T}\right)\right] \label{14.5.4} \end{align}
Equation $\ref{14.5.4}$ is the equation of a straight line,
$y = mx + b \nonumber$
where $y = \ln k$ and $x = 1/T$. This means that a plot of $\ln k$ versus $1/T$ is a straight line with a slope of $−E_a/R$ and an intercept of $\ln A$. In fact, we need to measure the reaction rate at only two temperatures to estimate $E_a$.
Knowing the $E_a$ at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining $E_a$ from reaction rates measured at several temperatures is illustrated in Example $1$.
A Video Discussing The Arrhenius Equation: The Arrhenius Equation(opens in new window) [youtu.be]
Example $1$: Chirping Tree Crickets
Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping ($f$) as a function of temperature ($T$). Use the data in the following table, along with the graph of ln[chirping rate] versus $1/T$ to calculate $E_a$ for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F).
Chirping Tree Crickets Frequency Table
Frequency (f; chirps/min) ln f T (K) 1/T (K)
200 5.30 299 3.34 × 10−3
179 5.19 298 3.36 × 10−3
158 5.06 296 3.38 × 10−3
141 4.95 294 3.40 × 10−3
126 4.84 293 3.41 × 10−3
112 4.72 292 3.42 × 10−3
100 4.61 290 3.45 × 10−3
89 4.49 289 3.46 × 10−3
79 4.37 287 3.48 × 10−3
Given: chirping rate at various temperatures
Asked for: activation energy and chirping rate at specified temperature
Strategy:
1. From the plot of $\ln f$ versus $1/T$, calculate the slope of the line (−Ea/R) and then solve for the activation energy.
2. Express Equation \ref{14.5.4} in terms of k1 and T1 and then in terms of k2 and T2.
3. Subtract the two equations; rearrange the result to describe k2/k1 in terms of T2 and T1.
4. Using measured data from the table, solve the equation to obtain the ratio k2/k1. Using the value listed in the table for k1, solve for k2.
Solution
A If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of $\ln f$ versus $1/T$ should give a straight line (Figure $6$).
Also, the slope of the plot of $\ln f$ versus $1/T$ should be equal to $−E_a/R$. We can use the two endpoints in Figure $6$ to estimate the slope:
\begin{align*}\textrm{slope}&=\dfrac{\Delta\ln f}{\Delta(1/T)} \[4pt] &=\dfrac{5.30-4.37}{3.34\times10^{-3}\textrm{ K}^{-1}-3.48\times10^{-3}\textrm{ K}^{-1}} \[4pt] &=\dfrac{0.93}{-0.14\times10^{-3}\textrm{ K}^{-1}} \[4pt] &=-6.6\times10^3\textrm{ K}\end{align*} \nonumber
A computer best-fit line through all the points has a slope of −6.67 × 103 K, so our estimate is very close. We now use it to solve for the activation energy:
\begin{align*} E_{\textrm a} &=-(\textrm{slope})(R) \[4pt] &=-(-6.6\times10^3\textrm{ K})\left(\dfrac{8.314 \textrm{ J}}{\mathrm{K\cdot mol}}\right)\left(\dfrac{\textrm{1 KJ}}{\textrm{1000 J}}\right) \[4pt] &=\dfrac{\textrm{55 kJ}}{\textrm{mol}} \end{align*} \nonumber
B If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use Equation \ref{14.5.4} to express the known rate constant ($k_1$) at the first temperature ($T_1$) as follows:
$\ln k_1=\ln A-\dfrac{E_{\textrm a}}{RT_1} \nonumber$
Similarly, we can express the unknown rate constant ($k_2$) at the second temperature ($T_2$) as follows:
$\ln k_2=\ln A-\dfrac{E_{\textrm a}}{RT_2} \nonumber$
C These two equations contain four known quantities (Ea, T1, T2, and k1) and two unknowns (A and k2). We can eliminate A by subtracting the first equation from the second:
\begin{align*} \ln k_2-\ln k_1 &=\left(\ln A-\dfrac{E_{\textrm a}}{RT_2}\right)-\left(\ln A-\dfrac{E_{\textrm a}}{RT_1}\right) \[4pt] &=-\dfrac{E_{\textrm a}}{RT_2}+\dfrac{E_{\textrm a}}{RT_1} \end{align*} \nonumber
Then
$\ln \dfrac{k_2}{k_1}=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \nonumber$
D To obtain the best prediction of chirping rate at 308 K (T2), we try to choose for T1 and k1 the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for T1 = 296 K, where f = 158, and using the $E_a$ calculated previously,
\begin{align*} \ln\dfrac{k_{T_2}}{k_{T_1}} &=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \[4pt] &=\dfrac{55\textrm{ kJ/mol}}{8.314\textrm{ J}/(\mathrm{K\cdot mol})}\left(\dfrac{1000\textrm{ J}}{\textrm{1 kJ}}\right)\left(\dfrac{1}{296 \textrm{ K}}-\dfrac{1}{\textrm{308 K}}\right) \[4pt] &=0.87 \end{align*} \nonumber
Thus k308/k296 = 2.4 and k308 = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute.
Exercise $\PageIndex{1A}$
The equation for the decomposition of $\ce{NO2}$ to $\ce{NO}$ and $\ce{O2}$ is second order in $\ce{NO2}$:
$\ce{2NO2(g) → 2NO(g) + O2(g)} \nonumber$
Data for the reaction rate as a function of temperature are listed in the following table. Calculate $E_a$ for the reaction and the rate constant at 700 K.
Data for the reaction rate as a function of temperature
T (K) k (M−1·s−1)
592 522
603 755
627 1700
652 4020
656 5030
Answer
$E_a$ = 114 kJ/mol; k700= 18,600 M−1·s−1 = 1.86 × 104 M−1·s−1.
Exercise $\PageIndex{1B}$
What $E_a$ results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C?
Answer
about 51 kJ/mol
A Video Discussing Graphing Using the Arrhenius Equation: Graphing Using the Arrhenius Equation (opens in new window) [youtu.be] (opens in new window)
Summary
For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation. The Arrhenius equation is $k=Ae^{-E_{\Large a}/RT}$. A minimum energy (activation energy,v$E_a$) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the Ea, the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: $k=Ae^{-E_{\Large a}/RT}$. A plot of the natural logarithm of k versus 1/T is a straight line with a slope of −Ea/R. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/15%3A_Chemical_Kinetics/15.06%3A_The_Effect_of_Temperature_on_Reaction_Rate.txt |
Learning Objectives
• To determine the individual steps of a simple reaction.
One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.
In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:
$\ce{2C8H18 (l) + 25O2(g) -> 16CO2 (g) + 18H2O(g)} \label{14.6.1}$
For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction, involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.
The overall sequence of elementary reactions is the mechanism of the reaction.
Molecularity and the Rate-Determining Step
To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.
$\ce{NO2(g) + CO(g) -> NO(g) + CO2 (g)} \label{14.6.2}$
From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of $\ce{NO2}$ with a molecule of $\ce{CO}$ that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:
$rate = k[\ce{NO2}]^2 \label{14.6.3}$
The fact that the reaction is second order in $[\ce{NO2}]$ and independent of $[\ce{CO}]$ tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be
$rate = k[\ce{NO2}][\ce{CO}]. \nonumber$
The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:
two-step mechanism
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\textrm{slow}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{elementary reaction}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\rightarrow\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{elementary reaction}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$ $\textrm{overall reaction}$
According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The $\ce{NO3}$ molecule is an intermediate in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.
The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.
Using Molecularity to Describe a Rate Law
The molecularity of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)
Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (Table $1$). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is
$rate = k[A]. \nonumber$
For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure $1$. For a bimolecular elementary reaction of the form A + B → products, the general rate law is
$rate = k[A][B]. \nonumber$
Table $1$: Common Types of Elementary Reactions and Their Rate Laws
Elementary Reaction Molecularity Rate Law Reaction Order
A → products unimolecular rate = k[A] first
2A → products bimolecular rate = k[A]2 second
A + B → products bimolecular rate = k[A][B] second
2A + B → products termolecular rate = k[A]2[B] third
A + B + C → products termolecular rate = k[A][B][C] third
For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law cannot be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step).
Identifying the Rate-Determining Step
Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining step, that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.
Look at the rate laws for each elementary reaction in our example as well as for the overall reaction.
rate laws for each elementary reaction in our example as well as for the overall reaction.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\mathrm{k_1}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{rate}=k_1[\mathrm{NO_2}]^2\textrm{ (predicted)}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{rate}=k_2[\mathrm{NO_3}][\mathrm{CO}]\textrm{ (predicted)}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\xrightarrow{k}\mathrm{NO}+\mathrm{CO_2}$ $\textrm{rate}=k[\mathrm{NO_2}]^2\textrm{ (observed)}$
The experimentally determined rate law for the reaction of $NO_2$ with $CO$ is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so $k$ for the overall reaction must equal $k_1$. That is, NO3 is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.
Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.
Example $1$: A Reaction with an Intermediate
In an alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
alternative mechanism for the reaction of $\ce{NO2}$ with $\ce{CO}$ with $\ce{N2O4}$ appearing as an intermediate.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{k_1}\mathrm{N_2O_4}$
$\textrm{step 2}$ $\underline{\mathrm{N_2O_4}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO}+\mathrm{NO_2}+\mathrm{CO_2}}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$
Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = k[NO2]2)?
Given: elementary reactions
Asked for: rate law for each elementary reaction and overall rate law
Strategy:
1. Determine the rate law for each elementary reaction in the reaction.
2. Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.
Solution
A The rate law for step 1 is rate = k1[NO2]2; for step 2, it is rate = k2[N2O4][CO].
B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k1[NO2]2. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4 as an intermediate, and the one described previously, with NO3 as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3 and N2O4, directly.
Exercise $1$
Iodine monochloride ($\ce{ICl}$) reacts with $\ce{H2}$ as follows:
$\ce{2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s)} \nonumber$
The experimentally determined rate law is $rate = k[\ce{ICl}][\ce{H2}]$. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: $\ce{HI}$ is an intermediate.)
Answer
Solutions to Exercise 14.6.1
$\textrm{step 1}$ $\mathrm{ICl}+\mathrm{H_2}\xrightarrow{k_1}\mathrm{HCl}+\mathrm{HI}$ $\mathrm{rate}=k_1[\mathrm{ICl}][\mathrm{H_2}]\,(\textrm{slow})$
$\textrm{step 2}$ $\underline{\mathrm{HI}+\mathrm{ICl}\xrightarrow{k_2}\mathrm{HCl}+\mathrm{I_2}}$ $\mathrm{rate}=k_2[\mathrm{HI}][\mathrm{ICl}]\,(\textrm{fast})$
$\textrm{sum}$ $\mathrm{2ICl}+\mathrm{H_2}\rightarrow\mathrm{2HCl}+\mathrm{I_2}$
This mechanism is consistent with the experimental rate law if the first step is the rate-determining step.
Example $2$ : Nitrogen Oxide Reacting with Molecular Hydrogen
Assume the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process:
the reaction between $\ce{NO}$ and $\ce{H_2}$ occurs via a three-step process
$\textrm{step 1}$ $\mathrm{NO}+\mathrm{NO}\xrightarrow{k_1}\mathrm{N_2O_2}$ $\textrm{(fast)}$
$\textrm{step 2}$ $\mathrm{N_2O_2}+\mathrm{H_2}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{H_2O}$ $\textrm{(slow)}$
$\textrm{step 3}$ $\mathrm{N_2O}+\mathrm{H_2}\xrightarrow{k_3}\mathrm{N_2}+\mathrm{H_2O}$ $\textrm{(fast)}$
Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction:
$\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \tag{observed}$
Answer
• Step 1: $rate = k_1[\ce{NO}]^2$
• Step 2: $rate = k_2[\ce{N_2O_2}][\ce{H_2}]$
• Step 3: $rate = k_3[\ce{N_2O}][\ce{H_2}]$
The overall reaction is then
$\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber$
• Rate Determining Step : #2
• Yes, because the rate of formation of $[\ce{N_2O_2}] = k_1[\ce{NO}]^2$. Substituting $k_1[\ce{NO}]^2$ for $[\ce{N_2O_2}]$ in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where $k = k_1k_2$.
Reaction Mechanism (Slow step followed by fast step): Reaction Mechanism (Slow step Followed by Fast Step)(opens in new window) [youtu.be] (opens in new window)
Summary
A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity, the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/15%3A_Chemical_Kinetics/15.07%3A_Reaction_Mechanisms.txt |
Learning Objectives
• To understand how catalysts increase the reaction rate and the selectivity of chemical reactions.
Catalysts are substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower Ea, but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst (Figure $1$). Nevertheless, because of its lower Ea, the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes.
A catalyst affects Ea, not ΔE.
Heterogeneous Catalysis
In heterogeneous catalysis, the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency.
An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in Figure $2$, the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H2 is substantially lower on the catalyst surface.
Figure $2$ shows a process called hydrogenation, in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter.
Several important examples of industrial heterogeneous catalytic reactions are in Table $1$. Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface.
Table $1$: Some Commercially Important Reactions that Employ Heterogeneous Catalysts
Commercial Process Catalyst Initial Reaction Final Commercial Product
contact process V2O5 or Pt 2SO2 + O2 → 2SO3 H2SO4
Haber process Fe, K2O, Al2O3 N2 + 3H2 → 2NH3 NH3
Ostwald process Pt and Rh 4NH3 + 5O2 → 4NO + 6H2O HNO3
water–gas shift reaction Fe, Cr2O3, or Cu CO + H2O → CO2 + H2 H2 for NH3, CH3OH, and other fuels
steam reforming Ni CH4 + H2O → CO + 3H2 H2
methanol synthesis ZnO and Cr2O3 CO + 2H2 → CH3OH CH3OH
Sohio process bismuth phosphomolybdate $\mathrm{CH}_2\textrm{=CHCH}_3+\mathrm{NH_3}+\mathrm{\frac{3}{2}O_2}\rightarrow\mathrm{CH_2}\textrm{=CHCN}+\mathrm{3H_2O}$ $\underset{\textrm{acrylonitrile}}{\mathrm{CH_2}\textrm{=CHCN}}$
catalytic hydrogenation Ni, Pd, or Pt RCH=CHR′ + H2 → RCH2—CH2R′ partially hydrogenated oils for margarine, and so forth
Homogeneous Catalysis
In homogeneous catalysis, the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds (Table $2$), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis.
Table $2$: Some Commercially Important Reactions that Employ Homogeneous Catalysts
Commercial Process Catalyst Reactants Final Product
Union Carbide [Rh(CO)2I2] CO + CH3OH CH3CO2H
hydroperoxide process Mo(VI) complexes CH3CH=CH2 + R–O–O–H
hydroformylation Rh/PR3 complexes RCH=CH2 + CO + H2 RCH2CH2CHO
adiponitrile process Ni/PR3complexes 2HCN + CH2=CHCH=CH2 NCCH2CH2CH2CH2CN used to synthesize nylon
olefin polymerization (RC5H5)2ZrCl2 CH2=CH2 –(CH2CH2–)n: high-density polyethylene
Enzymes
Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a substrate.
Because enzymes can increase reaction rates by enormous factors (up to 1017 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37 °C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water (Figure $3$).
Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research.
Summary
Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/15%3A_Chemical_Kinetics/15.08%3A_Catalysis.txt |
• 16.1: Fetal Hemoglobin and Equilibrium
• 16.2: The Concept of Dynamic Equilibrium
At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached.
• 16.3: The Equilibrium Constant (K)
The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants. For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.
• 16.4: Expressing the Equilibrium Constant in Terms of Pressure
An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium.
• 16.5: Heterogenous Equilibria - Reactions Involving Solids and Liquids
When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid.
• 16.6: The Reaction Quotient- Predicting the Direction of Change
The reaction Quotient has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, Q=K . Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve are nonequilibrium states.
• 16.7: Finding Equilibrium Concentrations
Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture. When an equilibrium constant is calculated from equilibrium concentrations, concentrations or partial pressures are use into the equilibrium constant expression.
• 16.8: Le Châtelier’s Principle- How a System at Equilibrium Responds to Disturbances
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Châtelier's principle: The system will respond in a way that counteracts the disturbance. Adding a catalyst affects the reaction rates but does not alter equilibrium.
16: Chemical Equilibrium
Learning Objectives
• To understand what is meant by chemical equilibrium.
In the last chapter, we discussed the principles of chemical kinetics, which deal with the rate of change, or how quickly a given chemical reaction occurs. We now turn our attention to the extent to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are reversible to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches chemical equilibrium, the point at which the composition of the system no longer changes with time.
Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide ($\ce{N_2O_4}$) to nitrogen dioxide ($\ce{NO_2}$). You may recall that $\ce{NO_2}$ is responsible for the brown color we associate with smog. When a sealed tube containing solid $\ce{N_2O_4}$ (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of $\ce{NO_2}$ appears (Figure $1$). The reaction can be followed visually because the product ($\ce{NO_2}$) is colored, whereas the reactant ($\ce{N_2O_4}$) is colorless:
$\underset{colorless }{\ce{N2O4 (g)}} \ce{ <=>[k_f][k_r] } \underset{red-brown }{\ce{2NO2(g)}}\label{Eq1}$
The double arrow indicates that both the forward reaction
$\ce{N2O4 (g) ->[k_f] 2NO2(g)} \label{eq1B}$
and reverse reaction
$\ce{2NO2(g) ->[k_r] N2O4 (g) } \label{eq1C}$
occurring simultaneously (i.e, the reaction is reversible). However, this does not necessarily mean the system is equilibrium as the following chapter demonstrates.
Figure $2$ shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of $\ce{NO_2}$ were zero, then it increases as the concentration of $\ce{N_2O_4}$ decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no $\ce{N_2O_4}$ but an initial $\ce{NO_2}$ concentration twice the initial concentration of $\ce{N_2O_4}$ (Figure $\PageIndex{2a}$), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure $\PageIndex{2b}$). Thus equilibrium can be approached from either direction in a chemical reaction.
Figure $3$ shows the forward and reverse reaction rates for a sample that initially contains pure $\ce{NO_2}$. Because the initial concentration of $\ce{N_2O_4}$ is zero, the forward reaction rate (dissociation of $\ce{N_2O_4}$) is initially zero as well. In contrast, the reverse reaction rate (dimerization of $\ce{NO_2}$) is initially very high ($2.0 \times 10^6\, M/s$), but it decreases rapidly as the concentration of $\ce{NO_2}$ decreases. As the concentration of $\ce{N_2O_4}$ increases, the rate of dissociation of $\ce{N_2O_4}$ increases—but more slowly than the dimerization of $\ce{NO_2}$—because the reaction is only first order in $\ce{N_2O_4}$ (rate = $k_f[N_2O_4]$, where $k_f$ is the rate constant for the forward reaction in Equations $\ref{Eq1}$ and $\ref{eq1B}$). Eventually, the forward and reverse reaction rates become identical, $k_f = k_r$, and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium.
The rate of dimerization of $\ce{NO_2}$ (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of $\ce{N_2O_4}$ is zero, the rate of the dissociation reaction (forward reaction) at $t = 0$ is also zero. As the dimerization reaction proceeds, the $\ce{N_2O_4}$ concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of $\ce{N_2O_4}$ and $\ce{NO_2}$ no longer change.
At equilibrium, the forward reaction rate is equal to the reverse reaction rate.
Example $1$
The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation:
$2A \rightleftharpoons B \nonumber$
where the blue circles are $A$ and the purple ovals are $B$. Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium?
Given: three reaction systems
Asked for: relative time to reach chemical equilibrium
Strategy:
Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium.
Solution:
In systems 1 and 3, the concentration of A decreases from $t_0$ through $t_2$ but is the same at both $t_2$ and $t_3$. Thus systems 1 and 3 are at equilibrium by $t_3$. In system 2, the concentrations of A and B are still changing between $t_2$ and $t_3$, so system 2 may not yet have reached equilibrium by $t_3$. Thus system 2 took the longest to reach chemical equilibrium.
Exercise $1$
In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium?
Answer
system 2
A Video Introduction to Dynamic Equilibrium: Introduction to Dynamic Equilibrium(opens in new window) [youtu.be]
Summary
At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/16%3A_Chemical_Equilibrium/16.02%3A_The_Concept_of_Dynamic_Equilibrium.txt |
Learning Objectives
• To know the relationship between the equilibrium constant and the rate constants for the forward and reverse reactions.
• To write an equilibrium constant expression for any reaction.
Because an equilibrium state is achieved when the forward reaction rate equals the reverse reaction rate, under a given set of conditions there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction (represented by rate constants). We can show this relationship using the decomposition reaction of $\ce{N_2O_4}$ to $\ce{NO2}$. Both the forward and reverse reactions for this system consist of a single elementary reaction, so the reaction rates are as follows:
$\text{forward rate} = k_f[\ce{N2O4}] \nonumber$
and
$\text{reverse rate} = k_r[\ce{NO2}]^2\nonumber$
At equilibrium, the forward rate equals the reverse rate (definition of equilibrium):
$k_f[\ce{N2O4}] = k_r[\ce{NO2}]^2 \label{Eq3}$
so
$\dfrac{k_f}{k_r}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{Eq4}$
The ratio of the rate constants gives us a new constant, the equilibrium constant ($K$), which is defined as follows:
$K=\dfrac{k_f}{k_r} \label{Eq5}$
Hence there is a fundamental relationship between chemical kinetics and chemical equilibrium: under a given set of conditions, the composition of the equilibrium mixture is determined by the magnitudes of the rate constants for the forward and the reverse reactions.
The equilibrium constant is equal to the rate constant for the forward reaction divided by the rate constant for the reverse reaction.
A Video for Determining the Equilibrium Expression: Determining the Equilibrium Expression(opens in new window) [youtu.be]
Table $1$ lists the initial and equilibrium concentrations from five different experiments using the reaction system described by Equation $\ref{Eq3}$. At equilibrium the magnitude of the quantity $[\ce{NO2}]^2/[\ce{N2O4}]$ is essentially the same for all five experiments. In fact, no matter what the initial concentrations of $\ce{NO2}$ and $\ce{N2O4}$ are, at equilibrium the quantity $[\ce{NO2}]^2/[\ce{N2O4}]$ will always be $6.53 \pm 0.03 \times 10^{-3}$ at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions. That is, at a given temperature, the equilibrium constant for a reaction always has the same value, even though the specific concentrations of the reactants and products vary depending on their initial concentrations.
Table $1$: Initial and Equilibrium Concentrations for $\ce{NO2}:\ce{N2O4}$ Mixtures at 25°C
Initial Concentrations Concentrations at Equilibrium
Experiment [$\ce{N2O4}$] (M) [$\ce{NO2}$] (M) [$\ce{N2O4}$] (M) [$\ce{NO2}$] (M) $K = [\ce{NO2}]^2/[\ce{N2O4}]$
1 0.0500 0.0000 0.0417 0.0165 $6.54 \times 10^{-3}$
2 0.0000 0.1000 0.0417 0.0165 $6.54 \times 10^{-3}$
3 0.0750 0.0000 0.0647 0.0206 $6.56 \times 10^{-3}$
4 0.0000 0.0750 0.0304 0.0141 $6.54 \times 10^{-3}$
5 0.0250 0.0750 0.0532 0.0186 $6.50 \times 10^{-3}$
Developing an Equilibrium Constant Expression
In 1864, the Norwegian chemists Cato Guldberg (1836–1902) and Peter Waage (1833–1900) carefully measured the compositions of many reaction systems at equilibrium. They discovered that for any reversible reaction of the general form
$aA+bB \rightleftharpoons cC+dD \label{Eq6}$
where A and B are reactants, C and D are products, and a, b, c, and d are the stoichiometric coefficients in the balanced chemical equation for the reaction, the ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced chemical equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced chemical equation) is always a constant under a given set of conditions. This relationship is known as the law of mass action (or law of chemical equilibrium) and can be stated as follows:
$K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Eq7}$
where $K$ is the equilibrium constant for the reaction. Equation $\ref{Eq6}$ is called the equilibrium equation, and the right side of Equation $\ref{Eq7}$ is called the equilibrium constant expression. The relationship shown in Equation $\ref{Eq7}$ is true for any pair of opposing reactions regardless of the mechanism of the reaction or the number of steps in the mechanism.
The equilibrium constant can vary over a wide range of values. The values of $K$ shown in Table $2$, for example, vary by 60 orders of magnitude. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than $10^3$ indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between $H_2$ and $Cl_2$ to produce $HCl$, which has an equilibrium constant of $1.6 \times 10^{33}$ at 300 K. Because $H_2$ is a good reductant and $Cl_2$ is a good oxidant, the reaction proceeds essentially to completion. In contrast, values of $K$ less than $10^{-3}$ indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants.
Table $2$: Equilibrium Constants for Selected Reactions*
Reaction Temperature (K) Equilibrium Constant (K)
*Equilibrium constants vary with temperature. The K values shown are for systems at the indicated temperatures.
$S_{(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}$ 300 $4.4 \times 10^{53}$
$2H_{2(g)}+O_{2(g)} \rightleftharpoons 2H2O_{(g)}$ 500 $2.4 \times 10^{47}$
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}$ 300 $1.6 \times 10^{33}$
$H_{2(g)}+Br_{2(g)} \rightleftharpoons 2HBr_{(g)}$ 300 $4.1 \times 10^{18}$
$2NO_{(g)}+O_{2(g)} \rightleftharpoons 2NO_{2(g)}$ 300 $4.2 \times 10^{13}$
$3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)}$ 300 $2.7 \times 10^{8}$
$H_{2(g)}+D_{2(g)} \rightleftharpoons 2HD_{(g)}$ 100 $1.92$
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$ 300 $2.9 \times 10^{-1}$
$I_{2(g)} \rightleftharpoons 2I_{(g)}$ 800 $4.6 \times 10^{-7}$
$Br_{2(g)} \rightleftharpoons 2Br_{(g)}$ 1000 $4.0 \times 10^{-7}$
$Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}$ 1000 $1.8 \times 10^{-9}$
$F_{2(g)} \rightleftharpoons 2F_{(g)}$ 500 $7.4 \times 10^{-13}$
Effective vs. True Concentrations
You will also notice in Table $2$ that equilibrium constants have no units, even though Equation $\ref{Eq7}$ suggests that the units of concentration might not always cancel because the exponents may vary. In fact, equilibrium constants are calculated using “effective concentrations,” or activities, of reactants and products, which are the ratios of the measured concentrations to a standard state of 1 M. As shown in Equation $\ref{Eq8}$, the units of concentration cancel, which makes $K$ unitless as well:
$\dfrac{[A]_{measured}}{[A]_{standard\; state}}=\dfrac{\cancel{M}}{\cancel{M}} = \dfrac{\cancel{\frac{mol}{L}}}{\cancel{\frac{mol}{L}}} \label{Eq8}$
Because equilibrium constants are calculated using “effective concentrations” relative to a standard state of 1 M, values of K are unitless.
Many reactions have equilibrium constants between 1000 and 0.001 ($10^3 \ge K \ge 10^{-3}$), neither very large nor very small. At equilibrium, these systems tend to contain significant amounts of both products and reactants, indicating that there is not a strong tendency to form either products from reactants or reactants from products. An example of this type of system is the reaction of gaseous hydrogen and deuterium, a component of high-stability fiber-optic light sources used in ocean studies, to form $\ce{HD}$:
$\ce{H2(g) + D2(g) <=> 2HD(g)} \nonumber$
The equilibrium constant expression for this reaction is
$K= \dfrac{[HD]^2}{[H_2][D_2]} \nonumber$
with $K$ varying between 1.9 and 4 over a wide temperature range (100–1000 K). Thus an equilibrium mixture of $H_2$, $D_2$, and $HD$ contains significant concentrations of both product and reactants.
Figure $3$ summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants $\rightleftharpoons$ products. Because there is a direct relationship between the kinetics of a reaction and the equilibrium concentrations of products and reactants (Equations $\ref{Eq8}$ and $\ref{Eq7}$), when $k_f \gg k_r$, $K$ is a large number, and the concentration of products at equilibrium predominate. This corresponds to an essentially irreversible reaction. Conversely, when $k_f \ll k_r$, $K$ is a very small number, and the reaction produces almost no products as written. Systems for which $k_f ≈ k_r$ have significant concentrations of both reactants and products at equilibrium.
A large value of the equilibrium constant $K$ means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium.
Example $1$: Equilibrium Constant Expressions
Write the equilibrium constant expression for each reaction.
• $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$
• $\ce{CO(g) + 1/2 O2(g) <=> CO2(g)}$
• $\ce{2CO2(g) <=> 2CO(g) + O2(g)}$
Given: balanced chemical equations
Asked for: equilibrium constant expressions
Strategy:
Refer to Equation $\ref{Eq7}$. Place the arithmetic product of the concentrations of the products (raised to their stoichiometric coefficients) in the numerator and the product of the concentrations of the reactants (raised to their stoichiometric coefficients) in the denominator.
Solution:
The only product is ammonia, which has a coefficient of 2. For the reactants, $\ce{N2}$ has a coefficient of 1 and $\ce{H2}$ has a coefficient of 3. The equilibrium constant expression is as follows:
$\dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3} \nonumber$
The only product is carbon dioxide, which has a coefficient of 1. The reactants are $\ce{CO}$, with a coefficient of 1, and $\ce{O2}$, with a coefficient of $\frac{1}{2}$. Thus the equilibrium constant expression is as follows:
$\dfrac{[\ce{CO2}]}{[\ce{CO}][\ce{O2}]^{1/2}} \nonumber$
This reaction is the reverse of the reaction in part b, with all coefficients multiplied by 2 to remove the fractional coefficient for $\ce{O2}$. The equilibrium constant expression is therefore the inverse of the expression in part b, with all exponents multiplied by 2
$\dfrac{[\ce{CO}]^2[\ce{O2}]}{[\ce{CO2}]^2} \nonumber$
Exercise $1$
Write the equilibrium constant expression for each reaction.
1. $\ce{N2O(g) <=> N2(g) + 1/2O2(g)}$
2. $\ce{2C8H18(g) + 25O2(g) <=> 16CO2(g) + 18H2O(g)}$
3. $\ce{H2(g) + I2(g) <=> 2HI(g)}$
Answer a
$K=\dfrac{[N_2][O_2]^{1/2}}{[N_2O]}$
Answer b
$K=\dfrac{[CO_2]^{16}[H_2O]^{18}}{[C_8H_{18}]^2[O_2]^{25}}$
Answer c
$K=\dfrac{[HI]^2}{[H_2][I_2]}$
Example $2$
Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants.
1. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\;\;\; K_{(700K)}=54$
2. $2CO_{2(g)} \rightleftharpoons 2CO_{(g)}+O_{2(g)}\;\;\; K_{(1200K)}=3.1 \times 10^{-18}$
3. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}\;\;\; K_{(613K)}=97$
4. $2O_{3(g)} \rightleftharpoons 3O_{2(g)} \;\;\; K_{(298 K)}=5.9 \times 10^{55}$
Given: systems and values of $K$
Asked for: composition of systems at equilibrium
Strategy:
Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both.
Solution:
1. Only system 4 has $K \gg 10^3$, so at equilibrium it will consist of essentially only products.
2. System 2 has $K \ll 10^{-3}$, so the reactants have little tendency to form products under the conditions specified; thus, at equilibrium the system will contain essentially only reactants.
3. Both systems 1 and 3 have equilibrium constants in the range $10^3 \ge K \ge 10^{-3}$, indicating that the equilibrium mixtures will contain appreciable amounts of both products and reactants.
Exercise $2$
Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation:
$\ce{3H2(g) + N2(g) <=> 2NH3(g)} \nonumber$
Values of the equilibrium constant at various temperatures were reported as
• $K_{25°C} = 3.3 \times 10^8$,
• $K_{177°C} = 2.6 \times 10^3$, and
• $K_{327°C} = 4.1$.
1. At which temperature would you expect to find the highest proportion of $H_2$ and $N_2$ in the equilibrium mixture?
2. Assuming that the reaction rates are fast enough so that equilibrium is reached quickly, at what temperature would you design a commercial reactor to operate to maximize the yield of ammonia?
Answer a
327°C, where $K$ is smallest
Answer b
25°C
Video which Discusses What Does K Tell us About a Reaction?: What Does K Tell us About a Reaction?(opens in new window) [youtu.be]
Variations in the Form of the Equilibrium Constant Expression
Because equilibrium can be approached from either direction in a chemical reaction, the equilibrium constant expression and thus the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. For example, if we write the reaction described in Equation $\ref{Eq6}$ in reverse, we obtain the following:
$cC+dD \rightleftharpoons aA+bB \label{Eq10}$
The corresponding equilibrium constant $K′$ is as follows:
$K'=\dfrac{[A]^a[B]^b}{[C]^c[D]^d} \label{Eq11}$
This expression is the inverse of the expression for the original equilibrium constant, so $K′ = 1/K$. That is, when we write a reaction in the reverse direction, the equilibrium constant expression is inverted. For instance, the equilibrium constant for the reaction $\ce{N2O4 <=> 2NO2}$ is as follows:
$K=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{Eq12}$
but for the opposite reaction, $2 NO_2 \rightleftharpoons N_2O_4$, the equilibrium constant K′ is given by the inverse expression:
$K'=\dfrac{[\ce{N2O4}]}{[\ce{NO2}]^2} \label{Eq13}$
Consider another example, the formation of water:
$\ce{2H2(g) + O2(g) <=> 2H2O(g)}. \nonumber$
Because $\ce{H2}$ is a good reductant and $\ce{O2}$ is a good oxidant, this reaction has a very large equilibrium constant ($K = 2.4 \times 10^{47}$ at 500 K). Consequently, the equilibrium constant for the reverse reaction, the decomposition of water to form $\ce{O2}$ and $\ce{H2}$, is very small: $K′ = 1/K = 1/(2.4 \times 10^{47}) = 4.2 \times 10^{-48}$. As suggested by the very small equilibrium constant, and fortunately for life as we know it, a substantial amount of energy is indeed needed to dissociate water into $\ce{H2}$ and $\ce{O2}$.
The equilibrium constant for a reaction written in reverse is the inverse of the equilibrium constant for the reaction as written originally.
Writing an equation in different but chemically equivalent forms also causes both the equilibrium constant expression and the magnitude of the equilibrium constant to be different. For example, we could write the equation for the reaction
$\ce{2NO2 <=> N2O4} \nonumber$
as
$\ce{NO2 <=> 1/2 N2O4} \nonumber$
with the equilibrium constant K″ is as follows:
$K′′=\dfrac{[\ce{N2O4}]^{1/2}}{[\ce{NO2}]} \label{Eq14}$
The values for K′ (Equation $\ref{Eq13}$) and K″ are related as follows:
$K′′=(K')^{1/2}=\sqrt{K'} \label{Eq15}$
In general, if all the coefficients in a balanced chemical equation were subsequently multiplied by $n$, then the new equilibrium constant is the original equilibrium constant raised to the $n^{th}$ power.
A Video Discussing Relationships Involving Equilibrium Constants: Relationships Involving Equilibrium Constants(opens in new window) [youtu.be] (opens in new window)
Example $3$: The Haber Process
At 745 K, K is 0.118 for the following reaction:
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
What is the equilibrium constant for each related reaction at 745 K?
1. $\ce{2NH3(g) <=> N2(g) + 3H2(g)}$
2. $\ce{1/2 N2(g) + 3/2 H2(g) <=> NH3(g)}$
Given: balanced equilibrium equation, $K$ at a given temperature, and equations of related reactions
Asked for: values of $K$ for related reactions
Strategy:
Write the equilibrium constant expression for the given reaction and for each related reaction. From these expressions, calculate $K$ for each reaction.
Solution:
The equilibrium constant expression for the given reaction of $N_{2(g)}$ with $H_{2(g)}$ to produce $NH_{3(g)}$ at 745 K is as follows:
$K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}=0.118 \nonumber$
This reaction is the reverse of the one given, so its equilibrium constant expression is as follows:
$K'=\dfrac{1}{K}=\dfrac{[N_2][H_2]^3}{[NH_3]^2}=\dfrac{1}{0.118}=8.47 \nonumber$
In this reaction, the stoichiometric coefficients of the given reaction are divided by 2, so the equilibrium constant is calculated as follows:
$K′′=\dfrac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}=K^{1/2}=\sqrt{K}=\sqrt{0.118} = 0.344 \nonumber$
Exercise
At 527°C, the equilibrium constant for the reaction
$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \nonumber$
is $7.9 \times 10^4$. Calculate the equilibrium constant for the following reaction at the same temperature:
$\ce{SO3(g) <=> SO2(g) + 1/2 O2(g)} \nonumber$
Answer
$3.6 \times 10^{-3}$
Equilibrium Constant Expressions for Systems that Contain Gases
For reactions that involve species in solution, the concentrations used in equilibrium calculations are usually expressed in moles/liter. For gases, however, the concentrations are usually expressed in terms of partial pressures rather than molarity, where the standard state is 1 atm of pressure. The symbol $K_p$ is used to denote equilibrium constants calculated from partial pressures. For the general reaction $aA+bB \rightleftharpoons cC+dD$, in which all the components are gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products and reactants (each raised to its coefficient in the chemical equation):
$K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \label{Eq16}$
Thus $K_p$ for the decomposition of $N_2O_4$ (Equation 15.1) is as follows:
$K_p=\dfrac{(P_{NO_2})^2}{P_{N_2O_4}} \label{Eq17}$
Like $K$, $K_p$ is a unitless quantity because the quantity that is actually used to calculate it is an “effective pressure,” the ratio of the measured pressure to a standard state of 1 bar (approximately 1 atm), which produces a unitless quantity. The “effective pressure” is called the fugacity, just as activity is the effective concentration.
Because partial pressures are usually expressed in atmospheres or mmHg, the molar concentration of a gas and its partial pressure do not have the same numerical value. Consequently, the numerical values of $K$ and $K_p$ are usually different. They are, however, related by the ideal gas constant ($R$) and the absolute temperature ($T$):
$\color{red} K_p = K(RT)^{Δn} \label{Eq18}$
where $K$ is the equilibrium constant expressed in units of concentration and $Δn$ is the difference between the numbers of moles of gaseous products and gaseous reactants ($n_p − n_r$). The temperature is expressed as the absolute temperature in Kelvin. According to Equation $\ref{Eq18}$, $K_p = K$ only if the moles of gaseous products and gaseous reactants are the same (i.e., $Δn = 0$). For the decomposition of $N_2O_4$, there are 2 mol of gaseous product and 1 mol of gaseous reactant, so $Δn = 1$. Thus, for this reaction,
$K_p = K(RT)^1 = KRT \nonumber$
Example $4$: The Haber Process (again)
The equilibrium constant for the reaction of nitrogen and hydrogen to give ammonia is 0.118 at 745 K. The balanced equilibrium equation is as follows:
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
What is $K_p$ for this reaction at the same temperature?
Given: equilibrium equation, equilibrium constant, and temperature
Asked for: $K_p$
Strategy:
Use the coefficients in the balanced chemical equation to calculate $Δn$. Then use Equation $\ref{Eq18}$ to calculate $K$ from $K_p$.
Solution:
This reaction has 2 mol of gaseous product and 4 mol of gaseous reactants, so $\Delta{n} = (2 − 4) = −2$. We know $K$, and $T = 745\; K$. Thus, from Equation $\ref{Eq15}$, we have the following:
\begin{align*} K_p &=K(RT)^{-2} \[4pt] &=\dfrac{K}{(RT)^2} \[4pt] &=\dfrac{0.118}{\{ [0.08206(L \cdot atm)/(mol \cdot K)][745\; K]\}^2} \[4pt] &=3.16 \times 10^{-5} \end{align*} \nonumber
Because $K_p$ is a unitless quantity, the answer is $K_p = 3.16 \times 10^{-5}$.
Exercise $4$
Calculate $K_p$ for the reaction
$\ce{2SO2(g) + O2(g) \rightleftharpoons 2SO3(g)} \nonumber$
at 527°C, if $K = 7.9 \times 10^4$ at this temperature.
Answer
$K_p = 1.2 \times 10^3$
Video Discussing Converting Kc to Kp: Converting Kc to Kp(opens in new window) [youtu.be]
Equilibrium Constant Expressions for the Sums of Reactions
Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known. The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions.
To illustrate this procedure, let’s consider the reaction of $\ce{N2}$ with $\ce{O2}$ to give $\ce{NO2}$. This reaction is an important source of the $\ce{NO2}$ that gives urban smog its typical brown color. The reaction normally occurs in two distinct steps. In the first reaction (step 1), $\ce{N2}$ reacts with $\ce{O2}$ at the high temperatures inside an internal combustion engine to give $\ce{NO}$. The released $\ce{NO}$ then reacts with additional $\ce{O2}$ to give $\ce{NO2}$ (step 2). The equilibrium constant for each reaction at 100°C is also given.
$\ce{N2(g) + O2(g) <=> 2NO(g)}\;\; K_1=2.0 \times 10^{-25} \label{step 1}$
$\ce{2NO(g) + O2(g) <=> 2NO2(g)}\;\;\;K_2=6.4 \times 10^9 \label{step 2}$
Summing reactions (step 1) and (step 2) gives the overall reaction of $N_2$ with $O_2$:
$\ce{N2(g) + 2O2(g) <=> 2NO2(g)} \;\;\;K_3=? \label{overall reaction 3}$
The equilibrium constant expressions for the reactions are as follows:
$K_1=\dfrac{[NO]^2}{[N_2][O_2]}\;\;\; K_2=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\;\;\; K_3=\dfrac{[NO_2]^2}{[N_2][O_2]^2} \nonumber$
What is the relationship between $K_1$, $K_2$, and $K_3$, all at 100°C? The expression for $K_1$ has $[NO]^2$ in the numerator, the expression for $K_2$ has $[NO]^2$ in the denominator, and $[NO]^2$ does not appear in the expression for $K_3$. Multiplying $K_1$ by $K_2$ and canceling the $[NO]^2$ terms,
$K_1K_2=\dfrac{\cancel{[NO]^2}}{[N_2][O_2]} \times \dfrac{[NO_2]^2}{\cancel{[NO]^2}[O_2]}=\dfrac{[NO_2]^2}{[N_2][O_2]^2}=K_3 \nonumber$
Thus the product of the equilibrium constant expressions for $K_1$ and $K_2$ is the same as the equilibrium constant expression for $K_3$:
$K_3 = K_1K_2 = (2.0 \times 10^{-25})(6.4 \times 10^9) = 1.3 \times 10^{-15} \nonumber$
The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions. In contrast, recall that according to Hess’s Law, $ΔH$ for the sum of two or more reactions is the sum of the ΔH values for the individual reactions.
To determine $K$ for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.
Example $6$
The following reactions occur at 1200°C:
1. $CO_{(g)}+3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)} \;\;\;K_1=9.17 \times 10^{-2}$
2. $CH_{4(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+4H_{2(g})\;\;\; K_2=3.3 \times 10^4$
Calculate the equilibrium constant for the following reaction at the same temperature.
1. $CO_{(g)}+2H_2S_{(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)}\;\;\; K_3=?$
Given: two balanced equilibrium equations, values of $K$, and an equilibrium equation for the overall reaction
Asked for: equilibrium constant for the overall reaction
Strategy:
Arrange the equations so that their sum produces the overall equation. If an equation had to be reversed, invert the value of $K$ for that equation. Calculate $K$ for the overall equation by multiplying the equilibrium constants for the individual equations.
Solution:
The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2:
$CO_{(g)}+ \cancel{3H_{2(g)}} \rightleftharpoons \cancel{CH_{4(g)}} + H_2O_{(g)} \nonumber$
$\cancel{CH_{4(g)}} +2H_2S_{(g)} \rightleftharpoons CS_{2(g)} + \cancel{3H_{2(g)}} + H_{2(g)} \nonumber$
$CO_{(g)} + 3H_{2(g)} \rightleftharpoons CS_{2(g)}+H_2O_{(g)}+H_{2(g)} \nonumber$
The values for $K_1$ and $K_2$ are given, so it is straightforward to calculate $K_3$:
$K_3 = K_1K_2 = (9.17 \times 10^{-2})(3.3 \times 10^4) = 3.03 \times 10^3 \nonumber$
Exercise $6$
In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide. In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.
1. $\frac{1}{8}S_{8(s)}+O_{2(g)} \rightleftharpoons SO_{2(g)}\;\;\; K_1=4.4 \times 10^{53}$
2. $SO_{2(g)}+\frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_2=2.6 \times 10^{12}$
3. $\frac{1}{8}S_{8(s)}+\frac{3}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}\;\;\; K_3=?$
Answer
$K_3 = 1.1 \times 10^{66}$
Summary
The ratio of the rate constants for the forward and reverse reactions at equilibrium is the equilibrium constant ($K$), a unitless quantity. The composition of the equilibrium mixture is therefore determined by the magnitudes of the forward and reverse rate constants at equilibrium. Under a given set of conditions, a reaction will always have the same $K$. For a system at equilibrium, the law of mass action relates $K$ to the ratio of the equilibrium concentrations of the products to the concentrations of the reactants raised to their respective powers to match the coefficients in the equilibrium equation. The ratio is called the equilibrium constant expression. When a reaction is written in the reverse direction, $K$ and the equilibrium constant expression are inverted. For gases, the equilibrium constant expression can be written as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to a power matching its coefficient in the chemical equation. An equilibrium constant calculated from partial pressures ($K_p$) is related to $K$ by the ideal gas constant ($R$), the temperature ($T$), and the change in the number of moles of gas during the reaction. An equilibrium system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium. When a reaction can be expressed as the sum of two or more reactions, its equilibrium constant is equal to the product of the equilibrium constants for the individual reactions.
• The law of mass action describes a system at equilibrium in terms of the concentrations of the products and the reactants.
• For a system involving one or more gases, either the molar concentrations of the gases or their partial pressures can be used.
• Definition of equilibrium constant in terms of forward and reverse rate constants: $K=\dfrac{k_f}{k_r} \nonumber$
• Equilibrium constant expression (law of mass action): $K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber$
• Equilibrium constant expression for reactions involving gases using partial pressures: $K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \nonumber$
• Relationship between $K_p$ and $K$: $K_p = K(RT)^{Δn} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/16%3A_Chemical_Equilibrium/16.03%3A_The_Equilibrium_Constant_%28K%29.txt |
Learning Objectives
• To understand how different phases affect equilibria.
When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid.
As noted in the previous section, the equilibrium constant expression is actually a ratio of activities. To simplify the calculations in general chemistry courses, the activity of each substance in the reaction is often approximated using a ratio of the molarity of a substance compared to the standard state of that substance. For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, the ratio of the molarity to the standard state for substances that are liquids or solids always has a value of 1. For example, for a compound such as CaF2(s), the term going into the equilibrium expression is [CaF2]/[CaF2] which cancels to unity. Thus, when the activities of the solids and liquids (including solvents) are incorporated into the equilibrium expression, they do not change the value.
Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:
$\ce{CO2(g) + C(s) \rightleftharpoons 2CO(g)} \label{Eq14.4.1}$
The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:
$K=\dfrac{a_{\ce{CO}}^2}{a_{\ce{CO2}}a_{C}}=\dfrac{[\ce{CO}]^2}{[\ce{CO2}][1]}=\dfrac{[\ce{CO}]^2}{[\ce{CO_2}]}\label{Eq14.4.2}$
The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:
$K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.3}$
Incorporating all the constant values into $K′$ or $K_p$ allows us to focus on the substances whose concentrations change during the reaction.
Although the activities of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of $\ce{CO}$ and $\ce{CO_2}$, the system described in Equation $\ref{Eq14.4.1}$ will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure $1$, it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium.
Example $1$
Write each expression for $K$, incorporating all constants, and $K_p$ for the following equilibrium reactions.
1. $\ce{PCl3(l) + Cl2(g) <=> PCl5(s)}$
2. $\ce{Fe3O4(s) + 4H2(g) <=> 3Fe(s) + 4H2O(g)}$
Given: balanced equilibrium equations.
Asked for: expressions for $K$ and $K_p$.
Strategy:
Find $K$ by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express $K_p$ as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation.
Solution
This reaction contains a pure solid ($PCl_5$) and a pure liquid ($PCl_3$). Their activities are equal to 1, so when incorporated into the equilibrium constant expression, they do not change the value. So
$K=\dfrac{1}{(1)[Cl_2]} \nonumber$
and
$K_p=\dfrac{1}{(1)P_{Cl_2}} \nonumber$
This reaction contains two pure solids ($Fe_3O_4$ and $Fe$), which are each assigned a value of 1 in the equilibrium constant expressions:
$K=\dfrac{(1)[H_2O]^4}{(1)[H_2]^4} \nonumber$
and
$K_p=\dfrac{(1)(P_{H_2O})^4}{(1)(P_{H_2})^4} \nonumber$
Exercise $1$
Write the expressions for $K$ and $K_p$ for the following reactions.
1. $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$
2. $\underset{glucose}{\ce{C6H12O6(s)}} + \ce{6O2(g) <=> 6CO2(g) + 6H2O(g)}$
Answer a
$K = [\ce{CO_2}]$ and $K_p = P_{\ce{CO_2}}$
Answer b
$K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$
For reactions carried out in solution, the solvent is assumed to be pure, and therefore is assigned an activity equal to 1 in the equilibrium constant expression. The activities of the solutes are approximated by their molarities. The result is that the equilibrium constant expressions appear to only depend upon the concentrations of the solutes.
The activities of pure solids, pure liquids, and solvents are defined as having a value of '1'. Often, it is said that these activities are "left out" of equilibrium constant expressions. This is an unfortunate use of words. The activities are not "left out" of equilibrium constant expressions. Rather, because they have a value of '1', they do not change the value of the equilibrium constant when they are multiplied together with the other terms. The activities of the solutes are approximated by their molarities.
Summary
An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium.
Contributors and Attributions
• Anonymous
• Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/16%3A_Chemical_Equilibrium/16.04%3A_Expressing_the_Equilibrium_Constant_in_Terms_of_Pressure.txt |
Learning Objectives
• To understand how different phases affect equilibria.
When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid, the system is a homogeneous equilibrium. In such situations, the concentrations of the reactants and products can vary over a wide range. In contrast, a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium, such as the reaction of a gas with a solid or liquid.
As noted in the previous section, the equilibrium constant expression is actually a ratio of activities. To simplify the calculations in general chemistry courses, the activity of each substance in the reaction is often approximated using a ratio of the molarity of a substance compared to the standard state of that substance. For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid. Because the molar concentrations of pure liquids and solids normally do not vary greatly with temperature, the ratio of the molarity to the standard state for substances that are liquids or solids always has a value of 1. For example, for a compound such as CaF2(s), the term going into the equilibrium expression is [CaF2]/[CaF2] which cancels to unity. Thus, when the activities of the solids and liquids (including solvents) are incorporated into the equilibrium expression, they do not change the value.
Consider the following reaction, which is used in the final firing of some types of pottery to produce brilliant metallic glazes:
$\ce{CO2(g) + C(s) \rightleftharpoons 2CO(g)} \label{Eq14.4.1}$
The glaze is created when metal oxides are reduced to metals by the product, carbon monoxide. The equilibrium constant expression for this reaction is as follows:
$K=\dfrac{a_{\ce{CO}}^2}{a_{\ce{CO2}}a_{C}}=\dfrac{[\ce{CO}]^2}{[\ce{CO2}][1]}=\dfrac{[\ce{CO}]^2}{[\ce{CO_2}]}\label{Eq14.4.2}$
The equilibrium constant for this reaction can also be written in terms of the partial pressures of the gases:
$K_p=\dfrac{(P_{CO})^2}{P_{CO_2}} \label{Eq14.4.3}$
Incorporating all the constant values into $K′$ or $K_p$ allows us to focus on the substances whose concentrations change during the reaction.
Although the activities of pure liquids or solids are not written explicitly in the equilibrium constant expression, these substances must be present in the reaction mixture for chemical equilibrium to occur. Whatever the concentrations of $\ce{CO}$ and $\ce{CO_2}$, the system described in Equation $\ref{Eq14.4.1}$ will reach chemical equilibrium only if a stoichiometric amount of solid carbon or excess solid carbon has been added so that some is still present once the system has reached equilibrium. As shown in Figure $1$, it does not matter whether 1 g or 100 g of solid carbon is present; in either case, the composition of the gaseous components of the system will be the same at equilibrium.
Example $1$
Write each expression for $K$, incorporating all constants, and $K_p$ for the following equilibrium reactions.
1. $\ce{PCl3(l) + Cl2(g) <=> PCl5(s)}$
2. $\ce{Fe3O4(s) + 4H2(g) <=> 3Fe(s) + 4H2O(g)}$
Given: balanced equilibrium equations.
Asked for: expressions for $K$ and $K_p$.
Strategy:
Find $K$ by writing each equilibrium constant expression as the ratio of the concentrations of the products and reactants, each raised to its coefficient in the chemical equation. Then express $K_p$ as the ratio of the partial pressures of the products and reactants, each also raised to its coefficient in the chemical equation.
Solution
This reaction contains a pure solid ($PCl_5$) and a pure liquid ($PCl_3$). Their activities are equal to 1, so when incorporated into the equilibrium constant expression, they do not change the value. So
$K=\dfrac{1}{(1)[Cl_2]} \nonumber$
and
$K_p=\dfrac{1}{(1)P_{Cl_2}} \nonumber$
This reaction contains two pure solids ($Fe_3O_4$ and $Fe$), which are each assigned a value of 1 in the equilibrium constant expressions:
$K=\dfrac{(1)[H_2O]^4}{(1)[H_2]^4} \nonumber$
and
$K_p=\dfrac{(1)(P_{H_2O})^4}{(1)(P_{H_2})^4} \nonumber$
Exercise $1$
Write the expressions for $K$ and $K_p$ for the following reactions.
1. $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$
2. $\underset{glucose}{\ce{C6H12O6(s)}} + \ce{6O2(g) <=> 6CO2(g) + 6H2O(g)}$
Answer a
$K = [\ce{CO_2}]$ and $K_p = P_{\ce{CO_2}}$
Answer b
$K=\dfrac{[CO_2]^6[H_2O]^6}{[O_2]^6}$ and $K_p=\dfrac{(P_{CO_2})^6(P_{H_2O})^6}{(P_{O_2})^6}$
For reactions carried out in solution, the solvent is assumed to be pure, and therefore is assigned an activity equal to 1 in the equilibrium constant expression. The activities of the solutes are approximated by their molarities. The result is that the equilibrium constant expressions appear to only depend upon the concentrations of the solutes.
The activities of pure solids, pure liquids, and solvents are defined as having a value of '1'. Often, it is said that these activities are "left out" of equilibrium constant expressions. This is an unfortunate use of words. The activities are not "left out" of equilibrium constant expressions. Rather, because they have a value of '1', they do not change the value of the equilibrium constant when they are multiplied together with the other terms. The activities of the solutes are approximated by their molarities.
Summary
An equilibrated system that contains products and reactants in a single phase is a homogeneous equilibrium; a system whose reactants, products, or both are in more than one phase is a heterogeneous equilibrium.
Contributors and Attributions
• Anonymous
• Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/16%3A_Chemical_Equilibrium/16.05%3A_Heterogenous_Equilibria_-_Reactions_Involving_Solids_and_Liquids.txt |
Learning Objectives
• To predict in which direction a reaction will proceed.
We previously saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination.
The Reaction Quotient
To determine whether a system has reached equilibrium, chemists use a Quantity called the reaction Quotient ($Q$). The expression for the reaction Quotient has precisely the same form as the equilibrium constant expression, except that $Q$ may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction:
$aA+bB \rightleftharpoons cC+dD \nonumber$
the reaction quotient is defined as follows:
$Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{15.6.1}$
To understand how information is obtained using a reaction Quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide,
$\ce{N2O4(g) <=> 2NO2(g)} \nonumber$
for which $K = 4.65 \times 10^{−3}$ at 298 K. We can write $Q$ for this reaction as follows:
$Q=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{15.6.2}$
The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of $Q$ were calculated for each. Each experiment begins with different proportions of product and reactant:
Table $1$: Equilibrium Experiment data
Experiment $[\ce{NO2}]\; (M)$ $[\ce{N2O4}]\; (M)$ $Q = \dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$
1 0 0.0400 $\dfrac{0^2}{0.0400}=0$
2 0.0600 0 $\dfrac{(0.0600)^2}{0}=\text{undefined}$
3 0.0200 0.0600 $\dfrac{(0.0200)^2}{0.0600}=6.67 \times 10^{−3}$
As these calculations demonstrate, $Q$ can have any numerical value between 0 and infinity (undefined); that is, $Q$ can be greater than, less than, or equal to $K$.
Comparing the magnitudes of $Q$ and $K$ enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes $Q$ approach $K$:
• If $Q = K$, for example, then the system is already at equilibrium, and no further change in the composition of the system will occur unless the conditions are changed.
• If $Q < K$, then the ratio of the concentrations of products to the concentrations of reactants is less than the ratio at equilibrium. Therefore, the reaction will proceed to the right as written, forming products at the expense of reactants.
• If $Q > K$, then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium, so the reaction will proceed to the left as written, forming reactants at the expense of products.
These points are illustrated graphically in Figure $1$.
If $Q < K$, the reaction will proceed to the right as written. If $Q > K$, the reaction will proceed to the left as written. If $Q = K$, then the system is at equilibrium.
A Video Discussing Using the Reaction Quotient (Q): Using the Reaction Quotient (Q) (opens in new window) [youtu.be]
Example $1$
At elevated temperatures, methane ($CH_4$) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction:
$\ce{CH4(g) + H2O(g) <=> CO(g) + 3H2(g)} \nonumber$
$K = 2.4 \times 10^{−4}$ at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If $1.2 \times 10^{−2}$ mol of $CH_4$, 8.0 × 10−3 mol of $H_2O$, $1.6 \times 10^{−2}$ mol of $CO$, and $6.0 \times 10^{−3}$ mol of $H_2$ are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce $\ce{CO}$ and $\ce{H_2}$ or to the left to form $\ce{CH_4}$ and $\ce{H_2O}$?
Given: balanced chemical equation, $K$, amounts of reactants and products, and volume
Asked for: direction of reaction
Strategy:
1. Calculate the molar concentrations of the reactants and the products.
2. Use Equation $\ref{15.6.1}$ to determine $Q$. Compare $Q$ and $K$ to determine in which direction the reaction will proceed.
Solution:
A We must first find the initial concentrations of the substances present. For example, we have $1.2 \times 10^{−2} mol$ of $\ce{CH_4}$ in a 2.0 L container, so
$[\ce{CH4}]=\dfrac{1.2\times 10^{−2} \, \text{mol}}{2.0\; \text{L}}=6.0 \times 10^{−3} M \nonumber$
We can calculate the other concentrations in a similar way:
• $[\ce{H2O}] = 4.0 \times 10^{−3} M$,
• $[\ce{CO}] = 8.0 \times 10^{−3} M$, and
• $[\ce{H_2}] = 3.0 \times 10^{−3} M$.
B We now compute $Q$ and compare it with $K$:
\begin{align*} Q&=\dfrac{[\ce{CO}][\ce{H_2}]^3}{[\ce{CH_4}][\ce{H_2O}]} \[4pt] &=\dfrac{(8.0 \times 10^{−3})(3.0 \times 10^{−3})^3}{(6.0\times 10^{−3})(4.0 \times 10^{−3})} \[4pt] &=9.0 \times 10^{−6} \end{align*} \nonumber
Because $K = 2.4 \times 10^{−4}$, we see that $Q < K$. Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming $\ce{H2}$ and $\ce{CO}$ at the expense of $\ce{H_2O}$ and $\ce{CH4}$.
Exercise $2$
In the water–gas shift reaction introduced in Example $1$, carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen:
$\ce{CO(g) + H_2O(g) <=> CO2(g) + H2(g)} \nonumber$
$K = 0.64$ at 900 K. If 0.010 mol of both $\ce{CO}$ and $\ce{H_2O}$, 0.0080 mol of $\ce{CO_2}$, and 0.012 mol of $\ce{H_2}$ are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written?
Answer
$Q = 0.96$. Since (Q > K), so the reaction will proceed to the left, and $CO$ and $H_2O$ will form.
Predicting the Direction of a Reaction with a Graph
By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which $Q = K$. Such a graph allows us to predict what will happen to a reaction when conditions change so that $Q$ no longer equals $K$, such as when a reactant concentration or a product concentration is increased or decreased.
Reaction 1
Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation:
$\ce{PbCO3(s) <=> PbO(s) + CO2(g)} \label{15.6.3}$
Because $\ce{PbCO_3}$ and $\ce{PbO}$ are solids, the equilibrium constant is simply
$K = [\ce{CO_2}]. \nonumber$
At a given temperature, therefore, any system that contains solid $\ce{PbCO_3}$ and solid $\ce{PbO}$ will have exactly the same concentration of $\ce{CO_2}$ at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in Figure $3$, which shows a plot of $[\ce{CO_2}]$ versus the amount of $\ce{PbCO_3}$ added. Initially, the added $\ce{PbCO_3}$ decomposes completely to $\ce{CO_2}$ because the amount of $\ce{PbCO_3}$ is not sufficient to give a $\ce{CO_2}$ concentration equal to $K$. Thus the left portion of the graph represents a system that is not at equilibrium because it contains only $\ce{CO2(g)}$ and $\ce{PbO(s)}$. In contrast, when just enough $\ce{PbCO_3}$ has been added to give $[CO_2] = K$, the system has reached equilibrium, and adding more $\ce{PbCO_3}$ has no effect on the $\ce{CO_2}$ concentration: the graph is a horizontal line.
Thus any $\ce{CO_2}$ concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough $\ce{PbCO_3}$ and $\ce{PbO}$ are present. For example, the point labeled A in Figure $2$ lies above the horizontal line, so it corresponds to a $[\ce{CO_2}]$ that is greater than the equilibrium concentration of $\ce{CO_2}$ (i.e., $Q > K$). To reach equilibrium, the system must decrease $[\ce{CO_2}]$, which it can do only by reacting $\ce{CO_2}$ with solid $\ce{PbO}$ to form solid $\ce{PbCO_3}$. Thus the reaction in Equation $\ref{15.6.3}$ will proceed to the left as written, until $[\ce{CO_2}] = K$. Conversely, the point labeled B in Figure $2$ lies below the horizontal line, so it corresponds to a $[\ce{CO_2}]$ that is less than the equilibrium concentration of $\ce{CO_2}$ (i.e., $Q < K$). To reach equilibrium, the system must increase $[\ce{CO_2}]$, which it can do only by decomposing solid $\ce{PbCO_3}$ to form $\ce{CO_2}$ and solid $\ce{PbO}$. The reaction in Equation \ref{15.6.3} will therefore proceed to the right as written, until $[\ce{CO_2}] = K$.
Reaction 2
In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor:
$\ce{CdO(s) + H2(g) <=> Cd(s) + H_2O(g)} \label{15.6.4}$
and the equilibrium constant is
$K = \dfrac{[\ce{H_2O}]}{[\ce{H_2}]}. \nonumber$
If $[\ce{H_2O}]$ is doubled at equilibrium, then $[\ce{H2}]$ must also be doubled for the system to remain at equilibrium. A plot of $[\ce{H_2O}]$ versus $[\ce{H_2}]$ at equilibrium is a straight line with a slope of $K$ (Figure $3$). Again, only those pairs of concentrations of $\ce{H_2O}$ and $\ce{H_2}$ that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in Equation $\ref{15.6.4}$ will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point A in Figure $3$ lies below the line, indicating that the $[\ce{H_2O}]/[\ce{H_2}]$ ratio is less than the ratio of an equilibrium mixture (i.e., $Q < K$). Thus the reaction in Equation \ref{15.6.4} will proceed to the right as written, consuming $\ce{H_2}$ and producing $\ce{H_2O}$, which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point B in Figure $3$ lies above the line, indicating that the $[\ce{H_2O}]/[\ce{H_2}]$ ratio is greater than the ratio of an equilibrium mixture ($Q > K$). Thus the reaction in Equation $\ref{15.6.4}$ will proceed to the left as written, consuming $\ce{H_2O}$ and producing $\ce{H_2}$, which causes the concentration ratio to move down and to the right toward the equilibrium line.
Reaction 3
In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures:
$\ce{ NH4I(s) <=> NH3(g) + HI(g)} \label{15.6.5}$
For this system, $K$ is equal to the product of the concentrations of the two products:
$K = [\ce{NH_3}][\ce{HI}]. \nonumber$
If we double the concentration of $\ce{NH3}$, the concentration of $\ce{HI}$ must decrease by approximately a factor of 2 to maintain equilibrium, as shown in Figure $4$. As a result, for a given concentration of either $\ce{HI}$ or $\ce{NH_3}$, only a single equilibrium composition that contains equal concentrations of both $\ce{NH_3}$ and $\ce{HI}$ is possible, for which
$[\ce{NH_3}] = [\ce{HI}] = \sqrt{K}. \nonumber$
Any point that lies below and to the left of the equilibrium curve (such as point A in Figure $4$) corresponds to $Q < K$, and the reaction in Equation $\ref{15.6.5}$ will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point B in Figure $\ref{15.6.5}$) corresponds to $Q > K$, and the reaction in Equation $\ref{15.6.5}$ will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium.
Summary
The reaction Quotient ($Q$) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction. The reaction Quotient ($Q$ or $Q_p$) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, $Q = K$. Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/16%3A_Chemical_Equilibrium/16.06%3A_The_Reaction_Quotient-_Predicting_the_Direction_of_Change.txt |
Learning Objectives
• To solve quantitative problems involving chemical equilibriums.
There are two fundamental kinds of equilibrium problems:
1. those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and
2. those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems.
Calculating an Equilibrium Constant from Equilibrium Concentrations
We saw in the exercise in Example 6 in Section 15.2 that the equilibrium constant for the decomposition of $CaCO_{3(s)}$ to $CaO_{(s)}$ and $CO_{2(g)}$ is $K = [CO_2]$. At 800°C, the concentration of $CO_2$ in equilibrium with solid $CaCO_3$ and $CaO$ is $2.5 \times 10^{-3}\; M$. Thus K at 800°C is $2.5 \times 10^{-3}$. (Remember that equilibrium constants are unitless.)
A more complex example of this type of problem is the conversion of n-butane, an additive used to increase the volatility of gasoline, into isobutane (2-methylpropane).
This reaction can be written as follows:
$\ce{n-butane_{(g)} \rightleftharpoons isobutane_{(g)}} \label{Eq1}$
and the equilibrium constant $K = [\text{isobutane}]/[\text{n-butane}]$. At equilibrium, a mixture of n-butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M n-butane. Substituting these concentrations into the equilibrium constant expression,
$K=\dfrac{[\textit{isobutane}]}{[\textit{n-butane}]}=0.041\; M = 2.6 \label{Eq2}$
Thus the equilibrium constant for the reaction as written is 2.6.
Example $1$
The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid:
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}\nonumber$
A mixture of $SO_2$ and $O_2$ was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained
• $5.0 \times 10^{-2}\; M\; SO_3$,
• $3.5 \times 10^{-3}\; M\; O_2$, and
• $3.0 \times 10^{-3}\; M\; SO_2$.
Calculate $K$ and $K_p$ at this temperature.
Given: balanced equilibrium equation and composition of equilibrium mixture
Asked for: equilibrium constant
Strategy
Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain $K$.
Solution
Substituting the appropriate equilibrium concentrations into the equilibrium constant expression,
$K=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]}=\dfrac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})}=7.9 \times 10^4 \nonumber$
To solve for $K_p$, we use the relationship derived previously
$K_p = K(RT)^{Δn} \nonumber$
where $Δn = 2 − 3 = −1$:
$K_p=K(RT)^{Δn}\nonumber$
$K_p=7.9 \times 10^4 [(0.08206\; L⋅atm/mol⋅K)(800 K)]^{−1}\nonumber$
$K_p=1.2 \times 10^3\nonumber$
Exercise $1$
Hydrogen gas and iodine react to form hydrogen iodide via the reaction
$2 NOCl_{(g)} \leftrightharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
A mixture of $H_2$ and $I_2$ was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained
• $1.37\times 10^{−2}\; M\; HI$,
• $6.47 \times 10^{−3}\; M\; H_2$, and
• $5.94 \times 10^{-4}\; M\; I_2$.
Calculate $K$ and $K_p$ for this reaction.
Answer
$K = 48.8$ and $K_p = 48.8$
Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example $2$ shows one way to do this.
Example $2$
A 1.00 mol sample of $NOCl$ was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of $Cl_2$. Calculate $K$ at this temperature. The equation for the decomposition of $NOCl$ to $NO$ and $Cl_2$ is as follows:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
Given: balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium
Asked for: $K$
Strategy:
1. Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations).
2. Calculate all possible initial concentrations from the data given and insert them in the table.
3. Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table.
4. Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction.
Solution
A The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows:
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\nonumber$
To obtain the concentrations of $NOCl$, $NO$, and $Cl_2$ at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations.
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial
Change
Final
B Initially, the system contains 1.00 mol of $NOCl$ in a 2.00 L container. Thus $[NOCl]_i = 1.00\; mol/2.00\; L = 0.500\; M$. The initial concentrations of $NO$ and $Cl_2$ are $0\; M$ because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of $Cl_2$ in a 2.00 L container, so $[Cl_2]_f = 0.056 \;mol/2.00 \;L = 0.028\; M$. We insert these values into the following table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial 0.500 0 0
Change
Final 0.028
C We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of $Cl_2$, the substance for which initial and final concentrations are known:
$Δ[Cl_2] = 0.028 \;M_{(final)} − 0.00\; M_{(initial)}] = +0.028\; M\nonumber$
According to the coefficients in the balanced chemical equation, 2 mol of $NO$ are produced for every 1 mol of $Cl_2$, so the change in the $NO$ concentration is as follows:
$Δ[NO]=\left(\dfrac{0.028\; \cancel{mol \;Cl_2}}{ L}\right)\left(\dfrac{2\; mol\; NO}{1 \cancel{\;mol \;Cl_2}}\right)=0.056\; M\nonumber$
Similarly, 2 mol of $NOCl$ are consumed for every 1 mol of $Cl_2$ produced, so the change in the $NOCl$ concentration is as follows:
$Δ[NOCl]= \left(\dfrac{0.028\; \cancel{mol\; Cl_2}}{L}\right) \left(\dfrac{−2\; mol \;NOCl}{1\; \cancel{mol\; Cl_2}} \right) = -0.056 \;M\nonumber$
We insert these values into our table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl]$ $[NO]$ $[Cl_2]$
Initial 0.500 0 0
Change −0.056 +0.056 +0.028
Final 0.028
D We sum the numbers in the $[NOCl]$ and $[NO]$ columns to obtain the final concentrations of $NO$ and $NOCl$:
$[NO]_f = 0.000\; M + 0.056 \;M = 0.056\; M\nonumber$
$[NOCl]_f = 0.500\; M + (−0.056\; M) = 0.444 M\nonumber$
We can now complete the table:
$2 NOCl_{(g)} \rightleftharpoons 2NO_{(g)}+Cl_{2(g)}\nonumber$
ICE $[NOCl] \([NO]$ $[Cl_2]$
initial 0.500 0 0
change −0.056 +0.056 +0.028
final 0.444 0.056 0.028
We can now calculate the equilibrium constant for the reaction:
$K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}=\dfrac{(0.056)^2(0.028)}{(0.444)^2}=4.5 \times 10^{−4}\nonumber$
Exercise $2$
The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia ($NH_3$) by reacting $0.1248\; M \;H_2$ and $0.0416\; M \;N_2$ at about 500°C. At equilibrium, the mixture contained 0.00272 M $NH_3$. What is $K$ for the reaction
$N_2+3H_2 \rightleftharpoons 2NH_3\nonumber$
at this temperature? What is $K_p$?
Answer
$K = 0.105$ and $K_p = 2.61 \times 10^{-5}$
A Video Disucssing Using ICE Tables to find Kc: Using ICE Tables to find Kc(opens in new window) [youtu.be]
Calculating Equilibrium Concentrations from the Equilibrium Constant
To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of n-butane to isobutane (Equation $\ref{Eq1}$), for which K = 2.6 at 25°C. If we begin with a 1.00 M sample of n-butane, we can determine the concentration of n-butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example $2$.
$\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)} \nonumber$
ICE $[\text{n-butane}_{(g)} ]$ $[\text{isobutane}_{(g)}]$
Initial
Change
Final
The initial concentrations of the reactant and product are both known: [n-butane]i = 1.00 M and [isobutane]i = 0 M. We need to calculate the equilibrium concentrations of both n-butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the change in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as $+x$, then the change in the concentration of n-butane is Δ[n-butane] = $−x$. This is because the balanced chemical equation for the reaction tells us that 1 mol of n-butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone.
$\text{n-butane}_{(g)} \rightleftharpoons \text{isobutane}_{(g)} \nonumber$
ICE $[\text{n-butane}_{(g)} ]$ $[\text{isobutane}_{(g)}]$
Initial 1.00 0
Change $−x$ $+x$
Final $(1.00 − x)$ $(0 + x) = x$
Substituting the expressions for the final concentrations of n-butane and isobutane from the table into the equilibrium equation,
$K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\dfrac{x}{1.00−x}=2.6 \nonumber$
Rearranging and solving for $x$,
$x=2.6(1.00−x)= 2.6−2.6x \nonumber$
$x+2.6x= 2.6 \nonumber$
$x=0.72 \nonumber$
We obtain the final concentrations by substituting this $x$ value into the expressions for the final concentrations of n-butane and isobutane listed in the table:
$[\text{n-butane}]_f = (1.00 − x) M = (1.00 − 0.72) M = 0.28\; M \nonumber$
$[\text{isobutane}]_f = (0.00 + x) M = (0.00 + 0.72) M = 0.72\; M \nonumber$
We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same $K$ that we used in the calculation:
$K=\dfrac{[\text{isobutane}]}{[\text{n-butane}]}=\left(\dfrac{0.72\; \cancel{M}}{0.28\;\cancel{M}}\right) =2.6 \nonumber$
This is the same $K$ we were given, so we can be confident of our results.
Example $3$ illustrates a common type of equilibrium problem that you are likely to encounter.
Example $3$: The water–gas shift reaction
The water–gas shift reaction is important in several chemical processes, such as the production of H2 for fuel cells. This reaction can be written as follows:
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\nonumber$
$K = 0.106$ at 700 K. If a mixture of gases that initially contains 0.0150 M $H_2$ and 0.0150 M $CO_2$ is allowed to equilibrate at 700 K, what are the final concentrations of all substances present?
Given: balanced equilibrium equation, $K$, and initial concentrations
Asked for: final concentrations
Strategy:
1. Construct a table showing what is known and what needs to be calculated. Define $x$ as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of $x$. From the values in the table, calculate the final concentrations.
2. Write the equilibrium equation for the reaction. Substitute appropriate values from the ICE table to obtain $x$.
3. Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain $K$.
Solution
A The initial concentrations of the reactants are $[H_2]_i = [CO_2]_i = 0.0150\; M$. Just as before, we will focus on the change in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of $H_2O$ as $x$, then $Δ[H_2O] = +x$. We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of $x$. For example, 1 mol of $CO$ is produced for every 1 mol of $H_2O$, so the change in the $CO$ concentration can be expressed as $Δ[CO] = +x$. Similarly, for every 1 mol of $H_2O$ produced, 1 mol each of $H_2$ and $CO_2$ are consumed, so the change in the concentration of the reactants is $Δ[H_2] = Δ[CO_2] = −x$. We enter the values in the following table and calculate the final concentrations.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\nonumber$
ICE $[H_2]$ $[CO_2]$ $[H_2O]$ $[CO]$
Initial 0.0150 0.0150 0 0
Change $−x$ $−x$ $+x$ $+x$
Final $(0.0150 − x)$ $(0.0150 − x)$ $x$ $x$
B We can now use the equilibrium equation and the given $K$ to solve for $x$:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(x)(x)}{(0.0150−x)(0.0150−x}=\dfrac{x^2}{(0.0150−x)^2}=0.106\nonumber$
We could solve this equation with the quadratic formula, but it is far easier to solve for $x$ by recognizing that the left side of the equation is a perfect square; that is,
$\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106\nonumber$
Taking the square root of the middle and right terms,
$\dfrac{x}{(0.0150−x)} =(0.106)^{1/2}=0.326\nonumber$
$x =(0.326)(0.0150)−0.326x\nonumber$
$1.326x=0.00489\nonumber$
$x =0.00369=3.69 \times 10^{−3}\nonumber$
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f=[H_2]_i+Δ[H_2]=(0.0150−0.00369) \;M=0.0113\; M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=(0.0150−0.00369)\; M=0.0113\; M$
• $[H_2O]_f=[H_2O]_i+Δ[H_2O]=(0+0.00369) \;M=0.00369\; M$
• $[CO]_f=[CO]_i+Δ[CO]=(0+0.00369)\; M=0.00369 \;M$
We can check our work by inserting the calculated values back into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107\nonumber$
To two significant figures, this K is the same as the value given in the problem, so our answer is confirmed.
Exercise $3$
Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation:
$H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}\nonumber$
$K = 54$ at 425°C. If 0.172 M $H_2$ and $I_2$ are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture?
Answer
• $[HI]_f = 0.270 \;M$
• $[H_2]_f = [I_2]_f = 0.037\; M$
In Example $3$, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example $4$.
Example $4$
In the water–gas shift reaction shown in Example $3$, a sample containing 0.632 M CO2 and 0.570 M $H_2$ is allowed to equilibrate at 700 K. At this temperature, $K = 0.106$. What is the composition of the reaction mixture at equilibrium?
Given: balanced equilibrium equation, concentrations of reactants, and $K$
Asked for: composition of reaction mixture at equilibrium
Strategy:
1. Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations ($x) and the final concentrations. 2. Write the equilibrium constant expression for the reaction. Substitute the known K value and the final concentrations to solve for \(x$.
3. Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain $K$.
Solution
A $[CO_2]_i = 0.632\; M$ and $[H_2]_i = 0.570\; M$. Again, $x$ is defined as the change in the concentration of $H_2O$: $Δ[H_2O] = +x$. Because 1 mol of $CO$ is produced for every 1 mol of $H_2O$, the change in the concentration of $CO$ is the same as the change in the concentration of H2O, so Δ[CO] = +x. Similarly, because 1 mol each of $H_2$ and $CO_2$ are consumed for every 1 mol of $H_2O$ produced, $Δ[H_2] = Δ[CO_2] = −x$. The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium.
$H_{2(g)}+CO_{2(g)} \rightleftharpoons H_2O_{(g)}+CO_{(g)}\nonumber$
ICE $H_{2(g)}$ $CO_{2(g)}$ $H_2O_{(g)}$ $CO_{(g)}$
Initial 0.570 0.632 0 0
Change $−x$ $−x$ $+x$ $+x$
Final $(0.570 − x)$ $(0.632 − x)$ $x$ $x$
B We can now use the equilibrium equation and the known $K$ value to solve for $x$:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106\nonumber$
In contrast to Example $3$, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator:
$x^2 = 0.106(0.360 − 1.202x + x^2)\nonumber$
Collecting terms on one side of the equation,
$0.894x^2 + 0.127x − 0.0382 = 0\nonumber$
This equation can be solved using the quadratic formula:
$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)}\nonumber$
$x =0.148 \text{ and } −0.290\nonumber$
Only the answer with the positive value has any physical significance, so $Δ[H_2O] = Δ[CO] = +0.148 M$, and $Δ[H_2] = Δ[CO_2] = −0.148\; M$.
C The final concentrations of all species in the reaction mixture are as follows:
• $[H_2]_f[ = [H_2]_i+Δ[H_2]=0.570 \;M −0.148\; M=0.422 M$
• $[CO_2]_f =[CO_2]_i+Δ[CO_2]=0.632 \;M−0.148 \;M=0.484 M$
• $[H_2O]_f =[H_2O]_i+Δ[H_2O]=0\; M+0.148\; M =0.148\; M$
• $[CO]_f=[CO]_i+Δ[CO]=0 M+0.148\;M=0.148 M$
We can check our work by substituting these values into the equilibrium constant expression:
$K=\dfrac{[H_2O][CO]}{[H_2][CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107\nonumber$
Because $K$ is essentially the same as the value given in the problem, our calculations are confirmed.
Exercise $4$
The exercise in Example $1$ showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which $K = 54$ at 425°C. If a sample containing 0.200 M $H_2$ and 0.0450 M $I_2$ is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture?
Answer
• $[H_I]_f = 0.0882\; M$
• $[H_2]_f = 0.156\; M$
• $[I_2]_f = 9.2 \times 10^{−4} M$
In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small ($K ≤ 10^{−3}$) or very large ($K ≥ 10^3$), which means that the change in the concentration (defined as $x$) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example $5$.
Example $5$
Atmospheric nitrogen and oxygen react to form nitric oxide:
$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber$
with $K_p = 2.0 \times 10^{−31}$ at 25°C.
What is the partial pressure of NO in equilibrium with $N_2$ and $O_2$ in the atmosphere (at 1 atm, $P_{N_2} = 0.78\; atm$ and $P_{O_2} = 0.21\; atm$?
Given: balanced equilibrium equation and values of $K_p$, $P_{O_2}$, and $P_{N_2}$
Asked for: partial pressure of NO
Strategy:
1. Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances.
2. Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration ($x). 3. Calculate the partial pressure of \(NO$. Check your answer by substituting values into the equilibrium equation and solving for $K$.
Solution
A Because we are given Kp and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of $O_2$ is 0.21 atm and that of $N_2$ is 0.78 atm. If we define the change in the partial pressure of $NO$ as $2x$, then the change in the partial pressure of $O_2$ and of $N_2$ is $−x$ because 1 mol each of $N_2$ and of $O_2$ is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium.
$N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber$
ICE $P_{N_2}$ $P_{O_2}$ $P_{NO}$
Initial 0.78 0.21 0
Change $−x$ $−x$ $+2x$
Final $(0.78 − x)$ $(0.21 − x)$ $2x$
B Substituting these values into the equation for the equilibrium constant,
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31}\nonumber$
In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the $x$ value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, $(0.78 − x) = 0.78$ and $(0.21 − x) = 0.21$. Substituting these expressions into our original equation,
$\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31\nonumber} \nonumber$
$\dfrac{4x^2}{0.16} =2.0 \times10^{−31}\nonumber$
$x^2=\dfrac{0.33 \times 10^{−31}}{4}\nonumber$
$x^=9.1 \times 10^{−17}\nonumber$
C Substituting this value of $x$ into our expressions for the final partial pressures of the substances,
• $P_{NO}=2x \; atm=1.8 \times 10^{−16} \;atm$
• $P_{N_2}=(0.78−x) \;atm=0.78 \;atm$
• $P_{O_2}=(0.21−x) \;atm=0.21\; atm$
From these calculations, we see that our initial assumption regarding $x$ was correct: given two significant figures, $2.0 \times 10^{−16}$ is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if $x$ is less than about 5% of the total, or $10^{−3} > K > 10^3$, then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic $NO$, an ingredient of smog, does not form from atmospheric concentrations of $N_2$ and $O_2$ to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation:
$K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 \times 10^{−31}\nonumber$
The final $K_p$ agrees with the value given at the beginning of this example.
Exercise $5$
Under certain conditions, oxygen will react to form ozone, as shown in the following equation:
$3O_{2(g)} \rightleftharpoons 2O_{3(g)}\nonumber$
with $K_p = 2.5 \times 10^{−59}$ at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere ($P_{O_2}=0.21\; atm$)?
Answer
$4.8 \times 10^{−31} \;atm$
Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large ($K \geq 10^3$). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the products side of the reaction rather than the reactants side. This approach is illustrated in Example $6$.
Example $6$
The chemical equation for the reaction of hydrogen with ethylene ($C_2H_4$) to give ethane ($C_2H_6$) is as follows:
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}\nonumber$
with $K = 9.6 \times 10^{18}$ at 25°C. If a mixture of 0.200 M $H_2$ and 0.155 M $C_2H_4$ is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture?
Given: balanced chemical equation, $K$, and initial concentrations of reactants
Asked for: equilibrium concentrations
Strategy:
1. Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations.
2. Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for $x$ (the change in concentration).
3. Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation.
Solution:
A From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example $5$. If we define $−x$ as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is $+x$. The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions.
$H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)}\nonumber$
IACE $[H_{2(g)}]$ $[C_2H_{4(g)}]$ $[C_2H_{6(g)}]$
Initial 0.200 0.155 0
Assuming 100% reaction 0.045 0 0.155
Change $+x$ $+x$ $−x$
Final $(0.045 + x)$ $(0 + x)$ $(0.155 − x)$
B Substituting values into the equilibrium constant expression,
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18}\nonumber$
Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus $x$ is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified ($(0.045 + x)$ = 0.045 and $(0.155 − x) = 0.155$) as follows:
$K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18}\nonumber$
$x=3.6 \times 10^{−19}\nonumber$
C The small $x$ value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table:
• $[C_2H_6]_f = (0.155 − x)\; M = 0.155 \; M$
• $[C_2H_4]_f = x\; M = 3.6 \times 10^{−19} M$
• $[H_2]_f = (0.045 + x) \;M = 0.045 \;M$
We can verify our calculations by substituting the final concentrations into the equilibrium constant expression:
$K=\dfrac{[C_2H_6]}{[H_2][C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18}\nonumber$
This $K$ value agrees with our initial value at the beginning of the example.
Exercise $6$
Hydrogen reacts with chlorine gas to form hydrogen chloride:
$H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}\nonumber$
with $K_p = 4.0 \times 10^{31}$ at 47°C. If a mixture of 0.257 M $H_2$ and 0.392 M $Cl_2$ is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture?
Answer
$[H_2]_f = 4.8 \times 10^{−32}\; M$ $[Cl_2]_f = 0.135\; M$ $[HCl]_f = 0.514\; M$
A Video Discussing Using ICE Tables to find Eq. Concentrations & Kc: Using ICE Tables to find Eq. Concentrations & Kc(opens in new window) [youtu.be]
Summary
Various methods can be used to solve the two fundamental types of equilibrium problems: (1) those in which we calculate the concentrations of reactants and products at equilibrium and (2) those in which we use the equilibrium constant and the initial concentrations of reactants to determine the composition of the equilibrium mixture. When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/16%3A_Chemical_Equilibrium/16.07%3A_Finding_Equilibrium_Concentrations.txt |
Learning Objectives
• Describe the ways in which an equilibrium system can be stressed
• Predict the response of a stressed equilibrium using Le Chatelier’s principle
As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient ($Q$) is equal to the equilibrium constant ($K$). We next address what happens when a system at equilibrium is disturbed so that $Q$ is no longer equal to $K$. If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of $Q$ will no longer equal the value of $K$. To re-establish equilibrium, the system will either shift toward the products (if $(Q \leq K)$ or the reactants (if $(Q \geq K)$ until $Q$ returns to the same value as $K$. This process is described by Le Chatelier's principle.
Le Chatelier's principle
When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in $Q$; the reaction will shift to re-establish $Q = K$.
Predicting the Direction of a Reversible Reaction
Le Chatelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of $Q$ and $K$ for the system to predict the changes.
A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium.
The stress on the system in Figure $1$ is the reduction of the equilibrium concentration of SCN (lowering the concentration of one of the reactants would cause $Q$ to be larger than K). As a consequence, Le Chatelier's principle leads us to predict that the concentration of Fe(SCN)2+ should decrease, increasing the concentration of SCN part way back to its original concentration, and increasing the concentration of Fe3+ above its initial equilibrium concentration.
The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} \label{15.7.1a}$
$K_c=\mathrm{50.0 \; at\; 400°C} \label{15.7.1b}$
The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with $\mathrm{[H_2] = [I_2]} = 0.221\; M$ and $\ce{[HI]} = 1.563 \;M$ is at equilibrium; for this mixture, $Q_c = K_c = 50.0$. If $\ce{H_2}$ is introduced into the system so quickly that its concentration doubles before it begins to react (new $\ce{[H_2]} = 0.442\; M$), the reaction will shift so that a new equilibrium is reached, at which
• $\ce{[H_2]} = 0.374\; M$,
• $\ce{[I_2]} = 0.153\; M$, and
• $\ce{[HI]} = 1.692\; M$.
This gives:
\begin{align*} Q_c &=\mathrm{\dfrac{[HI]^2}{[H_2][I_2]}} \[4pt] &=\dfrac{(1.692)^2}{(0.374)(0.153)} \[4pt] &= 50.0 =K_c \label{15.7.2} \end{align*}
We have stressed this system by introducing additional $\ce{H_2}$. The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess $\ce{H_2}$, reducing the amount of uncombined $\ce{I_2}$, and forming additional $\ce{HI}$. Le Chatelier’s Principle (Changing Concentrations):
A Video Discussing Le Chatelier’s Principle (Changing Concentrations): Le Chatelier’s Principle (Changing Concentrations)(opens in new window) [youtu.be] (opens in new window)
Effect of Change in Pressure on Equilibrium
Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for Kc) or partial pressure (for KP). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.
As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure.
Consider what happens when we increase the pressure on a system in which $\ce{NO}$, $\ce{O_2}$, and $\ce{NO_2}$ are at equilibrium:
$\ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g)} \label{15.7.3}$
The formation of additional amounts of $\ce{NO2}$ decreases the total number of molecules in the system because each time two molecules of $\ce{NO_2}$ form, a total of three molecules of $\ce{NO}$ and $\ce{O_2}$ are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of $\ce{NO_2}$ into $\ce{NO}$ and $\ce{O_2}$, which tends to restore the pressure.
Now consider this reaction:
$\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g)} \label{15.7.4}$
Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.
Le Chatelier’s Principle (Changes in Pressure or Volume):
Effect of Change in Temperature on Equilibrium
Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle.
When hydrogen reacts with gaseous iodine, heat is evolved.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{15.7.5}$
Because this reaction is exothermic, we can write it with heat as a product.
$\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} + \text{heat} \label{15.7.6}$
Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of H2 and I2 and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide.
When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium constant: At the new equilibrium the concentration of HI has increased and the concentrations of H2 and I2 decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.
Temperature affects the equilibrium between $\ce{NO_2}$ and $\ce{N_2O_4}$ in this reaction
$\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{15.7.7}$
The positive ΔH value tells us that the reaction is endothermic and could be written
$\ce{heat}+\ce{N_2O4(g) \rightleftharpoons 2NO2(g)} \label{15.7.8}$
At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown $\ce{NO_2}$ molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless $\ce{N_2O_4}$ increases, and the concentration of brown $\ce{NO_2}$ decreases, causing the brown color to fade.
The overview of how different disturbances affect the reaction equilibrium properties is tabulated in Table $1$.
Table $1$: Effects of Disturbances of Equilibrium and $K$
Disturbance Observed Change as Equilibrium is Restored Direction of Shift Effect on K
reactant added added reactant is partially consumed toward products none
product added added product is partially consumed toward reactants none
decrease in volume/increase in gas pressure pressure decreases toward side with fewer moles of gas none
increase in volume/decrease in gas pressure pressure increases toward side with more moles of gas none
temperature increase heat is absorbed toward products for endothermic, toward reactants for exothermic changes
temperature decrease heat is given off toward reactants for endothermic, toward products for exothermic changes
Example $1$
Write an equilibrium constant expression for each reaction and use this expression to predict what will happen to the concentration of the substance in bold when the indicated change is made if the system is to maintain equilibrium.
1. $2HgO_{(s)} \rightleftharpoons 2Hg_{(l)} + \mathbf{O}_{2(g)}$: the amount of HgO is doubled.
2. $NH_4HS_{(s)} \rightleftharpoons \mathbf{NH}_{3(g)} + H_2S_{(g)}$: the concentration of $H_2S$ is tripled.
3. $\textbf{n-butane}_{(g)} \rightleftharpoons isobutane_{(g)}$: the concentration of isobutane is halved.
Given: equilibrium systems and changes
Asked for: equilibrium constant expressions and effects of changes
Strategy:
Write the equilibrium constant expression, remembering that pure liquids and solids do not appear in the expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made.
Solution:
Because $HgO_{(s)}$ and $Hg_{(l)}$ are pure substances, they do not appear in the equilibrium constant expression. Thus, for this reaction, $K = [O_2]$. The equilibrium concentration of $O_2$ is a constant and does not depend on the amount of $HgO$ present. Hence adding more $HgO$ will not affect the equilibrium concentration of $O_2$, so no compensatory change is necessary.
$NH_4HS$ does not appear in the equilibrium constant expression because it is a solid. Thus $K = [NH_3][H_2S]$, which means that the concentrations of the products are inversely proportional. If adding $H_2S$ triples the $H_2S$ concentration, for example, then the $NH_3$ concentration must decrease by about a factor of 3 for the system to remain at equilibrium so that the product of the concentrations equals $K$.
For this reaction, $K = \frac{[isobutane]}{[\textit{n-butane}]}$, so halving the concentration of isobutane means that the n-butane concentration must also decrease by about half if the system is to maintain equilibrium.
Exercise $1$
Write an equilibrium constant expression for each reaction. What must happen to the concentration of the substance in bold when the indicated change occurs if the system is to maintain equilibrium?
1. $\ce{HBr (g) + NaH (s) \rightleftharpoons NaBr (s)} + \mathbf{H_2(g)}$: the concentration of $\ce{HBr}$ is decreased by a factor of 3.
2. $\ce{6Li (s)} + \mathbf{N_2(g)} \ce{ \rightleftharpoons 2Li3N(s)}$: the amount of $\ce{Li}$ is tripled.
3. $\mathbf{SO_2(g)} + \ce{ Cl2(g) \rightleftharpoons SO2Cl2(l)}$: the concentration of $\ce{Cl2}$ is doubled.
Answer a
$K = \dfrac{[H_2]}{[HBr]}$; $[H_2]$ must decrease by about a factor of 3.
Answer b
$K = \dfrac{1}{[N_2]}$; solid lithium does not appear in the equilibrium constant expression, so no compensatory change is necessary.
Answer c
$K = \dfrac{1}{[SO_2][Cl_2]}$; $[SO_2]$ must decrease by about half.
Le Chatelier’s Principle (Changes in Temperature):
Catalysts Do Not Affect Equilibrium
As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations. The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \label{15.7.9}$
A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year. Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry.
Fritz Haber
Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements (Equation \ref{15.7.9}). The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate.
Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008. The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen ($\ce{N_2}$) is nutritionally unavailable to a majority of plants due the tremendous stability of the nitrogen-nitrogen triple bond. Therefore, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation). Legumes achieve this conversion at ambient temperature by exploiting bacteria equipped with suitable enzymes.
In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.”1 Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.
Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.
Summary
Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side.
Footnotes
1. 1 Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764.
Glossary
Le Chatelier's principle
when a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance
position of equilibrium
concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe conditions before a disturbance)
stress
change to a reaction's conditions that may cause a shift in the equilibrium | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/16%3A_Chemical_Equilibrium/16.08%3A_Le_Chateliers_Principle-_How_a_System_at_Equilibrium_Responds_to_Disturbances.txt |
• 17.1: Batman’s Basic Blunder
• 17.2: The Nature of Acids and Bases
In chemistry, acids and bases have been defined differently by three sets of theories: One is the Arrhenius definition defined above, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (H+) ions while bases produce hydroxide (OH-) ions in solution. The other two definitions are discussed in detail alter in the chapter and include the Brønsted-Lowry definition and the Lewis theory.
• 17.3: Definitions of Acids and Bases
A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base.
• 17.4: Acid Strength and Molecular Structure
Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an H+H+ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of H+H+ , making the conjugate acid a stronger acid.
• 17.5: Acid Strength and the Acid Ionization Constant (Ka)
Acid–base reactions always contain two conjugate acid–base pairs. Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases.
• 17.6: Autoionization of Water and pH
Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion ( H3O+ ). The autoionization of liquid water produces OH− and H3O+ ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as Kw=[H3O+][OH−] . At 25°C, Kw is 1.01×10−14 ; hence pH+pOH=pKw=14.00 .
• 17.7: Finding the [H3O+] and pH of Strong and Weak Acid Solutions
Acid–base reactions always contain two conjugate acid–base pairs. Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases.
• 17.8: Finding the [OH-] and pH of Strong and Weak Base Solutions
• 17.9: The Acid-Base Properties of Ions and Salts
A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion ( A−A− ), the conjugate acid of a weak base as the cation ( BH+ ), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction.
• 17.10: Polyprotic Acids
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of Ka of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations sequentially.
• 17.11: Lewis Acids and Bases
Lewis proposed that the electron pair is the dominant actor in acid-base chemistry. An Lewis acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons. A Lewis base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared. Lewis acis/base theory is a powerful tool for describing many chemical reactions used in organic and inorganic chemistry.
17: Acids and Bases
Acids and bases have been known for a long time. When Robert Boyle characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.
Acids and bases are common solutions that exist everywhere. Almost every liquid that we encounter in our daily lives consists of acidic and basic properties, with the exception of water. They have completely different properties and are able to neutralize to form H2O, which will be discussed later in a subsection. Acids and bases can be defined by their physical and chemical observations (Table $1$).
ACIDS BASES
Table $1$: General Properties of Acids and Bases
produce a piercing pain in a wound. give a slippery feel.
taste sour. taste bitter.
are colorless when placed in phenolphthalein (an indicator). are pink when placed in phenolphthalein (an indicator).
are red on blue litmus paper (a pH indicator). are blue on red litmus paper (a pH indicator).
have a pH<7. have a pH>7.
produce hydrogen gas when reacted with metals.
produce carbon dioxide when reacted with carbonates.
Common examples: Lemons, oranges, vinegar, urine, sulfuric acid, hydrochloric acid Common Examples: Soap, toothpaste, bleach, cleaning agents, limewater, ammonia water, sodium hydroxide.
Acids and bases in aqueous solutions will conduct electricity because they contain dissolved ions. Therefore, acids and bases are electrolytes. Strong acids and bases will be strong electrolytes. Weak acids and bases will be weak electrolytes. This affects the amount of conductivity.
The Arrhenius Definition of Acids and Bases
In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. The Arrhenius definition of acid-base reactions is a development of the "hydrogen theory of acids". It was used to provide a modern definition of acids and bases, and followed from Arrhenius's work with Friedrich Wilhelm Ostwald in establishing the presence of ions in aqueous solution in 1884. This led to Arrhenius receiving the Nobel Prize in Chemistry in 1903.
An Arrhenius acid is a compound that increases the concentration of $H^+$ ions that are present when added to water. These $H^+$ ions form the hydronium ion ($H_3O^+$) when they combine with water molecules. This process is represented in a chemical equation by adding H2O to the reactants side.
$HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \nonumber$
In this reaction, hydrochloric acid ($HCl$) dissociates into hydrogen ($H^+$) and chlorine ($Cl^-$) ions when dissolved in water, thereby releasing H+ ions into solution. Formation of the hydronium ion equation:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)} \nonumber$
The Arrhenius definitions of acidity and alkalinity are restricted to aqueous solutions and refer to the concentration of the solvated ions. Under this definition, pure $H_2SO_4$ or $HCl$ dissolved in toluene are not acidic, despite the fact that both of these acids will donate a proton to toluene. In addition, under the Arrhenius definition, a solution of sodium amide ($NaNH_2$) in liquid ammonia is not alkaline, despite the fact that the amide ion ($NH^−_2$) will readily deprotonate ammonia. Thus, the Arrhenius definition can only describe acids and bases in an aqueous environment.
Limitation of the Arrhenius Definition of Acids and Bases
The Arrhenius definition can only describe acids and bases in an aqueous environment.
In chemistry, acids and bases have been defined differently by three sets of theories: One is the Arrhenius definition defined above, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen ($H^+$) ions while bases produce hydroxide ($OH^-$) ions in solution. The other two definitions are discussed in detail alter in the chapter and include the Brønsted-Lowry definition the defines acids as substances that donate protons ($H^+$) whereas bases are substances that accept protons and the Lewis theory of acids and bases states that acids are electron pair acceptors while bases are electron pair donors.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.02%3A_The_Nature_of_Acids_and_Bases.txt |
Learning Objectives
• Identify acids, bases, and conjugate acid-base pairs according to the Brønsted-Lowry definition
• Write equations for acid and base ionization reactions
• Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations
• Describe the acid-base behavior of amphiprotic substances
Acids and bases have been known for a long time. When Robert Boyle characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2), and interact with alkalis to form neutral substances. In 1815, Humphry Davy contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.
Previously, we defined acids and bases as Arrhenius did: An acid is a compound that dissolves in water to yield hydronium ions ($H_3O^+$) and a base as a compound that dissolves in water to yield hydroxide ions ($\ce{OH-}$). This definition is not wrong; it is simply limited. We extended the definition of an acid or a base using the more general definition proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, $\ce{H^+}$. A proton is what remains when a normal hydrogen atom, $\ce{^1_1H}$, loses an electron. A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base). In a subsequent chapter of this text we will introduce the most general model of acid-base behavior introduced by the American chemist G. N. Lewis.
Acids may be compounds such as $HCl$ or $H_2SO_4$, organic acids like acetic acid ($\ce{CH_3COOH}$) or ascorbic acid (vitamin C), or $H_2O$. Anions (such as $\ce{HSO_4^-}$, $\ce{H_2PO_4^-}$, $\ce{HS^-}$, and $\ce{HCO_3^-}$) and cations (such as $\ce{H_3O^+}$, $\ce{NH_4^+}$, and $\ce{[Al(H_2O)_6]^{3+}}$) may also act as acids. Bases fall into the same three categories. Bases may be neutral molecules (such as $\ce{H_2O}$, $\ce{NH_3}$, and $\ce{CH_3NH_2}$), anions (such as $\ce{OH^-}$, $\ce{HS^-}$, $\ce{HCO_3^-}$, $\ce{CO_3^{2−}}$, $\ce{F^-}$, and $\ce{PO_4^{3−}}$), or cations (such as $\ce{[Al(H_2O)_5OH]^{2+}}$). The most familiar bases are ionic compounds such as $\ce{NaOH}$ and $\ce{Ca(OH)_2}$, which contain the hydroxide ion, $\ce{OH^-}$. The hydroxide ion in these compounds accepts a proton from acids to form water:
$\ce{H^+ + OH^- \rightarrow H_2O} \label{16.2.1}$
We call the product that remains after an acid donates a proton the conjugate base of the acid. This species is a base because it can accept a proton (to re-form the acid):
$\text{acid} \rightleftharpoons \text{proton} + \text{conjugate base}\label{16.2.2a}$
$\ce{HF \rightleftharpoons H^+ + F^-} \label{16.2.2b}$
$\ce{H_2SO_4 \rightleftharpoons H^+ + HSO_4^{−}}\label{16.2.2c}$
$\ce{H_2O \rightleftharpoons H^+ + OH^-}\label{16.2.2d}$
$\ce{HSO_4^- \rightleftharpoons H^+ + SO_4^{2−}}\label{16.2.2e}$
$\ce{NH_4^+ \rightleftharpoons H^+ + NH_3} \label{16.2.2f}$
We call the product that results when a base accepts a proton the base’s conjugate acid. This species is an acid because it can give up a proton (and thus re-form the base):
$\text{base} + \text{proton} \rightleftharpoons \text{conjugate acid} \label{16.2.3a}$
$\ce{OH^- +H^+ \rightleftharpoons H2O}\label{16.2.3b}$
$\ce{H_2O + H^+ \rightleftharpoons H3O+}\label{16.2.3c}$
$\ce{NH_3 +H^+ \rightleftharpoons NH4+}\label{16.2.3d}$
$\ce{S^{2-} +H^+ \rightleftharpoons HS-}\label{16.2.3e}$
$\ce{CO_3^{2-} +H^+ \rightleftharpoons HCO3-}\label{16.2.3f}$
$\ce{F^- +H^+ \rightleftharpoons HF} \label{16.2.3g}$
In these two sets of equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, $\ce{OH^-}$, and the conjugate acid of ammonia, $\ce{NH4+}$:
The reaction between a Brønsted-Lowry acid and water is called acid ionization. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:
When we add a base to water, a base ionization reaction occurs in which protons are transferred from water molecules to base molecules. For example, adding pyridine to water yields hydroxide ions and pyridinium ions:
" height="167" width="612" src="/@api/deki/files/56740/CNX_Chem_14_01_NH3_img.jpg">
Notice that both these ionization reactions are represented as equilibrium processes. The relative extent to which these acid and base ionization reactions proceed is an important topic treated in a later section of this chapter. In the preceding paragraphs we saw that water can function as either an acid or a base, depending on the nature of the solute dissolved in it. In fact, in pure water or in any aqueous solution, water acts both as an acid and a base. A very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions:
This type of reaction, in which a substance ionizes when one molecule of the substance reacts with another molecule of the same substance, is referred to as autoionization. Pure water undergoes autoionization to a very slight extent. Only about two out of every $10^9$ molecules in a sample of pure water are ionized at 25 °C. The equilibrium constant for the ionization of water is called the ion-product constant for water (Kw):
$\ce{H_2O}_{(l)}+\ce{H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)}+H\ce{O^-}_{(aq)}\;\;\; K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{16.2.4}$
The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C, Kw has a value of $1.0 \times 10^{−14}$. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 °C, the value for $K_\ce{w}$ is approximately $5.1 \times 10^{−13}$, roughly 100-times larger than the value at 25 °C.
A Video Discussing Conjugate Acid-Base Pairs: Conjugate Acid-Base Pairs [youtu.be]
Example $1$: Ion Concentrations in Pure Water
What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C?
Solution
The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, $\ce{[H_3O^+]} = \ce{[OH^- ]}$. At 25 °C:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]}=\ce{[H_3O^+]^2+}=\ce{[OH^- ]^2+}=1.0 \times 10^{−14} \nonumber$
So:
$\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M \nonumber$
The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal $1.0 \times 10^{−7}\; M$.
Exercise $1$
The ion product of water at 80 °C is $2.4 \times 10^{−13}$. What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C?
Answer
$\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7}\; M$
It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium determined by the autoionization reaction but it does shift the relative concentrations of $\ce{[OH^-]}$ and $\ce{[H_3O^+]}$. Example 16.2.2 demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations.
A Video Describing the Self-Ionization of Water (Kw): Self-Ionization of Water (Kw) [youtu.be]
Example $2$: The Inverse Proportionality of $\ce{[H_3O^+]}$ and $\ce{[OH^- ]}$
A solution of carbon dioxide in water has a hydronium ion concentration of $2.0 \times 10^{−6}\; M$. What is the concentration of hydroxide ion at 25 °C?
Solution
We know the value of the ion-product constant for water at 25 °C:
$\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)} \nonumber$
$K_\ce{w}=\ce{[H3O+][OH^- ]}=1.0 \times 10^{−14} \nonumber$
Thus, we can calculate the missing equilibrium concentration.
Rearrangement of the Kw expression yields that $[\ce{OH^- }]$ is directly proportional to the inverse of [H3O+]:
$[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9} \nonumber$
The hydroxide ion concentration in water is reduced to $5.0 \times 10^{−9}\: M$ as the hydrogen ion concentration increases to $2.0 \times 10^{−6}\; M$. This is expected from Le Chatelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the $\ce{[OH^- ]}$ is reduced relative to that in pure water.
A check of these concentrations confirms that our arithmetic is correct:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]}=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber$
Exercise $2$
What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?
Answer
$\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber$
Amphiprotic Species
Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. Another term used to describe such species is amphoteric, which is a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here:
$\ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{CO^{2-}}_{3(aq)} + \ce{H_3O^+}_{(aq)} \label{16.2.5a}$
$ \ce{HCO^-}_{3(aq)} + \ce{H_2O}_{(l)} \rightleftharpoons \ce{H_2CO}_{3(aq)} + \ce{OH^-}_{(aq)} \label{16.2.5b}$
Example $3$: The Acid-Base Behavior of an Amphoteric Substance
Write separate equations representing the reaction of $\ce{HSO3-}$
1. as an acid with $\ce{OH^-}$
2. as a base with HI
Solution
1. $HSO_{3(aq)}^- + OH_{(aq)}^- \rightleftharpoons SO_{3(aq)}^{2-} + H_2O_{(l)}$
2. $HSO_{3(aq)}^- + HI_{(aq)} \rightleftharpoons H_2SO_{3(aq)}+ I_{(aq)}^-$
Example $4$
Write separate equations representing the reaction of $\ce{H2PO4-}$
1. as a base with HBr
2. as an acid with $\ce{OH^-}$
Answer
1. $H_2PO_{4(aq)}^- + HBr_{(aq)} \rightleftharpoons H_3PO_{4(aq)} + Br^-_{(aq)}$
2. $H_2PO_{4(aq)}^- + OH^-_{(aq)} \rightleftharpoons HPO_{4(aq)}^{2-} + H_2O_{(l)}$
Summary
A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H3O+, and the hydroxide ion, $\ce{OH^-}$ when it undergoes autoionization:
$\ce{2 H_2O}_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} \nonumber$
The ion product of water, Kw is the equilibrium constant for the autoionization reaction:
$K_\ce{w}=\mathrm{[H_2O^+][OH^- ]=1.0 \times 10^{−14} \; at\; 25°C} \nonumber$
Key Equations
• $K_{\ce w} = \ce{[H3O+][OH^- ]} = 1.0 \times 10^{−14}\textrm{ (at 25 °C)} \nonumber$
Glossary
acid ionization
reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid
amphiprotic
species that may either gain or lose a proton in a reaction
amphoteric
species that can act as either an acid or a base
autoionization
reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions
base ionization
reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base
Brønsted-Lowry acid
proton donor
Brønsted-Lowry base
proton acceptor
conjugate acid
substance formed when a base gains a proton
conjugate base
substance formed when an acid loses a proton
ion-product constant for water (Kw)
equilibrium constant for the autoionization of water | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.03%3A_Definitions_of_Acids_and_Bases.txt |
Learning Objectives
• To understand how molecular structure affects the strength of an acid or base.
We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule.
Bond Strengths
In general, the stronger the $\ce{A–H}$ or $\ce{B–H^+}$ bond, the less likely the bond is to break to form $H^+$ ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides:
Relative Acid Strength HF HCl HBr HI
H–X Bond Energy (kJ/mol) 570 432 366 298
pKa 3.20 −6.1 −8.9 −9.3
The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of Group 16 elements is as follows, with $pK_a$ values in parentheses:
$H_2O (14.00 = pK_w) < H_2S (7.05) < H_2Se (3.89) < H_2Te (2.6) \label{1}$
Stability of the Conjugate Base
Whether we write an acid–base reaction as $AH \rightleftharpoons A^−+H^+$ or as $BH^+ \rightleftharpoons B+H^+$, the conjugate base ($A^−$ or $B$) contains one more lone pair of electrons than the parent acid ($AH$ or $BH^+$). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of $H^+$ and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses:
$CH_4 (~50) \ll NH_3 (~36) < H_2O (14.00) < HF (3.20) \label{2}$
Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of $CH_4$ is $CH_3^−$, and the conjugate base of $HF$ is $F^−$. Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the $F^−$ ion than carbon can stabilize the negative charge in the CH3− ion. Consequently, $\ce{HF}$ has a greater tendency to dissociate to form $H^+$ and $F^−$ than does methane to form $H^+$ and $CH_3^−$, making HF a much stronger acid than $CH_4$.
The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula $HE$, as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form $E^−$ and $H^+$. Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table.
Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table.
the strongest acid Known: The hydrohelium Cation
The stornger acid, the weaker the covalent bond to a hydrogen atom. So the strongest acid possible is the molecule with the weakest bond. That is the hydrohelium (1+) cation, $\ce{HeH^{+}}$, which is a positively charged ion formed by the reaction of a proton with a helium atom in the gas phase. It was first produced in the laboratory in 1925 and is isoelectronic with molecular hydrogen (\ce{H2}}). It is the strongest known acid, with a proton affinity of 177.8 kJ/mol.
Ball and stick model of the hydrohelium ion. (CC BY-SA 3.0; CCoil).
$\ce{HeH^{+}}$ cannot be prepared in a condensed phase, as it would protonate any anion, molecule or atom with which it were associated. However it is possible to estimate a hypothetical aqueous acidity using Hess's law:
HHe+(g) H+(g) + He(g) +178 kJ/mol
HHe+(aq) HHe+(g) +973 kJ/mol
H+(g) H+(aq) −1530 kJ/mol
He(g) He(aq) +19 kJ/mol
HHe+(aq) H+(aq) + He(aq) −360 kJ/mol
A free energy change of dissociation of −360 kJ/mol is equivalent to a pKa of −63.
It has been suggested that $\ce{HeH^{+}}$ should occur naturally in the interstellar medium, but it has not yet been detected.
Inductive Effects
Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium:
$HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \label{3}$
The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms:
HOX Electronegativity of X pKa
HOCl 3.0 7.40
HOBr 2.8 8.55
HOI 2.5 10.5
As the electronegativity of $X$ increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as $H^+$.
The acidity of oxoacids, with the general formula $HOXO_n$ (with $n$ = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom $X$. As shown in Figure $1$, the $K_a$ values of the oxoacids of chlorine increase by a factor of about $10^4$ to $10^6$ with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base.
Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound.
Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure $1$ show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure $1$ and Figure $2$, blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from $HClO$ to $HClO_4$ (also written as $HOClO_3$, while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as $H^+$ ions, thereby increasing the strength of the acid.
At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure $2$, the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms.
Electron delocalization in the conjugate base increases acid strength.
The electrostatic potential plots in Figure $2$ demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO− is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in $ClO_4^+$, indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion ($ClO_4^−$), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion ($OCl^−$), the negative charge is largely localized on a single oxygen atom (Figure $2$). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known.
As the number of terminal oxygen atoms increases, the number of resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic.
Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, $H_3PO_4$ is a weak acid, $H_2SO_4$ is a strong acid, and $HClO_4$ is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to $Cl$, which causes electrons to be drawn from oxygen to the central atom, weakening the $\ce{O–H}$ bond and increasing the strength of the oxoacid.
Careful inspection of the data in Table $1$ shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid $(H_2CO_3$) were a discrete molecule with the structure $\ce{(HO)_2C=O}$, it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid ($H_3PO_4$), for which pKa1 = 2.16. Instead, the tabulated value of $pK_{a1}$ for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see, however, $H_2CO_3$ is only a minor component of the aqueous solutions of $CO_2$ that are referred to as carbonic acid. Similarly, if phosphorous acid ($H_3PO_3$) actually had the structure $(HO)_3P$, it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as $HOCl$ (pKa = 7.40). In fact, the $pK_{a1}$ for phosphorous acid is 1.30, and the structure of phosphorous acid is $\ce{(HO)_2P(=O)H}$ with one H atom directly bonded to P and one $\ce{P=O}$ bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as $H_3PO_4$. Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen.
Table $1$: Values of pKa for Selected Polyprotic Acids and Bases
*$H_2CO_3$ and $H_2SO_3$ are at best minor components of aqueous solutions of $CO_{2(g)}$ and $SO_{2(g)}$, respectively, but such solutions are commonly referred to as containing carbonic acid and sulfurous acid, respectively.
Polyprotic Acids Formula $pK_{a1}$ $pK_{a2}$ $pK_{a3}$
carbonic acid* “$H_2CO_3$” 6.35 10.33
citric acid $HO_2CCH-2C(OH)(CO_2H)CH_2CO_2H$ 3.13 4.76 6.40
malonic acid $HO-2CCH_2CO_2H$ 2.85 5.70
oxalic acid $HO_2CCO_2H$ 1.25 3.81
phosphoric acid $H_3PO_4$ 2.16 7.21 12.32
phosphorous acid $H_3PO_3$ 1.3 6.70
succinic acid $HO_2CCH_2CH_2CO_2H$ 4.21 5.64
sulfuric acid $H_2SO_4$ −2.0 1.99
sulfurous acid* “$H_2SO_3$” 1.85 7.21
Polyprotic Bases Formula $pK_{b1}$ $pK_{b2}$
ethylenediamine $H_2N(CH_2)_2NH_2$ 4.08 7.14
piperazine $HN(CH_2CH_2)_2NH$ 4.27 8.67
propylenediamine $H_2N(CH_2)_3NH_2$ 3.45 5.12
Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives:
$pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \nonumber$
As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the $\ce{–CH_3}$ group of acetic acid by a $\ce{–CF_3}$ group results in about a 10,000-fold increase in acidity!
Example $1$
Arrange the compounds of each series in order of increasing acid or base strength.
1. sulfuric acid [$H_2SO_4$, or $(HO)_2SO_2$], fluorosulfonic acid ($FSO_3H$, or $FSO_2OH$), and sulfurous acid [$H_2SO_3$, or $(HO)_2SO$]
2. ammonia ($NH_3$), trifluoramine ($NF_3$), and hydroxylamine ($NH_2OH$)
The structures are shown here.
Given: series of compounds
Asked for: relative acid or base strengths
Strategy:
Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution.
Solution:
Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, $FSO_3H$ is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids:
$pKa H_2SO_3 1.85<H_2SO_4^{−2} < FSO_3H−10 \nonumber$
The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an $H^+$ ion. Thus $NF_3$ is predicted to be a much weaker base than $NH_3$. Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in $NH_3$ by $OH$ will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured $pK_b$ values:
$pK_bNF_3—<<NH_2OH 8.06<NH_3 4.75 \nonumber$
Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has never been measured.
Exercise $1$
Arrange the compounds of each series in order of
1. decreasing acid strength: $H_3PO_4$, $CH_3PO_3H_2$, and $HClO_3$.
2. increasing base strength: $CH_3S^−$, $OH^−$, and $CF_3S^−$.
Answer a
$HClO-3 > CH_3PO_3H_2 > H_3PO_4$
Answer a
$CF_3S^− < CH_3S^− < OH^−$
Summary
Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an $H^+$ ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of $H^+$, making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an $\ce{O–H}$ bond and allow hydrogen to be more easily lost as $H^+$ ions.
• Anonymous | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.04%3A_Acid_Strength_and_Molecular_Structure.txt |
Learning Objectives
• To know the relationship between acid or base strength and the magnitude of $K_a$, $K_b$, $pK_a$, and $pK_b$.
• To understand the leveling effect.
The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate base, is as follows:
$HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}$
The equilibrium constant for this dissociation is as follows:
$K=\dfrac{[H_3O^+][A^−]}{[H_2O][HA]} \label{16.5.2}$
As we noted earlier, because water is the solvent, it has an activity equal to 1, so the $[H_2O]$ term in Equation $\ref{16.5.2}$ is actually the $\textit{a}_{H_2O}$, which is equal to 1.
Again, for simplicity, $H_3O^+$ can be written as $H^+$ in Equation $\ref{16.5.3}$.
$HA_{(aq)} \rightleftharpoons H^+_{(aq)}+A^−_{(aq)} \label{16.5.3}$
Keep in mind, though, that free $H^+$ does not exist in aqueous solutions and that a proton is transferred to $H_2O$ in all acid ionization reactions to form hydronium ions, $H_3O^+$. The larger the $K_a$, the stronger the acid and the higher the $H^+$ concentration at equilibrium. Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of $H^+$ or $OH^−$, thus making them unitless. The values of $K_a$ for a number of common acids are given in Table $1$.
Table $1$: Values of $K_a$, $pK_a$, $K_b$, and $pK_b$ for Selected Acids ($HA$ and Their Conjugate Bases ($A^−$)
Acid $HA$ $K_a$ $pK_a$ $A^−$ $K_b$ $pK_b$
*The number in parentheses indicates the ionization step referred to for a polyprotic acid.
hydroiodic acid $HI$ $2 \times 10^{9}$ −9.3 $I^−$ $5.5 \times 10^{−24}$ 23.26
sulfuric acid (1)* $H_2SO_4$ $1 \times 10^{2}$ −2.0 $HSO_4^−$ $1 \times 10^{−16}$ 16.0
nitric acid $HNO_3$ $2.3 \times 10^{1}$ −1.37 $NO_3^−$ $4.3 \times 10^{−16}$ 15.37
hydronium ion $H_3O^+$ $1.0$ 0.00 $H_2O$ $1.0 \times 10^{−14}$ 14.00
sulfuric acid (2)* $HSO_4^−$ $1.0 \times 10^{−2}$ 1.99 $SO_4^{2−}$ $9.8 \times 10^{−13}$ 12.01
hydrofluoric acid $HF$ $6.3 \times 10^{−4}$ 3.20 $F^−$ $1.6 \times 10^{−11}$ 10.80
nitrous acid $HNO_2$ $5.6 \times 10^{−4}$ 3.25 $NO2^−$ $1.8 \times 10^{−11}$ 10.75
formic acid $HCO_2H$ $1.78 \times 10^{−4}$ 3.750 $HCO_2−$ $5.6 \times 10^{−11}$ 10.25
benzoic acid $C_6H_5CO_2H$ $6.3 \times 10^{−5}$ 4.20 $C_6H_5CO_2^−$ $1.6 \times 10^{−10}$ 9.80
acetic acid $CH_3CO_2H$ $1.7 \times 10^{−5}$ 4.76 $CH_3CO_2^−$ $5.8 \times 10^{−10}$ 9.24
pyridinium ion $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23 $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77
hypochlorous acid $HOCl$ $4.0 \times 10^{−8}$ 7.40 $OCl^−$ $2.5 \times 10^{−7}$ 6.60
hydrocyanic acid $HCN$ $6.2 \times 10^{−10}$ 9.21 $CN^−$ $1.6 \times 10^{−5}$ 4.79
ammonium ion $NH_4^+$ $5.6 \times 10^{−10}$ 9.25 $NH_3$ $1.8 \times 10^{−5}$ 4.75
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $OH^−$ $1.00$ 0.00
acetylene $C_2H_2$ $1 \times 10^{−26}$ 26.0 $HC_2^−$ $1 \times 10^{12}$ −12.0
ammonia $NH_3$ $1 \times 10^{−35}$ 35.0 $NH_2^−$ $1 \times 10^{21}$ −21.0
Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid:
$B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}$
The equilibrium constant for this reaction is the base ionization constant (Kb), also called the base dissociation constant:
$K_b= \frac{[BH^+][OH^−]}{[B]} \label{16.5.5}$
Once again, the activity of water has a value of 1, so water does not appear in the equilibrium constant expression. The larger the $K_b$, the stronger the base and the higher the $OH^−$ concentration at equilibrium. The values of $K_b$ for a number of common weak bases are given in Table $2$.
Table $2$: Values of $K_b$, $pK_b$, $K_a$, and $pK_a$ for Selected Weak Bases (B) and Their Conjugate Acids (BH+)
Base $B$ $K_b$ $pK_b$ $BH^+$ $K_a$ $pK_a$
*As in Table $1$.
hydroxide ion $OH^−$ $1.0$ 0.00* $H_2O$ $1.0 \times 10^{−14}$ 14.00
phosphate ion $PO_4^{3−}$ $2.1 \times 10^{−2}$ 1.68 $HPO_4^{2−}$ $4.8 \times 10^{−13}$ 12.32
dimethylamine $(CH_3)_2NH$ $5.4 \times 10^{−4}$ 3.27 $(CH_3)_2NH_2^+$ $1.9 \times 10^{−11}$ 10.73
methylamine $CH_3NH_2$ $4.6 \times 10^{−4}$ 3.34 $CH_3NH_3^+$ $2.2 \times 10^{−11}$ 10.66
trimethylamine $(CH_3)_3N$ $6.3 \times 10^{−5}$ 4.20 $(CH_3)_3NH^+$ $1.6 \times 10^{−10}$ 9.80
ammonia $NH_3$ $1.8 \times 10^{−5}$ 4.75 $NH_4^+$ $5.6 \times 10^{−10}$ 9.25
pyridine $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77 $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23
aniline $C_6H_5NH_2$ $7.4 \times 10^{−10}$ 9.13 $C_6H_5NH_3^+$ $1.3 \times 10^{−5}$ 4.87
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $H_3O^+$ $1.0^*$ 0.00
There is a simple relationship between the magnitude of $K_a$ for an acid and $K_b$ for its conjugate base. Consider, for example, the ionization of hydrocyanic acid ($HCN$) in water to produce an acidic solution, and the reaction of $CN^−$ with water to produce a basic solution:
$HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \label{16.5.6}$
$CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \label{16.5.7}$
The equilibrium constant expression for the ionization of HCN is as follows:
$K_a=\dfrac{[H^+][CN^−]}{[HCN]} \label{16.5.8}$
The corresponding expression for the reaction of cyanide with water is as follows:
$K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \label{16.5.9}$
If we add Equations $\ref{16.5.6}$ and $\ref{16.5.7}$, we obtain the following:
add Equations $\ref{16.5.6}$ and $\ref{16.5.7}$, we obtain
Reaction Equilibrium Constants
$\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}}$ $K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}$
$\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}}$ $K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}$
$H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)}$ $K=K_a \times K_b=[H^+][OH^−]$
In this case, the sum of the reactions described by $K_a$ and $K_b$ is the equation for the autoionization of water, and the product of the two equilibrium constants is $K_w$:
$K_aK_b = K_w \label{16.5.10}$
Thus if we know either $K_a$ for an acid or $K_b$ for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair.
Just as with $pH$, $pOH$, and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining $pK_a$ as follows:
$pKa = −\log_{10}K_a \label{16.5.11}$
$K_a=10^{−pK_a} \label{16.5.12}$
and $pK_b$ as
$pK_b = −\log_{10}K_b \label{16.5.13}$
$K_b=10^{−pK_b} \label{16.5.14}$
Similarly, Equation $\ref{16.5.10}$, which expresses the relationship between $K_a$ and $K_b$, can be written in logarithmic form as follows:
$pK_a + pK_b = pK_w \label{16.5.15}$
At 25 °C, this becomes
$pK_a + pK_b = 14.00 \label{16.5.16}$
The values of $pK_a$ and $pK_b$ are given for several common acids and bases in Tables $1$ and $2$, respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid ($HNO_2$), with a $pK_a$ of 3.25, is about a million times stronger acid than hydrocyanic acid (HCN), with a $pK_a$ of 9.21. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases.
The relative strengths of some common acids and their conjugate bases are shown graphically in Figure $1$. The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of $pK_a$. This order corresponds to decreasing strength of the conjugate base or increasing values of $pK_b$. At the bottom left of Figure $2$ are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base.
The conjugate base of a strong acid is a weak base and vice versa.
We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows:
$\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \nonumber$
In an acid–base reaction, the proton always reacts with the stronger base.
For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce $H_3O^+$ and $Cl^−$; only negligible amounts of $HCl$ molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}$
In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of $H_3O^+$ and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows:
$\ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- } \nonumber$
Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left:
$H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)} \nonumber$
All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base.
Example $1$: Butyrate and Dimethylammonium Ions
1. Calculate $K_b$ and $pK_b$ of the butyrate ion ($CH_3CH_2CH_2CO_2^−$). The $pK_a$ of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
2. Calculate $K_a$ and $pK_a$ of the dimethylammonium ion ($(CH_3)_2NH_2^+$). The base ionization constant $K_b$ of dimethylamine ($(CH_3)_2NH$) is $5.4 \times 10^{−4}$ at 25°C.
Given: $pK_a$ and $K_b$
Asked for: corresponding $K_b$ and $pK_b$, $K_a$ and $pK_a$
Strategy:
The constants $K_a$ and $K_b$ are related as shown in Equation $\ref{16.5.10}$. The $pK_a$ and $pK_b$ for an acid and its conjugate base are related as shown in Equations $\ref{16.5.15}$ and $\ref{16.5.16}$. Use the relationships pK = −log K and K = 10−pK (Equations $\ref{16.5.11}$ and $\ref{16.5.13}$) to convert between $K_a$ and $pK_a$ or $K_b$ and $pK_b$.
Solution:
We are given the $pK_a$ for butyric acid and asked to calculate the $K_b$ and the $pK_b$ for its conjugate base, the butyrate ion. Because the $pK_a$ value cited is for a temperature of 25°C, we can use Equation $\ref{16.5.16}$: $pK_a$ + $pK_b$ = pKw = 14.00. Substituting the $pK_a$ and solving for the $pK_b$,
$4.83+pK_b=14.00 \nonumber$
$pK_b=14.00−4.83=9.17 \nonumber$
Because $pK_b = −\log K_b$, $K_b$ is $10^{−9.17} = 6.8 \times 10^{−10}$.
In this case, we are given $K_b$ for a base (dimethylamine) and asked to calculate $K_a$ and $pK_a$ for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is $K_b$ rather than $pK_b$, we can use Equation $\ref{16.5.10}$: $K_aK_b = K_w$. Substituting the values of $K_b$ and $K_w$ at 25°C and solving for $K_a$,
$K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14} \nonumber$
$K_a=1.9 \times 10^{−11} \nonumber$
Because $pK_a$ = −log $K_a$, we have $pK_a = −\log(1.9 \times 10^{−11}) = 10.72$. We could also have converted $K_b$ to $pK_b$ to obtain the same answer:
$pK_b=−\log(5.4 \times 10^{−4})=3.27 \nonumber$
$pKa+pK_b=14.00 \nonumber$
$pK_a=10.73 \nonumber$
$K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11} \nonumber$
If we are given any one of these four quantities for an acid or a base ($K_a$, $pK_a$, $K_b$, or $pK_b$), we can calculate the other three.
Exercise $1$: Lactic Acid
Lactic acid ($CH_3CH(OH)CO_2H$) is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its $pK_a$ is 3.86 at 25°C. Calculate $K_a$ for lactic acid and $pK_b$ and $K_b$ for the lactate ion.
Answer
• $K_a = 1.4 \times 10^{−4}$ for lactic acid;
• $pK_b$ = 10.14 and
• $K_b = 7.2 \times 10^{−11}$ for the lactate ion
A Video Calculating pH in Strong Acid or Strong Base Solutions: Calculating pH in Strong Acid or Strong Base Solutions [youtu.be]
Solutions of Strong Acids and Bases: The Leveling Effect
You will notice in Table $1$ that acids like $H_2SO_4$ and $HNO_3$ lie above the hydronium ion, meaning that they have $pK_a$ values less than zero and are stronger acids than the $H_3O^+$ ion. Recall from Chapter 4 that the acidic proton in virtually all oxoacids is bonded to one of the oxygen atoms of the oxoanion. Thus nitric acid should properly be written as $HONO_2$. Unfortunately, however, the formulas of oxoacids are almost always written with hydrogen on the left and oxygen on the right, giving $HNO_3$ instead. In fact, all six of the common strong acids that we first encountered in Chapter 4 have $pK_a$ values less than zero, which means that they have a greater tendency to lose a proton than does the $H_3O^+$ ion. Conversely, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton-transfer equilibria for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the $H_3O^+$ ion and the conjugate base of the acid.
Although $K_a$ for $HI$ is about 108 greater than $K_a$ for $HNO_3$, the reaction of either $HI$ or $HNO_3$ with water gives an essentially stoichiometric solution of $H_3O^+$ and I− or $NO_3^−$. In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M $H_3O^+$, regardless of the identity of the strong acid. This phenomenon is called the leveling effect: any species that is a stronger acid than the conjugate acid of water ($H_3O^+$) is leveled to the strength of $H_3O^+$ in aqueous solution because $H_3O^+$ is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acid strengths.
One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than $H_2O$. Measurements of the conductivity of 0.1 M solutions of both HI and $HNO_3$ in acetic acid show that HI is completely dissociated, but $HNO_3$ is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than $HNO_3$. The relative order of acid strengths and approximate $K_a$ and $pK_a$ values for the strong acids at the top of Table $1$ were determined using measurements like this and different nonaqueous solvents.
In aqueous solutions, $H_3O^+$ is the strongest acid and $OH^−$ is the strongest base that can exist in equilibrium with $H_2O$.
The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than OH− is leveled to the strength of OH− because OH− is the strongest base that can exist in equilibrium with water. Salts such as $K_2O$, $NaOCH_3$ (sodium methoxide), and $NaNH_2$ (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table $2$, are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of $OH^−$ and the corresponding cation:
$K_2O_{(s)}+H_2O_{(l)} \rightarrow 2OH^−_{(aq)}+2K^+_{(aq)} \label{16.5.18}$
$NaOCH_{3(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+CH_3OH_{(aq)} \label{16.5.19}$
$NaNH_{2(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+NH_{3(aq)} \label{16.5.20}$
Other examples that you may encounter are potassium hydride ($KH$) and organometallic compounds such as methyl lithium ($CH_3Li$).
Polyprotic Acids and Bases
As you learned, polyprotic acids such as $H_2SO_4$, $H_3PO_4$, and $H_2CO_3$ contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus acid strength decreases with the loss of subsequent protons, and, correspondingly, the $pK_a$ increases. Consider $H_2SO_4$, for example:
$HSO^−_{4 (aq)} \ce{ <=>>} SO^{2−}_{4(aq)}+H^+_{(aq)} \;\;\; pK_a=-2 \nonumber$
The equilibrium in the first reaction lies far to the right, consistent with $H_2SO_4$ being a strong acid. In contrast, in the second reaction, appreciable quantities of both $HSO_4^−$ and $SO_4^{2−}$ are present at equilibrium.
For a polyprotic acid, acid strength decreases and the $pK_a$ increases with the sequential loss of each proton.
The hydrogen sulfate ion ($HSO_4^−$) is both the conjugate base of $H_2SO_4$ and the conjugate acid of $SO_4^{2−}$. Just like water, HSO4− can therefore act as either an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion ($SO_4^{2−}$) is a polyprotic base that is capable of accepting two protons in a stepwise manner:
$SO^{2−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} HSO^{−}_{4(aq)}+OH_{(aq)}^- \nonumber$
$HSO^{−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} H_2SO_{4(aq)}+OH_{(aq)}^- \label{16.6}$
Like any other conjugate acid–base pair, the strengths of the conjugate acids and bases are related by $pK_a$ + $pK_b$ = pKw. Consider, for example, the $HSO_4^−/ SO_4^{2−}$ conjugate acid–base pair. From Table $1$, we see that the $pK_a$ of $HSO_4^−$ is 1.99. Hence the $pK_b$ of $SO_4^{2−}$ is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak base, whereas $OH^−$ is a strong base, so the equilibrium shown in Equation $\ref{16.6}$ lies to the left. The $HSO_4^−$ ion is also a very weak base ($pK_a$ of $H_2SO_4$ = 2.0, $pK_b$ of $HSO_4^− = 14 − (−2.0) = 16$), which is consistent with what we expect for the conjugate base of a strong acid.
Example $2$
Predict whether the equilibrium for each reaction lies to the left or the right as written.
• $NH^+_{4(aq)}+PO^{3−}_{4(aq)} \rightleftharpoons NH_{3(aq)}+HPO^{2−}_{4(aq)}$
• $CH_3CH_2CO_2H_{(aq)}+CN^−_{(aq)} \rightleftharpoons CH_3CH_2CO^−_{2(aq)}+HCN_{(aq)}$
Given: balanced chemical equation
Asked for: equilibrium position
Strategy:
Identify the conjugate acid–base pairs in each reaction. Then refer to Tables $1$and$2$ and Figure $2$ to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acid–base pair.
Solution:
The conjugate acid–base pairs are $NH_4^+/NH_3$ and $HPO_4^{2−}/PO_4^{3−}$. According to Tables $1$ and $2$, $NH_4^+$ is a stronger acid ($pK_a = 9.25$) than $HPO_4^{2−}$ (pKa = 12.32), and $PO_4^{3−}$ is a stronger base ($pK_b = 1.68$) than $NH_3$ ($pK_b = 4.75$). The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{NH^+_{4(aq)}} + \underset{\text{stronger base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weaker base}}{NH_{3(aq)}} +\underset{\text{weaker acid}} {HPO^{2-}_{4(aq)}} \nonumber$
The conjugate acid–base pairs are $CH_3CH_2CO_2H/CH_3CH_2CO_2^−$ and $HCN/CN^−$. According to Table $1$, HCN is a weak acid (pKa = 9.21) and $CN^−$ is a moderately weak base (pKb = 4.79). Propionic acid ($CH_3CH_2CO_2H$) is not listed in Table $1$, however. In a situation like this, the best approach is to look for a similar compound whose acid–base properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid ($\ce{−CH_2CH_3}$ versus $\ce{−CH_3}$), so we might expect the two compounds to have similar acid–base properties. In particular, we would expect the $pK_a$ of propionic acid to be similar in magnitude to the $pK_a$ of acetic acid. (In fact, the $pK_a$ of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) Thus propionic acid should be a significantly stronger acid than $HCN$. Because the stronger acid forms the weaker conjugate base, we predict that cyanide will be a stronger base than propionate. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{CH_3CH_2CO_2H_{(aq)}} + \underset{\text{stronger base}}{CN^-_{(aq)}} \ce{<=>>} \underset{\text{weaker base}}{CH_3CH_2CO^-_{2(aq)}} +\underset{\text{weaker acid}} {HCN_{(aq)}} \nonumber$
Exercise $1$
Predict whether the equilibrium for each reaction lies to the left or the right as written.
1. $H_2O_{(l)}+HS^−_{(aq)} \rightleftharpoons OH^−_{(aq)}+H_2S_{(aq)}$
2. $HCO^−_{2(aq)}+HSO^−_{4(aq)} \rightleftharpoons HCO_2H_{(aq)}+SO^{2−}_{4(aq)}$
Answer a
left
Answer b
left
A Video Discussing Polyprotic Acids: Polyprotic Acids [youtu.be]
Summary
Acid–base reactions always contain two conjugate acid–base pairs. Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant ($K_a$). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant ($K_b$). For any conjugate acid–base pair, $K_aK_b = K_w$. Smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases. At 25°C, $pK_a + pK_b = 14.00$. Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. No acid stronger than $H_3O^+$ and no base stronger than $OH^−$ can exist in aqueous solution, leading to the phenomenon known as the leveling effect. Polyprotic acids (and bases) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest base.
Key Equations
• Acid ionization constant: $K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
• Base ionization constant: $K_b= \dfrac{[BH^+][OH^−]}{[B]} \nonumber$
• Relationship between $K_a$ and $K_b$ of a conjugate acid–base pair: $K_aK_b = K_w \nonumber$
• Definition of $pK_a$: $pKa = −\log_{10}K_a \nonumber$ $K_a=10^{−pK_a} \nonumber$
• Definition of $pK_b$: $pK_b = −\log_{10}K_b \nonumber$ $K_b=10^{−pK_b} \nonumber$
• Relationship between $pK_a$ and $pK_b$ of a conjugate acid–base pair: $pK_a + pK_b = pK_w \nonumber$ $pK_a + pK_b = 14.00 \; \text{at 25°C} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.05%3A_Acid_Strength_and_the_Acid_Ionization_Constant_%28Ka%29.txt |
Learning Objectives
• To understand the autoionization reaction of liquid water.
• To know the relationship among pH, pOH, and $pK_w$.
As you learned previously acids and bases can be defined in several different ways (Table $1$). Recall that the Arrhenius definition of an acid is a substance that dissociates in water to produce $H^+$ ions (protons), and an Arrhenius base is a substance that dissociates in water to produce $OH^−$ (hydroxide) ions. According to this view, an acid–base reaction involves the reaction of a proton with a hydroxide ion to form water. Although Brønsted and Lowry defined an acid similarly to Arrhenius by describing an acid as any substance that can donate a proton, the Brønsted–Lowry definition of a base is much more general than the Arrhenius definition. In Brønsted–Lowry terms, a base is any substance that can accept a proton, so a base is not limited to just a hydroxide ion. This means that for every Brønsted–Lowry acid, there exists a corresponding conjugate base with one fewer proton. Consequently, all Brønsted–Lowry acid–base reactions actually involve two conjugate acid–base pairs and the transfer of a proton from one substance (the acid) to another (the base). In contrast, the Lewis definition of acids and bases, focuses on accepting or donating pairs of electrons rather than protons. A Lewis base is an electron-pair donor, and a Lewis acid is an electron-pair acceptor.
Table $1$: Definitions of Acids and Bases
Definition Acids Bases
Arrhenius $H^+$ donor $OH^−$ donor
Brønsted–Lowry $H^+$ donor $H^+$ acceptor
Lewis electron-pair acceptor electron-pair donor
Because this chapter deals with acid–base equilibria in aqueous solution, our discussion will use primarily the Brønsted–Lowry definitions and nomenclature. Remember, however, that all three definitions are just different ways of looking at the same kind of reaction: a proton is an acid, and the hydroxide ion is a base—no matter which definition you use. In practice, chemists tend to use whichever definition is most helpful to make a particular point or understand a given system. If, for example, we refer to a base as having one or more lone pairs of electrons that can accept a proton, we are simply combining the Lewis and Brønsted–Lowry definitions to emphasize the characteristic properties of a base.
Acid–Base Properties of Water
Recall that because of its highly polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, when a strong acid such as HCl dissolves in water, it dissociates into chloride ions ($Cl^−$) and protons ($H^+$). The proton, in turn, reacts with a water molecule to form the hydronium ion ($H_3O^+$):
$\underset{acid}{HCl_{(aq)}} + \underset{base}{H_2O_{(l)}} \rightarrow \underset{acid}{H_3O^+_{(aq)}} + \underset{base}{Cl^-_{(aq)}} \label{16.3.1a}$
In this reaction, $HCl$ is the acid, and water acts as a base by accepting an $H^+$ ion. The reaction in Equation \ref{16.3.1a} is often written in a simpler form by removing $H_2O$ from each side:
$HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \label{16.3.1b}$
In Equation \ref{16.3.1b}, the hydronium ion is represented by $H^+$, although free $H^+$ ions do not exist in liquid water as this reaction demonstrates:
$H^+_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} \nonumber$
Water can also act as an acid, as shown in Equation \ref{16.3.2}. In this equilibrium reaction, $H_2O$ donates a proton to $NH_3$, which acts as a base:
$\underset{acid}{H_2O_{(l)}} + \underset{base}{NH_{3(aq)}} \rightleftharpoons \underset{acid}{NH^+_{4 (aq)}} + \underset{base}{OH^-_{(aq)}} \label{16.3.2}$
Water is thus termed amphiprotic, meaning that it can behave as either an acid or a base, depending on the nature of the other reactant. Notice that Equation $\ref{16.3.2}$ is an equilibrium reaction as indicated by the double arrow and hence has an equilibrium constant associated with it.
The Ion-Product Constant of Liquid Water
Because water is amphiprotic, one water molecule can react with another to form an $OH^−$ ion and an $H_3O^+$ ion in an autoionization process:
$2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+OH^−_{(aq)} \label{16.3.3}$
The equilibrium constant $K$ for this reaction can be written as follows:
$K_{a}=\dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2} \approx \frac{[H_{3}O^{+}][HO^{-}]}{(1)^{2}}=[H_{3}O^{+}][HO^{-}] \label{16.3.4}$
where $a$ is the activity of a species. Because water is the solvent, and the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The activity of each solute is approximated by the molarity of the solute.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Ka = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka = Keq($\textit{a}_{H_2O}$). Because $\textit{a}_{H_2O}$ = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.
In this reaction, one water molecule acts as an acid and one water molecule acts as a base. Thus, this reaction actually can be designated as the $K_a$ of water and as the $K_b$ of water. It is most common, however, to designate this reaction and the associated law of mass action as the $K_w$ of water:
$K_{w}=[H_{3}O^{+}][HO^{-}] \label{16.3.5}$
When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25 °C, the concentrations of the hydronium ion and the hydroxide ion are equal:
$[H_3O^+] = [OH^−] = 1.003 \times 10^{−7}\; M \label{16.3.6}$
Thus the number of dissociated water molecules is very small indeed, approximately 2 ppb.
Substituting the values for $[H_3O^+]$ and $[OH^−]$ at 25 °C into this expression
$K_w=(1.003 \times10^{−7})(1.003 \times 10^{−7})=1.006 \times 10^{−14} \label{16.3.7}$
Thus, to three significant figures, $K_w = 1.01 \times 10^{−14}$ at room temperature, and $K_w = 1.01 \times 10^{-14} = [H_3O^+][OH^-] \label{16.3.7b}$.
Like any other equilibrium constant, $K_w$ varies with temperature, ranging from $1.15 \times 10^{−15}$ at 0 °C to $4.99 \times 10^{−13}$ at 100 °C.
In pure water, the concentrations of the hydronium ion and the hydroxide ion are equal, and the solution is therefore neutral. If $[H_3O^+] > [OH^−]$, however, the solution is acidic, whereas if $[H_3O^+] < [OH^−]$, the solution is basic. For an aqueous solution, the $H_3O^+$ concentration is a quantitative measure of acidity: the higher the $H_3O^+$ concentration, the more acidic the solution. Conversely, the higher the $OH^−$ concentration, the more basic the solution. In most situations that you will encounter, the $H_3O^+$ and $OH^−$ concentrations from the dissociation of water are so small ($1.003 \times 10^{−7} M$) that they can be ignored in calculating the $H_3O^+$ or $OH^−$ concentrations of solutions of acids and bases, but this is not always the case.
A Video Describing the Self-Ionization of Water (Kw): Self-Ionization of Water (Kw): [youtu.be]
The Relationship among pH, pOH, and $pK_w$
The pH scale is a concise way of describing the $H_3O^+$ concentration and hence the acidity or basicity of a solution. Recall that pH and the $H^+$ ($H_3O^+$) concentration are related as follows:
$pH=−\log_{10}[H^+] \label{16.3.8}$
$[H^+]=10^{−pH} \label{16.3.9}$
Because the scale is logarithmic, a pH difference of 1 between two solutions corresponds to a difference of a factor of 10 in their hydronium ion concentrations. Recall also that the pH of a neutral solution is 7.00 ($[H_3O^+] = 1.0 \times 10^{−7}\; M$), whereas acidic solutions have pH < 7.00 (corresponding to $[H_3O^+] > 1.0 \times 10^{−7}$) and basic solutions have pH > 7.00 (corresponding to $[H_3O^+] < 1.0 \times 10^{−7}$).
Similar notation systems are used to describe many other chemical quantities that contain a large negative exponent. For example, chemists use an analogous pOH scale to describe the hydroxide ion concentration of a solution. The pOH and $[OH^−]$ are related as follows:
$pOH=−\log_{10}[OH^−] \label{16.3.10}$
$[OH^−]=10^{−pOH} \label{16.3.11}$
The constant $K_w$ can also be expressed using this notation, where $pK_w = −\log\; K_w$.
Because a neutral solution has $[OH^−] = 1.0 \times 10^{−7}$, the pOH of a neutral solution is 7.00. Consequently, the sum of the pH and the pOH for a neutral solution at 25 °C is 7.00 + 7.00 = 14.00. We can show that the sum of pH and pOH is equal to 14.00 for any aqueous solution at 25 °C by taking the negative logarithm of both sides of Equation \ref{16.3.6b}:
\begin{align} −\log_{10} K_w &= pK_w \[4pt] &=−\log([H_3O^{+}][OH^{−}]) \[4pt] &= (−\log[H_3O^{+}])+(−\log[OH^{−}])\[4pt] &= pH+pOH \label{16.3.12} \end{align}
Thus at any temperature, $pH + pOH = pK_w$, so at 25 °C, where $K_w = 1.0 \times 10^{−14}$, pH + pOH = 14.00. More generally, the pH of any neutral solution is half of the $pK_w$ at that temperature. The relationship among pH, pOH, and the acidity or basicity of a solution is summarized graphically in Figure $1$ over the common pH range of 0 to 14. Notice the inverse relationship between the pH and pOH scales.
For any neutral solution, pH + pOH = 14.00 (at 25 °C) with pH=pOH=7.
A Video Introduction to pH: Introduction to pH [youtu.be]
Example $1$
The Kw for water at 100 °C is $4.99 \times 10^{−13}$. Calculate $pK_w$ for water at this temperature and the pH and the pOH for a neutral aqueous solution at 100 °C. Report pH and pOH values to two decimal places.
Given: $K_w$
Asked for: $pK_w$, $pH$, and $pOH$
Strategy:
1. Calculate $pK_w$ by taking the negative logarithm of $K_w$.
2. For a neutral aqueous solution, $[H_3O^+] = [OH^−]$. Use this relationship and Equation \ref{16.3.7b} to calculate $[H_3O^+]$ and $[OH^−]$. Then determine the pH and the pOH for the solution.
Solution:
A
Because $pK_w$ is the negative logarithm of Kw, we can write
$pK_w = −\log K_w = −\log(4.99 \times 10^{−13}) = 12.302 \nonumber$
The answer is reasonable: $K_w$ is between $10^{−13}$ and $10^{−12}$, so $pK_w$ must be between 12 and 13.
B
Equation \ref{16.3.6b} shows that $K_w = [H_3O^+][OH^−]$. Because $[H_3O^+] = [OH^−]$ in a neutral solution, we can let $x = [H_3O^+] = [OH^−]$:
\begin{align*} K_w &=[H_3O^+][OH^−] \[4pt] &=(x)(x)=x^2 \[4pt] x&=\sqrt{K_w} \[4pt] &=\sqrt{4.99 \times 10^{−13}} \[4pt] &=7.06 \times 10^{−7}\; M \end{align*} \nonumber
Because $x$ is equal to both $[H_3O^+]$ and $[OH^−]$,
\begin{align*} pH &= pOH = −\log(7.06 \times 10^{−7}) \[4pt] &= 6.15 \,\,\, \text{(to two decimal places)} \end{align*} \nonumber
We could obtain the same answer more easily (without using logarithms) by using the $pK_w$. In this case, we know that $pK_w = 12.302$, and from Equation \ref{16.3.12}, we know that $pK_w = pH + pOH$. Because $pH = pOH$ in a neutral solution, we can use Equation \ref{16.3.12} directly, setting $pH = pOH = y$. Solving to two decimal places we obtain the following:
\begin{align*} pK_w &= pH + pOH \[4pt] &= y + y \[4pt] &= 2y \[4pt] y &=\dfrac{pK_w}{2} \[4pt] &=\dfrac{12.302}{2} \[4pt] &=6.15 =pH=pOH \end{align*} \nonumber
Exercise $1$
Humans maintain an internal temperature of about 37 °C. At this temperature, $K_w = 3.55 \times 10^{−14}$. Calculate $pK_w$ and the pH and the pOH of a neutral solution at 37 °C. Report pH and pOH values to two decimal places.
Answer
• $pK_w = 13.45$
• $pH = pOH = 6.73$
Summary
Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion ($H_3O^+$). The autoionization of liquid water produces $OH^−$ and $H_3O^+$ ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (Kw) and is defined as $K_w = [H_3O^+][OH^−]$. At 25 °C, $K_w$ is $1.01 \times 10^{−14}$; hence $pH + pOH = pK_w = 14.00$.
• For any neutral solution, $pH + pOH = 14.00$ (at 25 °C) and $pH = 1/2 pK_w$.
• Ion-product constant of liquid water: $K_w = [H_3O^+][OH^−] \nonumber$
• Definition of $pH$: $pH = −\log10[H^+] \nonumber$ or $[H^+] = 10^{−pH} \nonumber$
• Definition of $pOH$: $pOH = −\log_{10}[OH^+] \nonumber$ or $[OH^−] = 10^{−pOH} \nonumber$
• Relationship among $pH$, $pOH$, and $pK_w$: $pK_w= pH + pOH \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.06%3A_Autoionization_of_Water_and_pH.txt |
Learning Objectives
• To know the relationship between acid or base strength and the magnitude of $K_a$, $K_b$, $pK_a$, and $pK_b$.
• To understand the leveling effect.
The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate base, is as follows:
$HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}$
The equilibrium constant for this dissociation is as follows:
$K=\dfrac{[H_3O^+][A^−]}{[H_2O][HA]} \label{16.5.2}$
As we noted earlier, because water is the solvent, it has an activity equal to 1, so the $[H_2O]$ term in Equation $\ref{16.5.2}$ is actually the $\textit{a}_{H_2O}$, which is equal to 1.
Again, for simplicity, $H_3O^+$ can be written as $H^+$ in Equation $\ref{16.5.3}$.
$HA_{(aq)} \rightleftharpoons H^+_{(aq)}+A^−_{(aq)} \label{16.5.3}$
Keep in mind, though, that free $H^+$ does not exist in aqueous solutions and that a proton is transferred to $H_2O$ in all acid ionization reactions to form hydronium ions, $H_3O^+$. The larger the $K_a$, the stronger the acid and the higher the $H^+$ concentration at equilibrium. Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of $H^+$ or $OH^−$, thus making them unitless. The values of $K_a$ for a number of common acids are given in Table $1$.
Table $1$: Values of $K_a$, $pK_a$, $K_b$, and $pK_b$ for Selected Acids ($HA$ and Their Conjugate Bases ($A^−$)
Acid $HA$ $K_a$ $pK_a$ $A^−$ $K_b$ $pK_b$
*The number in parentheses indicates the ionization step referred to for a polyprotic acid.
hydroiodic acid $HI$ $2 \times 10^{9}$ −9.3 $I^−$ $5.5 \times 10^{−24}$ 23.26
sulfuric acid (1)* $H_2SO_4$ $1 \times 10^{2}$ −2.0 $HSO_4^−$ $1 \times 10^{−16}$ 16.0
nitric acid $HNO_3$ $2.3 \times 10^{1}$ −1.37 $NO_3^−$ $4.3 \times 10^{−16}$ 15.37
hydronium ion $H_3O^+$ $1.0$ 0.00 $H_2O$ $1.0 \times 10^{−14}$ 14.00
sulfuric acid (2)* $HSO_4^−$ $1.0 \times 10^{−2}$ 1.99 $SO_4^{2−}$ $9.8 \times 10^{−13}$ 12.01
hydrofluoric acid $HF$ $6.3 \times 10^{−4}$ 3.20 $F^−$ $1.6 \times 10^{−11}$ 10.80
nitrous acid $HNO_2$ $5.6 \times 10^{−4}$ 3.25 $NO2^−$ $1.8 \times 10^{−11}$ 10.75
formic acid $HCO_2H$ $1.78 \times 10^{−4}$ 3.750 $HCO_2−$ $5.6 \times 10^{−11}$ 10.25
benzoic acid $C_6H_5CO_2H$ $6.3 \times 10^{−5}$ 4.20 $C_6H_5CO_2^−$ $1.6 \times 10^{−10}$ 9.80
acetic acid $CH_3CO_2H$ $1.7 \times 10^{−5}$ 4.76 $CH_3CO_2^−$ $5.8 \times 10^{−10}$ 9.24
pyridinium ion $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23 $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77
hypochlorous acid $HOCl$ $4.0 \times 10^{−8}$ 7.40 $OCl^−$ $2.5 \times 10^{−7}$ 6.60
hydrocyanic acid $HCN$ $6.2 \times 10^{−10}$ 9.21 $CN^−$ $1.6 \times 10^{−5}$ 4.79
ammonium ion $NH_4^+$ $5.6 \times 10^{−10}$ 9.25 $NH_3$ $1.8 \times 10^{−5}$ 4.75
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $OH^−$ $1.00$ 0.00
acetylene $C_2H_2$ $1 \times 10^{−26}$ 26.0 $HC_2^−$ $1 \times 10^{12}$ −12.0
ammonia $NH_3$ $1 \times 10^{−35}$ 35.0 $NH_2^−$ $1 \times 10^{21}$ −21.0
Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid:
$B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}$
The equilibrium constant for this reaction is the base ionization constant (Kb), also called the base dissociation constant:
$K_b= \frac{[BH^+][OH^−]}{[B]} \label{16.5.5}$
Once again, the activity of water has a value of 1, so water does not appear in the equilibrium constant expression. The larger the $K_b$, the stronger the base and the higher the $OH^−$ concentration at equilibrium. The values of $K_b$ for a number of common weak bases are given in Table $2$.
Table $2$: Values of $K_b$, $pK_b$, $K_a$, and $pK_a$ for Selected Weak Bases (B) and Their Conjugate Acids (BH+)
Base $B$ $K_b$ $pK_b$ $BH^+$ $K_a$ $pK_a$
*As in Table $1$.
hydroxide ion $OH^−$ $1.0$ 0.00* $H_2O$ $1.0 \times 10^{−14}$ 14.00
phosphate ion $PO_4^{3−}$ $2.1 \times 10^{−2}$ 1.68 $HPO_4^{2−}$ $4.8 \times 10^{−13}$ 12.32
dimethylamine $(CH_3)_2NH$ $5.4 \times 10^{−4}$ 3.27 $(CH_3)_2NH_2^+$ $1.9 \times 10^{−11}$ 10.73
methylamine $CH_3NH_2$ $4.6 \times 10^{−4}$ 3.34 $CH_3NH_3^+$ $2.2 \times 10^{−11}$ 10.66
trimethylamine $(CH_3)_3N$ $6.3 \times 10^{−5}$ 4.20 $(CH_3)_3NH^+$ $1.6 \times 10^{−10}$ 9.80
ammonia $NH_3$ $1.8 \times 10^{−5}$ 4.75 $NH_4^+$ $5.6 \times 10^{−10}$ 9.25
pyridine $C_5H_5N$ $1.7 \times 10^{−9}$ 8.77 $C_5H_5NH^+$ $5.9 \times 10^{−6}$ 5.23
aniline $C_6H_5NH_2$ $7.4 \times 10^{−10}$ 9.13 $C_6H_5NH_3^+$ $1.3 \times 10^{−5}$ 4.87
water $H_2O$ $1.0 \times 10^{−14}$ 14.00 $H_3O^+$ $1.0^*$ 0.00
There is a simple relationship between the magnitude of $K_a$ for an acid and $K_b$ for its conjugate base. Consider, for example, the ionization of hydrocyanic acid ($HCN$) in water to produce an acidic solution, and the reaction of $CN^−$ with water to produce a basic solution:
$HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \label{16.5.6}$
$CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \label{16.5.7}$
The equilibrium constant expression for the ionization of HCN is as follows:
$K_a=\dfrac{[H^+][CN^−]}{[HCN]} \label{16.5.8}$
The corresponding expression for the reaction of cyanide with water is as follows:
$K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \label{16.5.9}$
If we add Equations $\ref{16.5.6}$ and $\ref{16.5.7}$, we obtain the following:
add Equations $\ref{16.5.6}$ and $\ref{16.5.7}$, we obtain
Reaction Equilibrium Constants
$\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}}$ $K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}$
$\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}}$ $K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}$
$H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)}$ $K=K_a \times K_b=[H^+][OH^−]$
In this case, the sum of the reactions described by $K_a$ and $K_b$ is the equation for the autoionization of water, and the product of the two equilibrium constants is $K_w$:
$K_aK_b = K_w \label{16.5.10}$
Thus if we know either $K_a$ for an acid or $K_b$ for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair.
Just as with $pH$, $pOH$, and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining $pK_a$ as follows:
$pKa = −\log_{10}K_a \label{16.5.11}$
$K_a=10^{−pK_a} \label{16.5.12}$
and $pK_b$ as
$pK_b = −\log_{10}K_b \label{16.5.13}$
$K_b=10^{−pK_b} \label{16.5.14}$
Similarly, Equation $\ref{16.5.10}$, which expresses the relationship between $K_a$ and $K_b$, can be written in logarithmic form as follows:
$pK_a + pK_b = pK_w \label{16.5.15}$
At 25 °C, this becomes
$pK_a + pK_b = 14.00 \label{16.5.16}$
The values of $pK_a$ and $pK_b$ are given for several common acids and bases in Tables $1$ and $2$, respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid ($HNO_2$), with a $pK_a$ of 3.25, is about a million times stronger acid than hydrocyanic acid (HCN), with a $pK_a$ of 9.21. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases.
The relative strengths of some common acids and their conjugate bases are shown graphically in Figure $1$. The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of $pK_a$. This order corresponds to decreasing strength of the conjugate base or increasing values of $pK_b$. At the bottom left of Figure $2$ are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base.
The conjugate base of a strong acid is a weak base and vice versa.
We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows:
$\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \nonumber$
In an acid–base reaction, the proton always reacts with the stronger base.
For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce $H_3O^+$ and $Cl^−$; only negligible amounts of $HCl$ molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow:
$HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}$
In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of $H_3O^+$ and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows:
$\ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- } \nonumber$
Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left:
$H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)} \nonumber$
All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base.
Example $1$: Butyrate and Dimethylammonium Ions
1. Calculate $K_b$ and $pK_b$ of the butyrate ion ($CH_3CH_2CH_2CO_2^−$). The $pK_a$ of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter.
2. Calculate $K_a$ and $pK_a$ of the dimethylammonium ion ($(CH_3)_2NH_2^+$). The base ionization constant $K_b$ of dimethylamine ($(CH_3)_2NH$) is $5.4 \times 10^{−4}$ at 25°C.
Given: $pK_a$ and $K_b$
Asked for: corresponding $K_b$ and $pK_b$, $K_a$ and $pK_a$
Strategy:
The constants $K_a$ and $K_b$ are related as shown in Equation $\ref{16.5.10}$. The $pK_a$ and $pK_b$ for an acid and its conjugate base are related as shown in Equations $\ref{16.5.15}$ and $\ref{16.5.16}$. Use the relationships pK = −log K and K = 10−pK (Equations $\ref{16.5.11}$ and $\ref{16.5.13}$) to convert between $K_a$ and $pK_a$ or $K_b$ and $pK_b$.
Solution:
We are given the $pK_a$ for butyric acid and asked to calculate the $K_b$ and the $pK_b$ for its conjugate base, the butyrate ion. Because the $pK_a$ value cited is for a temperature of 25°C, we can use Equation $\ref{16.5.16}$: $pK_a$ + $pK_b$ = pKw = 14.00. Substituting the $pK_a$ and solving for the $pK_b$,
$4.83+pK_b=14.00 \nonumber$
$pK_b=14.00−4.83=9.17 \nonumber$
Because $pK_b = −\log K_b$, $K_b$ is $10^{−9.17} = 6.8 \times 10^{−10}$.
In this case, we are given $K_b$ for a base (dimethylamine) and asked to calculate $K_a$ and $pK_a$ for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is $K_b$ rather than $pK_b$, we can use Equation $\ref{16.5.10}$: $K_aK_b = K_w$. Substituting the values of $K_b$ and $K_w$ at 25°C and solving for $K_a$,
$K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14} \nonumber$
$K_a=1.9 \times 10^{−11} \nonumber$
Because $pK_a$ = −log $K_a$, we have $pK_a = −\log(1.9 \times 10^{−11}) = 10.72$. We could also have converted $K_b$ to $pK_b$ to obtain the same answer:
$pK_b=−\log(5.4 \times 10^{−4})=3.27 \nonumber$
$pKa+pK_b=14.00 \nonumber$
$pK_a=10.73 \nonumber$
$K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11} \nonumber$
If we are given any one of these four quantities for an acid or a base ($K_a$, $pK_a$, $K_b$, or $pK_b$), we can calculate the other three.
Exercise $1$: Lactic Acid
Lactic acid ($CH_3CH(OH)CO_2H$) is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its $pK_a$ is 3.86 at 25°C. Calculate $K_a$ for lactic acid and $pK_b$ and $K_b$ for the lactate ion.
Answer
• $K_a = 1.4 \times 10^{−4}$ for lactic acid;
• $pK_b$ = 10.14 and
• $K_b = 7.2 \times 10^{−11}$ for the lactate ion
A Video Calculating pH in Strong Acid or Strong Base Solutions: Calculating pH in Strong Acid or Strong Base Solutions [youtu.be]
Solutions of Strong Acids and Bases: The Leveling Effect
You will notice in Table $1$ that acids like $H_2SO_4$ and $HNO_3$ lie above the hydronium ion, meaning that they have $pK_a$ values less than zero and are stronger acids than the $H_3O^+$ ion. Recall from Chapter 4 that the acidic proton in virtually all oxoacids is bonded to one of the oxygen atoms of the oxoanion. Thus nitric acid should properly be written as $HONO_2$. Unfortunately, however, the formulas of oxoacids are almost always written with hydrogen on the left and oxygen on the right, giving $HNO_3$ instead. In fact, all six of the common strong acids that we first encountered in Chapter 4 have $pK_a$ values less than zero, which means that they have a greater tendency to lose a proton than does the $H_3O^+$ ion. Conversely, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton-transfer equilibria for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the $H_3O^+$ ion and the conjugate base of the acid.
Although $K_a$ for $HI$ is about 108 greater than $K_a$ for $HNO_3$, the reaction of either $HI$ or $HNO_3$ with water gives an essentially stoichiometric solution of $H_3O^+$ and I− or $NO_3^−$. In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M $H_3O^+$, regardless of the identity of the strong acid. This phenomenon is called the leveling effect: any species that is a stronger acid than the conjugate acid of water ($H_3O^+$) is leveled to the strength of $H_3O^+$ in aqueous solution because $H_3O^+$ is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acid strengths.
One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than $H_2O$. Measurements of the conductivity of 0.1 M solutions of both HI and $HNO_3$ in acetic acid show that HI is completely dissociated, but $HNO_3$ is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than $HNO_3$. The relative order of acid strengths and approximate $K_a$ and $pK_a$ values for the strong acids at the top of Table $1$ were determined using measurements like this and different nonaqueous solvents.
In aqueous solutions, $H_3O^+$ is the strongest acid and $OH^−$ is the strongest base that can exist in equilibrium with $H_2O$.
The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than OH− is leveled to the strength of OH− because OH− is the strongest base that can exist in equilibrium with water. Salts such as $K_2O$, $NaOCH_3$ (sodium methoxide), and $NaNH_2$ (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table $2$, are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of $OH^−$ and the corresponding cation:
$K_2O_{(s)}+H_2O_{(l)} \rightarrow 2OH^−_{(aq)}+2K^+_{(aq)} \label{16.5.18}$
$NaOCH_{3(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+CH_3OH_{(aq)} \label{16.5.19}$
$NaNH_{2(s)}+H_2O_{(l)} \rightarrow OH^−_{(aq)}+Na^+_{(aq)}+NH_{3(aq)} \label{16.5.20}$
Other examples that you may encounter are potassium hydride ($KH$) and organometallic compounds such as methyl lithium ($CH_3Li$).
Polyprotic Acids and Bases
As you learned, polyprotic acids such as $H_2SO_4$, $H_3PO_4$, and $H_2CO_3$ contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus acid strength decreases with the loss of subsequent protons, and, correspondingly, the $pK_a$ increases. Consider $H_2SO_4$, for example:
$HSO^−_{4 (aq)} \ce{ <=>>} SO^{2−}_{4(aq)}+H^+_{(aq)} \;\;\; pK_a=-2 \nonumber$
The equilibrium in the first reaction lies far to the right, consistent with $H_2SO_4$ being a strong acid. In contrast, in the second reaction, appreciable quantities of both $HSO_4^−$ and $SO_4^{2−}$ are present at equilibrium.
For a polyprotic acid, acid strength decreases and the $pK_a$ increases with the sequential loss of each proton.
The hydrogen sulfate ion ($HSO_4^−$) is both the conjugate base of $H_2SO_4$ and the conjugate acid of $SO_4^{2−}$. Just like water, HSO4− can therefore act as either an acid or a base, depending on whether the other reactant is a stronger acid or a stronger base. Conversely, the sulfate ion ($SO_4^{2−}$) is a polyprotic base that is capable of accepting two protons in a stepwise manner:
$SO^{2−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} HSO^{−}_{4(aq)}+OH_{(aq)}^- \nonumber$
$HSO^{−}_{4 (aq)} + H_2O_{(aq)} \ce{ <=>>} H_2SO_{4(aq)}+OH_{(aq)}^- \label{16.6}$
Like any other conjugate acid–base pair, the strengths of the conjugate acids and bases are related by $pK_a$ + $pK_b$ = pKw. Consider, for example, the $HSO_4^−/ SO_4^{2−}$ conjugate acid–base pair. From Table $1$, we see that the $pK_a$ of $HSO_4^−$ is 1.99. Hence the $pK_b$ of $SO_4^{2−}$ is 14.00 − 1.99 = 12.01. Thus sulfate is a rather weak base, whereas $OH^−$ is a strong base, so the equilibrium shown in Equation $\ref{16.6}$ lies to the left. The $HSO_4^−$ ion is also a very weak base ($pK_a$ of $H_2SO_4$ = 2.0, $pK_b$ of $HSO_4^− = 14 − (−2.0) = 16$), which is consistent with what we expect for the conjugate base of a strong acid.
Example $2$
Predict whether the equilibrium for each reaction lies to the left or the right as written.
• $NH^+_{4(aq)}+PO^{3−}_{4(aq)} \rightleftharpoons NH_{3(aq)}+HPO^{2−}_{4(aq)}$
• $CH_3CH_2CO_2H_{(aq)}+CN^−_{(aq)} \rightleftharpoons CH_3CH_2CO^−_{2(aq)}+HCN_{(aq)}$
Given: balanced chemical equation
Asked for: equilibrium position
Strategy:
Identify the conjugate acid–base pairs in each reaction. Then refer to Tables $1$and$2$ and Figure $2$ to determine which is the stronger acid and base. Equilibrium always favors the formation of the weaker acid–base pair.
Solution:
The conjugate acid–base pairs are $NH_4^+/NH_3$ and $HPO_4^{2−}/PO_4^{3−}$. According to Tables $1$ and $2$, $NH_4^+$ is a stronger acid ($pK_a = 9.25$) than $HPO_4^{2−}$ (pKa = 12.32), and $PO_4^{3−}$ is a stronger base ($pK_b = 1.68$) than $NH_3$ ($pK_b = 4.75$). The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{NH^+_{4(aq)}} + \underset{\text{stronger base}}{PO^{3-}_{4(aq)}} \ce{<=>>} \underset{\text{weaker base}}{NH_{3(aq)}} +\underset{\text{weaker acid}} {HPO^{2-}_{4(aq)}} \nonumber$
The conjugate acid–base pairs are $CH_3CH_2CO_2H/CH_3CH_2CO_2^−$ and $HCN/CN^−$. According to Table $1$, HCN is a weak acid (pKa = 9.21) and $CN^−$ is a moderately weak base (pKb = 4.79). Propionic acid ($CH_3CH_2CO_2H$) is not listed in Table $1$, however. In a situation like this, the best approach is to look for a similar compound whose acid–base properties are listed. For example, propionic acid and acetic acid are identical except for the groups attached to the carbon atom of the carboxylic acid ($\ce{−CH_2CH_3}$ versus $\ce{−CH_3}$), so we might expect the two compounds to have similar acid–base properties. In particular, we would expect the $pK_a$ of propionic acid to be similar in magnitude to the $pK_a$ of acetic acid. (In fact, the $pK_a$ of propionic acid is 4.87, compared to 4.76 for acetic acid, which makes propionic acid a slightly weaker acid than acetic acid.) Thus propionic acid should be a significantly stronger acid than $HCN$. Because the stronger acid forms the weaker conjugate base, we predict that cyanide will be a stronger base than propionate. The equilibrium will therefore lie to the right, favoring the formation of the weaker acid–base pair:
$\underset{\text{stronger acid}}{CH_3CH_2CO_2H_{(aq)}} + \underset{\text{stronger base}}{CN^-_{(aq)}} \ce{<=>>} \underset{\text{weaker base}}{CH_3CH_2CO^-_{2(aq)}} +\underset{\text{weaker acid}} {HCN_{(aq)}} \nonumber$
Exercise $1$
Predict whether the equilibrium for each reaction lies to the left or the right as written.
1. $H_2O_{(l)}+HS^−_{(aq)} \rightleftharpoons OH^−_{(aq)}+H_2S_{(aq)}$
2. $HCO^−_{2(aq)}+HSO^−_{4(aq)} \rightleftharpoons HCO_2H_{(aq)}+SO^{2−}_{4(aq)}$
Answer a
left
Answer b
left
A Video Discussing Polyprotic Acids: Polyprotic Acids [youtu.be]
Summary
Acid–base reactions always contain two conjugate acid–base pairs. Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant ($K_a$). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant ($K_b$). For any conjugate acid–base pair, $K_aK_b = K_w$. Smaller values of $pK_a$ correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of $pK_b$ correspond to larger base ionization constants and hence stronger bases. At 25°C, $pK_a + pK_b = 14.00$. Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. No acid stronger than $H_3O^+$ and no base stronger than $OH^−$ can exist in aqueous solution, leading to the phenomenon known as the leveling effect. Polyprotic acids (and bases) lose (and gain) protons in a stepwise manner, with the fully protonated species being the strongest acid and the fully deprotonated species the strongest base.
Key Equations
• Acid ionization constant: $K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
• Base ionization constant: $K_b= \dfrac{[BH^+][OH^−]}{[B]} \nonumber$
• Relationship between $K_a$ and $K_b$ of a conjugate acid–base pair: $K_aK_b = K_w \nonumber$
• Definition of $pK_a$: $pKa = −\log_{10}K_a \nonumber$ $K_a=10^{−pK_a} \nonumber$
• Definition of $pK_b$: $pK_b = −\log_{10}K_b \nonumber$ $K_b=10^{−pK_b} \nonumber$
• Relationship between $pK_a$ and $pK_b$ of a conjugate acid–base pair: $pK_a + pK_b = pK_w \nonumber$ $pK_a + pK_b = 14.00 \; \text{at 25°C} \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.07%3A_Finding_the_H3O_and_pH_of_Strong_and_Weak_Acid_Solutions.txt |
Learning Objectives
• To recognize salts that will produce acidic, basic, or neutral solutions in water
• To understand the Lewis acidity of small, highly-charged metal ions in water
A neutralization reaction can be defined as the reaction of an acid and a base to produce a salt and water. That is, another cation, such as $Na^+$, replaces the proton on the acid. An example is the reaction of $CH_3CO_2H$, a weak acid, with $NaOH$, a strong base:
$\underset{acid}{CH_3CO_2H_{(l)}} +\underset{base}{NaOH_{(s)}} \overset{H_2O}{\longrightarrow} \underset{salt}{H_2OCH_3CO_2Na_{(aq)} }+\underset{water}{H_2O_{(l)}} \nonumber$
Depending on the acid–base properties of its component ions, however, a salt can dissolve in water to produce a neutral solution, a basic solution, or an acidic solution.
When a salt such as $NaCl$ dissolves in water, it produces $Na^+_{(aq)}$ and $Cl^−_{(aq)}$ ions. Using a Lewis approach, the $Na^+$ ion can be viewed as an acid because it is an electron pair acceptor, although its low charge and relatively large radius make it a very weak acid. The $Cl^−$ ion is the conjugate base of the strong acid $HCl$, so it has essentially no basic character. Consequently, dissolving $NaCl$ in water has no effect on the $pH$ of a solution, and the solution remains neutral.
Now let's compare this behavior to the behavior of aqueous solutions of potassium cyanide and sodium acetate. Again, the cations ($K^+$ and $Na^+$) have essentially no acidic character, but the anions ($CN^−$ and $CH_3CO_2^−$) are weak bases that can react with water because they are the conjugate bases of the weak acids $HCN$ and acetic acid, respectively.
$CN^-_{(aq)} + H_2O_{(l)} \ce{ <<=>} HCN_{(aq)} + OH^-_{(aq)} \nonumber$
$CH_3CO^2_{2(aq)} + H_2O_{(l)} \ce{<<=>} CH_3CO_2H_{(aq)} + OH^-_{(aq)} \nonumber$
Neither reaction proceeds very far to the right as written because the formation of the weaker acid–base pair is favored. Both $HCN$ and acetic acid are stronger acids than water, and hydroxide is a stronger base than either acetate or cyanide, so in both cases, the equilibrium lies to the left. Nonetheless, each of these reactions generates enough hydroxide ions to produce a basic solution. For example, the $pH$ of a 0.1 M solution of sodium acetate or potassium cyanide at 25°C is 8.8 or 11.1, respectively. From Table $1$ and Figure $1$, we can see that $CN^−$ is a stronger base ($pK_b = 4.79$) than acetate ($pK_b = 9.24$), which is consistent with $KCN$ producing a more basic solution than sodium acetate at the same concentration.
In contrast, the conjugate acid of a weak base should be a weak acid (Equation $\ref{16.2}$). For example, ammonium chloride and pyridinium chloride are salts produced by reacting ammonia and pyridine, respectively, with $HCl$. As you already know, the chloride ion is such a weak base that it does not react with water. In contrast, the cations of the two salts are weak acids that react with water as follows:
$NH^+_{4(aq)} + H_2O_{(l)} \ce{ <<=>} HH_{3(aq)} + H_3O^+_{(aq)} \label{16.2}$
$C_5H_5NH^+_{(aq)} + H_2O_{(l)} \ce{<<=>} C_5H_5NH_{(aq)} + H_3O^+_{(aq)} \label{16.3}$
Equation $\ref{16.2}$ indicates that $H_3O^+$ is a stronger acid than either $NH_4^+$ or $C_5H_5NH^+$, and conversely, ammonia and pyridine are both stronger bases than water. The equilibrium will therefore lie far to the left in both cases, favoring the weaker acid–base pair. The $H_3O^+$ concentration produced by the reactions is great enough, however, to decrease the $pH$ of the solution significantly: the $pH$ of a 0.10 M solution of ammonium chloride or pyridinium chloride at 25°C is 5.13 or 3.12, respectively. This is consistent with the information shown in Figure 16.2, indicating that the pyridinium ion is more acidic than the ammonium ion.
What happens with aqueous solutions of a salt such as ammonium acetate, where both the cation and the anion can react separately with water to produce an acid and a base, respectively? According to Figure 16.10, the ammonium ion will lower the $pH$, while according to Equation $\ref{16.3}$, the acetate ion will raise the $pH$. This particular case is unusual, in that the cation is as strong an acid as the anion is a base (pKa ≈ pKb). Consequently, the two effects cancel, and the solution remains neutral. With salts in which the cation is a stronger acid than the anion is a base, the final solution has a $pH$ < 7.00. Conversely, if the cation is a weaker acid than the anion is a base, the final solution has a $pH$ > 7.00.
Solutions of simple salts of metal ions can also be acidic, even though a metal ion cannot donate a proton directly to water to produce $H_3O^+$. Instead, a metal ion can act as a Lewis acid and interact with water, a Lewis base, by coordinating to a lone pair of electrons on the oxygen atom to form a hydrated metal ion (part (a) in Figure $1$). A water molecule coordinated to a metal ion is more acidic than a free water molecule for two reasons. First, repulsive electrostatic interactions between the positively charged metal ion and the partially positively charged hydrogen atoms of the coordinated water molecule make it easier for the coordinated water to lose a proton.
The magnitude of this effect depends on the following two factors (Figure $2$):
1. The charge on the metal ion. A divalent ion ($M^{2+}$) has approximately twice as strong an effect on the electron density in a coordinated water molecule as a monovalent ion ($M^+$) of the same radius.
2. The radius of the metal ion. For metal ions with the same charge, the smaller the ion, the shorter the internuclear distance to the oxygen atom of the water molecule and the greater the effect of the metal on the electron density distribution in the water molecule.
Thus aqueous solutions of small, highly charged metal ions, such as $Al^{3+}$ and $Fe^{3+}$, are acidic:
$[Al(H_2O)_6]^{3+}_{(aq)} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+}_{(aq)}+H^+_{(aq)} \label{16.36}$
The $[Al(H_2O)_6]^{3+}$ ion has a $pK_a$ of 5.0, making it almost as strong an acid as acetic acid. Because of the two factors described previously, the most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. A number of pairs of metal ions that lie on a diagonal line in the periodic table, such as $Li^+$ and $Mg^{2+}$ or $Ca^{2+}$ and $Y^{3+}$, have different sizes and charges, but similar charge-to-radius ratios. As a result, these pairs of metal ions have similar effects on the acidity of coordinated water molecules, and they often exhibit other significant similarities in chemistry as well.
Solutions of small, highly charged metal ions in water are acidic.
Reactions such as those discussed in this section, in which a salt reacts with water to give an acidic or basic solution, are often called hydrolysis reactions. Using a separate name for this type of reaction is unfortunate because it suggests that they are somehow different. In fact, hydrolysis reactions are just acid–base reactions in which the acid is a cation or the base is an anion; they obey the same principles and rules as all other acid–base reactions.
A hydrolysis reaction is an acid–base reaction.
Example $1$
Predict whether aqueous solutions of these compounds are acidic, basic, or neutral.
1. $\ce{KNO_3}$
2. $\ce{CrBr_3} cdot \ce{H_2O}$
3. $\ce{Na_2SO_4}$
Given: compound
Asked for: acidity or basicity of aqueous solution
Strategy:
1. Assess the acid–base properties of the cation and the anion. If the cation is a weak Lewis acid, it will not affect the $pH$ of the solution. If the cation is the conjugate acid of a weak base or a relatively highly charged metal cation, however, it will react with water to produce an acidic solution.
2. f the anion is the conjugate base of a strong acid, it will not affect the $pH$ of the solution. If, however, the anion is the conjugate base of a weak acid, the solution will be basic.
Solution:
a
1. The $K^+$ cation has a small positive charge (+1) and a relatively large radius (because it is in the fourth row of the periodic table), so it is a very weak Lewis acid.
2. The $NO_3−$ anion is the conjugate base of a strong acid, so it has essentially no basic character (Table 16.1). Hence neither the cation nor the anion will react with water to produce $H^+$ or $OH^−$, and the solution will be neutral.
b.
1. The $Cr^{3+}$ ion is a relatively highly charged metal cation that should behave similarly to the $Al^{3+}$ ion and form the $[Cr(H2O)_6]^{3+}$ complex, which will behave as a weak acid: $Cr(H_2O)_6]^{3+}_{(aq)} \ce{ <=>>} Cr(H_2O)_5(OH)]^{2+}_{(aq)} + H^+_{(aq)}\nonumber$
2. The $Br^−$ anion is a very weak base (it is the conjugate base of the strong acid $HBr$), so it does not affect the $pH$ of the solution. Hence the solution will be acidic.
c.
1. The $Na^+$ ion, like the $K^+$, is a very weak acid, so it should not affect the acidity of the solution.
2. In contrast, $SO_4^{2−}$ is the conjugate base of $HSO_4^−$, which is a weak acid. Hence the $SO_4^{2−}$ ion will react with water as shown in Figure 16.6 to give a slightly basic solution.
Exercise $1$
Predict whether aqueous solutions of the following are acidic, basic, or neutral.
1. $KI$
2. $Mg(ClO_4)_2$
3. $NaHS$
Answer a
neutral
Answer b
acidic
Answer c
basic (due to the reaction of $\ce{HS^{-}}$ with water to form $\ce{H_2S}$ and $\ce{OH^{-}}$)
Summary
A salt can dissolve in water to produce a neutral, a basic, or an acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion ($A^−$), the conjugate acid of a weak base as the cation ($BH^+$), or both. Salts that contain small, highly charged metal ions produce acidic solutions in water. The reaction of a salt with water to produce an acidic or a basic solution is called a hydrolysis reaction.
Key Takeaways
• Acid–base reactions always contain two conjugate acid–base pairs.
• Each acid and each base has an associated ionization constant that corresponds to its acid or base strength. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.09%3A_The_Acid-Base_Properties_of_Ions_and_Salts.txt |
Learning Objectives
• Extend previously introduced equilibrium concepts to acids and bases that may donate or accept more than one proton
We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as $\ce{HCl}$, $\ce{HNO3}$, and $\ce{HCN}$ that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:
$\ce{HCl}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{Cl-}(aq) \nonumber$
$\ce{HNO3}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{NO3-}(aq) \nonumber$
$\ce{HCN}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CN-}(aq) \nonumber$
Even though it contains four hydrogen atoms, acetic acid, $\ce{CH3CO2H}$, is also monoprotic because only the hydrogen atom from the carboxyl group ($\ce{-COOH}$) reacts with bases:
Similarly, monoprotic bases are bases that will accept a single proton.
Diprotic Acids
Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:
• The first ionization is
$\ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq) \nonumber$
with $K_{\ce a1} > 10^2;\: {complete\: dissociation}$.
• The second ionization is
$\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^{2−}}(aq) \nonumber$
with $K_{\ce a2}=1.2×10^{−2}$.
This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, $\ce{H2CO3}$, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.
• First Ionization
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \nonumber$
with
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=4.3×10^{−7} \nonumber$
The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.
• Second Ionization
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
with
$K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11} \nonumber$
$K_{\ce{H2CO3}}$ is larger than $K_{\ce{HCO3-}}$ by a factor of 104, so H2CO3 is the dominant producer of hydronium ion in the solution. This means that little of the $\ce{HCO3-}$ formed by the ionization of H2CO3 ionizes to give hydronium ions (and carbonate ions), and the concentrations of H3O+ and $\ce{HCO3-}$ are practically equal in a pure aqueous solution of H2CO3.
If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H3O+ and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.
Example $1$: Ionization of a Diprotic Acid
When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO2 reacts with water to form carbonic acid, H2CO3. What are $\ce{[H3O+]}$, $\ce{[HCO3- ]}$, and $\ce{[CO3^2- ]}$ in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \label{step1} \tag{equilibrium step 1}$
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11} \label{step2} \tag{equilibrium step 2}$
Solution
As indicated by the ionization constants, H2CO3 is a much stronger acid than $\ce{HCO3-}$, so $\ce{H2CO3}$ is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem:
1. Using the customary four steps, we determine the concentration of H3O+ and $\ce{HCO3-}$ produced by ionization of H2CO3.
2. Then we determine the concentration of $\ce{CO3^2-}$ in a solution with the concentration of H3O+ and $\ce{HCO3-}$ determined in (1).
To summarize:
1. First Ionization: Determine the concentrations of $\ce{H3O+}$ and $\ce{HCO3-}$.
Since \ref{step1} is has a much bigger $K_{a1}=4.3×10^{−7}$ than $K_{a2}=4.7×10^{−11}$ for \ref{step2}, we can safely ignore the second ionization step and focus only on the first step (but address it in next part of problem).
$\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \nonumber$
As for the ionization of any other weak acid:
An abbreviated table of changes and concentrations shows:
Abbreviated table of changes and concentrations
ICE Table $\ce{H2CO3}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{HCO3-}(aq)$
Initial (M) $0.033 \:M$ - $0$ $0$
Change (M) $- x$ - $+x$ $+x$
Equilibrium (M) $0.033 \:M - x$ - $x$ $x$
Substituting the equilibrium concentrations into the equilibrium constant gives us:
$K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+][HCO3- ]}{[H2CO3]}}=\dfrac{(x)(x)}{0.033−x}=4.3×10^{−7} \nonumber$
Solving the preceding equation making our standard assumptions gives:
$x=1.2×10^{−4} \nonumber$
Thus:
$\ce{[H2CO3]}=0.033\:M \nonumber$
$\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber$
2. Second Ionization: Determine the concentration of $CO_3^{2-}$ in a solution at equilibrium.
Since the \ref{step1} is has a much bigger $K_a$ than \ref{step2}, we can the equilibrium conditions calculated from first part of example as the initial conditions for an ICER Table for the \ref{step2}:
$\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber$
ICER Table for the \ref{step2}:
ICE Table $\ce{HCO3-}(aq)$ $\ce{H2O}(l)$ $\ce{H3O+}(aq)$ $\ce{CO3^2-}(aq)$
Initial (M) $1.2×10^{−4}\:M$ - $1.2×10^{−4}\:M$ $0$
Change (M) $- y$ - $+y$ $+y$
Equilibrium (M) $1.2×10^{−4}\:M - y$ - $1.2×10^{−4}\:M + y$ $y$
\begin{align*} K_{\ce{HCO3-}}&=\ce{\dfrac{[H3O+][CO3^2- ]}{[HCO3- ]}} \[4pt] &=\dfrac{(1.2×10^{−4}\:M + y) (y)}{(1.2×10^{−4}\:M - y)} \end{align*} \nonumber
To avoid solving a quadratic equation, we can assume $y \ll 1.2×10^{−4}\:M$ so
$K_{\ce{HCO3-}} = 4.7×10^{−11} \approx \dfrac{(1.2×10^{−4}\:M ) (y)}{(1.2×10^{−4}\:M)} \nonumber$
Rearranging to solve for $y$
$y \approx \dfrac{ (4.7×10^{−11})(1.2×10^{−4}\:M )}{ 1.2×10^{−4}\:M} \nonumber$
$[\ce{CO3^2-}]=y \approx 4.7×10^{−11} \nonumber$
To summarize:
In part 1 of this example, we found that the $\ce{H2CO3}$ in a 0.033-M solution ionizes slightly and at equilibrium $[\ce{H2CO3}] = 0.033\, M$, $[\ce{H3O^{+}}] = 1.2 × 10^{−4}$, and $\ce{[HCO3- ]}=1.2×10^{−4}\:M$. In part 2, we determined that $\ce{[CO3^2- ]}=5.6×10^{−11}\:M$.
Exercise $2$: Hydrogen Sulfide
The concentration of $H_2S$ in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate $\ce{[H3O+]}$, $\ce{[HS^{−}]}$, and $\ce{[S^{2−}]}$ in the solution:
$\ce{H2S}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HS-}(aq) \hspace{20px} K_{\ce a1}=8.9×10^{−8} \nonumber$
$\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_{\ce a2}=1.0×10^{−19} \nonumber$
Answer
$[\ce{H2S}] = 0.1 M$, $\ce{[H3O+]} = [HS^{−}] = 0.0001\, M$, $[S^{2−}] = 1 × 10^{−19}\, M$
We note that the concentration of the sulfide ion is the same as Ka2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).
Triprotic Acids
A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:
• The first ionization is
$\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \nonumber$
with $K_{\ce a1}=7.5×10^{−3}$.
• The second ionization is
$\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq) \nonumber$
with $K_{\ce a2}=6.2×10^{−8}$.
• The third ionization is
$\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \nonumber$
with $K_{\ce a3}=4.2×10^{−13}$.
As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 105 to 106. This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, the calculations can be broken down into a series of parts similar to those for diprotic acids.
Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:
$\ce{H2O}(l)+\ce{CO3^2-}(aq)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \nonumber$
and
$\ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq) \nonumber$
Summary
An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of Ka of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps.
Glossary
diprotic acid
acid containing two ionizable hydrogen atoms per molecule. A diprotic acid ionizes in two steps
diprotic base
base capable of accepting two protons. The protons are accepted in two steps
monoprotic acid
acid containing one ionizable hydrogen atom per molecule
stepwise ionization
process in which an acid is ionized by losing protons sequentially
triprotic acid
acid that contains three ionizable hydrogen atoms per molecule; ionization of triprotic acids occurs in three steps | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.10%3A_Polyprotic_Acids.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
• Write equations illustrating the behavior of a given non-aqueous acid-base system.
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. According to Lewis,
• An acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons.
• A base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Proton-Transfer Reactions Involve Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note carefully that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H2O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because $\ce{HF}$ is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
Acid-base Reactions without Transferring Protons
The major utility of the Lewis definition is that it extends the concept of acids and bases beyond the realm of proton transfer reactions. The classic example is the reaction of boron trifluoride with ammonia to form an adduct:
$\ce{BF_3 + NH_3 \rightarrow F_3B-NH_3}$
One of the most commonly-encountered kinds of Lewis acid-base reactions occurs when electron-donating ligands form coordination complexes with transition-metal ions.
Exercise $1$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $\ce{Al(OH)_3 + OH^{–} \rightarrow Al(OH)_4^–}$
2. $\ce{SnS_2 + S^{2–} \rightarrow SnS_3^{2–}}$
3. $\ce{Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}}$
4. $\ce{AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–}$
5. $\ce{Fe^{2+} + NO \rightarrow Fe(NO)^{2+}}$
6. $\ce{Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}}$
Applications to organic reaction mechanisms
Although organic chemistry is beyond the scope of these lessons, it is instructive to see how electron donors and acceptors play a role in chemical reactions. The following two diagrams show the mechanisms of two common types of reactions initiated by simple inorganic Lewis acids:
In each case, the species labeled "Complex" is an intermediate that decomposes into the products, which are conjugates of the original acid and base pairs. The electric charges indicated in the complexes are formal charges, but those in the products are "real".
In reaction 1, the incomplete octet of the aluminum atom in $\ce{AlCl3}$ serves as a better electron acceptor to the chlorine atom than does the isobutyl part of the base. In reaction 2, the pair of non-bonding electrons on the dimethyl ether coordinates with the electron-deficient boron atom, leading to a complex that breaks down by releasing a bromide ion.
Non-aqueous Protonic Acid-Base Systems
We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base solvent systems, of which the water system is only one (Table $1$). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water.
The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. However, many other familiar substances can also serve as the basis of protonic solvent systems as Table $1$ indicates:
Table $1$: Popular Solvent systems
solvent
autoprotolysis reaction
pKap
water 2 H2O → H3O+ + OH 14
ammonia 2 NH3 → NH4+ + NH2 33
acetic acid 2 CH3COOH → CH3COOH2+ + CH3COO 13
ethanol 2 C2H5OH → C2H5OH2+ + C2H5O 19
hydrogen peroxide 2 HO-OH → HO-OH2+ + HO-O 13
hydrofluoric acid 2 HF → H2F+ + F 10
sulfuric acid 2 H2SO4 → H3SO4+ + HSO4 3.5
One use of nonaqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H3O+ or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/17%3A_Acids_and_Bases/17.11%3A_Lewis_Acids_and_Bases.txt |
An antifreeze is an additive which lowers the freezing point of a water-based liquid. An antifreeze mixture is used to achieve freezing-point depression for cold environments and also achieves boiling-point elevation to allow higher coolant temperature. Freezing and boiling points are colligative properties of a solution, which depend on the concentration of the dissolved substance. Because water has good properties as a coolant, water plus antifreeze is used in internal combustion engines and other heat transfer applications. The purpose of antifreeze is to prevent a rigid enclosure from bursting due to expansion when water freezes. Commercially, both the additive (pure concentrate) and the mixture (diluted solution) are called antifreeze, depending on the context. Careful selection of an antifreeze can enable a wide temperature range in which the mixture remains in the liquid phase, which is critical to efficient heat transfer and the proper functioning of heat exchangers.
Ethylene glycol solutions became available in 1926 and were marketed as "permanent antifreeze" since the higher boiling points provided advantages for summertime use as well as during cold weather. They are used today for a variety of applications, including automobiles, but gradually being replaced by propylene glycol due to its lower toxicity.
When ethylene glycol is used in a system, it may become oxidized to five organic acids (formic, oxalic, glycolic, glyoxalic and acetic acid). Inhibited ethylene glycol antifreeze mixes are available, with additives that buffer the pH and reserve alkalinity of the solution to prevent oxidation of ethylene glycol and formation of these acids. Nitrites, silicates, theodin, borates and azoles may also be used to prevent corrosive attack on metal.
Ethylene glycol is poisonous to humans and other animals,[4][5] and should be handled carefully and disposed of properly. Its sweet taste can lead to accidental ingestion or allow its deliberate use as a murder weapon.[6][7][8] Ethylene glycol is difficult to detect in the body, and causes symptoms—including intoxication, severe diarrhea, and vomiting—that can be confused with other illnesses or diseases.[4][8] Its metabolism produces calcium oxalate, which crystallizes in the brain, heart, lungs, and kidneys, damaging them; depending on the level of exposure, accumulation of the poison in the body can last weeks or months before causing death, but death by acute kidney failure can result within 72 hours if the individual does not receive appropriate medical treatment for the poisoning.[4] Some ethylene glycol antifreeze mixtures contain an embittering agent, such as denatonium, to discourage accidental or deliberate consumption.
The toxic mechanism of ethylene glycol poisoning is mainly due to the metabolites of ethylene glycol. Initially it is metabolized by alcohol dehydrogenase to glycolaldehyde, which is then oxidized to glycolic acid.[7] The increase in metabolites may cause encephalopathy or cerebral edema.[13] The metabolic effects occur 12 to 36 hours post ingestion, causing primarily metabolic acidosis which is due mainly to accumulated glycolic acid. Additionally, as a side effect of the first two steps of metabolism, an increase in the blood concentration of lactic acid occurs contributing to lactic acidosis. The formation of acid metabolites also causes inhibition of other metabolic pathways, such as oxidative phosphorylation.[7]
The kidney toxicity of ethylene glycol occurs 24 to 72 hours post ingestion and is caused by a direct cytotoxic effect of glycolic acid. The glycolic acid is then metabolized to glyoxylic acid and finally to oxalic acid. Oxalic acid binds with calcium to form calcium oxalate crystals which may deposit and cause damage to many areas of the body including the brain, heart, kidneys, and lungs.[7] The most significant effect is accumulation of calcium oxalate crystals in the kidneys which causes kidney damage leading to oliguric or anuric acute kidney failure.[7] The rate-limiting step in this cascade is the conversion of glycolic to glyoxylic acid.[14] Accumulation of glycolic acid in the body is mainly responsible for toxicity.[15] | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/18%3A_Aqueous_Ionic_Equilibrium/18.01%3A_The_Danger_of_Antifreeze.txt |
Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
Introduction
The solubility products Ksp's are equilibrium constants in heterogeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}$
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
$\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}} \nonumber$
$\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}} \nonumber$
$\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}} \nonumber$
$\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}} \nonumber$
$\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}} \nonumber$
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M} \nonumber$
but
\begin{alignat}{3} \nonumber &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\ \nonumber & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\ \nonumber & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\ \nonumber & &&= && &&\mathrm{\:0.40\: M} \nonumber \end{alignat}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
Solution
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M} \nonumber$
Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example $2$: Solubility of Lead Chloride
Consider the lead(II) ion concentration in this saturated solution of PbCl2. The balanced reaction is
$PbCl_{2 (s)} \rightleftharpoons Pb^{2+} _{(aq)} + 2Cl^-_{(aq)} \nonumber$
Defining $s$ as the concentration of dissolved lead(II) chloride, then:
$[Pb^{2+}] = s \nonumber$
$[Cl^- ] = 2s \nonumber$
These values can be substituted into the solubility product expression, which can be solved for $s$:
$\begin{eqnarray} K_{sp} &=& [Pb^{2+}] [Cl^-]^2 \ &=& s \times (2s)^2 \ 1.7 \times 10^{-5} &=& 4s^3 \ s^3 &=& \frac{1.7 \times 10^{-5}}{4} \ &=& 4.25 \times 10^{-6} \ s &=& \sqrt[3]{4.25 \times 10^{-6}} \ &=& 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{eqnarray} \nonumber$The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression again:
$PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq) \nonumber$
What happens to that equilibrium if extra chloride ions are added? According to Le Châtelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
Example $3$
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
$[Pb^{2+}] = s \label{2}$
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.
In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
$[Cl^- ] = 0.100\; M \label{3}$
The rest of the mathematics looks like this:
$\begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 \ & = s \times (0.100)^2 \ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \nonumber$
therefore:
$\begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \label{4}$
Finally, compare that value with the simple saturated solution:
Original solution:
$[Pb^{2+}] = 0.0162 \, M \label{5}$
Solution in 0.100 M NaCl solution:
$[Pb^{2+}] = 0.0017 \, M \label{6}$
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
A Video Discussing Finding the Solubility of a Salt: Finding the Solubility of a Salt(opens in new window) [youtu.be]
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
The common ion effect of H3O+ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium. Now consider the common ion effect of $\ce{OH^{-}}$ on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Chatelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, $K_b=1.8 \times 10^{-5}$, does not change. The reaction is put out of balance, or equilibrium.
$Q_a = \frac{\ce{[NH_4^{+}][OH^{-}]}}{\ce{[NH3]}} \nonumber$
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing Q to decrease towards K.
Common Ion Effect on Solubility
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
$\ce{Ca3(PO4)2(s) <=> 3Ca^{2+}(aq) + 2PO^{3−}4(aq)} \label{17.4.1}$
As you will discover in more advanced chemistry courses, basic anions, such as S2−, PO43, and CO32, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate. The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{17.4.2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{17.4.2b}$
At 25°C and pH 7.00, $K_{sp}$ for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43 ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts vary dramatically for different compounds (Table E3). Although $K_{sp}$ is not a function of pH in Equation $\ref{17.4.2a}$, changes in pH can affect the solubility of a compound.
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that $K_{sp}$ is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. This dependency is another example of the common ion effect where adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2]. We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains
• $3 \times (1.14 \times 10^{−7}\; M) = 3.42 \times 10^{−7} M \; of \; Ca^{2+}$
• $2 \times (1.14 \times 10^{−7} M) = 2.28 \times 10^{−7} M \; of \; PO_4^{3−}$
according to the stoichiometry shown in Equation $\ref{17.4.2a}$ (neglecting hydrolysis to form HPO42). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation $\ref{17.4.2a}$ to proceed to the left, resulting in precipitation of Ca3(PO4)2. This will decrease the concentration of both Ca2+ and PO43 until Q = Ksp.
Note
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example $5$
Consider the reaction:
$PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq) \nonumber$
What happens to the solubility of PbCl2(s) when 0.1 M NaCl is added?
Solution
$K_{sp}=1.7 \times 10^{-5} \nonumber$
$Q_{sp}= 1.8 \times 10^{-5}\nonumber$
Identify the common ion: Cl-
Notice: Qsp > Ksp The addition of NaCl has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of PbCl2 is equivalent to the concentration of Pb2+ produced because they are in a 1:1 ratio.
Because Ksp for the reaction is 1.7×10-5, the overall reaction would be (s)(2s)2= 1.7×10-5. Solving the equation for s gives s= 1.62×10-2 M. The coefficient on Cl- is 2, so it is assumed that twice as much Cl- is produced as Pb2+, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because Na+ and Cl- are in a 1:1 ration in the ionic salt, NaCl. Therefore, the overall molarity of Cl- would be 2s + 0.1, with 2s referring to the contribution of the chloride ion from the dissociation of lead chloride.
$\begin{eqnarray} Q_{sp} &=& [Pb^{2+}][Cl^-]^2 \ 1.8 \times 10^{-5} &=& (s)(2s + 0.1)^2 \ s &=& [Pb^{2+}] \ &=& 1.8 \times 10^{-3} M \ 2s &=& [Cl^-] \ &\approx & 0.1 M \end{eqnarray} \nonumber$
Notice that the molarity of Pb2+ is lower when NaCl is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl-] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl2 is greater than the equilibrium constant because of the added Cl-. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
Exercise $5$
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
A Video Discussing the Common Ion Effect in Solubility Products: The Common Ion Effect in Solubility Products (opens in new window) [youtu.be]
Contributors and Attributions
• Emmellin Tung, Mahtab Danai (UCD)
• Jim Clark (ChemGuide)
• Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)
Learning Objectives
• To understand how adding a common ion affects the position of an acid–base equilibrium.
• To know how to use the Henderson-Hasselbalch approximation to calculate the pH of a buffer.
Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid ($HA$) and its conjugate base $(A^−$) or a weak base ($B$) and its conjugate acid ($BH^+$), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH.
The Common Ion Effect: Weak Acids Combined with Conjugate Bases
To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of $\ce{H^{+}}$). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. The dissociation reaction of acetic acid is as follows:
$\ce{CH3COOH (aq) <=> CH3COO^{−} (aq) + H^{+} (aq)} \label{Eq1}$
and the equilibrium constant expression is as follows:
$K_a=\dfrac{[\ce{H^{+}}][\ce{CH3COO^{-}}]}{[\ce{CH3CO2H}]} \label{Eq2}$
Sodium acetate ($\ce{CH_3CO_2Na}$) is a strong electrolyte that ionizes completely in aqueous solution to produce $\ce{Na^{+}}$ and $\ce{CH3CO2^{−}}$ ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation \ref{Eq1} will shift to the left, consuming some of the added $\ce{CH_3COO^{−}}$ and some of the $\ce{H^{+}}$ ions originally present in solution.
Because $\ce{Na^{+}}$ is a spectator ion, it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which $[\ce{H^{+}}]$ is less than the initial value. Because $[\ce{H^{+}}]$ has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH.
If we instead add a strong acid such as $\ce{HCl}$ to the system, $[\ce{H^{+}}]$ increases. Once again the equilibrium is temporarily disturbed, but the excess $\ce{H^{+}}$ ions react with the conjugate base ($\ce{CH_3CO_2^{−}}$), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [$\ce{CH_3CO_2^{−}}$] than before. In both cases, only the equilibrium composition has changed; the ionization constant $K_a$ for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case $\ce{CH3CO2^{−}}$, will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is via the common ion effect.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Example $1$
A 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized.
1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?
Given: solution concentration and pH, $pK_a$, and percent ionization of acid; final concentration of conjugate base or strong acid added
Asked for: pH and percent ionization of formic acid
Strategy:
1. Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.
2. Substitute the expressions for the final concentrations into the expression for Ka. Calculate $[\ce{H^{+}}]$ and the pH of the solution.
3. Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.
Solution:
A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The $\ce{Na^{+}}$ ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium:
$\ce{HCO2H (aq) <=> HCO^{−}2 (aq) + H^{+} (aq)} \nonumber$
The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.
Final Concentration
ICE $[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
Initial 0.150 $1.00 \times 10^{−7}$ 0.100
Change −x +x +x
Equilibrium (0.150 − x) x (0.100 + x)
B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so
\begin{align*} K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]} &=\dfrac{(x)(0.100+x)}{0.150−x} \[4pt] &\approx \dfrac{x(0.100)}{0.150} \[4pt] &\approx 10^{−3.75} \[4pt] &\approx 1.8 \times 10^{−4} \end{align*} \nonumber
Rearranging and solving for $x$,
\begin{align*} x &=(1.8 \times 10^{−4}) \times \dfrac{0.150 \;M}{ 0.100 \;M} \[4pt] &=2.7 \times 10^{−4}\[4pt] &=[H^+] \end{align*} \nonumber
The value of $x$ is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover,
$K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} \nonumber$
which is greater than $1.0 \times 10^{−6}$, so again, our assumption is justified. The final pH is:
$pH= −\log(2.7 \times 10^{−4}) = 3.57 \nonumber$
compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of $\ce{H^{+}}$ ions, driving the equilibrium to the left.
C Because $HCl$ is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.
$HCO_2H (aq) \leftrightharpoons H^+ (aq) +HCO^−_2 (aq) \nonumber$
initial concentrations, changes in concentration, and final concentrations
$[HCO_2H (aq) ]$ $[H^+ (aq) ]$ $[HCO^−_2 (aq) ]$
initial 0.150 0.200 0
change −x +x +x
final (0.150 − x) (0.200 + x) x
To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final $[\ce{HCO2^{-}}]$. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so
$K_a=\dfrac{[H^+][HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \nonumber$
Rearranging and solving for $x$,
\begin{align*} x &=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M} \[4pt] &=1.35 \times 10^{−4}=[HCO_2^−] \end{align*} \nonumber
Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:
$\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100\%=0.0900\% \nonumber$
Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding $\ce{H^{+}}$ ions drives the dissociation equilibrium to the left.
Exercise $1$
A 0.225 M solution of ethylamine ($\ce{CH3CH2NH2}$ with $pK_b = 3.19$) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:
1. the pH of the solution if enough solid ethylamine hydrochloride ($\ce{EtNH3Cl}$) is added to make the solution 0.100 M in $\ce{EtNH3^{+}}$
2. the percentage of ethylamine that is ionized if enough solid $\ce{NaOH}$ is added to the original solution to give a final concentration of 0.050 M $\ce{NaOH}$
Answer a
11.16
Answer b
1.3%
A Video Discussing the Common Ion Effect: The Common Ion Effecr(opens in new window) [youtu.be]
The Common Ion Effect: Weak Bases Combined with Conjugate Acids
Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base ($B$) and its conjugate acid ($BH^+$). The general equation for the ionization of a weak base is as follows:
$B (aq) +H_2O (l) \leftrightharpoons BH^+ (aq) +OH^− (aq) \label{Eq3}$
If the equilibrium constant for the reaction as written in Equation $\ref{Eq3}$ is small, for example $K_b = 10^{−5}$, then the equilibrium constant for the reverse reaction is very large: $K = \dfrac{1}{K_b} = 10^5$. Adding a strong base such as $OH^-$ to the solution therefore causes the equilibrium in Equation $\ref{Eq3}$ to shift to the left, consuming the added $OH^-$. As a result, the $OH^-$ ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the $OH^-$ ion concentration, the reaction will proceed to the left to counteract the stress.
If the $pK_b$ of the base is 5.0, the $pK_a$ of its conjugate acid is
$pK_a = pK_w − pK_b = 14.0 – 5.0 = 9.0. \nonumber$
Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows:
$BH^+ (aq) +H_2O (l) \leftrightharpoons B (aq) +H_3O^+ (aq) \label{Eq4}$
Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/Ka = 109. If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation $\ref{Eq4}$ shifts to the left. As a result, the $H^+$ ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb $H^+$ and $OH^-$ ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution.
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on $K$), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure $1$, when $NaOH$ is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M $NaOH$ to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of $NaOH$ solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the $NaOH$ solution results in only a relatively small change in pH.
Calculating the pH of a Buffer
The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the $pK_a$ or $pK_b$ of the weak acid or weak base. The procedure is analogous to that used in Example $1$ to calculate the pH of a solution containing known concentrations of formic acid and formate.
An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is $HA \leftrightharpoons H^+ + A^−$, for which the equilibrium constant expression is as follows:
$K_a=\dfrac{[H^+][A^-]}{[HA]} \label{Eq5}$
This equation can be rearranged as follows:
$[H^+]=K_a\dfrac{[HA]}{[A^−]} \label{Eq6}$
Taking the logarithm of both sides and multiplying both sides by −1,
\begin{align} −\log[H^+] &=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right) \[4pt] &=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \label{Eq7} \end{align}
Replacing the negative logarithms in Equation $\ref{Eq7}$,
$pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \label{Eq8}$
or, more generally,
$pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{Eq9}$
Equation $\ref{Eq8}$ and Equation $\ref{Eq9}$ are both forms of the Henderson-Hasselbalch approximation, named after the two early 20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch approximation may be limited to solutions whose concentrations are at least 100 times greater than their $K_a$ values.
There are three special cases where the Henderson-Hasselbalch approximation is easily interpreted without the need for calculations:
• $[base] = [acid]$: Under these conditions, $\dfrac{[base]}{[acid]} = 1 \nonumber$ in Equation \ref{Eq9}. Because $\log 1 = 0$, $pH = pK_a \nonumber$ regardless of the actual concentrations of the acid and base. Recall that this corresponds to the midpoint in the titration of a weak acid or a weak base.
• $[base]/[acid] = 10$: In Equation $\ref{Eq9}$, because $\log 10 = 1$, $pH = pK_a + 1. \nonumber$
• $[base]/[acid] = 100$: In Equation $\ref{Eq9}$, because $\log 100 = 2$, $pH = pK_a + 2. \nonumber$
Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = $pK_a$ − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit.
If [base] = [acid] for a buffer, then pH = $pK_a$. Changing this ratio by a factor of 10 either way changes the pH by ±1 unit.
Example $2$
What is the pH of a solution that contains
1. 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$? (The $pK_a$ of formic acid is 3.75.)
2. 0.0135 M $\ce{HCO2H}$ and 0.0215 M $\ce{HCO2Na}$?
3. 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The $pK_b$ of pyridine is 8.77.)
Given: concentration of acid, conjugate base, and $pK_a$; concentration of base, conjugate acid, and $pK_b$
Asked for: pH
Strategy:
Substitute values into either form of the Henderson-Hasselbalch approximation (Equations \ref{Eq8} or \ref{Eq9}) to calculate the pH.
Solution:
According to the Henderson-Hasselbalch approximation (Equation \ref{Eq8}), the pH of a solution that contains both a weak acid and its conjugate base is
$pH = pK_a + \log([A−]/[HA]). \nonumber$
A
Inserting the given values into the equation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.215}{0.135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result makes sense because the $[A^−]/[HA]$ ratio is between 1 and 10, so the pH of the buffer must be between the $pK_a$ (3.75) and $pK_a + 1$, or 4.75.
B
This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch approximation,
\begin{align*} pH &=3.75+\log\left(\dfrac{0.0215}{0.0135}\right) \[4pt] &=3.75+\log 1.593 \[4pt] &=3.95 \end{align*} \nonumber
This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).
C
In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion ($HPy^+$). We will therefore use Equation $\ref{Eq9}$, the more general form of the Henderson-Hasselbalch approximation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and $[acid] = [HPy^{+}] = 0.234\, M$. We also are given $pK_b = 8.77$ for pyridine, but we need $pK_a$ for the pyridinium ion. Recall from Equation 16.23 that the $pK_b$ of a weak base and the $pK_a$ of its conjugate acid are related:
$pK_a + pK_b = pK_w. \nonumber$
Thus $pK_a$ for the pyridinium ion is $pK_w − pK_b = 14.00 − 8.77 = 5.23$. Substituting this $pK_a$ value into the Henderson-Hasselbalch approximation,
\begin{align*} pH=pK_a+\log \left(\dfrac{[base]}{[acid]}\right) \[4pt] &=5.23+\log\left(\dfrac{0.119}{0.234}\right) \[4pt] & =5.23 −0.294 \[4pt] &=4.94 \end{align*} \nonumber
Once again, this result makes sense: the $[B]/[BH^+]$ ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the $pK_a$ (5.23) and $pK_a − 1$, or 4.23.
Exercise $2$
What is the pH of a solution that contains
1. 0.333 M benzoic acid and 0.252 M sodium benzoate?
2. 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?
The $pK_a$ of benzoic acid is 4.20, and the $pK_b$ of trimethylamine is also 4.20.
Answer a
4.08
Answer b
9.68
A Video Discussing Using the Henderson Hasselbalch Equation: Using the Henderson Hasselbalch Equation(opens in new window) [youtu.be] (opens in new window)
The Henderson-Hasselbalch approximation ((Equation $\ref{Eq8}$) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example $3$.
Example $3$
The buffer solution in Example $2$ contained 0.135 M $\ce{HCO2H}$ and 0.215 M $\ce{HCO2Na}$ and had a pH of 3.95.
1. What is the final pH if 5.00 mL of 1.00 M $HCl$ are added to 100 mL of this solution?
2. What is the final pH if 5.00 mL of 1.00 M $NaOH$ are added?
Given: composition and pH of buffer; concentration and volume of added acid or base
Asked for: final pH
Strategy:
1. Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example $1$. Then calculate the amount of acid or base added.
2. Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch approximation (Equation \ref{Eq9}) to obtain the pH.
Solution:
The added $\ce{HCl}$ (a strong acid) or $\ce{NaOH}$ (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.
A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer:
$100 \, \cancel{mL} \left( \dfrac{0.135 \, mmol\; \ce{HCO2H}}{\cancel{mL}} \right) = 13.5\, mmol\, \ce{HCO2H} \nonumber$
$100\, \cancel{mL } \left( \dfrac{0.215 \, mmol\; \ce{HCO2^{-}}}{\cancel{mL}} \right) = 21.5\, mmol\, \ce{HCO2^{-}} \nonumber$
The millimoles of $\ce{H^{+}}$ in 5.00 mL of 1.00 M $\ce{HCl}$ is as follows:
$5.00 \, \cancel{mL } \left( \dfrac{1.00 \,mmol\; \ce{H^{+}}}{\cancel{mL}} \right) = 5\, mmol\, \ce{H^{+}} \nonumber$
B Next, we construct a table of initial amounts, changes in amounts, and final amounts:
$\ce{HCO^{2−}(aq) + H^{+} (aq) <=> HCO2H (aq)} \nonumber$
initial amounts, changes in amounts, and final amounts:
$HCO^{2−} (aq)$ $H^+ (aq)$ $HCO_2H (aq)$
Initial 21.5 mmol 5.00 mmol 13.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 16.5 mmol ∼0 mmol 18.5 mmol
The final amount of $H^+$ in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final $[H^+]$ and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example $1$ or the Henderson–Hasselbach approximation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch approximation requires the concentrations of $HCO_2^−$ and $HCO_2H$, which can be calculated using the number of millimoles ($n$) of each and the total volume ($VT$). Substituting these values into the Henderson-Hasselbalch approximation (Equation $\ref{Eq9}$):
\begin{align*} pH &=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right) \[4pt] &=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right) \[4pt] &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \end{align*} \nonumber
Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So
\begin{align*} pH &=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right) \[4pt] &=3.75 −0.050=3.70 \end{align*} \nonumber
Once again, this result makes sense on two levels. First, the addition of $HCl$has decreased the pH from 3.95, as expected. Second, the ratio of $HCO_2^−$ to $HCO_2H$ is slightly less than 1, so the pH should be between the $pK_a$ and $pK_a$ − 1.
A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of $HCO_2H$ and 21.5 mmol of $HCO_2^−$. The number of millimoles of $OH^-$ in 5.00 mL of 1.00 M $NaOH$ is as follows:
B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.
$\ce{HCO2H (aq) + OH^{−} (aq) <=> HCO^{−}2 (aq) + H2O (l)} \nonumber$
initial amounts, changes in amounts, and final amounts
$HCO_2H (aq)$ $OH^−$ $HCO^−_2 (aq)$
Initial 13.5 mmol 5.00 mmol 21.5 mmol
Change −5.00 mmol −5.00 mmol +5.00 mmol
Final 8.5 mmol ∼0 mmol 26.5 mmol
The final amount of $OH^-$ in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both $HCO_2^−$ and $HCO_2H$ into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:
\begin{align*} pH &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \[4pt] &=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right) \[4pt] &=3.75+0.494 =4.24 \end{align*} \nonumber
Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the $pK_a$ and $pK_a$ + 1, as expected for a solution with a $HCO_2^−/HCO_2H$ ratio between 1 and 10.
Exercise $3$
The buffer solution from Example $2$ contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.
1. What is the final pH if 12.0 mL of 1.5 M $\ce{NaOH}$ are added to 250 mL of this solution?
2. What is the final pH if 12.0 mL of 1.5 M $\ce{HCl}$ are added?
Answer a
5.30
Answer b
4.42
Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch approximation, not their concentrations.
A Video Discussing the Change in pH with the Addition of a Strong Acid to a Buffer: The Change in pH with the Addition of a Strong Acid to a Buffer(opens in new window) [youtu.be]
The Change in pH with the Addition of a Strong Base to a Buffer:
The results obtained in Example $3$ and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of $HCl$ or $NaOH$ solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to $1.1 \times 10^{−4}$ M HCl). In this case, adding 5.00 mL of 1.00 M $HCl$ would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M $NaOH$ would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH.
The most effective buffers contain equal concentrations of an acid and its conjugate base.
A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure $2$ for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of $CH_3CO_2^−$ to $CH_3CO_2H$ from 1:1 reduces the buffer capacity of the solution.
A Video Discussing The Buffer Region: The Buffer Region (opens in new window) [youtu.be]
The Relationship between Titrations and Buffers
There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure $3$. As indicated by the labels, the region around $pK_a$ corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the $pK_a$ to approximately a pH value of 1 unit greater than the $pK_a$, which is why buffer solutions usually have a pH that is within ±1 pH units of the $pK_a$ of the acid component of the buffer.
This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water.
In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to $K_a$. In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to $K_b$. However, we can calculate either $K_a$ or $K_b$ from the other because they are related by $K_w$.
Blood: A Most Important Buffer
Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0.
Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $\ce{CO2}/\ce{HCO3^{−}}$ system, which dominates the buffering action of blood plasma.
The acid–base equilibrium in the $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is usually written as follows:
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{Eq10}$
with $K_a = 4.5 \times 10^{−7}$ and $pK_a = 6.35$ at 25°C. In fact, Equation $\ref{Eq10}$ is a grossly oversimplified version of the $\ce{CO2}/\ce{HCO3^{-}}$ system because a solution of $\ce{CO2}$ in water contains only rather small amounts of $H_2CO_3$. Thus Equation $\ref{Eq10}$ does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C.
As shown in Equation $\ref{Eq11}$, $\ce{CO2}$ is in equilibrium with $\ce{H2CO3}$, but the equilibrium lies far to the left, with an $\ce{H2CO3}/\ce{CO2}$ ratio less than 0.01 under most conditions:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{Eq11}$
with $K′ = 4.0 \times 10^{−3}$ at 37°C. The true $pK_a$ of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a $K_a$ of $2.0 \times 10^{−4}$, which makes it a much stronger acid than Equation \ref{Eq10} suggests. Adding Equation \ref{Eq10} and Equation \ref{Eq11} and canceling $\ce{H2CO3}$ from both sides give the following overall equation for the reaction of $\ce{CO2}$ with water to give a proton and the bicarbonate ion:
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{16.65a}$
with $K'=4.0 \times 10^{−3} (37°C)$
$\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65b}$
with $K_a=2.0 \times 10^{−4} (37°C)$
$\ce{CO2 (aq) + H2O (l) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65c}$
with $K=8.0 \times 10^{−7} (37°C)$
The $K$ value for the reaction in Equation \ref{16.65c} is the product of the true ionization constant for carbonic acid ($K_a$) and the equilibrium constant (K) for the reaction of $\ce{CO2 (aq)}$ with water to give carbonic acid. The equilibrium equation for the reaction of $\ce{CO2}$ with water to give bicarbonate and a proton is therefore
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{[\ce{CO2}]}=8.0 \times 10^{−7} \label{eq13}$
The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to Henry’s law,
$[\ce{CO2}]=k P_{\ce{CO2}} \nonumber$
where $k$ is the Henry’s law constant for $\ce{CO2}$, which is $3.0 \times 10^{−5} \;M/mmHg$ at 37°C. Substituting this expression for $[\ce{CO2}]$ in Equation \ref{eq13},
$K=\dfrac{[\ce{H^{+}}][\ce{HCO3^{-}}]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{\ce{CO2}})} \nonumber$
where $P_{\ce{CO2}}$ is in mmHg. Taking the negative logarithm of both sides and rearranging,
$pH=6.10+\log \left( \dfrac{ [\ce{HCO3^{−}}]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{\ce{CO2}}) } \right) \label{Eq15}$
Thus the pH of the solution depends on both the $\ce{CO2}$ pressure over the solution and $[\ce{HCO3^{−}}]$. Figure $4$ plots the relationship between pH and $[\ce{HCO3^{−}}]$ under physiological conditions for several different values of $P_{\ce{CO2}}$, with normal pH and $[\ce{HCO3^{−}}]$ values indicated by the dashed lines.
According to Equation \ref{Eq15}, adding a strong acid to the $\ce{CO2}/\ce{HCO3^{−}}$ system causes $[\ce{HCO3^{−}}]$ to decrease as $\ce{HCO3^{−}}$ is converted to $\ce{CO2}$. Excess $\ce{CO2}$ is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in $P_{\ce{CO2}}$. Because the change in $[\ce{HCO3^{−}}]/P_{CO_2}$ is small, Equation \ref{Eq15} predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the $\ce{OH^{-}}$ reacts with $\ce{CO2}$ to form $\ce{HCO3^{−}}$, but $\ce{CO2}$ is replenished by the body, again limiting the change in both $[\ce{HCO3^{−}}]/P_{\ce{CO2}}$ and pH. The $\ce{CO2}/\ce{HCO3^{−}}$ buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value.
If a passenger steps out of an airplane in Denver, Colorado, for example, the lower $P_{\ce{CO2}}$ at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and $[\ce{HCO3^{-}}]$. The increase in pH and decrease in $[\ce{HCO3^{−}}]$ in response to the decrease in $P_{\ce{CO2}}$ are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness.
A Video Summary of the pH Curve for a Strong Acid/Strong Base Titration:
Summary
Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid ($HA$) and its conjugate weak base ($A^−$). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the $K_a$ or $K_b$), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch approximation, which is valid for solutions whose concentrations are at least 100 times greater than their $K_a$ values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the $CO_2/HCO_3^−$ system, which dominates the buffering action of blood plasma. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/18%3A_Aqueous_Ionic_Equilibrium/18.02%3A_Buffers-_Solutions_That_Resist_pH_Change.txt |
Buffer Capacity
Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on \(K\)), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure \(1\), when \(NaOH\) is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH.
Figure \(1\): Effect of Buffer Concentration on the Capacity of a Buffer
A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M \(NaOH\) to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of \(NaOH\) solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the \(NaOH\) solution results in only a relatively small change in pH.
Selecting proper components for desired pH
Buffers function best when the pKa of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10). For example, we know the Ka for hydroflouric acid is 6.6 x 10-4 so its pKa= -log(6.6 x 10-4) = 3.18. So, a hydrofluoric acid buffer would work best in a buffer range of around pH = 3.18.
For the weak base ammonia (NH3), the value of Kb is 1.8x10-5, implying that the Ka for the dissociation of its conjugate acid, NH4+, is Kw/Kb=10-14/1.8x10-5 = 5.6x10-10. Thus, the pKa for NH4+ = 9.25, so buffers using NH4+/NH3 will work best around a pH of 9.25. (It's always the pKa of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pKb of the conjugate base, obviously.)
When the desired pH of a buffer solution is near the pKa of the conjugate acid being used (i.e., when the amounts of conjugate acid and conjugate base in solution are within about a factor of 10 of each other), the Henderson-Hasselbalch equation can be applied as a simple approximation of the solution pH, as we will see in the next section.
Example 1: HF Buffer
In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a pH of 3.0. We can use the Henderson-Hasselbalch approximation to calculate the necessary ratio of F- and HF.
\[pH = pKa + \log\dfrac{[Base]}{[Acid]}\]
\[3.0 = 3.18 + \log\dfrac{[Base]}{[Acid]}\]
\[\log\dfrac{[Base]}{[Acid]} = -0.18\]
\[\dfrac{[Base]}{[Acid]} = 10^{-0.18}\]
\[\dfrac{[Base]}{[Acid]} = 0.66\]
This is simply the ratio of the concentrations of conjugate base and conjugate acid we will need in our solution. However, what if we have 100 ml of 1 M HF and we want to prepare a buffer using NaF? How much Sodium Fluoride would we need to add in order to create a buffer at said pH (3.0)?
We know from our Henderson-Hasselbalch calculation that the ratio of our base/acid should be equal to 0.66. From a table of molar masses, such as a periodic table, we can calculate the molar mass of NaF to be equal to 41.99 g/mol. HF is a weak acid with a Ka = 6.6 x 10-4 and the concentration of HF is given above as 1 M. Using this information, we can calculate the amount of F- we need to add.
The dissociation reaction is:
\[HF_{(aq)} + H_2O_{(l)} \rightleftharpoons F^-_{(aq)} + H_3O^+_{(aq)}\]
We could use ICE tables to calculate the concentration of F- from HF dissociation, but, since Ka is so small, we can approximate that virtually all of the HF will remain undissociated, so the amount of F- in the solution from HF dissociation will be negligible. Thus, the [HF] is about 1 M and the [F-] is close to 0. This will be especially true once we have added more F-, the addition of which will even further suppress the dissociation of HF.
We want the ratio of Base/Acid to be 0.66, so we will need [Base]/1M = 0.66. Thus, [F-] should be about 0.66 M. For 100 mL of solution, then, we will want to add 0.066 moles (0.1 L x 0.66 M) of F-. Since we are adding NaF as our source of F-, and since NaF completely dissociates in water, we need 0.066 moles of NaF. Thus, 0.066 moles x 41.99 g/mol = 2.767 g.
Note that, since the conjugate acid and the conjugate base are both mixed into the same volume of solution in the buffer, the ratio of "Base/Acid" is the same whether we use a ratio of the "concentration of base over concentration of acid," OR a ratio of "moles of base over moles of acid." The pH of the solution does not, it turns out, depend on the volume! (This is only true so long as the solution does not get so dilute that the autoionization of water becomes an important source of H+ or OH-. Such dilute solutions are rarely used as buffers, however.)
Contributors
• Jose Pietri (UCD), Donald Land (UCD) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/18%3A_Aqueous_Ionic_Equilibrium/18.03%3A_Buffer_Effectiveness-_Buffer_Capacity_and_Buffer_Range.txt |
Learning Objectives
• To calculate the pH at any point in an acid–base titration.
In an acid–base titration, a buret is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve. The shape of the curve provides important information about what is occurring in solution during the titration.
Titrations of Strong Acids and Bases
Figure $\PageIndex{1a}$ shows a plot of the pH as 0.20 M $\ce{HCl}$ is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as $\ce{HCl}$ is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of $\ce{HCl}$ (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M $\ce{NaOH}$ is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of $\ce{NaOH}$ as shown in Figure $\PageIndex{1b}$. As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid.
Suppose that we now add 0.20 M $\ce{NaOH}$ to 50.0 mL of a 0.10 M solution of $\ce{HCl}$. Because $\ce{HCl}$ is a strong acid that is completely ionized in water, the initial $[H^+]$ is 0.10 M, and the initial pH is 1.00. Adding $\ce{NaOH}$ decreases the concentration of H+ because of the neutralization reaction (Figure $\PageIndex{2a}$):
$\ce{OH^{−} + H^{+} <=> H_2O}. \nonumber$
Thus the pH of the solution increases gradually. Near the equivalence point, however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the $\ce{H^{+}}$ ions originally present have been consumed. For the titration of a monoprotic strong acid ($\ce{HCl}$) with a monobasic strong base ($\ce{NaOH}$), we can calculate the volume of base needed to reach the equivalence point from the following relationship:
$moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \label{Eq1}$
If 0.20 M $\ce{NaOH}$ is added to 50.0 mL of a 0.10 M solution of $\ce{HCl}$, we solve for $V_b$:
$V_b(0.20 Me)=0.025 L=25 mL \nonumber$
At the equivalence point (when 25.0 mL of $\ce{NaOH}$ solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more $\ce{NaOH}$ produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M $NaOH$.
As shown in Figure $\PageIndex{2b}$, the titration of 50.0 mL of a 0.10 M solution of $\ce{NaOH}$ with 0.20 M $\ce{HCl}$ produces a titration curve that is nearly the mirror image of the titration curve in Figure $\PageIndex{2a}$. The pH is initially 13.00, and it slowly decreases as $\ce{HCl}$ is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M $\ce{HCl}$.
The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.
The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities.
Example $1$: Hydrochloric Acid
Calculate the pH of the solution after 24.90 mL of 0.200 M $\ce{NaOH}$ has been added to 50.00 mL of 0.100 M $\ce{HCl}$.
Given: volumes and concentrations of strong base and acid
Asked for: pH
Strategy:
1. Calculate the number of millimoles of $\ce{H^{+}}$ and $\ce{OH^{-}}$ to determine which, if either, is in excess after the neutralization reaction has occurred. If one species is in excess, calculate the amount that remains after the neutralization reaction.
2. Determine the final volume of the solution. Calculate the concentration of the species in excess and convert this value to pH.
Solution
A Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of $\ce{H^{+}}$ in 50.00 mL of 0.100 M $\ce{HCl}$ can be calculated as follows:
$50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \nonumber$
The number of millimoles of $\ce{NaOH}$ added is as follows:
$24.90 \cancel{mL} \left ( \dfrac{0.200 \;mmol \;NaOH}{\cancel{mL}} \right )= 4.98 \;mmol \;NaOH=4.98 \;mmol \;OH^{-} \nonumber$
Thus $\ce{H^{+}}$ is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of $\ce{OH^{-}}$ to the $\ce{HCl}$ solution. Because only 4.98 mmol of $OH^-$ has been added, the amount of excess $\ce{H^{+}}$ is 5.00 mmol − 4.98 mmol = 0.02 mmol of $H^+$.
B The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of $\ce{H^{+}}$ is as follows:
$\left [ H^{+} \right ]= \dfrac{0.02 \;mmol \;H^{+}}{74.90 \; mL}=3 \times 10^{-4} \; M \nonumber$
Hence,
$pH \approx −\log[\ce{H^{+}}] = −\log(3 \times 10^{-4}) = 3.5 \nonumber$
This is significantly less than the pH of 7.00 for a neutral solution.
Exercise $1$
Calculate the pH of a solution prepared by adding $40.00\; mL$ of $0.237\; M$ $HCl$ to $75.00\; mL$ of a $0.133 M$ solution of $NaOH$.
Answer
11.6
pH after the addition of 10 ml of Strong Base to a Strong Acid:
https://youtu.be/_cM1_-kdJ20 (opens in new window)
pH at the Equivalence Point in a Strong Acid/Strong Base Titration:
https://youtu.be/7POGDA5Ql2M
Titrations of Weak Acids and Bases
In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding $K_a$ or $K_b$. As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned previously, $[\ce{H^{+}}]$ of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its $pK_a$ and its concentration. Because only a fraction of a weak acid dissociates, $[\(\ce{H^{+}}]$ is less than $[\ce{HA}]$. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration.
Figure $\PageIndex{3a}$ shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M $\ce{NaOH}$ superimposed on the curve for the titration of 0.100 M $\ce{HCl}$ shown in part (a) in Figure $2$. Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the $\ce{HCl}$ solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the $\ce{HCl}$ titration; the magnitude of the pH change at the equivalence point depends on the $pK_a$ of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess $\ce{NaOH}$ present, regardless of whether the acid is weak or strong.
The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the $K_a$ or $K_b$.
The titration curve in Figure $\PageIndex{3a}$ was created by calculating the starting pH of the acetic acid solution before any $\ce{NaOH}$ is added and then calculating the pH of the solution after adding increasing volumes of $NaOH$. The procedure is illustrated in the following subsection and Example $2$ for three points on the titration curve, using the $pK_a$ of acetic acid (4.76 at 25°C; $K_a = 1.7 \times 10^{-5}$.
Calculating the pH of a Solution of a Weak Acid or a Weak Base
As explained discussed, if we know $K_a$ or $K_b$ and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a ICE table (i.e, initial concentrations, changes in concentrations, and final concentrations). In this situation, the initial concentration of acetic acid is 0.100 M. If we define $x$ as $[\ce{H^{+}}]$ due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows:
$\ce{CH3CO2H(aq) <=> H^{+}(aq) + CH3CO2^{−}} \nonumber$
table of concentrations for the ionization of 0.100 M acetic acid
ICE $[CH_3CO_2H]$ $[H^+]$ $[CH_3CO_2^−]$
initial 0.100 $1.00 \times 10^{−7}$ 0
change −x +x +x
final 0.100 − x x x
In this and all subsequent examples, we will ignore $[H^+]$ and $[OH^-]$ due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.45 and Equation 16.46 to check that this assumption is justified.
Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations),
\begin{align*} K_a &=\dfrac{[H^+][CH_3CO_2^-]}{[CH_3CO_2H]} \[4pt] &=\dfrac{(x)(x)}{0.100 - x} \[4pt] &\approx \dfrac{x^2}{0.100} \[4pt] &\approx 1.74 \times 10^{-5} \end{align*} \nonumber
Solving this equation gives $x = [H^+] = 1.32 \times 10^{-3}\; M$. Thus the pH of a 0.100 M solution of acetic acid is as follows:
$pH = −\log(1.32 \times 10^{-3}) = 2.879 \nonumber$
pH at the Start of a Weak Acid/Strong Base Titration: https://youtu.be/AtdBKfrfJNg
Calculating the pH during the Titration of a Weak Acid or a Weak Base
Now consider what happens when we add 5.00 mL of 0.200 M $\ce{NaOH}$ to 50.00 mL of 0.100 M $CH_3CO_2H$ (part (a) in Figure $3$). Because the neutralization reaction proceeds to completion, all of the $OH^-$ ions added will react with the acetic acid to generate acetate ion and water:
$CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \label{Eq2}$
All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation to determine $[\ce{H^{+}}]$ of the resulting solution.
Step 1
To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of $\ce{CH_3CO_2H}$ in the original solution and the amount of $\ce{OH^{-}}$ in the $\ce{NaOH}$ solution that was added. The acetic acid solution contained
$50.00 \; \cancel{mL} (0.100 \;mmol (\ce{CH_3CO_2H})/\cancel{mL} )=5.00\; mmol (\ce{CH_3CO_2H}) \nonumber$
The $\ce{NaOH}$ solution contained
5.00 mL=1.00 mmol $NaOH$
Comparing the amounts shows that $CH_3CO_2H$ is in excess. Because $OH^-$ reacts with $CH_3CO_2H$ in a 1:1 stoichiometry, the amount of excess $CH_3CO_2H$ is as follows:
5.00 mmol $CH_3CO_2H$ − 1.00 mmol $OH^-$ = 4.00 mmol $CH_3CO_2H$
Each 1 mmol of $OH^-$ reacts to produce 1 mmol of acetate ion, so the final amount of $CH_3CO_2^−$ is 1.00 mmol.
The stoichiometry of the reaction is summarized in the following ICE table, which shows the numbers of moles of the various species, not their concentrations.
$\ce{CH3CO2H(aq) + OH^{−} (aq) <=> CH3CO2^{-}(aq) + H2O(l)} \nonumber$
ICE table
ICE $[\ce{CH_3CO_2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH_3CO_2^{−}}]$
initial 5.00 mmol 1.00 mmol 0 mmol
change −1.00 mmol −1.00 mmol +1.00 mmol
final 4.00 mmol 0 mmol 1.00 mmol
This ICE table gives the initial amount of acetate and the final amount of $OH^-$ ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of $CH_3CO_2^−$ in equilibrium is insignificant compared to the amount of $OH^-$ added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of $OH^-$, but the amount of $OH^-$ due to the autoionization of water is insignificant compared to the amount of $OH^-$ added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem.
Step 2
To calculate $[\ce{H^{+}}]$ at equilibrium following the addition of $NaOH$, we must first calculate [$\ce{CH_3CO_2H}$] and $[\ce{CH3CO2^{−}}]$ using the number of millimoles of each and the total volume of the solution at this point in the titration:
$final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL \nonumber$ $\left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M \nonumber$ $\left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \nonumber$
Knowing the concentrations of acetic acid and acetate ion at equilibrium and $K_a$ for acetic acid ($1.74 \times 10^{-5}$), we can calculate $[H^+]$ at equilibrium:
$K_{a}=\dfrac{\left [ CH_{3}CO_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ CH_{3}CO_{2}H \right ]} \nonumber$
$\left [ H^{+} \right ]=\dfrac{K_{a}\left [ CH_{3}CO_{2}H \right ]}{\left [ CH_{3}CO_{2}^{-} \right ]} = \dfrac{\left ( 1.72 \times 10^{-5} \right )\left ( 7.27 \times 10^{-2} \;M\right )}{\left ( 1.82 \times 10^{-2} \right )}= 6.95 \times 10^{-5} \;M \nonumber$
Calculating $−\log[\ce{H^{+}}]$ gives
$pH = −\log(6.95 \times 10^{−5}) = 4.158. \nonumber$
Comparing the titration curves for $\ce{HCl}$ and acetic acid in Figure $\PageIndex{3a}$, we see that adding the same amount (5.00 mL) of 0.200 M $\ce{NaOH}$ to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for $\ce{HCl}$ (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example $2$, we calculate another point for constructing the titration curve of acetic acid.
pH Before the Equivalence Point of a Weak Acid/Strong Base Titration:
https://youtu.be/znpwGCsefXc
Example $2$
What is the pH of the solution after 25.00 mL of 0.200 M $\ce{NaOH}$ is added to 50.00 mL of 0.100 M acetic acid?
Given: volume and molarity of base and acid
Asked for: pH
Strategy:
1. Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$. Determine which species, if either, is present in excess.
2. Tabulate the results showing initial numbers, changes, and final numbers of millimoles.
3. If excess acetate is present after the reaction with $\ce{OH^{-}}$, write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present.
4. Calculate $K_b$ using the relationship $K_w = K_aK_b$. Calculate [OH−] and use this to calculate the pH of the solution.
Solution
A Ignoring the spectator ion ($Na^+$), the equation for this reaction is as follows:
$CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \nonumber$
The initial numbers of millimoles of $OH^-$ and $CH_3CO_2H$ are as follows:
25.00 mL(0.200 mmol OH−mL=5.00 mmol $OH-$
$50.00\; mL (0.100 CH_3CO_2 HL=5.00 mmol \; CH_3CO_2H \nonumber$
The number of millimoles of $OH^-$ equals the number of millimoles of $CH_3CO_2H$, so neither species is present in excess.
B Because the number of millimoles of $OH^-$ added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form.
$CH_3CO_2H_{(aq)}+OH^-_{(aq)} \rightleftharpoons CH_3CO_2^{-}(aq)+H_2O(l) \nonumber$
results of the neutralization reaction
ICE $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$ $[\ce{CH3CO2^{−}}]$
initial 5.00 mmol 5.00 mmol 0 mmol
change −5.00 mmol −5.00 mmol +5.00 mmol
final 0 mmol 0 mmol 5.00 mmol
C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction:
$[\ce{CH_3CO_2}]=\dfrac{5.00 \;mmol \; CH_3CO_2^{-}}{(50.00+25.00) \; mL}=6.67\times 10^{-2} \; M \nonumber$
The equilibrium reaction of acetate with water is as follows:
$\ce{CH_3CO^{-}2(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-} (aq)} \nonumber$
The equilibrium constant for this reaction is
$K_b = \dfrac{K_w}{K_a} \label{16.18}$
where $K_a$ is the acid ionization constant of acetic acid. We therefore define x as $[\ce{OH^{−}}]$ produced by the reaction of acetate with water. Here is the completed table of concentrations:
$H_2O_{(l)}+CH_3CO^−_{2(aq)} \rightleftharpoons CH_3CO_2H_{(aq)} +OH^−_{(aq)} \nonumber$
completed table of concentrations
$[\ce{CH3CO2^{−}}]$ $[\ce{CH3CO2H}]$ $[\ce{OH^{−}}]$
initial 0.0667 0 1.00 × 10−7
change −x +x +x
final (0.0667 − x) x x
D We can obtain $K_b$ by substituting the known values into Equation \ref{16.18}:
$K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10} \label{16.23}$
Substituting the expressions for the final values from the ICE table into Equation \ref{16.23} and solving for $x$:
\begin{align*} \dfrac{x^{2}}{0.0667} &= 5.80 \times 10^{-10} \[4pt] x &= \sqrt{(5.80 \times 10^{-10})(0.0667)} \[4pt] &= 6.22 \times 10^{-6}\end{align*} \nonumber
Thus $[OH^{−}] = 6.22 \times 10^{−6}\, M$ and the pH of the final solution is 8.794 (Figure $\PageIndex{3a}$). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce $\ce{OH^{-}}$.
Exercise $2$
Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M $\ce{HCl}$ solution to 125.0 mL of a 0.150 M solution of ammonia. The $pK_b$ of ammonia is 4.75 at 25°C.
Answer
9.23
As shown in part (b) in Figure $3$, the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid.
The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure $4$ illustrates the shape of titration curves as a function of the $pK_a$ or the $pK_b$. As the acid or the base being titrated becomes weaker (its $pK_a$ or $pK_b$ becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point.
One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in Figures $\PageIndex{4a}$ and $\PageIndex{4b}$ for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A−]. Recall that the ionization constant for a weak acid is as follows:
$K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \nonumber$
If $[HA] = [A^−]$, this reduces to $K_a = [H_3O^+]$. Taking the negative logarithm of both sides,
$−\log K_a = −\log[H_3O+] \nonumber$
From the definitions of $pK_a$ and pH, we see that this is identical to
$pK_a = pH \label{16.52}$
Thus the pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid, as indicated in part (a) in Figure $4$ for the weakest acid where we see that the midpoint for $pK_a$ = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base).
The pH at the midpoint of the titration of a weak acid is equal to the $pK_a$ of the weak acid.
Titrations of Polyprotic Acids or Bases
When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the $pK_a$ values are separated by at least three $pK_a$ units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid $H_3PO_4$ with $\ce{NaOH}$ is illustrated in Figure $5$ and shows two well-defined steps: the first midpoint corresponds to $pK_a$1, and the second midpoint corresponds to $pK_a$2. Because HPO42 is such a weak acid, $pK_a$3 has such a high value that the third step cannot be resolved using 0.100 M $\ce{NaOH}$ as the titrant.
The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure $5$. The initial pH is high, but as acid is added, the pH decreases in steps if the successive $pK_b$ values are well separated. Table E1 lists the ionization constants and $pK_a$ values for some common polyprotic acids and bases.
Example $3$
Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M $\ce{NaOH}$ solution to 100.0 mL of a 0.0510 M solution of oxalic acid ($\ce{HO_2CCO_2H}$), a diprotic acid (abbreviated as $\ce{H2ox}$). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion ($\ce{O2CCO2^{2−}}$, abbreviated $\ce{ox^{2-}}$).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as $\ce{Ca^{2+}}$ are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids.
Given: volume and concentration of acid and base
Asked for: pH
Strategy:
1. Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution.
2. Calculate the concentrations of all the species in the final solution. Determine $\ce{[H{+}]}$ and convert this value to pH.
Solution:
A Table E5 gives the $pK_a$ values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present:
$100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \nonumber$
$55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \nonumber$
The strongest acid ($H_2ox$) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of $OH^-$ to react with Hox−, forming ox2 and H2O. The reactions can be written as follows:
$\underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \nonumber$
$\underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \nonumber$
In tabular form,
Solutions to Example 17.3.3
$\ce{H2ox}$ $\ce{OH^{-}}$ $\ce{Hox^{−}}$ $\ce{ox^{2−}}$
initial 5.10 mmol 6.60 mmol 0 mmol 0 mmol
change (step 1) −5.10 mmol −5.10 mmol +5.10 mmol 0 mmol
final (step 1) 0 mmol 1.50 mmol 5.10 mmol 0 mmol
change (step 2) −1.50 mmol −1.50 mmol +1.50 mmol
final 0 mmol 0 mmol 3.60 mmol 1.50 mmol
B The equilibrium between the weak acid ($\ce{Hox^{-}}$) and its conjugate base ($\ce{ox^{2-}}$) in the final solution is determined by the magnitude of the second ionization constant, $K_{a2} = 10^{−3.81} = 1.6 \times 10^{−4}$. To calculate the pH of the solution, we need to know $\ce{[H^{+}]}$, which is determined using exactly the same method as in the acetic acid titration in Example $2$:
$\text{final volume of solution} = 100.0\, mL + 55.0\, mL = 155.0 \,mL \nonumber$
Thus the concentrations of $\ce{Hox^{-}}$ and $\ce{ox^{2-}}$ are as follows:
$\left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \nonumber$
$\left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \nonumber$
We can now calculate [H+] at equilibrium using the following equation:
$K_{a2} =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \nonumber$
Rearranging this equation and substituting the values for the concentrations of $\ce{Hox^{−}}$ and $\ce{ox^{2−}}$,
$\left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \nonumber$
So
$pH = -\log\left [ H^{+} \right ]= -\log\left ( 3.7 \times 10^{-4} \right )= 3.43 \nonumber$
This answer makes chemical sense because the pH is between the first and second $pK_a$ values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than $pK_{a1}$), but we added only enough to titrate less than half of the second, less acidic proton, with $pK_{a2}$. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to $pK_{a2}$.
Exercise $3$: Piperazine
Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine ($pK_{b1}$ = 4.27, $pK_{b2}$ = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M $\ce{HCl}$ (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine.
Answer
pH=4.9
Indicators
In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful.
We can describe the chemistry of indicators by the following general equation:
$\ce{ HIn (aq) <=> H^{+}(aq) + In^{-}(aq)} \nonumber$
where the protonated form is designated by $\ce{HIn}$ and the conjugate base by $\ce{In^{−}}$. The ionization constant for the deprotonation of indicator $\ce{HIn}$ is as follows:
$K_{In} =\dfrac{ [\ce{H^{+}} ][ \ce{In^{-}}]}{[\ce{HIn}]} \label{Eq3}$
The $pK_{in}$ (its $pK_a$) determines the pH at which the indicator changes color.
Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH. Similarly, Hydrangea macrophylla flowers can be blue, red, pink, light purple, or dark purple depending on the soil pH (Figure $6$). Acidic soils will produce blue flowers, whereas alkaline soils will produce pinkish flowers.
Irrespective of the origins, a good indicator must have the following properties:
• The color change must be easily detected.
• The color change must be rapid.
• The indicator molecule must not react with the substance being titrated.
• To minimize errors, the indicator should have a $pK_{in}$ that is within one pH unit of the expected pH at the equivalence point of the titration.
Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure $7$ shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.
It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see later, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units.
We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure $8$. This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. The pH ranges over which two common indicators (methyl red, $pK_{in} = 5.0$, and phenolphthalein, $pK_{in} = 9.5$) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the $\ce{HCl}$ titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of $\ce{NaOH}$ has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the $\ce{HCl}$ titration, the phenolphthalein indicator will turn pink when about 50 mL of $\ce{NaOH}$ has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of $\ce{NaOH}$ will therefore cause the methyl red indicator to change color, resulting in a huge error.
The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M $NaOH$. Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the $pK_a$ of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point.
In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used.
The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure $9$).
pH Indicators: pH Indicators(opens in new window) [youtu.be]
Summary and Takeaway
Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the $pK_a$, and the $pK_b$ of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the $pK_a$ of the weak acid or the $pK_b$ of the weak base. Thus titration methods can be used to determine both the concentration and the $pK_a$ (or the $pK_b$) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/18%3A_Aqueous_Ionic_Equilibrium/18.04%3A_Titrations_and_pH_Curves.txt |
Learning Objectives
• To calculate the solubility of an ionic compound from its Ksp
We begin our discussion of solubility and complexation equilibria—those associated with the formation of complex ions—by developing quantitative methods for describing dissolution and precipitation reactions of ionic compounds in aqueous solution. Just as with acid–base equilibria, we can describe the concentrations of ions in equilibrium with an ionic solid using an equilibrium constant expression.
The Solubility Product
When a slightly soluble ionic compound is added to water, some of it dissolves to form a solution, establishing an equilibrium between the pure solid and a solution of its ions. For the dissolution of calcium phosphate, one of the two main components of kidney stones, the equilibrium can be written as follows, with the solid salt on the left:
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \label{Eq1}$
As you will discover in Section 17.4 and in more advanced chemistry courses, basic anions, such as S2−, PO43, and CO32, react with water to produce OH and the corresponding protonated anion. Consequently, their calculated molarities, assuming no protonation in aqueous solution, are only approximate.
The equilibrium constant for the dissolution of a sparingly soluble salt is the solubility product (Ksp) of the salt. Because the concentration of a pure solid such as Ca3(PO4)2 is a constant, it does not appear explicitly in the equilibrium constant expression. The equilibrium constant expression for the dissolution of calcium phosphate is therefore
$K=\dfrac{[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2}{[\mathrm{Ca_3(PO_4)_2}]} \label{Eq2a}$
$[\mathrm{Ca_3(PO_4)_2}]K=K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2 \label{Eq2b}$
At 25°C and pH 7.00, Ksp for calcium phosphate is 2.07 × 10−33, indicating that the concentrations of Ca2+ and PO43 ions in solution that are in equilibrium with solid calcium phosphate are very low. The values of Ksp for some common salts are listed in Table $1$, which shows that the magnitude of Ksp varies dramatically for different compounds. Although Ksp is not a function of pH in Equations $\ref{Eq2a}$ and $\ref{Eq2b}$, changes in pH can affect the solubility of a compound as discussed later.
As with any K, the concentration of a pure solid does not appear explicitly in Ksp.
Table $1$: Solubility Products for Selected Ionic Substances at 25°C
Solid Color $K_{sp}$ Solid Color $K_{sp}$
*These contain the Hg22+ ion.
Acetates Iodides
Ca(O2CCH3)2·3H2O white 4 × 10−3 Hg2I2* yellow 5.2 × 10−29
Bromides PbI2 yellow 9.8 × 10−9
AgBr off-white 5.35 × 10−13 Oxalates
Hg2Br2* yellow 6.40 × 10−23 Ag2C2O4 white 5.40 × 10−12
Carbonates MgC2O4·2H2O white 4.83 × 10−6
CaCO3 white 3.36 × 10−9 PbC2O4 white 4.8 × 10−10
PbCO3 white 7.40 × 10−14 Phosphates
Chlorides Ag3PO4 white 8.89 × 10−17
AgCl white 1.77 × 10−10 Sr3(PO4)2 white 4.0 × 10−28
Hg2Cl2* white 1.43 × 10−18 FePO4·2H2O pink 9.91 × 10−16
PbCl2 white 1.70 × 10−5 Sulfates
Chromates Ag2SO4 white 1.20 × 10−5
CaCrO4 yellow 7.1 × 10−4 BaSO4 white 1.08 × 10−10
PbCrO4 yellow 2.8 × 10−13 PbSO4 white 2.53 × 10−8
Fluorides Sulfides
BaF2 white 1.84 × 10−7 Ag2S black 6.3 × 10−50
PbF2 white 3.3 × 10−8 CdS yellow 8.0 × 10−27
Hydroxides PbS black 8.0 × 10−28
Ca(OH)2 white 5.02 × 10−6 ZnS white 1.6 × 10−24
Cu(OH)2 pale blue 1 × 10−14
Mn(OH)2 light pink 1.9 × 10−13
Cr(OH)3 gray-green 6.3 × 10−31
Fe(OH)3 rust red 2.79 × 10−39
Definition of a Solubility Product: Definition of a Solubility Product(opens in new window) [youtu.be]
Solubility products are determined experimentally by directly measuring either the concentration of one of the component ions or the solubility of the compound in a given amount of water. However, whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, $K_{sp}$, like $K$, is defined in terms of the molar concentrations of the component ions.
Example $1$
Calcium oxalate monohydrate [Ca(O2CCO2)·H2O, also written as CaC2O4·H2O] is a sparingly soluble salt that is the other major component of kidney stones [along with Ca3(PO4)2]. Its solubility in water at 25°C is 7.36 × 10−4 g/100 mL. Calculate its Ksp.
Given: solubility in g/100 mL
Asked for: Ksp
Strategy:
1. Write the balanced dissolution equilibrium and the corresponding solubility product expression.
2. Convert the solubility of the salt to moles per liter. From the balanced dissolution equilibrium, determine the equilibrium concentrations of the dissolved solute ions. Substitute these values into the solubility product expression to calculate Ksp.
Solution
A We need to write the solubility product expression in terms of the concentrations of the component ions. For calcium oxalate monohydrate, the balanced dissolution equilibrium and the solubility product expression (abbreviating oxalate as ox2) are as follows:
$\mathrm{Ca(O_2CCO_2)}\cdot\mathrm{H_2O(s)}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{^-O_2CCO_2^-(aq)}+\mathrm{H_2O(l)}\hspace{5mm}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}][\mathrm{ox^{2-}}]$
Neither solid calcium oxalate monohydrate nor water appears in the solubility product expression because their concentrations are essentially constant.
B Next we need to determine [Ca2+] and [ox2] at equilibrium. We can use the mass of calcium oxalate monohydrate that dissolves in 100 mL of water to calculate the number of moles that dissolve in 100 mL of water. From this we can determine the number of moles that dissolve in 1.00 L of water. For dilute solutions, the density of the solution is nearly the same as that of water, so dissolving the salt in 1.00 L of water gives essentially 1.00 L of solution. Because each 1 mol of dissolved calcium oxalate monohydrate dissociates to produce 1 mol of calcium ions and 1 mol of oxalate ions, we can obtain the equilibrium concentrations that must be inserted into the solubility product expression. The number of moles of calcium oxalate monohydrate that dissolve in 100 mL of water is as follows:
$\dfrac{7.36\times10^{-4}\textrm{ g}}{146.1\textrm{ g/mol}}=5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2)\cdot H_2O}$
The number of moles of calcium oxalate monohydrate that dissolve in 1.00 L of the saturated solution is as follows:
$\left(\dfrac{5.04\times10^{-6}\textrm{ mol }\mathrm{Ca(O_2CCO_2\cdot)H_2O}}{\textrm{100 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1.00 L}}\right)=5.04\times10^{-5}\textrm{ mol/L}=5.04\times10^{-5}\textrm{ M}$
Because of the stoichiometry of the reaction, the concentration of Ca2+ and ox2 ions are both 5.04 × 10−5 M. Inserting these values into the solubility product expression,
$K_{sp} = [Ca^{2+}][ox^{2−}] = (5.04 \times 10^{−5})(5.04 \times10^{−5}) = 2.54 \times 10^{−9} \nonumber$
In our calculation, we have ignored the reaction of the weakly basic anion with water, which tends to make the actual solubility of many salts greater than the calculated value.
Exercise $1$: Calcite
One crystalline form of calcium carbonate (CaCO3) is "calcite", found as both a mineral and a structural material in many organisms. Calcite is found in the teeth of sea urchins. The urchins create depressions in limestone that they can settle in by grinding the rock with their teeth. Limestone, however, also consists of calcite, so how can the urchins grind the rock without also grinding their teeth? Researchers have discovered that the teeth are shaped like needles and plates and contain magnesium. The concentration of magnesium increases toward the tip, which contributes to the hardness. Moreover, each tooth is composed of two blocks of the polycrystalline calcite matrix that are interleaved near the tip. This creates a corrugated surface that presumably increases grinding efficiency. Toolmakers are particularly interested in this approach to grinding.
The solubility of calcite in water is 0.67 mg/100 mL. Calculate its Ksp.
Answer
4.5 × 10−9
The reaction of weakly basic anions with H2O tends to make the actual solubility of many salts higher than predicted.
Finding Ksp from Ion Concentrations: Finding Ksp from Ion Concentrations(opens in new window) [youtu.be]
Tabulated values of Ksp can also be used to estimate the solubility of a salt with a procedure that is essentially the reverse of the one used in Example $1$. In this case, we treat the problem as a typical equilibrium problem and set up a table of initial concentrations, changes in concentration, and final concentrations (ICE Tables), remembering that the concentration of the pure solid is essentially constant.
Example $2$
We saw that the Ksp for Ca3(PO4)2 is 2.07 × 10−33 at 25°C. Calculate the aqueous solubility of Ca3(PO4)2 in terms of the following:
1. the molarity of ions produced in solution
2. the mass of salt that dissolves in 100 mL of water at 25°C
Given: Ksp
Asked for: molar concentration and mass of salt that dissolves in 100 mL of water
Strategy:
1. Write the balanced equilibrium equation for the dissolution reaction and construct a table showing the concentrations of the species produced in solution. Insert the appropriate values into the solubility product expression and calculate the molar solubility at 25°C.
2. Calculate the mass of solute in 100 mL of solution from the molar solubility of the salt. Assume that the volume of the solution is the same as the volume of the solvent.
Solution:
1. A The dissolution equilibrium for Ca3(PO4)2 (Equation $\ref{Eq2a}$) is shown in the following ICE table. Because we are starting with distilled water, the initial concentration of both calcium and phosphate ions is zero. For every 1 mol of Ca3(PO4)2 that dissolves, 3 mol of Ca2+ and 2 mol of PO43 ions are produced in solution. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] will be +3x, and the change in [PO43] will be +2x. We can insert these values into the table.
Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43(aq)
Solutions to Example 17.4.2
Ca3(PO4)2 [Ca2+] [PO43]
initial pure solid 0 0
change +3x +2x
final pure solid 3x 2x
Although the amount of solid Ca3(PO4)2 changes as some of it dissolves, its molar concentration does not change. We now insert the expressions for the equilibrium concentrations of the ions into the solubility product expression (Equation 17.2):
\begin{align}K_{\textrm{sp}}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2&=(3x)^3(2x)^2 \2.07\times10^{-33}&=108x^5 \1.92\times10^{-35}&=x^5 \1.14\times10^{-7}\textrm{ M}&=x\end{align}
This is the molar solubility of calcium phosphate at 25°C. However, the molarity of the ions is 2x and 3x, which means that [PO43] = 2.28 × 10−7 and [Ca2+] = 3.42 × 10−7.
1. B To find the mass of solute in 100 mL of solution, we assume that the density of this dilute solution is the same as the density of water because of the low solubility of the salt, so that 100 mL of water gives 100 mL of solution. We can then determine the amount of salt that dissolves in 100 mL of water:
$\left(\dfrac{1.14\times10^{-7}\textrm{ mol}}{\textrm{1 L}}\right)\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}} \right )\left(\dfrac{310.18 \textrm{ g }\mathrm{Ca_3(PO_4)_2}}{\textrm{1 mol}}\right)=3.54\times10^{-6}\textrm{ g }\mathrm{Ca_3(PO_4)_2}$
Exercise $2$
The solubility product of silver carbonate (Ag2CO3) is 8.46 × 10−12 at 25°C. Calculate the following:
1. the molarity of a saturated solution
2. the mass of silver carbonate that will dissolve in 100 mL of water at this temperature
Answer
1. 1.28 × 10−4 M
2. 3.54 mg
Finding the Solubility of a Salt: Finding the Solubility of a Salt (opens in new window) [youtu.be]
The Ion Product
The ion product (Q) of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient (Q) discussed for gaseous equilibria. Whereas Ksp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations.
The ion product Q is analogous to the reaction quotient Q for gaseous equilibria.
As summarized in Figure $1$, there are three possible conditions for an aqueous solution of an ionic solid:
• Q < Ksp. The solution is unsaturated, and more of the ionic solid, if available, will dissolve.
• Q = Ksp. The solution is saturated and at equilibrium.
• Q > Ksp. The solution is supersaturated, and ionic solid will precipitate.
The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.
Example $3$
We mentioned that barium sulfate is used in medical imaging of the gastrointestinal tract. Its solubility product is 1.08 × 10−10 at 25°C, so it is ideally suited for this purpose because of its low solubility when a “barium milkshake” is consumed by a patient. The pathway of the sparingly soluble salt can be easily monitored by x-rays. Will barium sulfate precipitate if 10.0 mL of 0.0020 M Na2SO4 is added to 100 mL of 3.2 × 10−4 M BaCl2? Recall that NaCl is highly soluble in water.
Given: Ksp and volumes and concentrations of reactants
Asked for: whether precipitate will form
Strategy:
1. Write the balanced equilibrium equation for the precipitation reaction and the expression for Ksp.
2. Determine the concentrations of all ions in solution when the solutions are mixed and use them to calculate the ion product (Q).
3. Compare the values of Q and Ksp to decide whether a precipitate will form.
Solution
A The only slightly soluble salt that can be formed when these two solutions are mixed is BaSO4 because NaCl is highly soluble. The equation for the precipitation of BaSO4 is as follows:
$BaSO_{4(s)} \rightleftharpoons Ba^{2+}_{(aq)} + SO^{2−}_{4(aq)} \nonumber$
The solubility product expression is as follows:
Ksp = [Ba2+][SO42] = 1.08×10−10
B To solve this problem, we must first calculate the ion product—Q = [Ba2+][SO42]—using the concentrations of the ions that are present after the solutions are mixed and before any reaction occurs. The concentration of Ba2+ when the solutions are mixed is the total number of moles of Ba2+ in the original 100 mL of BaCl2 solution divided by the final volume (100 mL + 10.0 mL = 110 mL):
$\textrm{moles Ba}^{2+}=\textrm{100 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{3.2\times10^{-4}\textrm{ mol}}{\textrm{1 L}} \right )=3.2\times10^{-5}\textrm{ mol Ba}^{2+}$
$[\mathrm{Ba^{2+}}]=\left(\dfrac{3.2\times10^{-5}\textrm{ mol Ba}^{2+}}{\textrm{110 mL}}\right)\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=2.9\times10^{-4}\textrm{ M Ba}^{2+}$
Similarly, the concentration of SO42 after mixing is the total number of moles of SO42 in the original 10.0 mL of Na2SO4 solution divided by the final volume (110 mL):
$\textrm{moles SO}_4^{2-}=\textrm{10.0 mL}\left(\dfrac{\textrm{1 L}}{\textrm{1000 mL}}\right)\left(\dfrac{\textrm{0.0020 mol}}{\textrm{1 L}}\right)=2.0\times10^{-5}\textrm{ mol SO}_4^{2-}$
$[\mathrm{SO_4^{2-}}]=\left(\dfrac{2.0\times10^{-5}\textrm{ mol SO}_4^{2-}}{\textrm{110 mL}} \right )\left(\dfrac{\textrm{1000 mL}}{\textrm{1 L}}\right)=1.8\times10^{-4}\textrm{ M SO}_4^{2-}$
We can now calculate Q:
Q = [Ba2+][SO42] = (2.9×10−4)(1.8×10−4) = 5.2×10−8
C We now compare Q with the Ksp. If Q > Ksp, then BaSO4 will precipitate, but if Q < Ksp, it will not. Because Q > Ksp, we predict that BaSO4 will precipitate when the two solutions are mixed. In fact, BaSO4 will continue to precipitate until the system reaches equilibrium, which occurs when [Ba2+][SO42] = Ksp = 1.08 × 10−10.
Exercise $3$
The solubility product of calcium fluoride (CaF2) is 3.45 × 10−11. If 2.0 mL of a 0.10 M solution of NaF is added to 128 mL of a 2.0 × 10−5M solution of Ca(NO3)2, will CaF2 precipitate?
Answer
yes (Q = 4.7 × 10−11 > Ksp)
Determining if a Precipitate forms (The Ion Product): Determining if a Precipitate forms (The Ion Product)(opens in new window) [youtu.be]
The Common Ion Effect and Solubility
The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that Ksp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. Adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Chatelier’s principle. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion. The exceptions generally involve the formation of complex ions, which is discussed later.
Consider, for example, the effect of adding a soluble salt, such as CaCl2, to a saturated solution of calcium phosphate [Ca3(PO4)2]. We have seen that the solubility of Ca3(PO4)2 in water at 25°C is 1.14 × 10−7 M (Ksp = 2.07 × 10−33). Thus a saturated solution of Ca3(PO4)2 in water contains 3 × (1.14 × 10−7 M) = 3.42 × 10−7 M Ca2+ and 2 × (1.14 × 10−7 M) = 2.28 × 10−7 M PO43, according to the stoichiometry shown in Equation $\ref{Eq1}$ (neglecting hydrolysis to form HPO42 as described in Chapter 16). If CaCl2 is added to a saturated solution of Ca3(PO4)2, the Ca2+ ion concentration will increase such that [Ca2+] > 3.42 × 10−7 M, making Q > Ksp. The only way the system can return to equilibrium is for the reaction in Equation $\ref{Eq1}$ to proceed to the left, resulting in precipitation of Ca3(PO4)2. This will decrease the concentration of both Ca2+ and PO43 until Q = Ksp.
The common ion effect usually decreases the solubility of a sparingly soluble salt.
Example $4$
Calculate the solubility of calcium phosphate [Ca3(PO4)2] in 0.20 M CaCl2.
Given: concentration of CaCl2 solution
Asked for: solubility of Ca3(PO4)2 in CaCl2 solution
Strategy:
1. Write the balanced equilibrium equation for the dissolution of Ca3(PO4)2. Tabulate the concentrations of all species produced in solution.
2. Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca3(PO4)2.
Solution
A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca3(PO4)2 in moles per liter, then the change in [Ca2+] is once again +3x, and the change in [PO43] is +2x. We can insert these values into the ICE table.
$Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)} \nonumber$
Solutions to Example 17.4.4
Ca3(PO4)2 [Ca2+] [PO43]
initial pure solid 0.20 0
change +3x +2x
final pure solid 0.20 + 3x 2x
B The Ksp expression is as follows:
Ksp = [Ca2+]3[PO43]2 = (0.20 + 3x)3(2x)2 = 2.07×10−33
Because Ca3(PO4)2 is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the Ksp expression as follows:
\begin{align*}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33} \x^2&=6.5\times10^{-32} \x&=2.5\times10^{-16}\textrm{ M}\end{align*} \nonumber
This value is the solubility of Ca3(PO4)2 in 0.20 M CaCl2 at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Chatelier’s principle. With one exception, this example is identical to Example $2$—here the initial [Ca2+] was 0.20 M rather than 0.
Exercise $4$
Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10−12 at 25°C.
Answer
2.9 × 10−6 M (versus 1.3 × 10−4 M in pure water)
The Common Ion Effect in Solubility Products: The Common Ion Effect in Solubility Products(opens in new window) [youtu.be]
Summary
The solubility product (Ksp) is used to calculate equilibrium concentrations of the ions in solution, whereas the ion product (Q) describes concentrations that are not necessarily at equilibrium. The equilibrium constant for a dissolution reaction, called the solubility product (Ksp), is a measure of the solubility of a compound. Whereas solubility is usually expressed in terms of mass of solute per 100 mL of solvent, Ksp is defined in terms of the molar concentrations of the component ions. In contrast, the ion product (Q) describes concentrations that are not necessarily equilibrium concentrations. Comparing Q and Ksp enables us to determine whether a precipitate will form when solutions of two soluble salts are mixed. Adding a common cation or common anion to a solution of a sparingly soluble salt shifts the solubility equilibrium in the direction predicted by Le Chatelier’s principle. The solubility of the salt is almost always decreased by the presence of a common ion. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/18%3A_Aqueous_Ionic_Equilibrium/18.05%3A_Solubility_Equilibria_and_the_Solubility_Product_Constant.txt |
Learning Objectives
• Calculate ion concentrations to maintain a heterogeneous equilibrium.
• Calculate pH required to precipitate a metal hydroxide.
• Design experiments to separate metal ions in a solution of mixtures of metals.
A mixture of metal ions in a solution can be separated by precipitation with anions such as $\ce{Cl-}$, $\ce{Br-}$, $\ce{SO4^2-}$, $\ce{CO3^2-}$, $\ce{S^2-}$, $\ce{Cr2O4^2-}$, $\ce{PO4^2-}$, $\ce{OH-}$ etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions. There are no definite dividing lines between insoluble salts, sparingly soluble, and soluble salts, but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large.
What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws.
• All nitrates are soluble. The singly charged large $\ce{NO3-}$ ions form salts with high solubilities. So do $\ce{ClO4-}$, $\ce{ClO3-}$, $\ce{NO2-}$, $\ce{HCOO-}$, and $\ce{CH3COO-}$.
• All chlorides, bromides, and iodides are soluble except those of $\ce{Ag+}$, $\ce{Hg2^2+}$, and $\ce{Pb^2+}$. $\ce{CaF2}$, $\ce{BaF2}$, and $\ce{PbF2}$ are also insoluble.
• All sulfates are soluble, except those of $\ce{Ba^2+}$, $\ce{Sr^2+}$, and $\ce{Pb^2+}$. The doubly charged sulfates are usually less soluble than halides and nitrates.
• Most singly charge cations $\ce{K+}$, $\ce{Na+}$, $\ce{NH4+}$ form soluble salts. However, $\ce{K3Co(NO2)6}$ and $\ce{(NH4)3Co(NO2)6}$ are insoluble.
These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks.
Chemical Separation of Metal Ions
Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The Ksp values for various salts are valuable information, and some data are given in Table E3. In the first two examples, we show how barium and strontium can be separated as chromate.
Example $1$
The Ksp for strontium chromate is $3.6 \times 10^{-5}$ and the Ksp for barium chromate is $1.2 \times10^{-10}$. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?
Solution
Since the Ksp for barium chromate is smaller, we know that $\ce{BaCrO4}$ will form a precipitate first as $\ce{[CrO4^2- ]}$ increases so that Qsp for $\ce{BaCrO4}$ also increases from zero to Ksp of $\ce{BaCrO4}$, at which point, $\ce{BaCrO4}$ precipitates. As $\ce{[CrO4^2- ]}$ increases, $\ce{[Ba^2+]}$ decreases. Further increase of $\ce{[CrO4^2- ]}$ till Qsp for $\ce{SrCrO4}$ increases to Ksp of $\ce{SrCrO4}$; it then precipitates.
Let us write the equilibrium equations and data down to help us think. Let $x$ be the concentration of chromate to precipitate $\ce{Sr^2+}$, and $y$ be that to precipitate $\ce{Ba^2+}$:
$\ce{SrCrO4(s) \rightarrow Sr^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
According to the definition of Ksp we have we have $K_{\ce{sp}} = (0.30)(x) = 3.6 \times 10^{-5}$. Solving for $x$ gives
$x = \dfrac{3.6 \times 10^{-5}}{0.30} = 1.2 \times 10^{-4} M \nonumber$
Further, let $y$ be the concentration of chromate to precipitate precipitate $\ce{Ba^2+}$:
$\ce{BaCrO4(s) \rightarrow Ba^{2+}(aq) + CrO4^{2-}(aq)} \nonumber$
with $K_{\ce{sp}} = (0.30)(y) = 1.2 \times 10^{-10}$. Solving for $y$ gives
$y = \dfrac{1.2 \times 10^{-10}}{0.30} = 4.0 \times 10^{-10} \;M \nonumber$
The Ksp's for the two salts indicate $\ce{BaCrO4}$ to be much less soluble, and it will precipitate before any $\ce{SrCrO4}$ precipitates. If chromate concentration is maintained less than $1.2 \times 10^{-4} M$, then all $\ce{Sr^2+}$ ions will remain in the solution.
Discussion
In reality, controling the increase of $\ce{[CrO4^2- ]}$ is very difficult.
Example $2$
The Ksp for strontium chromate is $3.6\times 10^{-5}$ and the Ksp for barium chromate is $1.2\times 10^{-10}$. Potassium chromate is added a small amount at a time to first precipitate $\ce{BaCrO4}$. Calculate $\ce{[Ba^2+]}$ when the first trace of $\ce{SrCrO4}$ precipitate starts to form in a solution that contains 0.30 M each of $\ce{Ba^2+}$ and $\ce{Sr^2+}$ ions.
Solution
From the solution given in Example $1$, $\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M$ when $\ce{SrCrO4}$ starts to form. At this concentration, the $\ce{[Ba^2+]}$ is estimated at $3.6 \times 10^{-4} = 1.2\times 10^{-10}$.
The Ksp of $\ce{BaCrO4}$.
Thus,
$\ce{[Ba^2+]} = 3.33 \times 10^{-7}\, M \nonumber$
Very small indeed, compared to 0.30. In the fresh precipitate of $\ce{SrCrO4}$, the molar ratio of $\ce{SrCrO4}$ to $\ce{BaCrO4}$ is
$\dfrac{0.30}{3.33 \times 10^{-7}} = 9.0 \times 10^{5}. \nonumber$
Hence, the amount of $\ce{Ba^2+}$ ion in the solid is only $1 \times 10^{-6}$ (i.e., 1 ppm) of all metal ions, providing that all the solid was removed when
$\ce{[CrO4^{2-}]} = 3.6 \times 10^{-4} M. \nonumber$
Discussion
The calculation shown here indicates that the separation of $\ce{Sr}$ and $\ce{Ba}$ is pretty good. In practice, an impurity level of 1 ppm is a very small value.
Example $3$
What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task?
Solution
The Ksp's for salts of silver and lead are required. We list the Ksp's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts Ksp.
Solutions to Example 17.6.3
Salt Ksp Salt Ksp
$\ce{AgCl}$ $1.8 \times 10^{-10}$ $\ce{Ag2SO4}$ $1.4\times 10^{-5}$
$\ce{Hg2Cl2}$ $1.3\times 10^{-18}$ $\ce{BaSO4}$ $1.1\times 10^{-10}$
$\ce{PbCl2}$ $1.7 \times 10^{-5}$ $\ce{CaSO4}$ $2.4\times 10^{-5}$
$\ce{PbSO4}$ $6.3\times 10^{-7}$
$\ce{SrSO4}$ $3.2\times 10^{-7}$
Because the Ksp's $\ce{AgCl}$ and $\ce{PbCl2}$ are very different, chloride, $\ce{Cl-}$, apppears a good choice of negative ions for their separation.
The literature also indicates that $\ce{PbCl2}$ is rather soluble in warm water, and by heating the solution to 350 K (80oC), you can keep $\ce{Pb^2+}$ ions in solution and precipitate $\ce{AgCl}$ as a solid. The solubility of $\ce{AgCl}$ is very small even at high temperatures.
Discussion
Find more detailed information about the solubility of lead chloride as a function of temperature.
Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the $\ce{[Pb^2+]}$ when $\ce{Ag2SO4}$ begins to precipitate in a solution that contains 0.10 M $\ce{Ag+}$?
The Separation of Two Ions by a Difference in Solubility: The Separation of Two Ions by a Difference in Solubility(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/18%3A_Aqueous_Ionic_Equilibrium/18.06%3A_Precipitation.txt |
Learning Objectives
• To be introduced to complex ions, including ligands.
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H2O)6]3+ ion.
A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu2+ or Ru3+, have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions.
As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu2+ ion {[Cu(H2O)6]2+}. Because it is a stronger base than H2O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH3)4(H2O)2]2+ ion. Formation of the [Cu(NH3)4(H2O)2]2+ complex is accompanied by a dramatic color change, as shown in Figure $1$. The solution changes from the light blue of [Cu(H2O)6]2+ to the blue-violet characteristic of the [Cu(NH3)4(H2O)2]2+ ion.
Figure $1$: The Formation of Complex Ions. An aqueous solution of $\ce{CuSO4}$ consists of hydrated Cu2+ ions in the form of pale blue [Cu(H2O)6]2+ (left). The addition of aqueous ammonia to the solution results in the formation of the intensely blue-violet [Cu(NH3)4(H2O)2]2+ ions, usually written as [Cu(NH3)4]2+ ion (right) because ammonia, a stronger base than H2O, replaces water molecules from the hydrated Cu2+ ion. For a more complete description, see www.youtube.com/watch?v=IQNcLH6OZK0.
The Formation Constant
The replacement of water molecules from [Cu(H2O)6]2+ by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu2+ for simplicity, we can write the equilibrium reactions as follows:
\begin{align}\mathrm{Cu^{2+}(aq)}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)]^{2+}_{(aq)}}\hspace{5mm}K_1 \ \mathrm{[Cu(NH_3)]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}\hspace{3mm}K_2 \ \mathrm{[Cu(NH_3)_2]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons\mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}\hspace{3mm}K_3 \ \mathrm{[Cu(NH_3)_3]^{2+}_{(aq)}}+\mathrm{NH_{3(aq)}}&\rightleftharpoons \mathrm{[Cu(NH_3)_4]^{2+}_{(aq)}}\hspace{3mm}K_4 \end{align} \label{17.3.1}
The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu2+ ion contains six H2O ligands, but the complex ion that is produced contains only four $\ce{NH_3}$ ligands, not six.
$Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{17.3.2}$
The equilibrium constant for the formation of the complex ion from the hydrated ion is called the formation constant (Kf). The equilibrium constant expression for Kf has the same general form as any other equilibrium constant expression. In this case, the expression is as follows:
$K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{17.3.3}$
The formation constant (Kf) has the same general form as any other equilibrium constant expression.
Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu2+(aq) ion is represented as Cu2+ for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, Kf), the more stable the product. With Kf = 2.1 × 1013, the [Cu(NH3)4(H2O)2]2+ complex ion is very stable. The formation constants for some common complex ions are listed in Table $1$.
Table $1$: Common Complex Ions
Complex Ion Equilibrium Equation Kf
*Reported values are overall formation constants. Source: Data from Lange’s Handbook of Chemistry, 15th ed. (1999).
Ammonia Complexes [Ag(NH3)2]+ Ag+ + 2NH3 ⇌ [Ag(NH3)2]+ 1.1 × 107
[Cu(NH3)4]2+ Cu2+ + 4NH3 ⇌ [Cu(NH3)4]2+ 2.1 × 1013
[Ni(NH3)6]2+ Ni2+ + 6NH3 ⇌ [Ni(NH3)6]2+ 5.5 × 108
Cyanide Complexes [Ag(CN)2] Ag+ + 2CN ⇌ [Ag(CN)2] 1.1 × 1018
[Ni(CN)4]2− Ni2+ + 4CN ⇌ [Ni(CN)4]2− 2.2 × 1031
[Fe(CN)6]3− Fe3+ + 6CN ⇌ [Fe(CN)6]3− 1 × 1042
Hydroxide Complexes [Zn(OH)4]2− Zn2+ + 4OH ⇌ [Zn(OH)4]2− 4.6 × 1017
[Cr(OH)4] Cr3+ + 4OH ⇌ [Cr(OH)4] 8.0 × 1029
Halide Complexes [HgCl4]2− Hg2+ + 4Cl ⇌ [HgCl4]2− 1.2 × 1015
[CdI4]2− Cd2+ + 4I ⇌ [CdI4]2− 2.6 × 105
[AlF6]3− Al3+ + 6F ⇌ [AlF6]3− 6.9 × 1019
Other Complexes [Ag(S2O3)2]3− Ag+ + 2S2O32− ⇌ [Ag(S2O3)2]3− 2.9 × 1013
[Fe(C2O4)3]3− Fe3+ + 3C2O42− ⇌ [Fe(C2O4)3]3− 2.0 × 1020
Example $1$
If 12.5 g of $\ce{Cu(NO3)2•6H2O}$ is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu2+(aq)?
Given: mass of Cu2+ salt and volume and concentration of ammonia solution
Asked for: equilibrium concentration of Cu2+(aq)
Strategy:
1. Calculate the initial concentration of Cu2+ due to the addition of copper(II) nitrate hexahydrate. Use the stoichiometry of the reaction shown in Equation $\ref{17.3.2}$ to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations of all species in solution.
2. Substitute the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) to calculate the equilibrium concentration of Cu2+(aq).
Solution
Adding an ionic compound that contains Cu2+ to an aqueous ammonia solution will result in the formation of [Cu(NH3)4]2+(aq), as shown in Equation $\ref{17.3.2}$. We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible.
A The initial concentration of Cu2+ from the amount of added copper nitrate prior to any reaction is as follows:
$12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M}$
Because the stoichiometry of the reaction is four NH3 to one Cu2+, the amount of NH3 required to react completely with the Cu2+ is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 1013), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH3)4]2+ in the first step is complete and allow some of it to dissociate into Cu2+ and NH3 until equilibrium has been reached. If we define x as the amount of Cu2+ produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH3)4]2+ is −x, and the change in the concentration of ammonia is +4x, as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations.
$\ce{Cu^{2+} + 4NH3 <=> [Cu(NH3)4]^{2+}} \nonumber$
[Cu2+] [NH3] [[Cu(NH3)4]2+]
initial 0.0846 1.00 0
after complete reaction 0 0.66 0.0846
change +x +4x x
final x 0.66 + 4x 0.0846 − x
B Substituting the final concentrations into the expression for the formation constant (Equation $\ref{17.3.3}$) and assuming that x << 0.0846, which allows us to remove x from the sum and difference,
\begin{align*}K_\textrm f&=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}}][\mathrm{NH_3}]^4}=\dfrac{0.0846-x}{x(0.66+4x)^4}\approx\dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \x&=2.1\times10^{-14}\end{align*}
The value of x indicates that our assumption was justified. The equilibrium concentration of Cu2+(aq) in a 1.00 M ammonia solution is therefore 2.1 × 10−14 M.
Exercise $1$
The ferrocyanide ion {[Fe(CN)6]4−} is very stable, with a Kf of 1 × 1035. Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K4[Fe(CN)6].
Answer
2 × 10−6 M
The Effect of the Formation of Complex Ions on Solubility
What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a Ksp of 5.35 × 10−13 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt.
The reaction for the dissolution of silver bromide is as follows:
$AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{17.3.4a}$
with
$K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{17.3.4b}$
The equilibrium lies far to the left, and the equilibrium concentrations of Ag+ and Br ions are very low (7.31 × 10−7 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S2O32−). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate.
The reaction of Ag+ with thiosulfate is as follows:
$Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{17.3.5a}$
with
$K_f = 2.9 \times 10^{13} \label{17.3.5b}$
The magnitude of the equilibrium constant indicates that almost all Ag+ ions in solution will be immediately complexed by thiosulfate to form [Ag(S2O3)2]3−. We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together:
\begin{align}\mathrm{AgBr(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=5.35\times10^{-13} \ \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}\hspace{3mm}K_\textrm f&=2.9\times10^{13} \ \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)}\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=15\end{align} \label{17.3.6}
Comparing K with Ksp shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 1013. The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled.
If a complex ion has a large Kf, the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts.
Example $2$
Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl2). Calculate the solubility of AgCl in each situation:
1. in pure water
2. in 1.0 M KCl solution, ignoring the formation of any complex ions
3. the same solution as in part (b) except taking the formation of complex ions into account, assuming that AgCl2 is the only Ag+ complex that forms in significant concentrations
At 25°C, Ksp = 1.77 × 10−10 for AgCl and Kf = 1.1 × 105 for AgCl2.
Given: Ksp of AgCl, Kf of AgCl2, and KCl concentration
Asked for: solubility of AgCl in water and in KCl solution with and without the formation of complex ions
Strategy:
1. Write the solubility product expression for AgCl and calculate the concentration of Ag+ and Cl in water.
2. Calculate the concentration of Ag+ in the KCl solution.
3. Write balanced chemical equations for the dissolution of AgCl and for the formation of the AgCl2 complex. Add the two equations and calculate the equilibrium constant for the overall equilibrium.
4. Write the equilibrium constant expression for the overall reaction. Solve for the concentration of the complex ion.
Solution
1. A If we let x equal the solubility of AgCl, then at equilibrium [Ag+] = [Cl] = x M. Substituting this value into the solubility product expression,
Ksp = [Ag+][Cl] = (x)(x) = x2 =1.77×10−10
x = 1.33×10−5
Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10−5 M.
1. B If x equals the solubility of AgCl in the KCl solution, then at equilibrium [Ag+] = x M and [Cl] = (1.0 + x) M. Substituting these values into the solubility product expression and assuming that x << 1.0,
Ksp = [Ag+][Cl] = (x)(1.0 + x) ≈ x(1.0) = 1.77×10−10 = x
If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water.
1. C To account for the effects of the formation of complex ions, we must first write the equilibrium equations for both the dissolution and the formation of complex ions. Adding the equations corresponding to Ksp and Kf gives us an equation that describes the dissolution of AgCl in a KCl solution. The equilibrium constant for the reaction is therefore the product of Ksp and Kf:
\begin{align}\mathrm{AgCl(s)}\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Cl^-(aq)}\hspace{3mm}K_{\textrm{sp}}&=1.77\times10^{-10} \ \mathrm{Ag^+(aq)}+\mathrm{2Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K_\textrm f&=1.1\times10^{5} \ \mathrm{AgCl(s)}+\mathrm{Cl^{-}}\rightleftharpoons\mathrm{[AgCl_2]^{-}}\hspace{3mm}K&=K_{\textrm{sp}}K_{\textrm f}=1.9\times10^{-5}\end{align}
D If we let x equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl2] = x and [Cl] = 1.0 − x. Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that x << 1.0,
$K=\dfrac{[\mathrm{AgCl_2^-}]}{[\mathrm{Cl^-}]}=\dfrac{x}{1.0-x}\approx1.9\times10^{-5}=x$
That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10−5 M solution of the AgCl2 complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 105 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.)
Exercise $2$
Calculate the solubility of mercury(II) iodide (HgI2) in each situation:
1. pure water
2. a 3.0 M solution of NaI, assuming [HgI4]2− is the only Hg-containing species present in significant amounts
Ksp = 2.9 × 10−29 for HgI2 and Kf = 6.8 × 1029 for [HgI4]2−.
Answer a
1.9 × 10−10 M
Answer a
1.4 M
Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca2+ and Mg2+, which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O3POPO34−, or P2O74−) or triphosphate (P3O105−) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large:
$\ce{Ca^{2+} (aq) + O3POPO^{4−}4(aq) \rightleftharpoons [Ca(O3POPO3)]^{2−} (aq)} \label{17.3.7a}$
with
$K_f = 4\times 10^4\label{17.3.7b}$
However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research.
Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na+, but it contains fewer dissolved minerals.
Figure $2$: An MRI Image of the Heart, Arteries, and Veins. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues.
Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the 1H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately (Figure $2$). One of the most important metal ions for this application is Gd3+, which with seven unpaired electrons is highly paramagnetic. Because Gd3+(aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA5− (diethylene triamine pentaacetic acid), whose fully protonated form is shown here.
Summary
The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large Kf. A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant (Kf). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/18%3A_Aqueous_Ionic_Equilibrium/18.07%3A_Complex_Ion_Equilibria.txt |
• 19.1: Energy Spreads Out
• 19.2: Spontaneous and Nonspontaneous Processes
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system.
• 19.3: Entropy and the Second Law of Thermodynamics
Entropy (S) is a state function whose value increases with an increase in the number of available microstates.For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases.
• 19.4: Predicting Entropy and Entropy Changes for Chemical Reactions
Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system.
• 19.5: Heat Transfer and Changes in the Entropy of the Surroundings
• 19.6: Gibbs Energy
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written.
• 19.7: Free Energy Changes in Chemical Reactions- Calculating
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written.
• 19.8: Free Energy Changes for Nonstandard States - The Relationship between and
For a reversible process (with no external work), the change in free energy can be expressed in terms of volume, pressure, entropy, and temperature. If ΔG° < 0, then K > 1, and products are favored over reactants. If ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then K = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
• 19.9: Free Energy and Equilibrium- Relating to the Equilibrium Constant (K)
For a reversible process (with no external work), the change in free energy can be expressed in terms of volume, pressure, entropy, and temperature. If ΔG° < 0, then K > 1, and products are favored over reactants. If ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then K = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
19: Free Energy and Thermodynamics
Learning Objectives
• Distinguish between spontaneous and nonspontaneous processes
• Describe the dispersal of matter and energy that accompanies certain spontaneous processes
In this section, consider the differences between two types of changes in a system: Those that occur spontaneously and those that occur only with the continuous input of energy. In doing so, we’ll gain an understanding as to why some systems are naturally inclined to change in one direction under certain conditions. We’ll also gain insight into how the spontaneity of a process affects the distribution of energy and matter within the system.
Spontaneous and Nonspontaneous Processes
Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is “driven” by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.
The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure $1$).
As another example, consider the conversion of diamond into graphite (Figure $2$).
$\ce{C(s, diamond)}⟶\ce{C(s, graphite)} \label{Eq1}$
The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow, and so for all practical purposes diamonds are indeed “forever.” Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.
Dispersal of Matter and Energy
As we extend our discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas on the left and a vacuum on the right (Figure $3$). When the valve is opened, the gas spontaneously expands to fill both flasks. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.
\begin{align} w&=−PΔV \[4pt]&=0 \,\,\, \mathrm{(P=0\: in\: a\: vaccum)} \label{Eq2} \end{align}
Note as well that since the system is isolated, no heat has been exchanged with the surroundings (q = 0). The first law of thermodynamics confirms that there has been no change in the system’s internal energy as a result of this process.
\begin{align} ΔU&=q+w \tag{First Law of Thermodynamics} \[4pt] &=0+0=0 \label{Eq3}\end{align}
The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the movement of the gas appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous process took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).
Now consider two objects at different temperatures: object X at temperature TX and object Y at temperature TY, with TX > TY (Figure $4$). When these objects come into contact, heat spontaneously flows from the hotter object (X) to the colder one (Y). This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y.
$q_\ce{X}<0 \hspace{20px} \ce{and} \hspace{20px} q_\ce{Y}=−q_\ce{X}>0 \label{Eq4}$
From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.
As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.
Example $1$: Redistribution of Matter during a Spontaneous Process
Describe how matter and energy are redistributed when the following spontaneous processes take place:
1. A solid sublimes.
2. A gas condenses.
3. A drop of food coloring added to a glass of water forms a solution with uniform color.
Solution
1. Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition. However, an input of energy from the surroundings ss required for the molecules to leave the solid phase and enter the gas phase.
2. Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after the gas-to-liquid transition. As the gas molecules move together to form the droplets of liquid, they form intermolecular forces and thus release energy to the surroundings.
3. The process in question is dilution. The food dye molecules initially occupy a much smaller volume (the drop of dye solution) than they occupy once the process is complete (in the full glass of water). The process therefore entails a greater dispersal of matter. The process may also yield a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop, zero in the water), and the final state of the system contains a single dye concentration throughout. This process can occur with out a change in energy because the molecules have kinetic energy relative to the temperature of the water, and so will be constantly in motion.
Exercise $1$
Describe how matter and energy are redistributed when you empty a canister of compressed air into a room.
Answer
This process entails both a greater and more uniform dispersal of matter as the compressed air in the canister is permitted to expand into the lower-pressure air of the room. The process also requires an input of energy to disrupt the intermolecular forces between the closely-spaced gas molecules that are originally compressed into the container. If you were to touch the nozzle of the canister, you would notice that it is cold because the exiting molecules are taking energy away from their surroundings, and the canister is part of the surroundings.
Summary
Chemical and physical processes have a natural tendency to occur in one direction under certain conditions. A spontaneous process occurs without the need for a continual input of energy from some external source, while a nonspontaneous process requires such. Systems undergoing a spontaneous process may or may not experience a gain or loss of energy, but they will experience a change in the way matter and/or energy is distributed within the system. In this section we have only discussed nuclear decay, physical changes of pure substances, and macroscopic events such as water flowing downhill. In the following sections we will discuss mixtures and chemical reactions, situations in which the description of sponteneity becomes more challenging.
Glossary
nonspontaneous process
process that requires continual input of energy from an external source
spontaneous change
process that takes place without a continuous input of energy from an external source | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/19%3A_Free_Energy_and_Thermodynamics/19.02%3A_Spontaneous_and_Nonspontaneous_Processes.txt |
Learning Objectives
• To understand the relationship between internal energy and entropy.
The first law of thermodynamics governs changes in the state function we have called internal energy ($U$). Changes in the internal energy (ΔU) are closely related to changes in the enthalpy (ΔH), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously.
Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved.
Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously.
For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH4NO3, NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., ΔHsoln > 0). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure $1$.
Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (S), a thermodynamic property of all substances that is proportional to their degree of "disorder". In Chapter 13, we introduced the concept of entropy in relation to solution formation. Here we further explore the nature of this state function and define it mathematically.
Entropy
Chemical and physical changes in a system may be accompanied by either an increase or a decrease in the disorder of the system, corresponding to an increase in entropy (ΔS > 0) or a decrease in entropy (ΔS < 0), respectively. As with any other state function, the change in entropy is defined as the difference between the entropies of the final and initial states: ΔS = Sf − Si.
When a gas expands into a vacuum, its entropy increases because the increased volume allows for greater atomic or molecular disorder. The greater the number of atoms or molecules in the gas, the greater the disorder. The magnitude of the entropy of a system depends on the number of microscopic states, or microstates, associated with it (in this case, the number of atoms or molecules); that is, the greater the number of microstates, the greater the entropy.
We can illustrate the concepts of microstates and entropy using a deck of playing cards, as shown in Figure $2$. In any new deck, the 52 cards are arranged by four suits, with each suit arranged in descending order. If the cards are shuffled, however, there are approximately 1068 different ways they might be arranged, which corresponds to 1068 different microscopic states. The entropy of an ordered new deck of cards is therefore low, whereas the entropy of a randomly shuffled deck is high. Card games assign a higher value to a hand that has a low degree of disorder. In games such as five-card poker, only 4 of the 2,598,960 different possible hands, or microstates, contain the highly ordered and valued arrangement of cards called a royal flush, almost 1.1 million hands contain one pair, and more than 1.3 million hands are completely disordered and therefore have no value. Because the last two arrangements are far more probable than the first, the value of a poker hand is inversely proportional to its entropy.
We can see how to calculate these kinds of probabilities for a chemical system by considering the possible arrangements of a sample of four gas molecules in a two-bulb container (Figure $3$). There are five possible arrangements: all four molecules in the left bulb (I); three molecules in the left bulb and one in the right bulb (II); two molecules in each bulb (III); one molecule in the left bulb and three molecules in the right bulb (IV); and four molecules in the right bulb (V). If we assign a different color to each molecule to keep track of it for this discussion (remember, however, that in reality the molecules are indistinguishable from one another), we can see that there are 16 different ways the four molecules can be distributed in the bulbs, each corresponding to a particular microstate. As shown in Figure $3$, arrangement I is associated with a single microstate, as is arrangement V, so each arrangement has a probability of 1/16. Arrangements II and IV each have a probability of 4/16 because each can exist in four microstates. Similarly, six different microstates can occur as arrangement III, making the probability of this arrangement 6/16. Thus the arrangement that we would expect to encounter, with half the gas molecules in each bulb, is the most probable arrangement. The others are not impossible but simply less likely.
There are 16 different ways to distribute four gas molecules between the bulbs, with each distribution corresponding to a particular microstate. Arrangements I and V each produce a single microstate with a probability of 1/16. This particular arrangement is so improbable that it is likely not observed. Arrangements II and IV each produce four microstates, with a probability of 4/16. Arrangement III, with half the gas molecules in each bulb, has a probability of 6/16. It is the one encompassing the most microstates, so it is the most probable.
Instead of four molecules of gas, let’s now consider 1 L of an ideal gas at standard temperature and pressure (STP), which contains 2.69 × 1022 molecules (6.022 × 1023 molecules/22.4 L). If we allow the sample of gas to expand into a second 1 L container, the probability of finding all 2.69 × 1022 molecules in one container and none in the other at any given time is extremely small, approximately $\frac{2}{2.69 \times 10^{22}}$. The probability of such an occurrence is effectively zero. Although nothing prevents the molecules in the gas sample from occupying only one of the two bulbs, that particular arrangement is so improbable that it is never actually observed. The probability of arrangements with essentially equal numbers of molecules in each bulb is quite high, however, because there are many equivalent microstates in which the molecules are distributed equally. Hence a macroscopic sample of a gas occupies all of the space available to it, simply because this is the most probable arrangement.
A disordered system has a greater number of possible microstates than does an ordered system, so it has a higher entropy. This is most clearly seen in the entropy changes that accompany phase transitions, such as solid to liquid or liquid to gas. As you know, a crystalline solid is composed of an ordered array of molecules, ions, or atoms that occupy fixed positions in a lattice, whereas the molecules in a liquid are free to move and tumble within the volume of the liquid; molecules in a gas have even more freedom to move than those in a liquid. Each degree of motion increases the number of available microstates, resulting in a higher entropy. Thus the entropy of a system must increase during melting (ΔSfus > 0). Similarly, when a liquid is converted to a vapor, the greater freedom of motion of the molecules in the gas phase means that ΔSvap > 0. Conversely, the reverse processes (condensing a vapor to form a liquid or freezing a liquid to form a solid) must be accompanied by a decrease in the entropy of the system: ΔS < 0.
Entropy (S) is a thermodynamic property of all substances that is proportional to their degree of disorder. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy.
Experiments show that the magnitude of ΔSvap is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔSvap. For instance, ΔSvap for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure $4$, the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na+ ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an NaCl solution, disruption of the crystalline NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so ΔSsoln > 0.
Dissolving NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive.
Example $1$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of NH3(g) or 1 mol of He(g), both at 25°C
2. 1 mol of Pb(s) at 25°C or 1 mol of Pb(l) at 800°C
Given: amounts of substances and temperature
Asked for: higher entropy
Strategy:
From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy.
Solution:
1. Both substances are gases at 25°C, but one consists of He atoms and the other consists of NH3 molecules. With four atoms instead of one, the NH3 molecules have more motions available, leading to a greater number of microstates. Hence we predict that the NH3 sample will have the higher entropy.
2. The nature of the atomic species is the same in both cases, but the phase is different: one sample is a solid, and one is a liquid. Based on the greater freedom of motion available to atoms in a liquid, we predict that the liquid sample will have the higher entropy.
Exercise $1$
Predict which substance in each pair has the higher entropy and justify your answer.
1. 1 mol of He(g) at 10 K and 1 atm pressure or 1 mol of He(g) at 250°C and 0.2 atm
2. a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm or a sample of 2 mol of NH3(g) at 25°C and 1 atm
Answer a
1 mol of He(g) at 250°C and 0.2 atm (higher temperature and lower pressure indicate greater volume and more microstates)
Answer a
a mixture of 3 mol of H2(g) and 1 mol of N2(g) at 25°C and 1 atm (more molecules of gas are present)
Reversible and Irreversible Changes
Changes in entropy (ΔS), together with changes in enthalpy (ΔH), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (Pext = 0) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change.
Because work done during the expansion of a gas depends on the opposing external pressure (w = - PextΔV), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev. Whether a process is reversible or irreversible, ΔU = q + w. Because U is a state function, the magnitude of ΔU does not depend on reversibility and is independent of the path taken. So
$ΔU = q_{rev} + w_{rev} = q_{irrev} + w_{irrev} \label{Eq1}$
Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: wrev ≥ wirrev.
In other words, ΔU for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (qrev) to define entropy quantitatively.
The Relationship between Internal Energy and Entropy
Because the quantity of heat transferred (qrev) is directly proportional to the absolute temperature of an object (T) (qrev ∝ T), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder (ΔS ∝ qrev). Combining these relationships for any reversible process,
$q_{\textrm{rev}}=T\Delta S\;\textrm{ and }\;\Delta S=\dfrac{q_{\textrm{rev}}}{T} \label{Eq2}$
Because the numerator (qrev) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is wrev = −PΔV, we can express Equation $\ref{Eq1}$ as follows:
\begin{align} ΔU &= q_{rev} + w_{rev} \[4pt] &= TΔS − PΔV \label{Eq3} \end{align}
Thus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the $PV$ work done.
To illustrate the use of Equation $\ref{Eq2}$ and Equation $\ref{Eq3}$, we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its $T$ constant (Figure $5$). The internal energy of the gas does not change because the temperature of the gas does not change; that is, $ΔU = 0$ and $q_{rev} = −w_{rev}$. During expansion, ΔV > 0, so the gas performs work on its surroundings:
$w_{rev} = −PΔV < 0. \nonumber$
According to Equation $\ref{Eq3}$, this means that qrev must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the system is therefore ΔSsys = +qrev/T, and the entropy change of the surroundings is
$ΔS_{surr} = −\dfrac{q_{rev}}{T}. \nonumber$
The corresponding change in entropy of the universe is then as follows:
\begin{align*} \Delta S_{\textrm{univ}} &=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}} \[4pt] &= \dfrac{q_{\textrm{rev}}}{T}+\left(-\dfrac{q_\textrm{rev}}{T}\right) \[4pt] &= 0 \label{Eq4} \end{align*}
Thus no change in ΔSuniv has occurred.
In the initial state (top), the temperatures of a gas and the surroundings are the same. During the reversible expansion of the gas, heat must be added to the gas to maintain a constant temperature. Thus the internal energy of the gas does not change, but work is performed on the surroundings. In the final state (bottom), the temperature of the surroundings is lower because the gas has absorbed heat from the surroundings during expansion.
Now consider the reversible melting of a sample of ice at 0°C and 1 atm. The enthalpy of fusion of ice is 6.01 kJ/mol, which means that 6.01 kJ of heat are absorbed reversibly from the surroundings when 1 mol of ice melts at 0°C, as illustrated in Figure $6$. The surroundings constitute a sample of low-density carbon foam that is thermally conductive, and the system is the ice cube that has been placed on it. The direction of heat flow along the resulting temperature gradient is indicated with an arrow. From Equation $\ref{Eq2}$, we see that the entropy of fusion of ice can be written as follows:
$\Delta S_{\textrm{fus}}=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{\Delta H_{\textrm{fus}}}{T} \label{Eq5}$
By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change.
In this case, ΔSfus = (6.01 kJ/mol)/(273 K) = 22.0 J/(mol•K) = ΔSsys. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so ΔSsurr = qrev/T = −(6.01 kJ/mol)/(273 K) = −22.0 J/(mol•K). Once again, we see that the entropy of the universe does not change:
ΔSuniv = ΔSsys + ΔSsurr = 22.0 J/(mol•K) − 22.0 J/(mol•K) = 0
In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process.
The Second Law of Thermodynamics
The entropy of the universe increases during a spontaneous process. It also increases during an observable non-spontaneous process.
As an example of an irreversible process, consider the entropy changes that accompany the spontaneous and irreversible transfer of heat from a hot object to a cold one, as occurs when lava spewed from a volcano flows into cold ocean water. The cold substance, the water, gains heat (q > 0), so the change in the entropy of the water can be written as ΔScold = q/Tcold. Similarly, the hot substance, the lava, loses heat (q < 0), so its entropy change can be written as ΔShot = −q/Thot, where Tcold and Thot are the temperatures of the cold and hot substances, respectively. The total entropy change of the universe accompanying this process is therefore
$\Delta S_{\textrm{univ}}=\Delta S_{\textrm{cold}}+\Delta S_{\textrm{hot}}=\dfrac{q}{T_{\textrm{cold}}}+\left(-\dfrac{q}{T_{\textrm{hot}}}\right) \label{Eq6}$
The numerators on the right side of Equation $\ref{Eq6}$ are the same in magnitude but opposite in sign. Whether ΔSuniv is positive or negative depends on the relative magnitudes of the denominators. By definition, Thot > Tcold, so −q/Thot must be less than q/Tcold, and ΔSuniv must be positive. As predicted by the second law of thermodynamics, the entropy of the universe increases during this irreversible process. Any process for which ΔSuniv is positive is, by definition, a spontaneous one that will occur as written. Conversely, any process for which ΔSuniv is negative will not occur as written but will occur spontaneously in the reverse direction. We see, therefore, that heat is spontaneously transferred from a hot substance, the lava, to a cold substance, the ocean water. In fact, if the lava is hot enough (e.g., if it is molten), so much heat can be transferred that the water is converted to steam (Figure $7$).
Example $2$: Tin Pest
Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon was argued to have plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and may have disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C.
1. What is ΔS for this process?
2. Which is the more highly ordered form of tin—white or gray?
Given: ΔH and temperature
Asked for: ΔS and relative degree of order
Strategy:
Use Equation $\ref{Eq2}$ to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure.
Solution
1. We know from Equation $\ref{Eq2}$ that the entropy change for any reversible process is the heat transferred (in joules) divided by the temperature at which the process occurs. Because the conversion occurs at constant pressure, and ΔH and ΔU are essentially equal for reactions that involve only solids, we can calculate the change in entropy for the reversible phase transition where qrev = ΔH. Substituting the given values for ΔH and temperature in kelvins (in this case, T = 13.2°C = 286.4 K),
$\Delta S=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{(-2.1\;\mathrm{kJ/mol})(1000\;\mathrm{J/kJ})}{\textrm{286.4 K}}=-7.3\;\mathrm{J/(mol\cdot K)}$
1. The fact that ΔS < 0 means that entropy decreases when white tin is converted to gray tin. Thus gray tin must be the more highly ordered structure.
Video $1$: Time lapse tin pest reaction.
Note: Whether failing buttons were indeed a contributing factor in the failure of the invasion remains disputed; critics of the theory point out that the tin used would have been quite impure and thus more tolerant of low temperatures. Laboratory tests provide evidence that the time required for unalloyed tin to develop significant tin pest damage at lowered temperatures is about 18 months, which is more than twice the length of Napoleon's Russian campaign. It is clear though that some of the regiments employed in the campaign had tin buttons and that the temperature reached sufficiently low values (at least -40 °C)
Exercise $2$
Elemental sulfur exists in two forms: an orthorhombic form (Sα), which is stable below 95.3°C, and a monoclinic form (Sβ), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with ΔH = 0.401 kJ/mol at 1 atm.
1. What is ΔS for this process?
2. Which is the more highly ordered form of sulfur—Sα or Sβ?
Answer a
1.09 J/(mol•K)
Answer b
Sα
Entropy: Entropy(opens in new window) [youtu.be]
Summary
For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases. $\Delta S=\frac{q_{\textrm{rev}}}{T} \nonumber$
A measure of the disorder of a system is its entropy (S), a state function whose value increases with an increase in the number of available microstates. A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/19%3A_Free_Energy_and_Thermodynamics/19.03%3A_Entropy_and_the_Second_Law_of_Thermodynamics.txt |
Learning Objectives
• To calculate entropy changes for a chemical reaction
We have seen that the energy given off (or absorbed) by a reaction, and monitored by noting the change in temperature of the surroundings, can be used to determine the enthalpy of a reaction (e.g. by using a calorimeter). Tragically, there is no comparable easy way to experimentally measure the change in entropy for a reaction. Suppose we know that energy is going into a system (or coming out of it), and yet we do not observe any change in temperature. What is going on in such a situation? Changes in internal energy, that are not accompanied by a temperature change, might reflect changes in the entropy of the system.
For example, consider water at °0C at 1 atm pressure
• This is the temperature and pressure condition where liquid and solid phases of water are in equilibrium (also known as the melting point of ice)
$\ce{H2O(s) \rightarrow H2O(l)} \label{19.4.1}$
• At such a temperature and pressure we have a situation (by definition) where we have some ice and some liquid water
• If a small amount of energy is input into the system the equilibrium will shift slightly to the right (i.e. in favor of the liquid state)
• Likewise if a small amount of energy is withdrawn from the system, the equilibrium will shift to the left (more ice)
However, in both of the above situations, the energy change is not accompanied by a change in temperature (the temperature will not change until we no longer have an equilibrium condition; i.e. all the ice has melted or all the liquid has frozen)
Since the quantitative term that relates the amount of heat energy input vs. the rise in temperature is the heat capacity, it would seem that in some way, information about the heat capacity (and how it changes with temperature) would allow us to determine the entropy change in a system. In fact, values for the "standard molar entropy" of a substance have units of J/mol K, the same units as for molar heat capacity.
Standard Molar Entropy, S0
The entropy of a substance has an absolute value of 0 entropy at 0 K.
• Standard molar entropies are listed for a reference temperature (like 298 K) and 1 atm pressure (i.e. the entropy of a pure substance at 298 K and 1 atm pressure). A table of standard molar entropies at 0K would be pretty useless because it would be 0 for every substance (duh!) Standard molar entropy values are listed for a variety of substances in Table T2.
• When comparing standard molar entropies for a substance that is either a solid, liquid or gas at 298 K and 1 atm pressure, the gas will have more entropy than the liquid, and the liquid will have more entropy than the solid
• Unlike enthalpies of formation, standard molar entropies of elements are not 0.
The entropy change in a chemical reaction is given by the sum of the entropies of the products minus the sum of the entropies of the reactants. As with other calculations related to balanced equations, the coefficients of each component must be taken into account in the entropy calculation (the n, and m, terms below are there to indicate that the coefficients must be accounted for):
$\Delta S^0 = \sum_n nS^0(products) - \sum_m mS^0(reactants) \nonumber$
Example $1$: Haber Process
Calculate the change in entropy associated with the Haber process for the production of ammonia from nitrogen and hydrogen gas.
$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \nonumber$
At 298K as a standard temperature:
• S0(NH3) = 192.5 J/mol K
• S0(H2) = 130.6 J/mol K
• S0(N2) = 191.5 J/mol K
Solution
From the balanced equation we can write the equation for ΔS0 (the change in the standard molar entropy for the reaction):
ΔS0 = 2*S0(NH3) - [S0(N2) + (3*S0(H2))]
ΔS0 = 2*192.5 - [191.5 + (3*130.6)]
ΔS0 = -198.3 J/mol K
It would appear that the process results in a decrease in entropy - i.e. a decrease in disorder. This is expected because we are decreasing the number of gas molecules. In other words the N2(g) used to float around independently of the H2 gas molecules. After the reaction, the two are bonded together and can't float around freely from one another. (I guess you can consider marriage as a negative entropy process!)
To calculate ΔS° for a chemical reaction from standard molar entropies, we use the familiar “products minus reactants” rule, in which the absolute entropy of each reactant and product is multiplied by its stoichiometric coefficient in the balanced chemical equation. Example $2$ illustrates this procedure for the combustion of the liquid hydrocarbon isooctane (C8H18; 2,2,4-trimethylpentane).
ΔS° for a reaction can be calculated from absolute entropy values using the same “products minus reactants” rule used to calculate ΔH°.
Example $2$: Combustion of Octane
Use the data in Table T2 to calculate ΔS° for the combustion reaction of liquid isooctane with O2(g) to give CO2(g) and H2O(g) at 298 K.
Given: standard molar entropies, reactants, and products
Asked for: ΔS°
Strategy:
Write the balanced chemical equation for the reaction and identify the appropriate quantities in Table T2. Subtract the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products, each multiplied by their appropriate stoichiometric coefficients, to obtain ΔS° for the reaction.
Solution:
The balanced chemical equation for the complete combustion of isooctane (C8H18) is as follows:
$\ce{C8H_{18}(l) + 25/2 O2(g) \rightarrow 8CO2(g) + 9H2O(g)} \nonumber$
We calculate ΔS° for the reaction using the “products minus reactants” rule, where m and n are the stoichiometric coefficients of each product and each reactant:
\begin{align*}\Delta S^\circ_{\textrm{rxn}}&=\sum mS^\circ(\textrm{products})-\sum nS^\circ(\textrm{reactants})
\ &=[8S^\circ(\mathrm{CO_2})+9S^\circ(\mathrm{H_2O})]-[S^\circ(\mathrm{C_8H_{18}})+\dfrac{25}{2}S^\circ(\mathrm{O_2})]
\ &=\left \{ [8\textrm{ mol }\mathrm{CO_2}\times213.8\;\mathrm{J/(mol\cdot K)}]+[9\textrm{ mol }\mathrm{H_2O}\times188.8\;\mathrm{J/(mol\cdot K)}] \right \}
\ &-\left \{[1\textrm{ mol }\mathrm{C_8H_{18}}\times329.3\;\mathrm{J/(mol\cdot K)}]+\left [\dfrac{25}{2}\textrm{ mol }\mathrm{O_2}\times205.2\textrm{ J}/(\mathrm{mol\cdot K})\right ] \right \} \ &=515.3\;\mathrm{J/K}\end{align*}
ΔS° is positive, as expected for a combustion reaction in which one large hydrocarbon molecule is converted to many molecules of gaseous products.
Exercise $2$
Use the data in Table T2 to calculate ΔS° for the reaction of H2(g) with liquid benzene (C6H6) to give cyclohexane (C6H12).
Answer
−361.1 J/K
Calculating the Entropy of Reaction using S: Calculating the Entropy of Reaction using S(opens in new window) [youtu.be] | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/19%3A_Free_Energy_and_Thermodynamics/19.04%3A_Predicting_Entropy_and_Entropy_Changes_for_Chemical_Reactions.txt |
Learning Objectives
• To understand the relationship between Gibbs free energy and work.
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if ΔSuniv > 0 for a process or a reaction, then the process will occur spontaneously as written. Conversely, if ΔSuniv < 0, a process cannot occur spontaneously; if ΔSuniv = 0, the system is at equilibrium. The sign of ΔSuniv is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using ΔSuniv requires that we calculate ΔS for both a system and its surroundings. This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
Gibbs Free Energy and the Direction of Spontaneous Reactions
The Gibbs free energy ($G$), often called simply free energy, was named in honor of J. Willard Gibbs (1838–1903), an American physicist who first developed the concept. It is defined in terms of three other state functions with which you are already familiar: enthalpy, temperature, and entropy:
$G = H- TS$
Because it is a combination of state functions, $G$ is also a state function.
J. Willard Gibbs (1839–1903)
Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Chatelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes.
The criterion for predicting spontaneity is based on ($ΔG$), the change in $G$, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as $G$. At constant temperature and pressure,
$ΔG = ΔH − TΔS \label{Eq2}$
where all thermodynamic quantities are those of the system. Recall that at constant pressure, $ΔH = q$, whether a process is reversible or irreversible, and TΔS = qrev. Using these expressions, we can reduce Equation $\ref{Eq2}$ to $ΔG = q − q_{rev}$. Thus ΔG is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, q = qrev and ΔG = 0. As we shall soon see, if $ΔG$ is zero, the system is at equilibrium, and there will be no net change.
What about processes for which ΔG ≠ 0? To understand how the sign of ΔG for a system determines the direction in which change is spontaneous, we can rewrite the relationship between $\Delta S$ and $q_{rev}$, discussed earlier.
$\Delta S= \dfrac{q_{rev}}{T} \nonumber$
with the definition of $\Delta H$ in terms of $q_{rev}$
$q_{rev} = ΔH \nonumber$
to obtain
$\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}$
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, $ΔS_{univ} > 0$, so substituting we obtain
\begin{align} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \[4pt] &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align} \nonumber
Multiplying both sides of the inequality by −T reverses the sign of the inequality; rearranging,
$ΔH_{sys}−TΔS_{sys}<0 \nonumber$
which is equal to $ΔG$ (Equation $\ref{Eq2}$). We can therefore see that for a spontaneous process, $ΔG < 0$.
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation $\ref{Eq2}$ allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes ($ΔH \ll 0$) that increase the disorder of a system ($ΔS_{sys} \gg 0$) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
• If $ΔG < 0$, the process occurs spontaneously.
• If $ΔG = 0$, the system is at equilibrium.
• If $ΔG > 0$, the process is not spontaneous as written but occurs spontaneously in the reverse direction.
To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔHvap (40.657 kJ/mol). For vaporizing 1 mol of water, $ΔH = 40,657; J$, so the process is highly endothermic. From the definition of ΔS (Equation $\ref{Eq3}$), we know that for 1 mol of water,
\begin{align*} \Delta S_{\textrm{vap}}&=\dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b} \[4pt] &=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}} \[4pt] &=\textrm{108.96 J/K} \end{align*} \nonumber
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
\begin{align*}\Delta G_{100^\circ\textrm C}&=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \[4pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{0 J}\end{align*} \nonumber
The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes
\begin{align*}\Delta G_{110^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \[4pt] &=-\textrm{1091 J}\end{align*} \nonumber
At 110°C, $ΔG < 0$, and vaporization is predicted to occur spontaneously and irreversibly.
We can also calculate $ΔG$ for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
\begin{align*}\Delta G_{90^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{1088 J}\end{align*} \nonumber
At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict.
Relating Enthalpy and Entropy changes under Equilibrium Conditions
$ΔG = 0$ only if $ΔH = TΔS$.
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation $\ref{Eq2}$), setting $ΔG = 0$, and solving for $T$,
0 J=40,657 J−T(108.96 J/K)
T=373.15 K
Thus $ΔG = 0$ at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, $ΔG$ is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, $ΔG$ is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure $1$ shows how the $ΔH$ and $TΔS$ terms vary with temperature for the vaporization of water. When the two lines cross, $ΔG = 0$, and $ΔH = TΔS$.
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
The Definition of Gibbs Free Energy: The Definition of Gibbs Free Energy (opens in new window) [youtu.be]
The Relationship between ΔG and Work
In the previous subsection, we learned that the value of ΔG allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of ΔG for a process provides other important information. The change in free energy (ΔG) is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): ΔG = wmax. To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation $\ref{Eq2}$. We can rearrange this equation as follows:
$ΔH = ΔG + TΔS \label{Eq4}$
This equation tells us that when energy is released during an exothermic process (ΔH < 0), such as during the combustion of a fuel, some of that energy can be used to do work (ΔG < 0), while some is used to increase the entropy of the universe (TΔS > 0). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. (For more information on entropy and reversibility, see the previous section). Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, ΔH will never equal ΔG because ΔS has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table $1$, the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
Table $1$: Approximate Thermodynamic Efficiencies of Various Devices
Device Energy Conversion Approximate Efficiency (%)
large electrical generator mechanical → electrical 99
chemical battery chemical → electrical 90
home furnace chemical → heat 65
small electric tool electrical → mechanical 60
space shuttle engine chemical → mechanical 50
mammalian liver cell chemical → chemical 30–50
spinach leaf cell light → chemical 30
internal combustion engine chemical → mechanical 25–30
fluorescent light electrical → light 20
solar cell light → electricity 10-20
incandescent light bulb electricity → light 5
yeast cell chemical → chemical 2–4
Standard Free-Energy Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation $\ref{Eq5}$:
$ΔG° = ΔH° − TΔS° \label{Eq5}$
If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive ΔG° means that the equilibrium constant is less than 1.
Example $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$\ce{ H2(g) + O2(g) \rightleftharpoons H2O2(l)}\nonumber$
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
• S°(H2O2) = 109.6 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
1. Calculate ΔS° from the absolute molar entropy values given.
2. Use Equation $\ref{Eq5}$, the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.
Solution
A To calculate ΔG° for the reaction, we need to know $ΔH^o$, $ΔS^o$, and $T$. We are given $ΔH^o$, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \nonumber \[4pt] &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \nonumber \[4pt] &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times205.2\;\mathrm{J/(mol\cdot K)}] \right \} \nonumber \[4pt]&=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*} \nonumber
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, $ΔS^o$ is very negative for this reaction.
B Substituting the appropriate quantities into Equation $\ref{Eq5}$,
\begin{align*}\Delta G^\circ &=\Delta H^\circ -T\Delta S^\circ \[4pt] &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}]\nonumber \[4pt] &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol} \[4pt] &=-120.31\textrm{ kJ/mol}\nonumber \end{align*} \nonumber
The negative value of $ΔG^o$ indicates that the reaction is spontaneous as written. Because $ΔS^o$ and $ΔH^o$ for this reaction have the same sign, the sign of $ΔG^o$ depends on the relative magnitudes of the $ΔH^o$ and $TΔS^o$ terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable $ΔS^o$ term and for the energy needed to break bonds in the reactants.
Exercise $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l)\nonumber . \nonumber$
Is the reaction spontaneous as written at 25°C?
Hint
At 25°C, the standard enthalpy change ($ΔH^o$) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are
• S°(N2H4) = 121.2 J/(mol•K),
• S°(N2) = 191.6 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Answer
149.5 kJ/mol
no, not spontaneous
Video Solution
Determining if a Reaction is Spontaneous: Determining if a Reaction is Spontaneous(opens in new window) [youtu.be] (opens in new window)
Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation ($ΔG^∘_f$)of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has $\Delta G^∘_f = 0$. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔHf) and the standard entropy of formation (ΔSf) using the definition of free energy:
$\Delta G^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}$
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
$ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum nΔ^o_{f} (reactants) \label{Eq7a}$
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
The "Products minus Reactants" Rule
The $ΔG^o$ of a reaction can be calculated from tabulated ΔGf values (Table T1) using the “products minus reactants” rule.
Example $2$
Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data:
• ΔG°f(isooctane) = −353.2 kJ/mol,
• ΔG°f(CO2) = −394.4 kJ/mol, and
• ΔG°f(H2O) = −237.1 kJ/mol. Is the reaction spontaneous as written?
Given: balanced chemical equation and values of ΔG°f for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain ΔGrxn, remembering that ΔG°f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
$\ce{C8H_{18}(l) + 25/2 O2 (g) \rightarrow 8CO2(g) + 9H2O(l)}\nonumber$
We are given ΔGf values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, ΔGf (O2) is zero. Using the “products minus reactants” rule,
\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \nonumber \[4pt] &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \nonumber\[4pt]&-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \nonumber \[4pt] &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}})\nonumber \end{align*} \nonumber
Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the ΔGf of the stable products: water and carbon dioxide.
Exercise $2$
Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
• ΔGf(benzene) = 124.5 kJ/mol
• ΔGf (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
Answer
92.8 kJ; no
Video Solution
Calculating Grxn using Gf: Calculating Grxn using Gf(opens in new window) [youtu.be]
Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither $ΔH$ nor $ΔS$ vary greatly with temperature.
Example $3$
Calculate (a) ΔG° and (b) ΔG300°C for the reaction N2(g)+3H2(g)⇌2NH3(g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
• S°(N2) = 191.6 J/(mol•K),
• S°(H2) = 130.7 J/(mol•K),
• S°(NH3) = 192.8 J/(mol•K), and
• ΔHf (NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and ΔHf for NH3
Asked for: ΔG° and ΔG at 300°C
Strategy:
1. Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔHf for any element in its standard state is zero.
2. Substitute the appropriate values into Equation $\ref{Eq5}$ to obtain ΔG° for the reaction.
3. Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation $\ref{Eq2}$ to obtain ΔG for the reaction at 300°C.
Solution
A To calculate ΔG° for the reaction using Equation $\ref{Eq5}$, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \nonumber\ &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \nonumber\ &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \}\nonumber\ &=-198.1\textrm{ J/K (per mole of N}_2)\end{align}\nonumber
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔHf (NH3) is given, and ΔHf is zero for both N2 and H2:
\begin{align}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})]\nonumber \ &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})]\nonumber \ &=-91.8\textrm{ kJ(per mole of N}_2)\nonumber\end{align}\nonumber
B Inserting the appropriate values into Equation $\ref{Eq5}$
$\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=-\textrm{32.7 kJ (per mole of N}_2)\nonumber$
C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation $\ref{Eq2}$:
\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C}) \[4pt] &=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ\nonumber \[4pt] &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \[4pt]&=21.7\textrm{ kJ (per mole of N}_2) \end{align*} \nonumber
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise $3$
Calculate
1. $ΔG°$ and
2. $ΔG_{750°C}$
for the following reaction
$\ce{ 2NO(g) + O2 (g) \rightleftharpoons 2NO2 (g)}\nonumber$
which is important in the formation of urban smog. Assume that $ΔH$ and $ΔS$ do not change between 25.0°C and 750°C and use these data:
• S°(NO) = 210.8 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K),
• S°(NO2) = 240.1 J/(mol•K),
• ΔHf(NO2) = 33.2 kJ/mol, and
• ΔHf (NO) = 91.3 kJ/mol.
Answer a
−72.5 kJ/mol of $O_2$
Answer b
33.8 kJ/mol of $O_2$
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in Equation $\ref{Eq5}$, as illustrated in Example $4$.
Example $4$
As you saw in Example $3$, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example 9.5.3 to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set ΔG° equal to zero in Equation $\ref{Eq5}$ and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example $3$, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to ΔG° = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation $\ref{Eq5}$ to solve for T:
\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \[4pt] \Delta H^\circ &=T\Delta S^\circ \[4pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*} \nonumber
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise $4$
As you found in the exercise in Example $3$, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer
792.6 K
Video Solution
Summary
• The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
• Free-energy change:
$ΔG = ΔH − TΔS\nonumber$
• Standard free-energy change:
$ΔG° = ΔH° − TΔS°\nonumber$
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔGf), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/19%3A_Free_Energy_and_Thermodynamics/19.06%3A_Gibbs_Energy.txt |
Learning Objectives
• To understand the relationship between Gibbs free energy and work.
One of the major goals of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously. We have developed one such criterion, the change in entropy of the universe: if ΔSuniv > 0 for a process or a reaction, then the process will occur spontaneously as written. Conversely, if ΔSuniv < 0, a process cannot occur spontaneously; if ΔSuniv = 0, the system is at equilibrium. The sign of ΔSuniv is a universally applicable and infallible indicator of the spontaneity of a reaction. Unfortunately, using ΔSuniv requires that we calculate ΔS for both a system and its surroundings. This is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient and is provided by a new state function: the Gibbs free energy.
Gibbs Free Energy and the Direction of Spontaneous Reactions
The Gibbs free energy ($G$), often called simply free energy, was named in honor of J. Willard Gibbs (1838–1903), an American physicist who first developed the concept. It is defined in terms of three other state functions with which you are already familiar: enthalpy, temperature, and entropy:
$G = H- TS$
Because it is a combination of state functions, $G$ is also a state function.
J. Willard Gibbs (1839–1903)
Born in Connecticut, Josiah Willard Gibbs attended Yale, as did his father, a professor of sacred literature at Yale, who was involved in the Amistad trial. In 1863, Gibbs was awarded the first engineering doctorate granted in the United States. He was appointed professor of mathematical physics at Yale in 1871, the first such professorship in the United States. His series of papers entitled “On the Equilibrium of Heterogeneous Substances” was the foundation of the field of physical chemistry and is considered one of the great achievements of the 19th century. Gibbs, whose work was translated into French by Le Chatelier, lived with his sister and brother-in-law until his death in 1903, shortly before the inauguration of the Nobel Prizes.
The criterion for predicting spontaneity is based on ($ΔG$), the change in $G$, at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as $G$. At constant temperature and pressure,
$ΔG = ΔH − TΔS \label{Eq2}$
where all thermodynamic quantities are those of the system. Recall that at constant pressure, $ΔH = q$, whether a process is reversible or irreversible, and TΔS = qrev. Using these expressions, we can reduce Equation $\ref{Eq2}$ to $ΔG = q − q_{rev}$. Thus ΔG is the difference between the heat released during a process (via a reversible or an irreversible path) and the heat released for the same process occurring in a reversible manner. Under the special condition in which a process occurs reversibly, q = qrev and ΔG = 0. As we shall soon see, if $ΔG$ is zero, the system is at equilibrium, and there will be no net change.
What about processes for which ΔG ≠ 0? To understand how the sign of ΔG for a system determines the direction in which change is spontaneous, we can rewrite the relationship between $\Delta S$ and $q_{rev}$, discussed earlier.
$\Delta S= \dfrac{q_{rev}}{T} \nonumber$
with the definition of $\Delta H$ in terms of $q_{rev}$
$q_{rev} = ΔH \nonumber$
to obtain
$\Delta S_{\textrm{surr}}=-\dfrac{\Delta H_{\textrm{sys}}}{T} \label{Eq3}$
Thus the entropy change of the surroundings is related to the enthalpy change of the system. We have stated that for a spontaneous reaction, $ΔS_{univ} > 0$, so substituting we obtain
\begin{align} \Delta S_{\textrm{univ}}&=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}}>0 \[4pt] &=\Delta S_{\textrm{sys}}-\dfrac{\Delta H_{\textrm{sys}}}{T}>0\end{align} \nonumber
Multiplying both sides of the inequality by −T reverses the sign of the inequality; rearranging,
$ΔH_{sys}−TΔS_{sys}<0 \nonumber$
which is equal to $ΔG$ (Equation $\ref{Eq2}$). We can therefore see that for a spontaneous process, $ΔG < 0$.
The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation $\ref{Eq2}$ allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes ($ΔH \ll 0$) that increase the disorder of a system ($ΔS_{sys} \gg 0$) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results:
• If $ΔG < 0$, the process occurs spontaneously.
• If $ΔG = 0$, the system is at equilibrium.
• If $ΔG > 0$, the process is not spontaneous as written but occurs spontaneously in the reverse direction.
To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔHvap (40.657 kJ/mol). For vaporizing 1 mol of water, $ΔH = 40,657; J$, so the process is highly endothermic. From the definition of ΔS (Equation $\ref{Eq3}$), we know that for 1 mol of water,
\begin{align*} \Delta S_{\textrm{vap}}&=\dfrac{\Delta H_{\textrm{vap}}}{T_\textrm b} \[4pt] &=\dfrac{\textrm{40,657 J}}{\textrm{373.15 K}} \[4pt] &=\textrm{108.96 J/K} \end{align*} \nonumber
Hence there is an increase in the disorder of the system. At the normal boiling point of water,
\begin{align*}\Delta G_{100^\circ\textrm C}&=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \[4pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{0 J}\end{align*} \nonumber
The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.
Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes
\begin{align*}\Delta G_{110^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \[4pt] &=-\textrm{1091 J}\end{align*} \nonumber
At 110°C, $ΔG < 0$, and vaporization is predicted to occur spontaneously and irreversibly.
We can also calculate $ΔG$ for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions:
\begin{align*}\Delta G_{90^\circ\textrm C}&=\Delta H-T\Delta S \[4pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})] \[4pt] &=\textrm{1088 J}\end{align*} \nonumber
At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG110°C = 1090 J = −ΔG90°C, as we would predict.
Relating Enthalpy and Entropy changes under Equilibrium Conditions
$ΔG = 0$ only if $ΔH = TΔS$.
We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation $\ref{Eq2}$), setting $ΔG = 0$, and solving for $T$,
0 J=40,657 J−T(108.96 J/K)
T=373.15 K
Thus $ΔG = 0$ at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, $ΔG$ is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, $ΔG$ is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure $1$ shows how the $ΔH$ and $TΔS$ terms vary with temperature for the vaporization of water. When the two lines cross, $ΔG = 0$, and $ΔH = TΔS$.
A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K.
The Definition of Gibbs Free Energy: The Definition of Gibbs Free Energy (opens in new window) [youtu.be]
The Relationship between ΔG and Work
In the previous subsection, we learned that the value of ΔG allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of ΔG for a process provides other important information. The change in free energy (ΔG) is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): ΔG = wmax. To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation $\ref{Eq2}$. We can rearrange this equation as follows:
$ΔH = ΔG + TΔS \label{Eq4}$
This equation tells us that when energy is released during an exothermic process (ΔH < 0), such as during the combustion of a fuel, some of that energy can be used to do work (ΔG < 0), while some is used to increase the entropy of the universe (TΔS > 0). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. (For more information on entropy and reversibility, see the previous section). Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, ΔH will never equal ΔG because ΔS has a positive value.
One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table $1$, the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%.
Table $1$: Approximate Thermodynamic Efficiencies of Various Devices
Device Energy Conversion Approximate Efficiency (%)
large electrical generator mechanical → electrical 99
chemical battery chemical → electrical 90
home furnace chemical → heat 65
small electric tool electrical → mechanical 60
space shuttle engine chemical → mechanical 50
mammalian liver cell chemical → chemical 30–50
spinach leaf cell light → chemical 30
internal combustion engine chemical → mechanical 25–30
fluorescent light electrical → light 20
solar cell light → electricity 10-20
incandescent light bulb electricity → light 5
yeast cell chemical → chemical 2–4
Standard Free-Energy Change
We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation $\ref{Eq5}$:
$ΔG° = ΔH° − TΔS° \label{Eq5}$
If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants.
A positive ΔG° means that the equilibrium constant is less than 1.
Example $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$\ce{ H2(g) + O2(g) \rightleftharpoons H2O2(l)}\nonumber$
At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are:
• S°(H2O2) = 109.6 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Is the reaction spontaneous as written?
Given: balanced chemical equation, ΔH° and S° for reactants and products
Asked for: spontaneity of reaction as written
Strategy:
1. Calculate ΔS° from the absolute molar entropy values given.
2. Use Equation $\ref{Eq5}$, the calculated value of ΔS°, and other data given to calculate ΔG° for the reaction. Use the value of ΔG° to determine whether the reaction is spontaneous as written.
Solution
A To calculate ΔG° for the reaction, we need to know $ΔH^o$, $ΔS^o$, and $T$. We are given $ΔH^o$, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule:
\begin{align*}\Delta S^\circ &=S^\circ(\mathrm{H_2O_2})-[S^\circ(\mathrm{O_2})+S^\circ(\mathrm{H_2})] \nonumber \[4pt] &=[\mathrm{1\;mol\;H_2O_2}\times109.6\;\mathrm{J/(mol\cdot K})] \nonumber \[4pt] &-\left \{ [\textrm{1 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]+[\textrm{1 mol O}_2\times205.2\;\mathrm{J/(mol\cdot K)}] \right \} \nonumber \[4pt]&=-226.3\textrm{ J/K }(\textrm{per mole of }\mathrm{H_2O_2}) \end{align*} \nonumber
As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, $ΔS^o$ is very negative for this reaction.
B Substituting the appropriate quantities into Equation $\ref{Eq5}$,
\begin{align*}\Delta G^\circ &=\Delta H^\circ -T\Delta S^\circ \[4pt] &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}]\nonumber \[4pt] &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol} \[4pt] &=-120.31\textrm{ kJ/mol}\nonumber \end{align*} \nonumber
The negative value of $ΔG^o$ indicates that the reaction is spontaneous as written. Because $ΔS^o$ and $ΔH^o$ for this reaction have the same sign, the sign of $ΔG^o$ depends on the relative magnitudes of the $ΔH^o$ and $TΔS^o$ terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable $ΔS^o$ term and for the energy needed to break bonds in the reactants.
Exercise $1$
Calculate the standard free-energy change ($ΔG^o$) at 25°C for the reaction
$2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l)\nonumber . \nonumber$
Is the reaction spontaneous as written at 25°C?
Hint
At 25°C, the standard enthalpy change ($ΔH^o$) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are
• S°(N2H4) = 121.2 J/(mol•K),
• S°(N2) = 191.6 J/(mol•K), and
• S°(H2) = 130.7 J/(mol•K).
Answer
149.5 kJ/mol
no, not spontaneous
Video Solution
Determining if a Reaction is Spontaneous: Determining if a Reaction is Spontaneous(opens in new window) [youtu.be] (opens in new window)
Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation ($ΔG^∘_f$)of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl2 gas at 298.15 K, for example, has $\Delta G^∘_f = 0$. The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔHf) and the standard entropy of formation (ΔSf) using the definition of free energy:
$\Delta G^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}$
Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule:
$ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum nΔ^o_{f} (reactants) \label{Eq7a}$
where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction.
The "Products minus Reactants" Rule
The $ΔG^o$ of a reaction can be calculated from tabulated ΔGf values (Table T1) using the “products minus reactants” rule.
Example $2$
Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data:
• ΔG°f(isooctane) = −353.2 kJ/mol,
• ΔG°f(CO2) = −394.4 kJ/mol, and
• ΔG°f(H2O) = −237.1 kJ/mol. Is the reaction spontaneous as written?
Given: balanced chemical equation and values of ΔG°f for isooctane, CO2, and H2O
Asked for: spontaneity of reaction as written
Strategy:
Use the “products minus reactants” rule to obtain ΔGrxn, remembering that ΔG°f for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written.
Solution
The balanced chemical equation for the reaction is as follows:
$\ce{C8H_{18}(l) + 25/2 O2 (g) \rightarrow 8CO2(g) + 9H2O(l)}\nonumber$
We are given ΔGf values for all the products and reactants except O2(g). Because oxygen gas is an element in its standard state, ΔGf (O2) is zero. Using the “products minus reactants” rule,
\begin{align*} \Delta G^\circ &=[8\Delta G^\circ_\textrm f(\mathrm{CO_2})+9\Delta G^\circ_\textrm f(\mathrm{H_2O})]-\left[1\Delta G^\circ_\textrm f(\mathrm{C_8H_{18}})+\dfrac{25}{2}\Delta G^\circ_\textrm f(\mathrm{O_2})\right] \nonumber \[4pt] &=[(\textrm{8 mol})(-394.4\textrm{ kJ/mol})+(\textrm{9 mol})(-237.1\textrm{ kJ/mol})] \nonumber\[4pt]&-\left [(\textrm{1 mol})(-353.2\textrm{ kJ/mol})+\left(\dfrac{25}{2}\;\textrm{mol}\right)(0 \textrm{ kJ/mol}) \right ] \nonumber \[4pt] &=-4935.9\textrm{ kJ }(\textrm{per mol of }\mathrm{C_8H_{18}})\nonumber \end{align*} \nonumber
Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the ΔGf of the stable products: water and carbon dioxide.
Exercise $2$
Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the following data
• ΔGf(benzene) = 124.5 kJ/mol
• ΔGf (cyclohexane) = 217.3 kJ/mol.
Is the reaction spontaneous as written?
Answer
92.8 kJ; no
Video Solution
Calculating Grxn using Gf: Calculating Grxn using Gf(opens in new window) [youtu.be]
Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having Keq < 1, to one that is, having a Keq > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered.
In the absence of a phase change, neither $ΔH$ nor $ΔS$ vary greatly with temperature.
Example $3$
Calculate (a) ΔG° and (b) ΔG300°C for the reaction N2(g)+3H2(g)⇌2NH3(g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data:
• S°(N2) = 191.6 J/(mol•K),
• S°(H2) = 130.7 J/(mol•K),
• S°(NH3) = 192.8 J/(mol•K), and
• ΔHf (NH3) = −45.9 kJ/mol.
Given: balanced chemical equation, temperatures, S° values, and ΔHf for NH3
Asked for: ΔG° and ΔG at 300°C
Strategy:
1. Convert each temperature to kelvins. Then calculate ΔS° for the reaction. Calculate ΔH° for the reaction, recalling that ΔHf for any element in its standard state is zero.
2. Substitute the appropriate values into Equation $\ref{Eq5}$ to obtain ΔG° for the reaction.
3. Assuming that ΔH and ΔS are independent of temperature, substitute values into Equation $\ref{Eq2}$ to obtain ΔG for the reaction at 300°C.
Solution
A To calculate ΔG° for the reaction using Equation $\ref{Eq5}$, we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule:
\begin{align}\Delta S^\circ_{\textrm{rxn}}&=2S^\circ(\mathrm{NH_3})-[S^\circ(\mathrm{N_2})+3S^\circ(\mathrm{H_2})] \nonumber\ &=[\textrm{2 mol NH}_3\times192.8\;\mathrm{J/(mol\cdot K)}] \nonumber\ &-\left \{[\textrm{1 mol N}_2\times191.6\;\mathrm{J/(mol\cdot K)}]+[\textrm{3 mol H}_2\times130.7\;\mathrm{J/(mol\cdot K)}]\right \}\nonumber\ &=-198.1\textrm{ J/K (per mole of N}_2)\end{align}\nonumber
We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔHf (NH3) is given, and ΔHf is zero for both N2 and H2:
\begin{align}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})]\nonumber \ &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})]\nonumber \ &=-91.8\textrm{ kJ(per mole of N}_2)\nonumber\end{align}\nonumber
B Inserting the appropriate values into Equation $\ref{Eq5}$
$\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=-\textrm{32.7 kJ (per mole of N}_2)\nonumber$
C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH300°C = H° and ΔS300°C = ΔS°) and insert the appropriate temperature (573 K) into Equation $\ref{Eq2}$:
\begin{align*}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C}) \[4pt] &=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ\nonumber \[4pt] &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J}) \[4pt]&=21.7\textrm{ kJ (per mole of N}_2) \end{align*} \nonumber
In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its Keq is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable.
Exercise $3$
Calculate
1. $ΔG°$ and
2. $ΔG_{750°C}$
for the following reaction
$\ce{ 2NO(g) + O2 (g) \rightleftharpoons 2NO2 (g)}\nonumber$
which is important in the formation of urban smog. Assume that $ΔH$ and $ΔS$ do not change between 25.0°C and 750°C and use these data:
• S°(NO) = 210.8 J/(mol•K),
• S°(O2) = 205.2 J/(mol•K),
• S°(NO2) = 240.1 J/(mol•K),
• ΔHf(NO2) = 33.2 kJ/mol, and
• ΔHf (NO) = 91.3 kJ/mol.
Answer a
−72.5 kJ/mol of $O_2$
Answer b
33.8 kJ/mol of $O_2$
The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in Equation $\ref{Eq5}$, as illustrated in Example $4$.
Example $4$
As you saw in Example $3$, the reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example 9.5.3 to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature.
Given: ΔH° and ΔS°
Asked for: temperature at which reaction changes from spontaneous to nonspontaneous
Strategy:
Set ΔG° equal to zero in Equation $\ref{Eq5}$ and solve for T, the temperature at which the reaction becomes nonspontaneous.
Solution
In Example $3$, we calculated that ΔH° is −91.8 kJ/mol of N2 and ΔS° is −198.1 J/K per mole of N2, corresponding to ΔG° = −32.7 kJ/mol of N2 at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation $\ref{Eq5}$ to solve for T:
\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \[4pt] \Delta H^\circ &=T\Delta S^\circ \[4pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*} \nonumber
This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K.
Exercise $4$
As you found in the exercise in Example $3$, ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous.
Answer
792.6 K
Video Solution
Summary
• The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for predicting the spontaneity of a reaction.
• Free-energy change:
$ΔG = ΔH − TΔS\nonumber$
• Standard free-energy change:
$ΔG° = ΔH° − TΔS°\nonumber$
We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔGf), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/19%3A_Free_Energy_and_Thermodynamics/19.07%3A_Free_Energy_Changes_in_Chemical_Reactions-_Calculating.txt |
Learning Objectives
• To know the relationship between free energy and the equilibrium constant.
We have identified three criteria for whether a given reaction will occur spontaneously:
1. $ΔS_{univ} > 0$,
2. $ΔG_{sys} < 0$ (applicable under constant temperature and constant pressure conditions), and
3. the relative magnitude of the reaction quotient $Q$ versus the equilibrium constant $K$.
Recall that if $Q < K$, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if $Q > K$, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If $Q = K$, then the system is at equilibrium, and no net reaction occurs. Table $1$ summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes.
Table $1$: Criteria for the Spontaneity of a Process as Written
Spontaneous Equilibrium Nonspontaneous*
*Spontaneous in the reverse direction.
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGsys < 0 ΔGsys = 0 ΔGsys > 0
Q < K Q = K Q > K
Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. In this section, we explore the relationship between the standard free energy of reaction ($ΔG^o$) and the equilibrium constant ($K$).
Free Energy and the Equilibrium Constant
Because $ΔH^o$ and $ΔS^o$ determine the magnitude and sign of $ΔG^o$ and also because $K$ is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of $ΔG^o$ and vice versa. "Free Energy", $ΔG$ is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating $ΔH$ from the equation for $ΔG$. The general relationship can be shown as follows (derivation not shown):
$\Delta G = V \Delta P − S \Delta T \label{18.29}$
If a reaction is carried out at constant temperature ($ΔT = 0$), then Equation $\ref{18.29}$ simplifies to
$\Delta{G} = V\Delta{P} \label{18.30}$
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the $V$ in Equation $\ref{18.30}$ by $nRT/P$ (where $n$ is the number of moles of gas and $R$ is the ideal gas constant) and express $\Delta{G}$ in terms of the initial and final pressures ($P_i$ and $P_f$, respectively):
\begin{align} \Delta G &=\left(\dfrac{nRT}{P}\right)\Delta P \[4pt] &=nRT\dfrac{\Delta P}{P} \[4pt] &=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31}\end{align}
If the initial state is the standard state with $P_i = 1 \,atm$, then the change in free energy of a substance when going from the standard state to any other state with a pressure $P$ can be written as follows:
$G − G^° = nRT\ln{P} \nonumber$
This can be rearranged as follows:
$G = G^° + nRT\ln {P} \label{18.32}$
As you will soon discover, Equation $\ref{18.32}$ allows us to relate $ΔG^o$ and $K_p$. Any relationship that is true for $K_p$ must also be true for $K$ because $K_p$ and $K$ are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
$aA+bB \rightleftharpoons cC+dD \label{18.33}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for $ΔG$:
\begin{align} \Delta{G} &=\sum_m G_{products}−\sum_n G_{reactants} \[4pt] &=(cG_C+dG_D)−(aG_A+bG_B) \label{18.34} \end{align}
Substituting Equation $\ref{18.32}$ for each term into Equation $\ref{18.34}$,
$\Delta G = [(cG^o_C+cRT \ln P_C)+(dG^o_D+dRT\ln P_D)]−[(aG^o_A+aRT\ln P_A)+(bG^o_B+bRT\ln P_B)] \nonumber$
Combining terms gives the following relationship between $ΔG$ and the reaction quotient $Q$:
\begin{align} \Delta G &=\Delta G^\circ+RT \ln\left(\dfrac{P^c_\textrm CP^d_\textrm D}{P^a_\textrm AP^b_\textrm B}\right) \[4pt] &=\Delta G^\circ+RT\ln Q \label{18.35} \end{align}
where $ΔG^o$ indicates that all reactants and products are in their standard states. For gases at equilibrium ($Q = K_p$), and as you’ve learned in this chapter, $ΔG = 0$ for a system at equilibrium. Therefore, we can describe the relationship between $ΔG^o$ and $K_p$ for gases as follows:
\begin{align} 0 &= ΔG^o + RT\ln K_p \label{18.36a} \[4pt] ΔG^o &= −RT\ln K_p \label{18.36b} \end{align}
If the products and reactants are in their standard states and $ΔG^o < 0$, then $K_p > 1$, and products are favored over reactants when the reaction is at equilibrium. Conversely, if $ΔG^o > 0$, then $K_p < 1$, and reactants are favored over products when the reaction is at equilibrium. If $ΔG^o = 0$, then $K_p = 1$, and neither reactants nor products are favored when the reaction is at equilibrium.
For a spontaneous process under standard conditions, $K_{eq}$ and $K_p$ are greater than 1.
Example $1$
$ΔG^o$ is −32.7 kJ/mol of N2 for the reaction
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate $ΔG$ for the same reaction under the following nonstandard conditions:
• $P_{\textrm N_2}$ = 2.00 atm,
• $P_{\textrm H_2}$ = 7.00 atm,
• $P_{\textrm{NH}_3}$ = 0.021 atm, and
• $T = 100 ^oC$.
Does the reaction proceed to the right, as written, or to the left to reach equilibrium?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether the reaction proceeds to the right or to the left to reach equilibrium
Strategy:
1. Using the values given and Equation $\ref{18.35}$, calculate $Q$.
2. Determine if $Q$ is >, <, or = to $K$
3. Substitute the values of $ΔG^o$ and $Q$ into Equation $\ref{18.35}$ to obtain $ΔG$ for the reaction under nonstandard conditions.
Solution:
A The relationship between $ΔG^o$ and $ΔG$ under nonstandard conditions is given in Equation $\ref{18.35}$. Substituting the partial pressures given, we can calculate $Q$:
\begin{align*} Q &=\dfrac{P^2_{\textrm{NH}_3}}{P_{\textrm N_2}P^3_{\textrm H_2}} \[4pt] &=\dfrac{(0.021)^2}{(2.00)(7.00)^3}=6.4 \times 10^{-7} \end{align*} \nonumber
B Because $ΔG^o$ is −, K must be a number greater than 1
C Substituting the values of $ΔG^o$ and $Q$ into Equation $\ref{18.35}$,
\begin{align*} \Delta G &=\Delta G^\circ+RT\ln Q \[4pt] &=-32.7\textrm{ kJ}+\left[(\textrm{8.314 J/K})(\textrm{373 K})\left(\dfrac{\textrm{1 kJ}}{\textrm{1000 J}}\right)\ln(6.4\times10^{-7})\right] \[4pt] &=-32.7\textrm{ kJ}+(-44\textrm{ kJ}) \[4pt] &=-77\textrm{ kJ/mol of N}_2 \end{align*} \nonumber
Because $ΔG < 0$ and $Q < K$ (because $Q < 1$), the reaction proceeds spontaneously to the right, as written, in order to reach equilibrium.
Exercise $1$
Calculate $ΔG$ for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, PNO = 0.0100 atm, $P_{\ce{O_2}}$ = 0.200 atm, and $P_{\ce{NO_2}} = 1.00 × 10^{−4} atm$. The value of $ΔG^o$ for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer
−92.9 kJ/mol of $\ce{O2}$; the reaction is spontaneous to the right as written. The reaction will proceed in the forward direction to reach equilibrium.
Example $2$
Calculate $K_p$ for the reaction of $\ce{H_2}$ with $\ce{N2}$ to give $\ce{NH3}$ at 25°C. $ΔG^o$ for this reaction is −32.7 kJ/mol of $\ce{N2}$.
Given: balanced chemical equation from Example $1$, $ΔG^o$, and temperature
Asked for: $K_p$
Strategy:
Substitute values for $ΔG^o$ and T (in kelvin) into Equation $\ref{18.36b}$ to calculate $K_p$, the equilibrium constant for the formation of ammonia.
Solution
In Example $1$, we used tabulated values of ΔGf to calculate $ΔG^o$ for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging Equation $\ref{18.36b}$,
\begin{align*} \Delta G^\circ &=-RT\ln K_\textrm p \[4pt] \dfrac{-\Delta G^\circ}{RT} &=\ln K_\textrm p \end{align*} \nonumber
Inserting the value of $ΔG^o$ and the temperature (25°C = 298 K) into this equation,
\begin{align*}\ln K_\textrm p &=-\dfrac{(-\textrm{32.7 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298 K})}=13.2 \[4pt] K_\textrm p &=5.4\times10^5\end{align*} \nonumber
Thus the equilibrium constant for the formation of ammonia at room temperature is product-favored. However, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise $3$
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. $ΔG^o$ for this reaction is −70.5 kJ/mol of $\ce{O2}$.
Answer
2.3 × 1012
Although $K_p$ is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant $K$ is defined in terms of the concentrations of the reactants and the products. The numerical magnitude of $K_p$ and $K$ are related:
$K_p = K(RT)^{Δn} \label{18.37}$
where $Δn$ is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, $Δn = 0$, so $K_p = K$. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation $\ref{18.36b}$ can be written in a more general form:
$ΔG° = −RT \ln K \label{18.38}$
Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation $\ref{18.38}$ for the difference between $K_p$ and $K$.
Non-Ideal Behavior
Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equation $\ref{18.38}$ with $ΔG^o = ΔH^o − TΔS^o$ provides insight into how the components of $ΔG^o$ influence the magnitude of the equilibrium constant:
\begin{align} ΔG° &= ΔH° − TΔS° \[4pt] &= −RT \ln K \label{18.39} \end{align}
Notice that $K$ becomes larger as $ΔS^o$ becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, $K$ increases as $ΔH^o$ decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum entropy and seek the lowest energy state possible.
Relating Grxn and Kp: Relating Grxn and Kp(opens in new window) [youtu.be]
Temperature Dependence of the Equilibrium Constant
The fact that $ΔG^o$ and $K$ are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation $\ref{18.39}$, which can be rearranged as follows:
$\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}$
Assuming $ΔH^o$ and $ΔS^o$ are temperature independent, for an exothermic reaction ($ΔH^o < 0$), the magnitude of $K$ decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of $K$ increases with increasing temperature. The quantitative relationship expressed in Equation $\ref{18.40}$ agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of $K$. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of $K$. Equation $\ref{18.40}$ also shows that the magnitude of $ΔH^o$ dictates how rapidly $K$ changes as a function of temperature. In contrast, the magnitude and sign of $ΔS^o$ affect the magnitude of $K$ but not its temperature dependence.
If we know the value of $K$ at a given temperature and the value of $ΔH^o$ for a reaction, we can estimate the value of $K$ at any other temperature, even in the absence of information on $ΔS^o$. Suppose, for example, that $K_1$ and $K_2$ are the equilibrium constants for a reaction at temperatures $T_1$ and $T_2$, respectively. Applying Equation $\ref{18.40}$ gives the following relationship at each temperature:
$\ln K_1 =\dfrac{-\Delta H^\circ}{RT_1}+\dfrac{\Delta S^\circ}{R} \nonumber$
and
$\ln K_2 =\dfrac{-\Delta H^\circ}{RT_2}+\dfrac{\Delta S^\circ}{R} \nonumber$
Subtracting $\ln K_1$ from $\ln K_2$,
$\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}$
Thus calculating $ΔH^o$ from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature ($K_1$) allow us to calculate the value of the equilibrium constant at any other temperature ($K_2$), assuming that $ΔH^o$ and $ΔS^o$ are independent of temperature.
Example $4$
The equilibrium constant for the formation of $\ce{NH3}$ from $\ce{H2}$ and $\ce{N2}$ at 25°C was calculated to be Kp = 5.4 × 105 in Example $3$. What is $K_p$ at 500°C? (Use the data from Example $1$.)
Given: balanced chemical equation, $ΔH^o°$, initial and final $T$, and $K_p$ at 25°C
Asked for: $K_p$ at 500°C
Strategy:
Convert the initial and final temperatures to kelvin. Then substitute appropriate values into Equation $\ref{18.41}$ to obtain $K_2$, the equilibrium constant at the final temperature.
Solution:
The value of $ΔH^o$ for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set T1 = 25°C = 298.K and T2 = 500°C = 773 K, then from Equation $\ref{18.41}$ we obtain the following:
\begin{align*}\ln\dfrac{K_2}{K_1}&=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \[4pt] &=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{\textrm{8.314 J/K}}\left(\dfrac{1}{\textrm{298 K}}-\dfrac{1}{\textrm{773 K}}\right)=-22.8 \[4pt] \dfrac{K_2}{K_1}&=1.3\times10^{-10} \[4pt] K_2&=(5.4\times10^5)(1.3\times10^{-10})=7.0\times10^{-5}\end{align*} \nonumber
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise $4$
In the exercise in Example $3$, you calculated Kp = 2.2 × 1012 for the reaction of NO with O2 to give NO2 at 25°C. Use the $ΔH^o_f$ values in the exercise in Example $1$ to calculate $K_p$ for this reaction at 1000°C.
Answer
5.6 × 10−4
The Van't Hoff Equation: The Van't Hoff Equation (opens in new window) [youtu.be]
Summary
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If $ΔG^o$ < 0, then K > 1, and products are favored over reactants at equilibrium. Conversely, if $ΔG^o$ > 0, then K < 1, and reactants are favored over products at equilibrium. If $ΔG^o$ = 0, then K=1, and neither reactants nor products are favored at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
Contributors and Attributions
• Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/19%3A_Free_Energy_and_Thermodynamics/19.08%3A_Free_Energy_Changes_for_Nonstandard_States_-_The_Relationship_between_and.txt |
Learning Objectives
• To know the relationship between free energy and the equilibrium constant.
We have identified three criteria for whether a given reaction will occur spontaneously:
1. $ΔS_{univ} > 0$,
2. $ΔG_{sys} < 0$ (applicable under constant temperature and constant pressure conditions), and
3. the relative magnitude of the reaction quotient $Q$ versus the equilibrium constant $K$.
Recall that if $Q < K$, then the reaction proceeds spontaneously to the right as written, resulting in the net conversion of reactants to products. Conversely, if $Q > K$, then the reaction proceeds spontaneously to the left as written, resulting in the net conversion of products to reactants. If $Q = K$, then the system is at equilibrium, and no net reaction occurs. Table $1$ summarizes these criteria and their relative values for spontaneous, nonspontaneous, and equilibrium processes.
Table $1$: Criteria for the Spontaneity of a Process as Written
Spontaneous Equilibrium Nonspontaneous*
*Spontaneous in the reverse direction.
ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0
ΔGsys < 0 ΔGsys = 0 ΔGsys > 0
Q < K Q = K Q > K
Because all three criteria are assessing the same thing—the spontaneity of the process—it would be most surprising indeed if they were not related. In this section, we explore the relationship between the standard free energy of reaction ($ΔG^o$) and the equilibrium constant ($K$).
Free Energy and the Equilibrium Constant
Because $ΔH^o$ and $ΔS^o$ determine the magnitude and sign of $ΔG^o$ and also because $K$ is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of $ΔG^o$ and vice versa. "Free Energy", $ΔG$ is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating $ΔH$ from the equation for $ΔG$. The general relationship can be shown as follows (derivation not shown):
$\Delta G = V \Delta P − S \Delta T \label{18.29}$
If a reaction is carried out at constant temperature ($ΔT = 0$), then Equation $\ref{18.29}$ simplifies to
$\Delta{G} = V\Delta{P} \label{18.30}$
Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important.
Assuming ideal gas behavior, we can replace the $V$ in Equation $\ref{18.30}$ by $nRT/P$ (where $n$ is the number of moles of gas and $R$ is the ideal gas constant) and express $\Delta{G}$ in terms of the initial and final pressures ($P_i$ and $P_f$, respectively):
\begin{align} \Delta G &=\left(\dfrac{nRT}{P}\right)\Delta P \[4pt] &=nRT\dfrac{\Delta P}{P} \[4pt] &=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31}\end{align}
If the initial state is the standard state with $P_i = 1 \,atm$, then the change in free energy of a substance when going from the standard state to any other state with a pressure $P$ can be written as follows:
$G − G^° = nRT\ln{P} \nonumber$
This can be rearranged as follows:
$G = G^° + nRT\ln {P} \label{18.32}$
As you will soon discover, Equation $\ref{18.32}$ allows us to relate $ΔG^o$ and $K_p$. Any relationship that is true for $K_p$ must also be true for $K$ because $K_p$ and $K$ are simply different ways of expressing the equilibrium constant using different units.
Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species:
$aA+bB \rightleftharpoons cC+dD \label{18.33}$
Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for $ΔG$:
\begin{align} \Delta{G} &=\sum_m G_{products}−\sum_n G_{reactants} \[4pt] &=(cG_C+dG_D)−(aG_A+bG_B) \label{18.34} \end{align}
Substituting Equation $\ref{18.32}$ for each term into Equation $\ref{18.34}$,
$\Delta G = [(cG^o_C+cRT \ln P_C)+(dG^o_D+dRT\ln P_D)]−[(aG^o_A+aRT\ln P_A)+(bG^o_B+bRT\ln P_B)] \nonumber$
Combining terms gives the following relationship between $ΔG$ and the reaction quotient $Q$:
\begin{align} \Delta G &=\Delta G^\circ+RT \ln\left(\dfrac{P^c_\textrm CP^d_\textrm D}{P^a_\textrm AP^b_\textrm B}\right) \[4pt] &=\Delta G^\circ+RT\ln Q \label{18.35} \end{align}
where $ΔG^o$ indicates that all reactants and products are in their standard states. For gases at equilibrium ($Q = K_p$), and as you’ve learned in this chapter, $ΔG = 0$ for a system at equilibrium. Therefore, we can describe the relationship between $ΔG^o$ and $K_p$ for gases as follows:
\begin{align} 0 &= ΔG^o + RT\ln K_p \label{18.36a} \[4pt] ΔG^o &= −RT\ln K_p \label{18.36b} \end{align}
If the products and reactants are in their standard states and $ΔG^o < 0$, then $K_p > 1$, and products are favored over reactants when the reaction is at equilibrium. Conversely, if $ΔG^o > 0$, then $K_p < 1$, and reactants are favored over products when the reaction is at equilibrium. If $ΔG^o = 0$, then $K_p = 1$, and neither reactants nor products are favored when the reaction is at equilibrium.
For a spontaneous process under standard conditions, $K_{eq}$ and $K_p$ are greater than 1.
Example $1$
$ΔG^o$ is −32.7 kJ/mol of N2 for the reaction
$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber$
This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate $ΔG$ for the same reaction under the following nonstandard conditions:
• $P_{\textrm N_2}$ = 2.00 atm,
• $P_{\textrm H_2}$ = 7.00 atm,
• $P_{\textrm{NH}_3}$ = 0.021 atm, and
• $T = 100 ^oC$.
Does the reaction proceed to the right, as written, or to the left to reach equilibrium?
Given: balanced chemical equation, partial pressure of each species, temperature, and ΔG°
Asked for: whether the reaction proceeds to the right or to the left to reach equilibrium
Strategy:
1. Using the values given and Equation $\ref{18.35}$, calculate $Q$.
2. Determine if $Q$ is >, <, or = to $K$
3. Substitute the values of $ΔG^o$ and $Q$ into Equation $\ref{18.35}$ to obtain $ΔG$ for the reaction under nonstandard conditions.
Solution:
A The relationship between $ΔG^o$ and $ΔG$ under nonstandard conditions is given in Equation $\ref{18.35}$. Substituting the partial pressures given, we can calculate $Q$:
\begin{align*} Q &=\dfrac{P^2_{\textrm{NH}_3}}{P_{\textrm N_2}P^3_{\textrm H_2}} \[4pt] &=\dfrac{(0.021)^2}{(2.00)(7.00)^3}=6.4 \times 10^{-7} \end{align*} \nonumber
B Because $ΔG^o$ is −, K must be a number greater than 1
C Substituting the values of $ΔG^o$ and $Q$ into Equation $\ref{18.35}$,
\begin{align*} \Delta G &=\Delta G^\circ+RT\ln Q \[4pt] &=-32.7\textrm{ kJ}+\left[(\textrm{8.314 J/K})(\textrm{373 K})\left(\dfrac{\textrm{1 kJ}}{\textrm{1000 J}}\right)\ln(6.4\times10^{-7})\right] \[4pt] &=-32.7\textrm{ kJ}+(-44\textrm{ kJ}) \[4pt] &=-77\textrm{ kJ/mol of N}_2 \end{align*} \nonumber
Because $ΔG < 0$ and $Q < K$ (because $Q < 1$), the reaction proceeds spontaneously to the right, as written, in order to reach equilibrium.
Exercise $1$
Calculate $ΔG$ for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, PNO = 0.0100 atm, $P_{\ce{O_2}}$ = 0.200 atm, and $P_{\ce{NO_2}} = 1.00 × 10^{−4} atm$. The value of $ΔG^o$ for this reaction is −72.5 kJ/mol of O2. Are products or reactants favored?
Answer
−92.9 kJ/mol of $\ce{O2}$; the reaction is spontaneous to the right as written. The reaction will proceed in the forward direction to reach equilibrium.
Example $2$
Calculate $K_p$ for the reaction of $\ce{H_2}$ with $\ce{N2}$ to give $\ce{NH3}$ at 25°C. $ΔG^o$ for this reaction is −32.7 kJ/mol of $\ce{N2}$.
Given: balanced chemical equation from Example $1$, $ΔG^o$, and temperature
Asked for: $K_p$
Strategy:
Substitute values for $ΔG^o$ and T (in kelvin) into Equation $\ref{18.36b}$ to calculate $K_p$, the equilibrium constant for the formation of ammonia.
Solution
In Example $1$, we used tabulated values of ΔGf to calculate $ΔG^o$ for this reaction (−32.7 kJ/mol of N2). For equilibrium conditions, rearranging Equation $\ref{18.36b}$,
\begin{align*} \Delta G^\circ &=-RT\ln K_\textrm p \[4pt] \dfrac{-\Delta G^\circ}{RT} &=\ln K_\textrm p \end{align*} \nonumber
Inserting the value of $ΔG^o$ and the temperature (25°C = 298 K) into this equation,
\begin{align*}\ln K_\textrm p &=-\dfrac{(-\textrm{32.7 kJ})(\textrm{1000 J/kJ})}{(\textrm{8.314 J/K})(\textrm{298 K})}=13.2 \[4pt] K_\textrm p &=5.4\times10^5\end{align*} \nonumber
Thus the equilibrium constant for the formation of ammonia at room temperature is product-favored. However, the rate at which the reaction occurs at room temperature is too slow to be useful.
Exercise $3$
Calculate Kp for the reaction of NO with O2 to give NO2 at 25°C. $ΔG^o$ for this reaction is −70.5 kJ/mol of $\ce{O2}$.
Answer
2.3 × 1012
Although $K_p$ is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant $K$ is defined in terms of the concentrations of the reactants and the products. The numerical magnitude of $K_p$ and $K$ are related:
$K_p = K(RT)^{Δn} \label{18.37}$
where $Δn$ is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, $Δn = 0$, so $K_p = K$. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation $\ref{18.36b}$ can be written in a more general form:
$ΔG° = −RT \ln K \label{18.38}$
Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation $\ref{18.38}$ for the difference between $K_p$ and $K$.
Non-Ideal Behavior
Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of activities or fugacities in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively.
Combining Equation $\ref{18.38}$ with $ΔG^o = ΔH^o − TΔS^o$ provides insight into how the components of $ΔG^o$ influence the magnitude of the equilibrium constant:
\begin{align} ΔG° &= ΔH° − TΔS° \[4pt] &= −RT \ln K \label{18.39} \end{align}
Notice that $K$ becomes larger as $ΔS^o$ becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, $K$ increases as $ΔH^o$ decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible.
The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum entropy and seek the lowest energy state possible.
Relating Grxn and Kp: Relating Grxn and Kp(opens in new window) [youtu.be]
Temperature Dependence of the Equilibrium Constant
The fact that $ΔG^o$ and $K$ are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation $\ref{18.39}$, which can be rearranged as follows:
$\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}$
Assuming $ΔH^o$ and $ΔS^o$ are temperature independent, for an exothermic reaction ($ΔH^o < 0$), the magnitude of $K$ decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of $K$ increases with increasing temperature. The quantitative relationship expressed in Equation $\ref{18.40}$ agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of $K$. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of $K$. Equation $\ref{18.40}$ also shows that the magnitude of $ΔH^o$ dictates how rapidly $K$ changes as a function of temperature. In contrast, the magnitude and sign of $ΔS^o$ affect the magnitude of $K$ but not its temperature dependence.
If we know the value of $K$ at a given temperature and the value of $ΔH^o$ for a reaction, we can estimate the value of $K$ at any other temperature, even in the absence of information on $ΔS^o$. Suppose, for example, that $K_1$ and $K_2$ are the equilibrium constants for a reaction at temperatures $T_1$ and $T_2$, respectively. Applying Equation $\ref{18.40}$ gives the following relationship at each temperature:
$\ln K_1 =\dfrac{-\Delta H^\circ}{RT_1}+\dfrac{\Delta S^\circ}{R} \nonumber$
and
$\ln K_2 =\dfrac{-\Delta H^\circ}{RT_2}+\dfrac{\Delta S^\circ}{R} \nonumber$
Subtracting $\ln K_1$ from $\ln K_2$,
$\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}$
Thus calculating $ΔH^o$ from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature ($K_1$) allow us to calculate the value of the equilibrium constant at any other temperature ($K_2$), assuming that $ΔH^o$ and $ΔS^o$ are independent of temperature.
Example $4$
The equilibrium constant for the formation of $\ce{NH3}$ from $\ce{H2}$ and $\ce{N2}$ at 25°C was calculated to be Kp = 5.4 × 105 in Example $3$. What is $K_p$ at 500°C? (Use the data from Example $1$.)
Given: balanced chemical equation, $ΔH^o°$, initial and final $T$, and $K_p$ at 25°C
Asked for: $K_p$ at 500°C
Strategy:
Convert the initial and final temperatures to kelvin. Then substitute appropriate values into Equation $\ref{18.41}$ to obtain $K_2$, the equilibrium constant at the final temperature.
Solution:
The value of $ΔH^o$ for the reaction obtained using Hess’s law is −91.8 kJ/mol of N2. If we set T1 = 25°C = 298.K and T2 = 500°C = 773 K, then from Equation $\ref{18.41}$ we obtain the following:
\begin{align*}\ln\dfrac{K_2}{K_1}&=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \[4pt] &=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{\textrm{8.314 J/K}}\left(\dfrac{1}{\textrm{298 K}}-\dfrac{1}{\textrm{773 K}}\right)=-22.8 \[4pt] \dfrac{K_2}{K_1}&=1.3\times10^{-10} \[4pt] K_2&=(5.4\times10^5)(1.3\times10^{-10})=7.0\times10^{-5}\end{align*} \nonumber
Thus at 500°C, the equilibrium strongly favors the reactants over the products.
Exercise $4$
In the exercise in Example $3$, you calculated Kp = 2.2 × 1012 for the reaction of NO with O2 to give NO2 at 25°C. Use the $ΔH^o_f$ values in the exercise in Example $1$ to calculate $K_p$ for this reaction at 1000°C.
Answer
5.6 × 10−4
The Van't Hoff Equation: The Van't Hoff Equation (opens in new window) [youtu.be]
Summary
For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and Kp, the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If $ΔG^o$ < 0, then K > 1, and products are favored over reactants at equilibrium. Conversely, if $ΔG^o$ > 0, then K < 1, and reactants are favored over products at equilibrium. If $ΔG^o$ = 0, then K=1, and neither reactants nor products are favored at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature.
Contributors and Attributions
• Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/19%3A_Free_Energy_and_Thermodynamics/19.09%3A_Free_Energy_and_Equilibrium-_Relating_to_the_Equilibrium_Constant_%28K%29.txt |
Learning Objectives
• To identify oxidation–reduction reactions in solution.
We described the defining characteristics of oxidation–reduction, or redox, reactions. Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses oxidation states, and a second is referred to as the half-reaction method.
Balancing Redox Equations Using Oxidation States
To balance a redox equation using the oxidation state method, we conceptually separate the overall reaction into two parts: an oxidation—in which the atoms of one element lose electrons—and a reduction—in which the atoms of one element gain electrons. Consider, for example, the reaction of Cr2+(aq) with manganese dioxide (MnO2) in the presence of dilute acid. Equation $\ref{20.2.1}$ is the net ionic equation for this reaction before balancing; the oxidation state of each element in each species has been assigned using the procedure described previously (in red above each element):
$\overset{\color{red}{+2}}{\ce{Cr^{2+}} ( aq }) + \overset{\color{red}{+4}}{\ce{Mn}} \overset{\color{red}{-2}}{\ce{O_2} ( aq )} + \overset{\color{red}{+1}} {\ce{H^{+}} ( aq )} \rightarrow \overset{\color{red}{+3}}{\ce{Cr^{3+}} ( aq )} + \overset{\color{red}{+2}}{\ce{Mn^{2+}}( aq )} + \overset{\color{red}{+1}} {\ce{H_2}} \overset{\color{red}{-2}} {\ce{O} (l)} \label{20.2.1}$
Notice that chromium is oxidized from the +2 to the +3 oxidation state, while manganese is reduced from the +4 to the +2 oxidation state. We can write an equation for this reaction that shows only the atoms that are oxidized and reduced (ignoring the oxygen and hydrogen atoms):
$\ce{Cr^{2+} + Mn^{4+} -> Cr^{3+} + Mn^{2+}} \label{20.2.2}$
The oxidation can be written as
$\underbrace{\ce{Cr^{2+} -> Cr^{3+} + e^{-}}}_{\text{oxidation with 1 electron lost}} \label{20.2.3}$
and the reduction as
$\underbrace{\ce{Mn^{4+} + 2e^{-} \rightarrow Mn^{2+}}}_{\text{reduction with 2 electrons gained}} \label{20.2.4}$
For the overall chemical equation to be balanced, the number of electrons lost by the reductant must equal the number gained by the oxidant. We must therefore multiply the oxidation and the reduction equations by appropriate coefficients to give us the same number of electrons in both. In this example, we must multiply the oxidation (Equation \ref{20.2.3}) by 2 to give
$\ce{2Cr^{2+} -> 2Cr^{3+} + 2e^{-}} \label{20.2.5}$
The number of electrons lost in the oxidation now equals the number of electrons gained in the reduction (Equation \ref{20.2.4}):
\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \ce{2e^{-}} \label{20.2.6} \[8pt] \ce{Mn^{4+}} + \ce{2e^{-}} &\rightarrow \ce{Mn^{2+}} \end{align*}
We then add the equations for the oxidation and the reduction and cancel the electrons on both sides of the equation, using the actual chemical forms of the reactants and products:
\begin{align*} \ce{2Cr^{2+}} &\rightarrow \ce{2Cr^{3+}} + \cancel{\ce{2e^{-}}} \[8pt] \ce{Mn^{4+}} + \cancel{\ce{2e^{-}}} &\rightarrow \ce{Mn^{2+}} \end{align*} \nonumber
to result in the balanced redox reaction (metals only)
$\ce{ Mn^{4+} +2Cr^{2+} \rightarrow 2Cr^{3+} + Mn^{2+}} \label{20.2.7}$
now we can add the non-redox active atoms back into the equation (ignoring water and hydronium for now)
$\ce{MnO2(aq) + 2Cr^{2+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.7b}$
In a balanced redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.
Although the electrons cancel and the metal atoms are balanced, the total charge on the left side of Equation \ref{20.2.7b} (+4) does not equal the charge on the right side (+8). Because the reaction is carried out in the presence of aqueous acid, we can add $\ce{H^{+}}$ as necessary to either side of the equation to balance the charge. By the same token, if the reaction were carried out in the presence of aqueous base, we could balance the charge by adding $\ce{OH^{−}}$ as necessary to either side of the equation to balance the charges.
In this case, adding four $\ce{H^{+}}$ ions to the left side of Equation \ref{20.2.7b} to give
$\ce{ MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq)} \label{20.2.8}$
Although the charges are now balanced in Equation \ref{20.2.8}, we have two oxygen atoms on the left side of the equation and none on the right. We can balance the oxygen atoms without affecting the overall charge balance by adding $\ce{H2O}$ as necessary to either side of the equation. Here, we need to add two $\ce{H2O}$ molecules to the right side of Equation \ref{20.2.8}:
$\ce{MnO2(s) + 2Cr^{2+}(aq) + 4H^{+}(aq) -> 2Cr^{3+}(aq) + Mn^{2+}(aq) + 2H_2O(l)} \label{20.2.9}$
Although we did not explicitly balance the hydrogen atoms, we can see by inspection that the overall chemical equation is now balanced with respect to all atoms and charge. All that remains is to check to make sure that we have not made a mistake. This procedure for balancing reactions is summarized below and illustrated in Example $1$ below.
Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method
1. Write the unbalanced chemical equation for the reaction, showing the reactants and the products.
2. Assign oxidation states to all atoms in the reactants and the products and determine which atoms change oxidation state.
3. Write separate equations for oxidation and reduction, showing (a) the atom(s) that is (are) oxidized and reduced and (b) the number of electrons accepted or donated by each.
4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons.
5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting the coefficients as necessary to give the numbers of atoms in step 4.
6. Add the two equations and cancel the electrons.
7. Balance the charge by adding $\ce{H^{+}}$ or $\ce{OH^{−}}$ ions as necessary for reactions in acidic or basic solution, respectively.
8. Balance the oxygen atoms by adding $\ce{H2O}$ molecules to one side of the equation.
9. Check to make sure that the equation is balanced in both atoms and total charges.
Example $1$: Balancing in Acid Solutions
Arsenic acid ($\ce{H3AsO4}$) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine ($\ce{AsH3}$, a highly toxic and unstable gas) and $\ce{Zn^{2+}(aq)}$. Balance the equation for this reaction using oxidation states:
$\ce{H3AsO4(aq) + Zn(s) -> AsH3(g) + Zn^{2+}(aq)} \nonumber$
Given: reactants and products in acidic solution
Asked for: balanced chemical equation using oxidation states
Strategy:
Follow the procedure given above for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced.
Solution:
1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step.
2. Assign oxidation states and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +5, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in $Zn^{2+}(aq)$ is +2: $H_3\overset{\color{red}{+5}}{As}O_4(aq) + \overset{\color{red}{0}}{Zn}(s) \rightarrow \overset{\color{red}{-3}}{As}H_3(g) + \overset{\color{red}{+2}}{Zn^{2+}}(aq) \nonumber$
3. Write separate equations for oxidation and reduction. The arsenic atom in H3AsO4 is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons:
$\underbrace{ \overset{\color{red}{+5}}{As} + 8e^- \rightarrow \overset{\color{red}{-3}}{As}}_{\text{Reduction with gain of 8 electrons}} \nonumber$
Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:
$\underbrace{ \overset{\color{red}{0}} {Zn} \rightarrow \overset{\color{red}{+2}} {Zn^{2+}} + 2e^- }_{\text{Oxidation with loss of 2 electrons}}\nonumber$
4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain \begin{align*} \overset{\color{red}{+5}}{\ce{As}} + \ce{8e^{-}} & \rightarrow \overset{\color{red}{-3}}{\ce{As}} \nonumber \ \overset{\color{red}{0}} {\ce{4Zn}} & \rightarrow \overset{\color{red}{+2}} {\ce{4Zn^{2+}}} + \ce{8e^{-}} \end{align*} \nonumber
5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting coefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives
• Reduction: $\ce{H3AsO4(aq) + 8e^{-} \rightarrow AsH3(g)} \nonumber$
• Oxidation: $\ce{4Zn(s) -> 4Zn^{2+}(aq) + 8e^{-}} \nonumber$
6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is $\ce{H3AsO4(aq) + 4Zn(s)} + \cancel{\ce{8e^{-}}} \rightarrow \ce{AsH3(g)} + \ce{4Zn^{2+}(aq)} + \cancel{\ce{8e^{-}}} \nonumber$ which then yields after canceling electrons $\ce{H3AsO4(aq) + 4Zn(s) \rightarrow AsH3(g) + 4Zn^{2+}(aq)} \nonumber$
7. Balance the charge by adding $\ce{H^{+}}$ or $\ce{OH^{−}}$ ions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H+ ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H+ ions to the left side gives a charge of +8 on both sides of the equation: $\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) \rightarrow AsH3(g) + 4Zn^{2+}(aq)} \nonumber$
8. Balance the oxygen atoms by adding $\ce{H2O}$ molecules to one side of the equation. There are 4 $\ce{O}$ atoms on the left side of the equation. Adding 4 $\ce{H2O}$ molecules to the right side balances the $\ce{O}$ atoms: $\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) \rightarrow AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber$ Although we have not explicitly balanced $\ce{H}$ atoms, each side of the equation has 11 $\ce{H}$ atoms.
9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is important to check that both the total number of atoms of each element and the total charges are the same on both sides of the equation:
• Atoms: $\ce{1As + 4Zn + 4O + 11H} \overset{\checkmark}{=} \ce{1As + 4Zn + 4O + 11H} \nonumber$
• Charge: $8(+1) \overset{\checkmark}{=} 4(+2) \nonumber$
The balanced chemical equation (both for charge and for atoms) for this reaction is therefore:
$\ce{H3AsO4(aq) + 4Zn(s) + 8H^{+}(aq) -> AsH3(g) + 4Zn^{2+}(aq) + 4H2O(l)} \nonumber$
Exercise $1$: Oxidizing Copper
Copper commonly occurs as the sulfide mineral $\ce{CuS}$. The first step in extracting copper from $\ce{CuS}$ is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to $\ce{NO}$. Balance the equation for this reaction using oxidation states:
$\ce{CuS(s) + H^{+}(aq) + NO^{-}3(aq) -> Cu^{2+}(aq) + NO(g) + SO^{2-}4(aq)} \nonumber$
Answer
$\ce{3CuS(s) + 8H^{+}(aq) + 8NO^{-}3(aq) \rightarrow 3Cu^{2+}(aq) + 8NO(g) + 3SO^{2-}4(aq) + 4H2O(l)} \nonumber$
Reactions in basic solutions are balanced in exactly the same manner. To make sure you understand the procedure, consider Example $2$.
Example $2$: Balancing in Basic Solution
The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows:
$\ce{Al(s) + H2O(l) \rightarrow [Al(OH)4]^{-}(aq) + H2(g)} \nonumber$
Balance this equation using oxidation states.
Given: reactants and products in a basic solution
Asked for: balanced chemical equation
Strategy:
Follow the procedure given above for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced.
Solution:
We will apply the same procedure used in Example $1$, but in a more abbreviated form.
1. The equation for the reaction is given, so we can skip this step.
2. The oxidation state of aluminum changes from 0 in metallic $Al$ to +3 in $\ce{[Al(OH)4]^{−}}$. The oxidation state of hydrogen changes from +1 in $\ce{H_2O}$ to 0 in $\ce{H2}$. Aluminum is oxidized, while hydrogen is reduced: $\overset{\color{red}{0}}{Al}_{(s)} + \overset{\color{red}{+1}}{H}_2 O_{(aq)} \rightarrow [ \overset{\color{red}{+3}}{Al} (OH)_4 ]^- _{(aq)} + \overset{\color{red}{0}}{H_2}_{(g)} \nonumber$
3. Write separate equations for oxidation and reduction.
• Reduction: $\overset{\color{red}{+1}}{H} + e^- \rightarrow \overset{\color{red}{0}}{H} \: (in\: H_2 ) \nonumber$
• Oxidation: $\overset{\color{red}{0}}{Al} \rightarrow \overset{\color{red}{+3}}{Al} + 3e^- \nonumber$
4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation:
• Reduction: $\ce{3H^{+} + 3e^{-} -> 3H^0}\: (in\: \ce{H2}) \nonumber$
• Oxidation: $\ce{Al^0 -> Al^{3+} + 3e^{-}} \nonumber$
5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of $\ce{H2O}$ contains two protons, in this case, $\ce{3H^{+}}$ corresponds to $\ce{3/2 H2O}$. Similarly, each molecule of hydrogen gas contains two H atoms, so $\ce{3H}$ corresponds to $\ce{3/2H2}$.
• Reduction: $\ce{3/2 H2O + 3e^{-} -> 3/2 H2} \nonumber$
• Oxidation: $\ce{Al -> [Al(OH)4]^{-} + 3e^{-}} \nonumber$
6. Adding the equations and canceling the electrons gives $\ce{Al} + \ce{3/2 H2O} + \cancel{\ce{3e^{-}}} \ce{->} \ce{[Al(OH)4]^{-}} + \ce{3/2 H2} + \cancel{\ce{3e^{-}}} \nonumber$ $\ce{Al} + \ce{3/2 H2O} \ce{->} \ce{[Al(OH)4]^{-}} + \ce{3/2 H2} \nonumber$ To remove the fractional coefficients, multiply both sides of the equation by 2: $\ce{2Al + 3H2O \rightarrow 2[Al(OH)4]^{-} + 3H2} \nonumber$
7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two $\ce{OH^{−}}$ ions to the left side: $\ce{2Al + 2OH^{-} + 3H2O -> 2[Al(OH)4]^{-} + 3H2} \nonumber$
8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H2O molecules to the left side: $\ce{2Al + 2OH^{-} + 6H2O -> 2[Al(OH)4]^{-} + 3H2} \nonumber$
9. Be sure the equation is balanced:
1. Atoms: $\ce{2Al + 8O + 14H} \overset{\checkmark}{=} \ce{2Al + 8O + 14H} \nonumber$
2. Charge: $(2)(0) + (2)(-1) + (6)(0) \overset{\checkmark}{=} (2)(-1) + (3)(0) \nonumber$
The balanced chemical equation is therefore
$\ce{ 2Al(s) + 2OH^{-}(aq) + 6H2O(l) \rightarrow 2[Al(OH)4]^{-}(aq) + 3H2(g)} \nonumber$
Thus 3 mol of $\ce{H2}$ gas are produced for every 2 mol of $\ce{Al}$ consumed.
Exercise $2$: Reducing Manganese in permanganate
The permanganate ion reacts with nitrite ion in basic solution to produce manganese (IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction.
Answer
$\ce{2MnO4^{-}(aq) + 3NO2^{-}(aq) + H2O(l) -> 2MnO2(s) + 3NO3^{-}(aq) + 2OH^{-}(aq)} \nonumber$
As suggested in Examples $1$ and $2$, a wide variety of redox reactions are possible in aqueous solutions. The identity of the products obtained from a given set of reactants often depends on both the ratio of oxidant to reductant and whether the reaction is carried out in acidic or basic solution, which is one reason it can be difficult to predict the outcome of a reaction. Because oxidation–reduction reactions in solution are so common and so important, however, chemists have developed two general guidelines for predicting whether a redox reaction will occur and the identity of the products:
1. Compounds of elements in high oxidation states (such as $\ce{ClO4^{−}}$, $\ce{NO3^{−}}$, $\ce{MnO4^{−}}$, $\ce{Cr2O7^{2−}}$, and $\ce{UF6}$) tend to act as oxidants and become reduced in chemical reactions.
2. Compounds of elements in low oxidation states (such as $\ce{CH4}$, $\ce{NH3}$, $\ce{H2S}$, and $\ce{HI}$) tend to act as reductants and become oxidized in chemical reactions.
When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur.
Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants.
Balancing a Redox Reaction in Acidic Conditions: Balancing a Redox Reaction in Acidic Conditions (opens in new window) [youtu.be]
Summary
Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the oxidation state method, in which the overall reaction is separated into an oxidation equation and a reduction equation. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.02%3A_Balancing_Oxidation-Reduction_Equations.txt |
Learning Objectives
• To understand the basics of voltaic cells
• To connect voltage from a voltaic cell to underlying redox chemistry
In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant is the substance that loses electrons and is oxidized in the process; the oxidant is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements.
Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactions, one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:
$\ce{Zn(s) + Br2(aq) \rightarrow Zn^{2+} (aq) + 2Br^{−} (aq)} \nonumber$
The half-reactions are as follows:
reduction half-reaction:
$\ce{Br2 (aq) + 2e^{−} \rightarrow 2Br^{−} (aq)} \nonumber$
oxidation half-reaction:
$\ce{Zn (s) \rightarrow Zn^{2+} (aq) + 2e^{−} }\nonumber$
Each half-reaction is written to show what is actually occurring in the system; $\ce{Zn}$ is the reductant in this reaction (it loses electrons), and $\ce{Br2}$ is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation $1$). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation.
In any redox reaction, the number of electrons lost by the oxidation reaction(s) equals the number of electrons gained by the reduction reaction(s).
In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cell.
There are two types of electrochemical cells: galvanic cells and electrolytic cells. Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction ($ΔG < 0$) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an electrolytic cell consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur ($ΔG > 0$). Both types contain two electrodes, which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure $1$). The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells.
Voltaic (Galvanic) Cells
To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu2+) to give copper metal and Zn2+ ion. The balanced chemical equation is as follows:
$\ce{Zn (s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu(s)} \label{20.3.4}$
We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work.
This same reaction can be carried out using the galvanic cell illustrated in Figure $\PageIndex{3a}$. To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of $\ce{Cu^{2+}}$ ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of $\ce{Zn^{2+}}$ ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a salt bridge, a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are $\ce{Na^{+}}$ or $\ce{K^{+}}$ and $\ce{NO3^{−}}$ or $\ce{SO4^{2−}}$, respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to $\ce{Zn^{2+}}$ ions at the zinc electrode (the anode), and $\ce{Cu^{2+}}$ ions are reduced to $\ce{Cu}$ metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of $\ce{Zn^{2+}}$ ions in the solution increases; simultaneously, the copper strip gains mass, and the concentration of $\ce{Cu^{2+}}$ ions in the solution decreases (Figure $\PageIndex{3b}$). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work.
The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the $\ce{Zn^{2+}}$ solution would increase as the zinc metal dissolves, and the total positive charge in the $\ce{Cu^{2+}}$ solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the $\ce{Zn^{2+}}$ solution and a flow of cations into the $\ce{Cu^{2+}}$ solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained.
A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The potential ($E_{cell}$) of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (Figure $\PageIndex{3a}$). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged.
Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example $1$.
A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction.
Example $1$
A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation:
$\ce{3Sn(s) + 2NO3^{-}(aq) + 8H^{+} (aq) \rightarrow 3Sn^{2+} (aq) + 2NO (g) + 4H2O (l)} \nonumber$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is the positive electrode and which is the negative electrode.
Given: galvanic cell and redox reaction
Asked for: half-reactions, identity of anode and cathode, and electrode assignment as positive or negative
Strategy:
1. Identify the oxidation half-reaction and the reduction half-reaction. Then identify the anode and cathode from the half-reaction that occurs at each electrode.
2. From the direction of electron flow, assign each electrode as either positive or negative.
Solution
A In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO2, with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows:
reduction: $\ce{NO3^{−} (aq) + 4H^{+}(aq) + 3e^{−} → NO(g) + 2H2O(l)} \nonumber$
oxidation: $\ce{Sn(s) → Sn^{2+}(aq) + 2e^{−}} \nonumber$
Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn2+.
Because the reduction reaction occurs at the Pt electrode, it is the cathode. Conversely, the oxidation reaction occurs at the tin electrode, so it is the anode.
B Electrons flow from the tin electrode through the wire to the platinum electrode, where they transfer to nitrate. The electric circuit is completed by the salt bridge, which permits the diffusion of cations toward the cathode and anions toward the anode. Because electrons flow from the tin electrode, it must be electrically negative. In contrast, electrons flow toward the Pt electrode, so that electrode must be electrically positive.
Exercise $1$
Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of $\ce{MnO_4^{−}}$ in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of $\ce{Sn^{2+}}$ in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation:
$\ce{2MnO^{−}4(aq) + 5Sn^{2+}(aq) + 16H^{+}(aq) \rightarrow 2Mn^{2+}(aq) + 5Sn^{4+}(aq) + 8H2O(l)} \nonumber$
For this galvanic cell,
1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is positive and which is negative.
Answer a
\begin{align*} \ce{MnO4^{−}(aq) + 8H^{+}(aq) + 5e^{−}} &→ \ce{Mn^{2+}(aq) + 4H2O(l)} \[4pt] \ce{Sn^{2+}(aq)} &→ \ce{Sn^{4+}(aq) + 2e^{−}} \end{align*} \nonumber
Answer b
The Pt electrode in the permanganate solution is the cathode; the one in the tin solution is the anode.
Answer c
The cathode (electrode in beaker that contains the permanganate solution) is positive, and the anode (electrode in beaker that contains the tin solution) is negative.
Electrochemical Cells: Electrochemical Cells(opens in new window) [youtu.be]
Constructing Cell Diagrams (Cell Notation)
Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the $\ce{Zn/Cu}$ cell shown in Figure $\PageIndex{3a}$ is written as follows:
Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the junction potential. One example of this type of galvanic cell is as follows:
$\ce{Pt(s)\, | \, H2(g) | HCl(aq, \, 1\,M)\,|\, AgCl(s) \,Ag(s)} \nonumber$
This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows:
cathode reaction:
$\ce{AgCl (s) + e^{−} \rightarrow Ag(s) + Cl^{−}(aq)} \nonumber$
anode reaction:
$\ce{ 1/2 H2(g) -> H^{+}(aq) + e^{-}} \nonumber$
overall:
$\ce{ AgCl(s) + 1/2H2(g) -> Ag(s) + Cl^{-} + H^{+}(aq)} \nonumber$
A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity.
Example $2$
Draw a cell diagram for the galvanic cell described in Example $1$. The balanced chemical reaction is as follows:
$\ce{3Sn(s) + 2NO^{−}3(aq) + 8H^{+}(aq) \rightarrow 3Sn^{2+}(aq) + 2NO(g) + 4H2O(l)} \nonumber$
Given: galvanic cell and redox reaction
Asked for: cell diagram
Strategy:
Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left.
Solution
The anode is the tin strip, and the cathode is the $\ce{Pt}$ electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus $\ce{Sn(s)∣Sn^{2+}(aq)}$. We could include $\ce{H2SO4(aq)}$ with the contents of the anode compartment, but the sulfate ion (as $\ce{HSO4^{−}}$) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction ($\ce{NO}$) and the $\ce{Pt}$ electrode. These are written as $\ce{HNO3(aq)∣NO(g)∣Pt(s)}$, with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge,
$\ce{Sn(s)\,|\,Sn^{2+}(aq)\,||\,HNO3(aq)\,|\,NO(g)\,|\,Pt_(s)} \nonumber$
The solution concentrations were not specified, so they are not included in this cell diagram.
Exercise $2$
Draw the cell diagram for the following reaction, assuming the concentration of $\ce{Ag^{+}}$ and $\ce{Mg^{2+}}$ are each 1 M:
$\ce{Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)} \nonumber$
Answer
$\ce{ Mg(s) \,|\,Mg^{2+}(aq, \;1 \,M )\,||\,Ag^+(aq, \;1\, M)\,|\,Ag(s)} \nonumber$
Cell Diagrams: Cell Diagrams(opens in new window) [youtu.be]
Summary
A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction to generate electricity, whereas an electrolytic cell consumes electrical energy from an external source to force a reaction to occur. Electrochemistry is the study of the relationship between electricity and chemical reactions. The oxidation–reduction reaction that occurs during an electrochemical process consists of two half-reactions, one representing the oxidation process and one the reduction process. The sum of the half-reactions gives the overall chemical reaction. The overall redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. An electric current is produced from the flow of electrons from the reductant to the oxidant. An electrochemical cell can either generate electricity from a spontaneous redox reaction or consume electricity to drive a nonspontaneous reaction. In a galvanic (voltaic) cell, the energy from a spontaneous reaction generates electricity, whereas in an electrolytic cell, electrical energy is consumed to drive a nonspontaneous redox reaction. Both types of cells use two electrodes that provide an electrical connection between systems that are separated in space. The oxidative half-reaction occurs at the anode, and the reductive half-reaction occurs at the cathode. A salt bridge connects the separated solutions, allowing ions to migrate to either solution to ensure the system’s electrical neutrality. A voltmeter is a device that measures the flow of electric current between two half-reactions. The potential of a cell, measured in volts, is the energy needed to move a charged particle in an electric field. An electrochemical cell can be described using line notation called a cell diagram, in which vertical lines indicate phase boundaries and the location of the salt bridge. Resistance to the flow of charge at a boundary is called the junction potential. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.03%3A_Voltaic_%28or_Galvanic%29_Cells-_Generating_Electricity_from_Spontaneous_Chemical_Reactions.txt |
Learning Objectives
• To use redox potentials to predict whether a reaction is spontaneous.
• To balance redox reactions using half-reactions.
In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper. Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work.
Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in Figure $1$ but instead of copper use a strip of cobalt metal and 1 M Co2+ in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V.
The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the standard cell potential (E°cell), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions, concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for non ideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C.
Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system.
Measuring Standard Electrode Potentials
It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured (this is analogous to measuring absolute enthalpies or free energies; recall that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram:
$Co_{(s)} ∣ Co^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M) ∣ Cu (s)\;\;\; E°=0.59\; V \label{20.4.1}$
This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu2+ in solution at the copper cathode.
All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances (Table P1). The standard cell potential (E°cell) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum:
$E°_{cell} = E°_{cathode} − E°_{anode} \label{20.4.2}$
In contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to Equation $\ref{20.4.2}$, when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell.
Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The standard hydrogen electrode (SHE) is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H+. The [H+] in solution is in equilibrium with H2 gas at a pressure of 1 atm at the Pt-solution interface (Figure $2$). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation:
$2H^+_{(aq)}+2e^− \rightleftharpoons H_{2(g)} \label{20.4.3}$
One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction.
Figure $3$ shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn2+ ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn2+, and H+ ions are reduced to H2 in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows:
$Zn_{(s)}∣Zn^{2+}_{(aq)}∥H^+(aq, 1 M)∣H_2(g, 1 atm)∣Pt_{(s)} \label{20.4.4}$
The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)}\;\;\; E°_{cathode}=0 V \label{20.4.5}$
• anode: $Zn_{(s)} \rightarrow Zn^{2+}_{(aq)}+2e^−\;\;\; E°_{anode}=−0.76\; V \label{20.4.6}$
• overall: $Zn_{(s)}+2H^+_{(aq)} \rightarrow Zn^{2+}_{(aq)}+H_{2(g)} \label{20.4.7}$
We then use Equation \ref{20.4.2} to calculate the cell potential
\begin{align*} E°_{cell} &=E°_{cathode}−E°_{anode}\[4pt] &=0.76\; V \end{align*} \nonumber
Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the standard electrode potential for that half-reaction. In this example, the standard reduction potential for Zn2+(aq) + 2e → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn2+, often called the Zn/Zn2+ redox couple, or the Zn/Zn2+ couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E°anode from E°cathode to obtain
$E°_{cell}: 0 \,V − (−0.76\, V) = 0.76\, V \nonumber$
Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential.
E° values do NOT depend on the stoichiometric coefficients for a half-reaction, because it is an intensive property.
The Standard Hydrogen Electrode (SHE): The Standard Hydrogen Electrode (SHE)(opens in new window) [youtu.be]
Standard Electrode Potentials
To measure the potential of the Cu/Cu2+ couple, we can construct a galvanic cell analogous to the one shown in Figure $3$ but containing a Cu/Cu2+ couple in the sample compartment instead of Zn/Zn2+. When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of $E°_{cell}$ indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn2+ couple. Hence the reactions that occur spontaneously, indicated by a positive $E°_{cell}$, are the reduction of Cu2+ to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H2 is oxidized to H+ at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu2+/Cu couple on the right:
$Pt_{(s)}∣H_2(g, 1 atm)∣H^+(aq, 1\; M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{20.4.8}$
The half-cell reactions and potentials of the spontaneous reaction are as follows:
• Cathode: $Cu^{2+}{(aq)} + 2e^− \rightarrow Cu_{(g)}\;\;\; E°_{cathode} = 0.34\; V \label{20.4.9}$
• Anode: $H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^−\;\;\; E°_{anode} = 0\; V \label{20.4.10}$
• Overall: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2H^+_{(aq)} + Cu_{(s)} \label{20.4.11}$
We then use Equation \ref{20.4.2} to calculate the cell potential
\begin{align*} E°_{cell} &= E°_{cathode}− E°_{anode} \[4pt] &= 0.34\; V \end{align*} \nonumber
Thus the standard electrode potential for the Cu2+/Cu couple is 0.34 V.
Electrode Potentials and ECell: Electrode and Potentials and Ecell(opens in new window) [youtu.be]
Balancing Redox Reactions Using the Half-Reaction Method
In Section 4.4, we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other.
We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas:
$Al_{(s)} + OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + H_{2(g)} \label{20.4.12}$
In this reaction, $Al_{(s)}$ is oxidized to Al3+, and H+ in water is reduced to H2 gas, which bubbles through the solution, agitating it and breaking up the clogs.
The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed Table P1, we find the corresponding half-reactions that describe the reduction of H+ ions in water to H2and the oxidation of Al to Al3+ in basic solution:
• reduction: $2H_2O_{(l)} + 2e^− \rightarrow 2OH^−_{(aq)} + H_{2(g)} \label{20.4.13}$
• oxidation: $Al_{(s)} + 4OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + 3e^− \label{20.4.14}$
The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution.
In Equation $\ref{20.4.13}$, two H+ ions gain one electron each in the reduction; in Equation $\ref{20.4.14}$, the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction (Equation $\ref{20.4.13}$) by 3 and the oxidation half-reaction (Equation $\ref{20.4.14}$) by 2 to give the same number of electrons in both half-reactions:
• reduction:
$6H_2O_{(l)} + 6e^− \rightarrow 6OH^−_{(aq)} + 3H_{2(g)} \label{20.4.15}$
• oxidation:
$2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 6e^− \label{20.4.16}$
Adding the two half-reactions,
$6H_2O_{(l)} + 2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−{4(aq)} + 3H_{2(g)} + 6OH^−_{(aq)} \label{20.4.17}$
Simplifying by canceling substances that appear on both sides of the equation,
$6H_2O_{(l)} + 2Al_{(s)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.18}$
We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced:
$2Al + 8O + 14H = 2Al + 8O + 14H \label{20.4.19}$
The atoms also balance, so Equation $\ref{20.4.18}$ is a balanced chemical equation for the redox reaction depicted in Equation $\ref{20.4.12}$.
The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction.
We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in Table P1, but instead focus on the atoms whose oxidation states change, as illustrated in the following steps:
Step 1: Write the reduction half-reaction and the oxidation half-reaction.
For the reaction shown in Equation $\ref{20.4.12}$, hydrogen is reduced from H+ in OH to H2, and aluminum is oxidized from Al° to Al3+:
• reduction:
$OH^−_{(aq)} \rightarrow H_{2(g)} \label{20.4.20}$
• oxidation:
$Al_{(s)} \rightarrow Al(OH)^−_{4(aq)} \label{20.4.21}$
Step 2: Balance the atoms by balancing elements other than O and H. Then balance O atoms by adding H2O and balance H atoms by adding H+.
Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction:
• reduction:
$OH^−_{(aq)} \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.22}$
• oxidation:
$Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} \label{20.4.23}$
Balancing H atoms by adding H+, we obtain the following:
• reduction:
$OH^−_{(aq)} + 3H^+_{(aq)} \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.24}$
• oxidation:
$Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} + 4H^+_{(aq)} \label{20.4.25}$
We have now balanced the atoms in each half-reaction, but the charges are not balanced.
Step 3: Balance the charges in each half-reaction by adding electrons.
Two electrons are gained in the reduction of H+ ions to H2, and three electrons are lost during the oxidation of Al° to Al3+:
• reduction:
$OH^−_{(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} + H_2O_{(l)} \label{20.4.26}$
• oxidation:
$Al_{(s)} + 4H_2O_{(l)} \rightarrow Al(OH)^−_{4(aq)} + 4H^+_{(aq)} + 3e^− \label{20.4.27}$
Step 4: Multiply the reductive and oxidative half-reactions by appropriate integers to obtain the same number of electrons in both half-reactions.
In this case, we multiply Equation $\ref{20.4.26}$ (the reductive half-reaction) by 3 and Equation $\ref{20.4.27}$ (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions:
• reduction:
$3OH^−_{(aq)} + 9H^+_{(aq)} + 6e^− \rightarrow 3H_{2(g)} + 3H_2O_{(l)} \label{20.4.28}$
• oxidation:
$2Al_{(s)} + 8H_2O_{(l)} \rightarrow 2Al(OH)^−_{4(aq)} + 8H^+_{(aq)} + 6e^− \label{20.4.29}$
Step 5: Add the two half-reactions and cancel substances that appear on both sides of the equation.
Adding and, in this case, canceling 8H+, 3H2O, and 6e,
$2Al_{(s)} + 5H_2O_{(l)} + 3OH^−_{(aq)} + H^+_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.30}$
We have three OH and one H+ on the left side. Neutralizing the H+ gives us a total of 5H2O + H2O = 6H2O and leaves 2OH on the left side:
$2Al_{(s)} + 6H_2O_{(l)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{20.4.31}$
Step 6: Check to make sure that all atoms and charges are balanced.
Equation $\ref{20.4.31}$ is identical to Equation $\ref{20.4.18}$, obtained using the first method, so the charges and numbers of atoms on each side of the equation balance.
Example $1$
In acidic solution, the redox reaction of dichromate ion ($\ce{Cr2O7^{2−}}$) and iodide ($\ce{I^{−}}$) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green $\ce{Cr^{3+}(aq)}$ complex and brown $\ce{I2(aq)}$ ions (Figure $4$):
$\ce{Cr2O7^{2−}(aq) + I^{−}(aq) -> Cr^{3+}(aq) + I2(aq)} \nonumber$
Balance this equation using half-reactions.
Given: redox reaction and Table P1
Asked for: balanced chemical equation using half-reactions
Strategy:
Follow the steps to balance the redox reaction using the half-reaction method.
Solution
From the standard electrode potentials listed in Table P1, we find the half-reactions corresponding to the overall reaction:
• reduction: $\ce{Cr2O7^{2−}(aq) + 14H^{+}(aq) + 6e^{−} -> 2Cr^{3+}(aq) + 7H2O(l)} \nonumber$
• oxidation: $\ce{2I^{−}(aq) -> I2(aq) + 2e^{−}} \nonumber$
Balancing the number of electrons by multiplying the oxidation reaction by 3,
• oxidation: $\ce{6I^{−}(aq) -> 3I2(aq) + 6e^{−}} \nonumber$
Adding the two half-reactions and canceling electrons,
$\ce{Cr2O^{2−}7(aq) + 14H^{+}(aq) + 6I^{−}(aq) -> 2Cr^{3+}(aq) + 7H2O(l) + 3I2(aq)} \nonumber$
We must now check to make sure the charges and atoms on each side of the equation balance:
\begin{align*} (−2) + 14 + (−6) &= +6 \[4pt] +6 &\overset{\checkmark}{=} +6 \end{align*} \nonumber
and atoms
$\ce{2Cr + 7O + 14H + 6I} \overset{\checkmark}{=} \ce{2Cr + 7O + 14H + 6I} \nonumber$
Both the charges and atoms balance, so our equation is balanced.
We can also use the alternative procedure, which does not require the half-reactions listed in Table P1.
Step 1: Chromium is reduced from $\ce{Cr^{6+}}$ in $\ce{Cr2O7^{2−}}$ to $\ce{Cr^{3+}}$, and $\ce{I^{−}}$ ions are oxidized to $\ce{I2}$. Dividing the reaction into two half-reactions,
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow Cr^{3+}_{(aq)} \nonumber$
• oxidation: $I^−_{(aq)} \rightarrow I_{2(aq)} \nonumber$
Step 2: Balancing the atoms other than oxygen and hydrogen,
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)} \nonumber$
• oxidation: $2I^−_{(aq)} \rightarrow I_{2(aq)} \nonumber$
We now balance the O atoms by adding H2O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step.
• reduction: $Cr_2O^{2−}_{7(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} \nonumber$
Next we balance the H atoms by adding H+ to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction.
• reduction: $Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} \nonumber$
Step 3: We must now add electrons to balance the charges. The reduction half-reaction (2Cr+6 to 2Cr+3) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I to I2) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge:
• reduction: Cr2O72(aq) + 14H+(aq) + 6e → 2Cr3+(aq) + 7H2O(l)
• oxidation: 2I(aq) → I2(aq) + 2e
Step 4: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3:
• oxidation: 6I(aq) → 3I2(s) + 6e
Step 5: Adding the two half-reactions and canceling substances that appear in both reactions,
$\ce{Cr2O7^{2−}(aq) + 14H^{+}(aq) + 6I^{−}(aq) → 2Cr^{3+}(aq) + 7H2O(l) + 3I2(aq)} \nonumber$
Step 6: This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance.
Exercise $1$
Copper is found as the mineral covellite ($\ce{CuS}$). The first step in extracting the copper is to dissolve the mineral in nitric acid ($\ce{HNO3}$), which oxidizes sulfide to sulfate and reduces nitric acid to $\ce{NO}$:
$\ce{CuS(s) + HNO3(aq) \rightarrow NO(g) + CuSO4(aq)} \nonumber$
Balance this equation using the half-reaction method.
Answer
$\ce{3CuS(s) + 8HNO3(aq) -> 8NO(g) + 3CuSO4(aq) + 4H2O(l)} \nonumber$
Calculating Standard Cell Potentials
The standard cell potential for a redox reaction (E°cell) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram:
$Zn{(s)}∣Zn^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{20.4.32}$
We know the values of E°anode for the reduction of Zn2+ and E°cathode for the reduction of Cu2+, so we can calculate $E°_{cell}$:
• cathode: $Cu^{2+}_{(aq)} + 2e^− \rightarrow Cu_{(s)} \;\;\; E°_{cathode} = 0.34\; V \label{20.4.33}$
• anode: $Zn_{(s)} \rightarrow Zn^{2+}(aq, 1 M) + 2e^−\;\;\; E°_{anode} = −0.76\; V \label{20.4.34}$
• overall: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{20.4.35}$
$E°_{cell} = E°_{cathode} − E°_{anode} = 1.10\; V \nonumber$
This is the same value that is observed experimentally. If the value of $E°_{cell}$ is positive, the reaction will occur spontaneously as written. If the value of $E°_{cell}$ is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction. As we shall see in Section 20.9, this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example $2$ and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples.
A positive $E°_{cell}$ means that the reaction will occur spontaneously as written. A negative $E°_{cell}$ means that the reaction will proceed spontaneously in the opposite direction.
Example $2$
A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl3, and the other contains a piece of nickel immersed in a 1 M solution of NiCl2. The half-reactions that occur when the compartments are connected are as follows:
cathode: Ni2+(aq) + 2e → Ni(s)
anode: Ga(s) → Ga3+(aq) + 3e
If the potential for the oxidation of Ga to Ga3+ is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni2+?
Given: galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions
Asked for: standard electrode potential of reaction occurring at the cathode
Strategy:
1. Write the equation for the half-reaction that occurs at the anode along with the value of the standard electrode potential for the half-reaction.
2. Use Equation $\ref{20.4.2}$ to calculate the standard electrode potential for the half-reaction that occurs at the cathode. Then reverse the sign to obtain the potential for the corresponding oxidation half-reaction under standard conditions.
Solution
A We have been given the potential for the oxidation of Ga to Ga3+ under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga3+(aq) + 3e → Ga(s), E°anode = −0.55 V.
B Using the value given for $E°_{cell}$ and the calculated value of E°anode, we can calculate the standard potential for the reduction of Ni2+ to Ni from Equation $\ref{20.4.2}$:
\begin{align*} E°_{cell} &= E°_{cathode} − E°_{anode} \[4pt] 0.27\, V &= E^o°_{cathhode} − (−0.55\, V) \[4pt] E^°_{cathode} &= −0.28 \,V \end{align*} \nonumber
This is the standard electrode potential for the reaction Ni2+(aq) + 2e → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni2+ under standard conditions, we must reverse the sign of E°cathode. Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry.
Exercise $2$
A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate $\ce{Hg(CH_3CO_2)_2}$ and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of $\ce{MgCl2}$. When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur:
• cathode: $\ce{Hg^{2+} (aq) + 2e^{−} → Hg(l)}$
• anode: $\ce{Mg(s) → Mg^{2+}(aq) + 2e^{−}}$
If the potential for the oxidation of $\ce{Mg}$ to $\ce{Mg^{2+}}$ is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the cathode?
Answer
0.85 V
Reference Electrodes and Measuring Concentrations
When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called indicator electrode, whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the reference electrode, must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode.
The potential of any reference electrode should not be affected by the properties of the solution to be analyzed, and it should also be physically isolated.
There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the silver–silver chloride electrode, which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows:
$Cl^−_{(aq)}∣AgCl_{(s)}∣Ag_{(s)} \label{20.4.36}$
$AgCl_{(s)}+e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)} \nonumber$
If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE.
A second common reference electrode is the saturated calomel electrode (SCE), which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg2Cl2; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in Figure $5$. Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows:
$Pt_{(s)} ∣ Hg_2Cl_{2(s)}∣KCl_{(aq, sat)} \label{20.4.37}$
$Hg_2Cl_{2(s)} + 2e^− \rightarrow 2Hg_{(l)} + 2Cl^−{(aq)} \label{20.4.38}$
At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential.
One of the most common uses of electrochemistry is to measure the H+ ion concentration of a solution. A glass electrode is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in Figure $5$. The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H+] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H+] as follows (recall that pH = −log[H+]):
$E_{glass} = E′ + (0.0591\; V \times \log[H^+]) = E′ − 0.0591\; V \times pH \label{20.4.39}$
The voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH.
Ion-selective electrodes are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in Figure $5$). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped $LaF_3$ as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in Table $1$.
Table $1$: Some Species Whose Aqueous Concentrations Can Be Measured Using Electrochemical Methods
Species Type of Sample
H+ laboratory samples, blood, soil, and ground and surface water
NH3/NH4+ wastewater and runoff water
K+ blood, wine, and soil
CO2/HCO3 blood and groundwater
F groundwater, drinking water, and soil
Br grains and plant extracts
I milk and pharmaceuticals
NO3 groundwater, drinking water, soil, and fertilizer
Summary
Redox reactions can be balanced using the half-reaction method. The standard cell potential is a measure of the driving force for the reaction. $E°_{cell} = E°_{cathode} − E°_{anode} \nonumber \] The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E°cell). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E°cell = E°cathode − E°anode). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed. If \(E°_{cell}$ is positive, the reaction will occur spontaneously under standard conditions. If $E°_{cell}$ is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.04%3A_Standard_Reduction_Potentials.txt |
Learning Objectives
• To understand the relationship between cell potential and the equilibrium constant.
• To use cell potentials to calculate solution concentrations.
Changes in reaction conditions can have a tremendous effect on the course of a redox reaction. For example, under standard conditions, the reaction of $\ce{Co(s)}$ with $\ce{Ni^{2+}(aq)}$ to form $\ce{Ni(s)}$ and $\ce{Co^{2+}(aq)}$ occurs spontaneously, but if we reduce the concentration of $\ce{Ni^{2+}}$ by a factor of 100, so that $\ce{[Ni^{2+}]}$ is 0.01 M, then the reverse reaction occurs spontaneously instead. The relationship between voltage and concentration is one of the factors that must be understood to predict whether a reaction will be spontaneous.
The Relationship between Cell Potential & Gibbs Energy
Electrochemical cells convert chemical energy to electrical energy and vice versa. The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the reductant to the oxidant during the course of a reaction. The resulting electric current is measured in coulombs (C), an SI unit that measures the number of electrons passing a given point in 1 s. A coulomb relates energy (in joules) to electrical potential (in volts). Electric current is measured in amperes (A); 1 A is defined as the flow of 1 C/s past a given point (1 C = 1 A·s):
$\dfrac{\textrm{1 J}}{\textrm{1 V}}=\textrm{1 C}=\mathrm{A\cdot s} \label{20.5.1}$
In chemical reactions, however, we need to relate the coulomb to the charge on a mole of electrons. Multiplying the charge on the electron by Avogadro’s number gives us the charge on 1 mol of electrons, which is called the faraday (F), named after the English physicist and chemist Michael Faraday (1791–1867):
\begin{align}F &=(1.60218\times10^{-19}\textrm{ C})\left(\dfrac{6.02214 \times 10^{23}\, J}{\textrm{1 mol e}^-}\right) \[4pt] &=9.64833212\times10^4\textrm{ C/mol e}^- \[4pt] &\simeq 96,485\, J/(\mathrm{V\cdot mol\;e^-})\end{align} \label{20.5.2}
The total charge transferred from the reductant to the oxidant is therefore $nF$, where $n$ is the number of moles of electrons.
Michael Faraday (1791–1867)
Faraday was a British physicist and chemist who was arguably one of the greatest experimental scientists in history. The son of a blacksmith, Faraday was self-educated and became an apprentice bookbinder at age 14 before turning to science. His experiments in electricity and magnetism made electricity a routine tool in science and led to both the electric motor and the electric generator. He discovered the phenomenon of electrolysis and laid the foundations of electrochemistry. In fact, most of the specialized terms introduced in this chapter (electrode, anode, cathode, and so forth) are due to Faraday. In addition, he discovered benzene and invented the system of oxidation state numbers that we use today. Faraday is probably best known for “The Chemical History of a Candle,” a series of public lectures on the chemistry and physics of flames.
The maximum amount of work that can be produced by an electrochemical cell ($w_{max}$) is equal to the product of the cell potential ($E^°_{cell}$) and the total charge transferred during the reaction ($nF$):
$w_{max} = nFE_{cell} \label{20.5.3}$
Work is expressed as a negative number because work is being done by a system (an electrochemical cell with a positive potential) on its surroundings.
The change in free energy ($\Delta{G}$) is also a measure of the maximum amount of work that can be performed during a chemical process ($ΔG = w_{max}$). Consequently, there must be a relationship between the potential of an electrochemical cell and $\Delta{G}$; this relationship is as follows:
$\Delta{G} = −nFE_{cell} \label{20.5.4}$
A spontaneous redox reaction is therefore characterized by a negative value of $\Delta{G}$ and a positive value of $E^°_{cell}$, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and $E^°_{cell}$ is as follows:
$\Delta{G^°} = −nFE^°_{cell} \label{20.5.5}$
A spontaneous redox reaction is characterized by a negative value of ΔG°, which corresponds to a positive value of E°cell.
Example $1$
Suppose you want to prepare elemental bromine from bromide using the dichromate ion as an oxidant. Using the data in Table P2, calculate the free-energy change (ΔG°) for this redox reaction under standard conditions. Is the reaction spontaneous?
Given: redox reaction
Asked for: $ΔG^o$ for the reaction and spontaneity
Strategy:
1. From the relevant half-reactions and the corresponding values of $E^o$, write the overall reaction and calculate $E^°_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Then use Equation \ref{20.5.5} to calculate $ΔG^o$. If $ΔG^o$ is negative, then the reaction is spontaneous.
A
As always, the first step is to write the relevant half-reactions and use them to obtain the overall reaction and the magnitude of $E^o$. From Table P2, we can find the reduction and oxidation half-reactions and corresponding $E^o$ values:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \ & \textrm{anode:} &\quad & \mathrm{2Br^{-}(aq)} \rightarrow \mathrm{Br_2(aq)} +\mathrm{2e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \end{align*} \nonumber
To obtain the overall balanced chemical equation, we must multiply both sides of the oxidation half-reaction by 3 to obtain the same number of electrons as in the reduction half-reaction, remembering that the magnitude of $E^o$ is not affected:
\begin{align*} & \textrm{cathode:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{14H^+(aq)}+\mathrm{6e^-}\rightarrow \mathrm{2Cr^{3+}(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cathode}} =\textrm{1.36 V} \[4pt] & \textrm{anode:} &\quad & \mathrm{6Br^{-}(aq)} \rightarrow \mathrm{3Br_2(aq)} +\mathrm{6e^-} &\quad & E^\circ_{\textrm{anode}} =\textrm{1.09 V} \[4pt] \hline & \textrm{overall:} &\quad & \mathrm{Cr_2O_7^{2-}(aq)} + \mathrm{6Br^{-}(aq)} + \mathrm{14H^+(aq)} \rightarrow \mathrm{2Cr^{3+}(aq)} + \mathrm{3Br_2(aq)} +\mathrm{7H_2O(l)} &\quad & E^\circ_{\textrm{cell}} =\textrm{0.27 V} \end{align*} \nonumber
B
We can now calculate ΔG° using Equation $\ref{20.5.5}$. Because six electrons are transferred in the overall reaction, the value of $n$ is 6:
\begin{align*}\Delta G^\circ &=-(n)(F)(E^\circ_{\textrm{cell}}) \[4pt] & =-(\textrm{6 mole})[96,485\;\mathrm{J/(V\cdot mol})(\textrm{0.27 V})] \& =-15.6 \times 10^4\textrm{ J} \ & =-156\;\mathrm{kJ/mol\;Cr_2O_7^{2-}} \end{align*} \nonumber
Thus $ΔG^o$ is −168 kJ/mol for the reaction as written, and the reaction is spontaneous.
Exercise $1$
Use the data in Table P2 to calculate $ΔG^o$ for the reduction of ferric ion by iodide:
$\ce{2Fe^{3+}(aq) + 2I^{−}(aq) → 2Fe^{2+}(aq) + I2(s)}\nonumber$
Is the reaction spontaneous?
Answer
−44 kJ/mol I2; yes
Relating G and Ecell: Relating G and Ecell(opens in new window) [youtu.be]
Potentials for the Sums of Half-Reactions
Although Table P2 list several half-reactions, many more are known. When the standard potential for a half-reaction is not available, we can use relationships between standard potentials and free energy to obtain the potential of any other half-reaction that can be written as the sum of two or more half-reactions whose standard potentials are available. For example, the potential for the reduction of $\ce{Fe^{3+}(aq)}$ to $\ce{Fe(s)}$ is not listed in the table, but two related reductions are given:
$\ce{Fe^{3+}(aq) + e^{−} -> Fe^{2+}(aq)} \;\;\;E^° = +0.77 V \label{20.5.6}$
$\ce{Fe^{2+}(aq) + 2e^{−} -> Fe(s)} \;\;\;E^° = −0.45 V \label{20.5.7}$
Although the sum of these two half-reactions gives the desired half-reaction, we cannot simply add the potentials of two reductive half-reactions to obtain the potential of a third reductive half-reaction because $E^o$ is not a state function. However, because $ΔG^o$ is a state function, the sum of the $ΔG^o$ values for the individual reactions gives us $ΔG^o$ for the overall reaction, which is proportional to both the potential and the number of electrons ($n$) transferred. To obtain the value of $E^o$ for the overall half-reaction, we first must add the values of $ΔG^o (= −nFE^o)$ for each individual half-reaction to obtain $ΔG^o$ for the overall half-reaction:
\begin{align*}\ce{Fe^{3+}(aq)} + \ce{e^-} &\rightarrow \ce\mathrm{Fe^{2+}(aq)} &\quad \Delta G^\circ &=-(1)(F)(\textrm{0.77 V})\[4pt] \ce{Fe^{2+}(aq)}+\ce{2e^-} &\rightarrow\ce{Fe(s)} &\quad\Delta G^\circ &=-(2)(F)(-\textrm{0.45 V})\[4pt] \ce{Fe^{3+}(aq)}+\ce{3e^-} &\rightarrow \ce{Fe(s)} &\quad\Delta G^\circ & =[-(1)(F)(\textrm{0.77 V})]+[-(2)(F)(-\textrm{0.45 V})] \end{align*} \nonumber
Solving the last expression for ΔG° for the overall half-reaction,
$\Delta{G^°} = F[(−0.77 V) + (−2)(−0.45 V)] = F(0.13 V) \label{20.5.9}$
Three electrons ($n = 3$) are transferred in the overall reaction, so substituting into Equation $\ref{20.5.5}$ and solving for $E^o$ gives the following:
\begin{align*}\Delta G^\circ & =-nFE^\circ_{\textrm{cell}} \[4pt] F(\textrm{0.13 V}) & =-(3)(F)(E^\circ_{\textrm{cell}}) \[4pt] E^\circ & =-\dfrac{0.13\textrm{ V}}{3}=-0.043\textrm{ V}\end{align*} \nonumber
This value of $E^o$ is very different from the value that is obtained by simply adding the potentials for the two half-reactions (0.32 V) and even has the opposite sign.
Values of $E^o$ for half-reactions cannot be added to give $E^o$ for the sum of the half-reactions; only values of $ΔG^o = −nFE^°_{cell}$ for half-reactions can be added.
The Relationship between Cell Potential & the Equilibrium Constant
We can use the relationship between $\Delta{G^°}$ and the equilibrium constant $K$, to obtain a relationship between $E^°_{cell}$ and $K$. Recall that for a general reaction of the type $aA + bB \rightarrow cC + dD$, the standard free-energy change and the equilibrium constant are related by the following equation:
$\Delta{G°} = −RT \ln K \label{20.5.10}$
Given the relationship between the standard free-energy change and the standard cell potential (Equation $\ref{20.5.5}$), we can write
$−nFE^°_{cell} = −RT \ln K \label{20.5.12}$
Rearranging this equation,
$E^\circ_{\textrm{cell}}= \left( \dfrac{RT}{nF} \right) \ln K \label{20.5.12B}$
For $T = 298\, K$, Equation $\ref{20.5.12B}$ can be simplified as follows:
\begin{align} E^\circ_{\textrm{cell}} &=\left(\dfrac{RT}{nF}\right)\ln K \[4pt] &=\left[ \dfrac{[8.314\;\mathrm{J/(mol\cdot K})(\textrm{298 K})]}{n[96,485\;\mathrm{J/(V\cdot mol)}]}\right]2.303 \log K \[4pt] &=\left(\dfrac{\textrm{0.0592 V}}{n}\right)\log K \label{20.5.13} \end{align}
Thus $E^°_{cell}$ is directly proportional to the logarithm of the equilibrium constant. This means that large equilibrium constants correspond to large positive values of $E^°_{cell}$ and vice versa.
Example $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of metallic lead with PbO2 in the presence of sulfate ions to give PbSO4 under standard conditions. (This reaction occurs when a car battery is discharged.) Report your answer to two significant figures.
Given: redox reaction
Asked for: $K$
Strategy:
1. Write the relevant half-reactions and potentials. From these, obtain the overall reaction and $E^o_{cell}$.
2. Determine the number of electrons transferred in the overall reaction. Use Equation $\ref{20.5.13}$ to solve for $\log K$ and then $K$.
Solution
A The relevant half-reactions and potentials from Table P2 are as follows:
\begin{align*} & \textrm {cathode:} & & \mathrm{PbO_2(s)}+\mathrm{SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}+\mathrm{2e^-}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cathode}=\textrm{1.69 V} \[4pt] & \textrm{anode:} & & \mathrm{Pb(s)}+\mathrm{SO_4^{2-}(aq)}\rightarrow\mathrm{PbSO_4(s)}+\mathrm{2e^-} & & E^\circ_\textrm{anode}=-\textrm{0.36 V} \[4pt] \hline & \textrm {overall:} & & \mathrm{Pb(s)}+\mathrm{PbO_2(s)}+\mathrm{2SO_4^{2-}(aq)}+\mathrm{4H^+(aq)}\rightarrow\mathrm{2PbSO_4(s)}+\mathrm{2H_2O(l)} & & E^\circ_\textrm{cell}=\textrm{2.05 V} \end{align*} \nonumber
B Two electrons are transferred in the overall reaction, so $n = 2$. Solving Equation $\ref{20.5.13}$ for log K and inserting the values of $n$ and $E^o$,
\begin{align*}\log K & =\dfrac{nE^\circ}{\textrm{0.0591 V}}=\dfrac{2(\textrm{2.05 V})}{\textrm{0.0591 V}}=69.37 \[4pt] K & =2.3\times10^{69}\end{align*} \nonumber
Thus the equilibrium lies far to the right, favoring a discharged battery (as anyone who has ever tried unsuccessfully to start a car after letting it sit for a long time will know).
Exercise $2$
Use the data in Table P2 to calculate the equilibrium constant for the reaction of $\ce{Sn^{2+}(aq)}$ with oxygen to produce $\ce{Sn^{4+}(aq)}$ and water under standard conditions. Report your answer to two significant figures. The reaction is as follows:
$\ce{2Sn^{2+}(aq) + O2(g) + 4H^{+}(aq) <=> 2Sn^{4+}(aq) + 2H2O(l)} \nonumber$
Answer
$5.7 \times 10^{72}$
Figure $1$ summarizes the relationships that we have developed based on properties of the system—that is, based on the equilibrium constant, standard free-energy change, and standard cell potential—and the criteria for spontaneity (ΔG° < 0). Unfortunately, these criteria apply only to systems in which all reactants and products are present in their standard states, a situation that is seldom encountered in the real world. A more generally useful relationship between cell potential and reactant and product concentrations, as we are about to see, uses the relationship between $\Delta{G}$ and the reaction quotient $Q$.
Electrode Potentials and ECell: Electrode Potentials and Ecell(opens in new window) [youtu.be]
Summary
A coulomb (C) relates electrical potential, expressed in volts, and energy, expressed in joules. The current generated from a redox reaction is measured in amperes (A), where 1 A is defined as the flow of 1 C/s past a given point. The faraday (F) is Avogadro’s number multiplied by the charge on an electron and corresponds to the charge on 1 mol of electrons. The product of the cell potential and the total charge is the maximum amount of energy available to do work, which is related to the change in free energy that occurs during the chemical process. Adding together the ΔG values for the half-reactions gives ΔG for the overall reaction, which is proportional to both the potential and the number of electrons (n) transferred. Spontaneous redox reactions have a negative ΔG and therefore a positive Ecell. Because the equilibrium constant K is related to ΔG, E°cell and K are also related. Large equilibrium constants correspond to large positive values of E°. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.05%3A_Cell_Potential_Gibbs_Energy_and_the_Equilibrium_Constant.txt |
Learning Objectives
• Relate cell potentials to Gibbs energy changes
• Use the Nernst equation to determine cell potentials at nonstandard conditions
• Perform calculations that involve converting between cell potentials, free energy changes, and equilibrium constants
The Nernst Equation enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the reaction quotient and allows the accurate determination of equilibrium constants (including solubility constants).
The Effect of Concentration on Cell Potential: The Nernst Equation
Recall that the actual free-energy change for a reaction under nonstandard conditions, $\Delta{G}$, is given as follows:
$\Delta{G} = \Delta{G°} + RT \ln Q \label{Eq1}$
We also know that $ΔG = −nFE_{cell}$ (under non-standard conditions) and $ΔG^o = −nFE^o_{cell}$ (under standard conditions). Substituting these expressions into Equation $\ref{Eq1}$, we obtain
$−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2}$
Dividing both sides of this equation by $−nF$,
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3}$
Equation $\ref{Eq3}$ is called the Nernst equation, after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium ($ΔG = 0$), then Equation $\ref{Eq3}$ reduces to Equation $\ref{Eq31}$ and $\ref{Eq32}$ because $Q = K$, and there is no net transfer of electrons (i.e., Ecell = 0).
$E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31}$
since
$E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32}$
Substituting the values of the constants into Equation $\ref{Eq3}$ with $T = 298\, K$ and converting to base-10 logarithms give the relationship of the actual cell potential (Ecell), the standard cell potential (E°cell), and the reactant and product concentrations at room temperature (contained in $Q$):
$E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4}$
The Power of the Nernst Equation
The Nernst Equation ($\ref{Eq3}$) can be used to determine the value of Ecell, and thus the direction of spontaneous reaction, for any redox reaction under any conditions.
Equation $\ref{Eq4}$ allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation $\ref{Eq4}$ that the cell potential changes by 0.0591/n V for each 10-fold change in the value of $Q$ because log 10 = 1.
Example $1$
The following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which means that ΔG° < 0):
$\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber$
Calculate $E_{cell}$ for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce4+] = 0.013 M, [Ce3+] = 0.60 M, [Cl] = 0.0030 M, $P_\mathrm{Cl_2}$ = 1.0 atm, and T = 25°C.
Given: balanced redox reaction, standard cell potential, and nonstandard conditions
Asked for: cell potential
Strategy:
Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions.
Solution
We can use the information given and the Nernst equation to calculate Ecell. Moreover, because the temperature is 25°C (298 K), we can use Equation $\ref{Eq4}$ instead of Equation $\ref{Eq3}$. The overall reaction involves the net transfer of two electrons:
$2Ce^{4+}_{(aq)} + 2e^− \rightarrow 2Ce^{3+}_{(aq)}\nonumber$
$2Cl^−_{(aq)} \rightarrow Cl_{2(g)} + 2e^−\nonumber$
so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl2, and the value of E°cell into Equation $\ref{Eq4}$,
\begin{align*}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ & =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \ & =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align*} \nonumber
Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture
Exercise $1$
Molecular oxygen will not oxidize $MnO_2$ to permanganate via the reaction
$\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber$
Calculate $E_{cell}$ for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, $P_\mathrm{O_2}$= 0.20 atm, [MNO4] = 1.0 × 10−4 M, and T = 25°C.
Answer
Ecell = −0.22 V; the reaction will not occur spontaneously.
Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows:
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5}$
The reaction quotient is therefore $Q = [Zn^{2+}]/[Cu^{2+}]$. Suppose that the cell initially contains 1.0 M Cu2+ and 1.0 × 10−6 M Zn2+. The initial voltage measured when the cell is connected can then be calculated from Equation $\ref{Eq4}$:
\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\ & =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6}
Thus the initial voltage is greater than E° because $Q<1$. As the reaction proceeds, [Zn2+] in the anode compartment increases as the zinc electrode dissolves, while [Cu2+] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn2+]/[Cu2+] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn2+] = [Cu2+], so Q = 1 and Ecell = E°cell. Beyond this point, [Zn2+] will continue to increase in the anode compartment, and [Cu2+] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in Ecell. When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn2+ and 1.0 × 10−6 M Cu2+), Q = 1.0 × 106, and the cell potential will be reduced to 0.92 V.
The variation of Ecell with $\log{Q}$ over this range is linear with a slope of −0.0591/n, as illustrated in Figure $1$. As the reaction proceeds still further, $Q$ continues to increase, and Ecell continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of $Q$ when Ecell = 0 is calculated as follows:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q=0 \ E^\circ &=\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \ \log Q &=\dfrac{E^\circ n}{\textrm{0.0591 V}}=\dfrac{(\textrm{1.10 V})(2)}{\textrm{0.0591 V}}=37.23 \ Q &=10^{37.23}=1.7\times10^{37}\end{align} \label{Eq7}
Recall that at equilibrium, $Q = K$. Thus the equilibrium constant for the reaction of Zn metal with Cu2+ to give Cu metal and Zn2+ is 1.7 × 1037 at 25°C.
The Nernst Equation: The Nernst Equation (opens in new window) [youtu.be]
Concentration Cells
A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO3 in one compartment and 1.0 M AgNO3 in the other. The cell diagram and corresponding half-reactions are as follows:
$\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8}$
cathode:
$\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9}$
anode:
$\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10}$
Overall
$\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11}$
As the reaction progresses, the concentration of $Ag^+$ will increase in the left (oxidation) compartment as the silver electrode dissolves, while the $Ag^+$ concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of $Ag(s)$ in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E°cell because E°cathode = −E°anode:
\begin{align*} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \[4pt] &=\textrm{0.12 V} \end{align*} \nonumber
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell. Finally, when the concentration of Ag+ is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (Ecell = 0).
Example $2$
Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl2 as the cathode, and a manganese electrode immersed in a 5.2 × 10−2 M solution of MnSO4 as the anode (T = 25°C).
Given: galvanic cell, identities of the electrodes, and solution concentrations
Asked for: voltage
Strategy:
1. Write the overall reaction that occurs in the cell.
2. Determine the number of electrons transferred. Substitute this value into the Nernst equation to calculate the voltage.
Solution
A This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO42) do not participate in the reaction, so their identity is not important. The overall reaction is as follows:
$\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber$
B For the reduction of Mn2+(aq) to Mn(s), n = 2. We substitute this value and the given Mn2+ concentrations into Equation $\ref{Eq4}$:
\begin{align*} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \[4pt] &=\textrm{0.047 V}\end{align*} \nonumber
Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution.
Exercise $2$
Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na2SO4 at pH 7.00. Both cells are in contact with the atmosphere, with $P_\mathrm{O_2}$ = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water:
$\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber$
What will be the potential when the circuit is closed?
Answer
0.41 V
Using Cell Potentials to Measure Solubility Products
Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products ($K_{sp}$) of sparingly soluble substances. As you learned previously, solubility products can be very small, with values of less than or equal to 10−30. Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods.
To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure $1$, which is designed to measure the solubility product of silver chloride:
$K_{sp} = [\ce{Ag^{+}}][\ce{Cl^{−}}]. \nonumber$
In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag+; the other compartment contains a silver wire inserted into a 1.0 M Cl solution saturated with AgCl. In this system, the Ag+ ion concentration in the first compartment equals Ksp. We can see this by dividing both sides of the equation for Ksp by [Cl] and substituting:
\begin{align*}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align*} \nonumber
The overall cell reaction is as follows:
Ag+(aq, concentrated) → Ag+(aq, dilute)
Thus the voltage of the concentration cell due to the difference in [Ag+] between the two cells is as follows:
\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]_\textrm{dilute}}{[\mathrm{Ag^+}]_\textrm{concentrated}}\right) \nonumber \[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align}
By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation $\ref{Eq122}$ for $K_{sp}$,
\begin{align*}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \[4pt] K_\textrm{sp} & =1.5\times10^{-10}\end{align*} \nonumber
Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt.
Example $3$: Solubility of lead(II) sulfate
To measure the solubility product of lead(II) sulfate (PbSO4) at 25°C, you construct a galvanic cell like the one shown in Figure $1$, which contains a 1.0 M solution of a very soluble Pb2+ salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na2SO4 saturated with PbSO4 in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is Ksp for PbSO4? Report your answer to two significant figures.
Given: galvanic cell, solution concentrations, electrodes, and voltage
Asked for: Ksp
Strategy:
1. From the information given, write the equation for Ksp. Express this equation in terms of the concentration of Pb2+.
2. Determine the number of electrons transferred in the electrochemical reaction. Substitute the appropriate values into Equation $\ref{Eq12}$ and solve for Ksp.
Solution
A You have constructed a concentration cell, with one compartment containing a 1.0 M solution of $\ce{Pb^{2+}}$ and the other containing a dilute solution of Pb2+ in 1.0 M Na2SO4. As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb2+ in the dilute solution to Ksp:
\begin{align*}[\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \ [\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align*} \nonumber
B The reduction of Pb2+ to Pb is a two-electron process and proceeds according to the following reaction:
Pb2+(aq, concentrated) → Pb2+(aq, dilute)
so
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \ \textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]_\textrm{dilute}}{[\mathrm{Pb^{2+}}]_\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \ -7.77 & =\log K_\textrm{sp} \ 1.7\times10^{-8} & =K_\textrm{sp}\end{align*} \nonumber
Exercise $3$
A concentration cell similar to the one described in Example $3$ contains a 1.0 M solution of lanthanum nitrate [La(NO3)3] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF3 in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the Ksp for LaF3? Report your answer to two significant figures.
Answer
5.7 × 10−17
Using Cell Potentials to Measure Concentrations
Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example $3$, in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to Ksp. Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H+] using a pH meter, as illustrated below.
Example $4$: Measuring pH
Suppose a galvanic cell is constructed with a standard Zn/Zn2+ couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H+] in the second compartment is unknown. The cell diagram is as follows:
$\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber$
What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C?
Given: galvanic cell, cell diagram, and cell potential
Asked for: pH of the solution
Strategy:
1. Write the overall cell reaction.
2. Substitute appropriate values into the Nernst equation and solve for −log[H+] to obtain the pH.
Solution
A Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H2 (note that Zn lies below H2 in Table P2):
Zn(s) + 2H2+(aq) → Zn2+(aq) + H2(g) E°=0.76 V
B By substituting the given values into the simplified Nernst equation (Equation $\ref{Eq4}$), we can calculate [H+] under nonstandard conditions:
\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \ \textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \ 16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \ 8.46 &=-\log[\mathrm{H^+}] \ 8.5 &=\mathrm{pH}\end{align*} \nonumber
Thus the potential of a galvanic cell can be used to measure the pH of a solution.
Exercise $4$
Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb2+ in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E°cathode = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows:
$Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber$
When the circuit is closed, the cell has a measured potential of 1.62 V. Use Table P2 to determine the concentration of Pb2+ in the groundwater.
Answer
$1.2 \times 10^{−9}\; M$
Summary
The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.06%3A_Cell_Potential_and_Concentration.txt |
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A battery (storage cell) is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells.
Batteries
There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell.
Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (Figure $1$). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells.
The major difference between batteries and the galvanic cells is that commercial typically batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. An obvious exception is the standard car battery which used solution phase chemistry.
Leclanché Dry Cell
The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the Leclanché dry cell is actually a “wet cell”: the electrolyte is an acidic water-based paste containing $MnO_2$, $NH_4Cl$, $ZnCl_2$, graphite, and starch (part (a) in Figure $1$). The half-reactions at the anode and the cathode can be summarized as follows:
• cathode (reduction):
$\ce{2MnO2(s) + 2NH^{+}4(aq) + 2e^{−} -> Mn2O3(s) + 2NH3(aq) + H2O(l)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{−}} \nonumber$
The $\ce{Zn^{2+}}$ ions formed by the oxidation of $\ce{Zn(s)}$ at the anode react with $\ce{NH_3}$ formed at the cathode and $\ce{Cl^{−}}$ ions present in solution, so the overall cell reaction is as follows:
• overall reaction:
$\ce{2MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)} \label{Eq3}$
The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the $\ce{MnO2}$ that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the $\ce{Zn}$ anode reacts spontaneously with $\ce{NH4Cl}$ in the electrolyte, causing the case to corrode and allowing the contents to leak out.
The alkaline battery is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows:
• cathode (reduction)
$\ce{2MnO2(s) + H2O(l) + 2e^{−} -> Mn2O3(s) + 2OH^{−}(aq)} \nonumber$
• anode (oxidation):
$\ce{Zn(s) + 2OH^{−}(aq) -> ZnO(s) + H2O(l) + 2e^{−}} \nonumber$
• overall reaction:
$\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)} \nonumber$
This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective.
Button Batteries
Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these "button" batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either $\ce{HgO}$ or $\ce{Ag2O}$ as the oxidant rather than $\ce{MnO2}$ in Figure $\PageIndex{1b}$).
The cathode, anode and overall reactions and cell output for these two types of button batteries are as follows (two half-reactions occur at the anode, but the overall oxidation half-reaction is shown):
• cathode (mercury battery): $\ce{HgO(s) + H2O(l) + 2e^{−} -> Hg(l) + 2OH^{−}(aq)} \nonumber$
• Anode (mercury battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• overall reaction (mercury battery): $\ce{Zn(s) + 2HgO(s) -> 2Hg(l) + ZnO(s)} \nonumber$ with $E_{cell} = 1.35 \,V$.
• cathode reaction (silver battery): $\ce{Ag2O(s) + H2O(l) + 2e^{−} -> 2Ag(s) + 2OH^{−}(aq)} \nonumber$
• anode (silver battery): $\ce{Zn + 2OH^{−} -> ZnO + H2O + 2e^{−}} \nonumber$
• Overall reaction (silver battery): $\ce{Zn(s) + 2Ag2O(s) -> 2Ag(s) + ZnO(s)} \nonumber$ with $E_{cell} = 1.6 \,V$.
The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as $\ce{Hg}$ and $\ce{Ag}$.
Lithium–Iodine Battery
None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries. One of the few commercially successful water-free batteries is the lithium–iodine battery. The anode is lithium metal, and the cathode is a solid complex of $I_2$. Separating them is a layer of solid $LiI$, which acts as the electrolyte by allowing the diffusion of Li+ ions. The electrode reactions are as follows:
• cathode (reduction):
$I_{2(s)} + 2e^− \rightarrow {2I^-}_{(LiI)}\label{Eq11}$
• anode (oxidation):
$2Li_{(s)} \rightarrow 2Li^+_{(LiI)} + 2e^− \label{Eq12}$
• overall:
$2Li_{(s)}+ I_{2(s)} \rightarrow 2LiI_{(s)} \label{Eq12a}$
with $E_{cell} = 3.5 \, V$
As shown in part (c) in Figure $1$, a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using $TiS_2$, for example, for the cathode.
Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next.
Nickel–Cadmium (NiCad) Battery
The nickel–cadmium, or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in Figure $2$, the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible.
The electrode reactions during the discharge of a $NiCad$ battery are as follows:
• cathode (reduction):
$2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{Eq13}$
• anode (oxidation):
$Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{Eq14}$
• overall:
$Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{Eq15}$
$E_{cell} = 1.4 V$
Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH)2 and 2Ni(OH)2], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium.
A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows:
$NiO(OH)_{(s)} + MH \rightarrow Ni(OH)_{2(s)} + M_{(s)} \label{Eq16}$
The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery.
Directive 2006/66/EC of the European Union prohibits the placing on the market of portable batteries that contain more than 0.002% of cadmium by weight. The aim of this directive was to improve "the environmental performance of batteries and accumulators"
Lead–Acid (Lead Storage) Battery
The lead–acid battery is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E°cell) times the number of cells.
As shown in Figure $3$, the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide ($PbO_2$). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M $H_2SO_4$). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows:
• cathode (reduction):
$PbO_{2(s)} + HSO^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq17}$
with $E^°_{cathode} = 1.685 \; V$
• anode (oxidation):
$Pb_{(s)} + HSO^−_{4(aq)} \rightarrow PbSO_{4(s) }+ H^+_{(aq)} + 2e^−\label{Eq18}$
with $E^°_{anode} = −0.356 \; V$
• overall:
$Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq19}$
and $E^°_{cell} = 2.041 \; V$
As the cell is discharged, a powder of $PbSO_4$ forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte. This is often done with the use of a hydrometer.
A hydrometer can be used to test the specific gravity of each cell as a measure of its state of charge (www.youtube.com/watch?v=SRcOqfL6GqQ).
When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and $PbSO_4$ is converted back to metallic lead and $PbO_2$. If the battery is recharged too vigorously, however, electrolysis of water can occur:
$2H_2O_{(l)} \rightarrow 2H_{2(g)} +O_{2 (g)} \label{EqX}$
This results in the evolution of potentially explosive hydrogen gas. The gas bubbles formed in this way can dislodge some of the $PbSO_4$ or $PbO_2$ particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse.
Fuel Cells
A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as US manned space vehicles.
These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H2(g) and O2(g), as illustrated in Figure $4$. The electrode reactions are as follows:
• cathode (reduction):
$O_{2(g)} + 4H^+ + 4e^− \rightarrow 2H_2O_{(g)} \label{Eq20}$
• anode (oxidation):
$2H_{2(g)} \rightarrow 4H^+ + 4e^− \label{Eq21}$
• overall:
$2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)} \label{Eq22}$
The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of $O_2 (g)$ at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of $O_2 (g)$.
Summary
Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.07%3A_Batteries-_Using_Chemistry_to_Generate_Electricity.txt |
Learning Objectives
• To understand electrolysis and describe it quantitatively.
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called electrolysis, occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications.
Electrolytic Cells
If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu2+ solution and the other electrode is cadmium metal immersed in a $\,1\; M\, Cd^{2+}$ solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd2+) and is the anode, while metallic copper will be deposited on the copper electrode (Cu2+ is reduced to Cu), which is the cathode (Figure $\PageIndex{1a}$).
The overall reaction is as follows:
$\ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber$
with $E°_{cell} = 0.74\; V$
This reaction is thermodynamically spontaneous as written ($ΔG^o < 0$):
\begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber
In this direction, the system is acting as a galvanic cell.
In an electrolytic cell, an external voltage is applied to drive a nonspontaneous reaction.
The reverse reaction, the reduction of Cd2+ by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd2+ is reduced) (Figure $\PageIndex{1b}$). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows:
• half-reaction at the cathode:
$\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3}$
with $E^°_{cathode} = −0.40 \, V$
• half-reaction at the anode:
$\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4}$
with $E^°_{anode} = 0.34 \, V$
• Overall Reaction:
$\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5}$
with $E^°_{cell} = −0.74 \: V$
Because $E^°_{cell} < 0$, the overall reaction—the reduction of $Cd^{2+}$ by $Cu$—clearly cannot occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table $1$.
Table $1$: Comparison of Galvanic and Electrolytic Cells
Property Galvanic Cell Electrolytic Cell
ΔG < 0 > 0
Ecell > 0 < 0
Electrode Process
anode oxidation oxidation
cathode reduction reduction
Sign of Electrode
anode +
cathode +
Electrolytic Reactions
At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten $\ce{NaCl}$, for example, and an electrical potential is applied, $\ce{Cl^{-}}$ is oxidized at the anode, and $\ce{Na^{+}}$ is reduced at the cathode. The overall reaction is as follows:
$\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6}$
This is the reverse of the formation of $\ce{NaCl}$ from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of $\ce{NaCl}$ (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten $\ce{NaCl}$ in a Downs cell (Figure $2$). In this specialized cell, $\ce{CaCl2}$ (melting point = 772°C) is first added to the $\ce{NaCl}$ to lower the melting point of the mixture to about 600°C, thereby lowering operating costs.
Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al2O3; melting point = 2054°C) and 95% cryolite (Na3AlF6; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO2 gas at the carbon anode. The overall reaction is as follows:
$\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7}$
Oxide ions react with oxidized carbon at the anode, producing CO2(g).
There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general.
1. The electrode potentials for molten salts are likely to be very different from the standard cell potentials listed in Table P2, which are compiled for the reduction of the hydrated ions in aqueous solutions under standard conditions.
2. Using a mixed salt system means there is a possibility of competition between different electrolytic reactions. When a mixture of NaCl and CaCl2 is electrolyzed, Cl is oxidized because it is the only anion present, but either Na+ or Ca2+ can be reduced. Conversely, in the Hall–Heroult process, only one cation is present that can be reduced (Al3+), but there are three species that can be oxidized: C, O2−, and F.
In the Hall–Heroult process, C is oxidized instead of O2− or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O2 or F2. Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl2 mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na+ more difficult to reduce. In fact, the reduction of Na+ to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome.
Example $1$
If a molten mixture of MgCl2 and KBr is electrolyzed, what products will form at the cathode and the anode, respectively?
Given: identity of salts
Asked for: electrolysis products
Strategy:
1. List all the possible reduction and oxidation products. Based on the electronegativity values shown in Figure 7.5, determine which species will be reduced and which species will be oxidized.
2. Identify the products that will form at each electrode.
Solution
A The possible reduction products are Mg and K, and the possible oxidation products are Cl2 and Br2. Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl2 is a stronger oxidant than Br2.
B Electrolysis will therefore produce Br2 at the anode and Mg at the cathode.
Exercise $1$
Predict the products if a molten mixture of AlBr3 and LiF is electrolyzed.
Answer
Br2 and Al
Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H2 and O2. However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H2SO4 or Na2SO4) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H2 and O2 (Figure $3$).
The reactions that occur are as follows:
• cathode: $2H^+_{(aq)} + 2e^− \rightarrow H_{2(g)} \;\;\; E^°_{cathode} = 0 V \label{20.9.8}$
• anode: $2H_2O_{(l)} → O_{2(g)} + 4H^+_{(aq)} + 4e^−\;\;\; E^°_{anode} = 1.23\; V \label{20.9.9}$
• overall: $2H_2O_{(l)} → O_{2(g)} + 2H_{2(g)}\;\;\; E^°_{cell} = −1.23 \;V \label{20.9.10}$
For a system that contains an electrolyte such as Na2SO4, which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H+] = [OH] = 1.0 × 10−7. Assuming that $P_\mathrm{O_2}$ = $P_\mathrm{H_2}$ = 1 atm, we can use the standard potentials to calculate E for the overall reaction:
\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11}
Thus Ecell is −1.23 V, which is the value of E°cell if the reaction is carried out in the presence of 1 M H+ rather than at pH 7.0.
In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an overvoltage, represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O2 can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F2 cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO3, SO42, PO43, are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis.
In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation.
Electroplating
In a process called electroplating, a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure $4$.
The half-reactions in electroplating a fork, for example, with silver are as follows:
• cathode (fork): $\ce{Ag^{+}(aq) + e^{−} -> Ag(s)} \quad E°_{cathode} = 0.80 V\ \nonumber$
• anode (silver bar): $\ce{Ag(s) -> Ag^{+}(aq) + e^{-}} \quad E°_{anode} = 0.80 V \nonumber$
The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because $E^o_{cell} = 0\, V$, it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating.
Quantitative Considerations
If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material.
The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction
$\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber$
1 mol of electrons reduces 1 mol of $\ce{Ag^{+}}$ to $\ce{Ag}$ metal. In contrast, in the reaction
$\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber$
1 mol of electrons reduces only 0.5 mol of $\ce{Cu^{2+}}$ to $\ce{Cu}$ metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge ($q$ in coulombs) transferred is the product of the current ($I$ in amperes) and the time ($t$, in seconds):
$q = I \times t \label{20.9.14}$
The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process.
For example, if a current of 0.60 A passes through an aqueous solution of $\ce{CuSO4}$ for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows:
\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \[4pt] &=\mathrm{220\;A\cdot s} \[4pt] &=\textrm{220 C} \end{align*} \nonumber
The number of moles of electrons transferred to $\ce{Cu^{2+}}$ is therefore
\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber
Because two electrons are required to reduce a single Cu2+ ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10−3 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks.
Example $2$
A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%?
Given: mass of metal, time, and efficiency
Asked for: current required
Strategy:
1. Calculate the number of moles of metal corresponding to the given mass transferred.
2. Write the reaction and determine the number of moles of electrons required for the electroplating process.
3. Use the definition of the faraday to calculate the number of coulombs required. Then convert coulombs to current in amperes.
Solution
A We must first determine the number of moles of Ag corresponding to 2.00 g of Ag:
$\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}$
B The reduction reaction is Ag+(aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver.
C Using the definition of the faraday,
coulombs = (1.85 × 102mol e)(96,485 C/mol e) = 1.78 × 103 C / mole
The current in amperes needed to deliver this amount of charge in 12.0 h is therefore
\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\ & =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber
Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A.
Exercise $2$
A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al2O3/Na3AlF6 mixture?
Answer
5.8 h
Electroplating: Electroplating(opens in new window) [youtu.be]
Summary
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time:
$q = I \times t \nonumber$
In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H2 and O2 from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.08%3A_Electrolysis-_Driving_Non-spontaneous_Chemical_Reactions_with_Electricity.txt |
Learning Objectives
• To understand the process of corrosion.
Corrosion is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated \$100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals.
Corrosion is a REDOX process.
Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both.
In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide ($\ce{Fe2O3 \cdot xH2O}$), commonly known as rust, that does not provide a tight protective film (Figure $1$). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil will not rust because of the absence of water even if the solvent is saturated with oxygen.
In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe2+; oxygen is reduced to water at the cathode. The relevant reactions are as follows:
• at cathode: $\ce{O2(g) + 4H^{+}(aq) + 4e^{−} -> 2H2O(l)} \nonumber$ with $E^o_{SRP}=1.23\; V$.
• at anode: $\ce{Fe(s) -> Fe^{2+}(aq) + 2e^{−}}\nonumber$ with $E^o_{SRP} = −0.45\; V$.
• overall: $\ce{2Fe(s) + O2(g) + 4H^{+}(aq) -> 2Fe^{2+}(aq) + 2H2O(l)} \label{Eq3}$ with $E^o_{cell} = 1.68\; V$.
The $\ce{Fe^{2+}}$ ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing $\ce{Fe^{3+}}$, as represented in the following equation:
$\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4}$
The sign and magnitude of $E^o_{cell}$ for the corrosion process (Equation $\ref{Eq3}$) indicate that there is a strong driving force for the oxidation of iron by O2 under standard conditions (1 M H+). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO2 with water to form H+ and HCO3 provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure $2$).
Prophylactic Protection
One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy).
As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^{2+}}$ (E° = −0.14 V) and Fe2+ (E° = −0.45 V) in Table P2 show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure $3$). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure.
Cathodic Protection
One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as $\ce{Zn}$ (E° = −0.76 V for $\ce{Zn^{2+} + 2e^{−} -> Zn}$) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows:
$\underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5}$
$\underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6}$
$\underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7}$
The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans.
In a similar strategy, sacrificial electrodes using magnesium, for example, are used to protect underground tanks or pipes (Figure $4$). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting.
Example $1$
Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin).
1. If the boat is immersed in seawater, what corrosion reaction will occur? What is $E^o°_{cell}$?
2. How could you prevent this corrosion from occurring?
Given: identity of metals
Asked for: corrosion reaction, $E^o°_{cell}$, and preventive measures
Strategy:
1. Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate $E^o°_{cell}$.
2. Based on the relative redox activity of various substances, suggest possible preventive measures.
Solution
1. A According to Table P2, both copper and tin are less active metals than iron (i.e., they have higher positive values of $E^o°_{cell}$ than iron). Thus if tin or copper is brought into electrical contact by seawater with iron in the presence of oxygen, corrosion will occur. We therefore anticipate that the bronze propeller will act as the cathode at which $\ce{O2}$ is reduced, and the iron screws will act as anodes at which iron dissolves:
\begin{align*} & \textrm{cathode:} & & \mathrm{O_2(s)} + \mathrm{4H^+(aq)}+\mathrm{4e^-}\rightarrow \mathrm{2H_2O(l)} & & E^\circ_{\textrm{cathode}} =\textrm{1.23 V} \ & \textrm{anode:} & & \mathrm{Fe(s)} \rightarrow \mathrm{Fe^{2+}} +\mathrm{2e^-} & & E^\circ_{\textrm{anode}} =-\textrm{0.45 V} \ & \textrm{overall:} & & \mathrm{2Fe(s)}+\mathrm{O_2(g)}+\mathrm{4H^+(aq)} \rightarrow \mathrm{2Fe^{2+}(aq)} +\mathrm{2H_2O(l)} & & E^\circ_{\textrm{overall}} =\textrm{1.68 V} \end{align*} \nonumber
Over time, the iron screws will dissolve, and the boat will fall apart.
1. B Possible ways to prevent corrosion, in order of decreasing cost and inconvenience, are as follows: disassembling the boat and rebuilding it with bronze screws; removing the boat from the water and storing it in a dry place; or attaching an inexpensive piece of zinc metal to the propeller shaft to act as a sacrificial electrode and replacing it once or twice a year. Because zinc is a more active metal than iron, it will act as the sacrificial anode in the electrochemical cell and dissolve (Equation $\ref{Eq7}$).
Exercise $1$
Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job.
1. Do you accept his proposal?
2. What else should you have the plumber do while at your home?
Answer a
Not unless you plan to sell the house very soon because the $\ce{Cu/Fe}$ pipe joints will lead to rapid corrosion.
Answer b
Any existing $\ce{Pb/Fe}$ joints should be examined carefully for corrosion of the iron pipes due to the $\ce{Pb–Fe}$ junction; the less active $\ce{Pb}$ will have served as the cathode for the reduction of $\ce{O2}$, promoting oxidation of the more active $\ce{Fe}$ nearby.
Summary
Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/20%3A_Electrochemistry/20.09%3A_Corrosion-_Undesirable_Redox_Reactions.txt |
• 21.1: Diagnosing Appendicitis
• 21.2: The Discovery of Radioactivity
Henri Becquerel, Marie Curie, and Pierre Curie shared the discovery of radioactivity.
• 21.3: Types of Radioactivity
Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved and the most common are protons, neutrons, positrons, alpha (α) particles, beta (β) particles (high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). As with chemical reactions, nuclear reactions are always balanced. When a nuclear reaction occurs, the total mass (number) and the total charge remain unchanged.
• 21.4: The Valley of Stability- Predicting the Type of Radioactivity
Many elements have at least one isotope whose atomic nucleus is stable indefinitely, but all elements have isotopes that are unstable and decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the nuclear decay results in changes within atomic nuclei.
• 21.5: Detecting Radioactivity
When alpha, beta or gamma particles collides with a target, some of the energy in the particle is transferred to the target, typically resulting in the promotion of an electron to an “excited state”. In many “targets”, especially gasses, this results in ionization, and alpha, beta and gamma radiation is broadly referred to as ionizing radiation. A Geiger counter (or Geiger-Müller counter) takes advantage of this in order to detect these particles.
• 21.6: The Kinetics of Radioactive Decay and Radiometric Dating
Unstable nuclei undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new stable nuclei sometimes via multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics and each radioisotope has its own half-life.
• 21.7: The Discovery of Fission- The Atomic Bomb and Nuclear Power
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons.
• 21.8: Converting Mass to Energy- Mass Defect and Nuclear Binding Energy
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect that corresponds to the nuclear binding energy.
• 21.9: Nuclear Fusion - The Power of the Sun
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect that corresponds to the nuclear binding energy.
• 21.10: Nuclear Transmutation and Transuranium Elements
It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way.
• 21.11: The Effects of Radiation on Life
The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues.
• 21.12: Radioactivity in Medicine and Other Applications
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US.
21: Radioactivity and Nuclear Chemistry
Learning Objectives
• List the most common emissions from naturally radioactive nuclei.
• Compare the energy released per gram of matter in nuclear reactions to that in chemical reactions.
• Express the relationship between nuclear stability and the nuclei's binding energy per nucleon ratio.
No one could have known in the 1800's that the discovery of the fascinating science and art form of photography would eventually lead to the splitting of the atom. The basis of photography is the fact that visible light causes certain chemical reactions. If the chemicals are spread thinly on a surface but protected from light by a covering, no reaction occurs. When the covering is removed, however, light acting on the chemicals causes them to darken. With millions of cameras in use today, we do not think of it as a strange phenomenon—but at the time of its discovery, photography was a strange and wonderful thing.
Even stranger was the discovery by Wilhelm Roentgen—that radiation other than visible light could expose photographic film. He found that film wrapped in dark paper would react when x-rays went through the paper and struck the film.
When Henri Becquerel heard about Roentgen's discovery, he wondered if his fluorescent minerals would give the same x-rays. Becquerel placed some of his rock crystals on top of a well-covered photographic plate and sat them in the sunlight. The sunlight made the crystals glow with a bright fluorescent light, but when Becquerel developed the film he was very disappointed. He found that only one of his minerals, a uranium salt, had fogged the photographic plate. He decided to try again, and this time, to leave them out in the sun for a longer period of time. Fortunately, the weather didn't cooperate, and Becquerel had to leave the crystals and film stored in a drawer for several cloudy days. Before continuing his experiments, Becquerel decided to check one of the photographic plates to make sure the chemicals were still good. To his amazement, he found that the plate had been exposed in spots where it had been near the uranium containing rocks, and some of these rocks had not been exposed to sunlight at all. In later experiments, Becquerel confirmed that the radiation from the uranium had no connection with light or fluorescence, but the amount of radiation was directly proportional to the concentration of uranium in the rock. Becquerel had discovered radioactivity.
The Curies and Radium
One of Becquerel's assistants, a young Polish scientist named Maria Sklowdowska (to become Marie Curie after she married Pierre Curie), became interested in the phenomenon of radioactivity. With her husband, she decided to find out if chemicals other than uranium were radioactive. The Austrian government was happy to send the Curies a ton of pitchblende from the mining region of Joachimstahl, because it was waste material that had to be disposed of anyway. The Curies wanted the pitchblende because it was the residue of uranium mining. From the ton of pitchblende, the Curies separated $0.10 \: \text{g}$ of a previously unknown element, radium, in the form of the compound radium chloride. This radium was many times more radioactive than uranium.
By 1902, the world was aware of a new phenomenon called radioactivity and of new elements which exhibited natural radioactivity. For this work, Becquerel and the Curies shared the 1903 Nobel Prize and for subsequent work; Marie Cure received a second Nobel Prize in 1911. She is the only person ever to receive two Nobel Prizes in science.
Further experiments provided information about the characteristics of the penetrating emissions from radioactive substances. It was soon discovered that there were three common types of radioactive emissions. Some of the radiation could pass easily through aluminum foil while some of the radiation was stopped by the foil. Some of the radiation could even pass through foil up to a centimeter thick. The three basic types of radiation were named alpha, beta, and gamma radiation. The actual composition of the three types of radiation was still not known.
Eventually, scientists were able to demonstrate experimentally that the alpha particle, $\alpha$, was a helium nucleus (a particle containing two protons and two neutrons), a beta particle, $\beta$, was a high speed electron, and gamma rays, $\gamma$, were a very high energy form of light (even higher energy than x-rays).
Unstable Nuclei May Disintegrate
A nucleus (with one exception, hydrogen-1) consists of some number of protons and neutrons pulled together in an extremely tiny volume. Since protons are positively charged and like charges repel, it is clear that protons cannot remain together in the nucleus unless there is a powerful force holding them there. The force which holds the nucleus together is generated by nuclear binding energy.
A nucleus with a large amount of binding energy per nucleon (proton or neutron) will be held together tightly and is referred to as stable. These nuclei do not break apart. When there is too little binding energy per nucleon, the nucleus will be less stable and may disintegrate (come apart). Such disintegration is referred to as natural radioactivity. It is also possible for scientists to smash nuclear particles together and cause nuclear reactions between normally stable nuclei. This disintegration is referred to as artificial radioactivity. None of the elements above #92 on the periodic table occur on earth naturally—they are all products of artificial (manmade) radioactivity.
When nuclei come apart, they come apart violently accompanied by a tremendous release of energy in the form of heat, light, and radiation. This energy comes from some of the nuclear binding energy. In nuclear changes, the energy involved comes from the nuclear binding energy. However, in chemical reactions, the energy comes from electrons moving energy levels. A typical nuclear change (such as fission) may involve millions of times more energy per atom changing compared to a chemical change (such as burning)!
Summary
• Henri Becquerel, Marie Curie, and Pierre Curie shared the discovery of radioactivity.
• Wikibooks | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.02%3A_The_Discovery_of_Radioactivity.txt |
Learning Objectives
• Write and balance nuclear equations
• To know the different kinds of radioactive decay.
• To balance a nuclear reaction.
Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of $\ce{^1_1H}$, neutrons. Recall that the number of protons in the nucleus is called the atomic number ($Z$) of the element, and the sum of the number of protons and the number of neutrons is the mass number ($A$). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide and identify it by the notation:
$\large \ce{^{A}_{Z}X} \label{Eq1a}$
where
• $X$ is the symbol for the element,
• $A$ is the mass number, and
• $Z$ is the atomic number.
Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, $\ce{^{14}_6C}$ is called “carbon-14.”
Protons and neutrons, collectively called nucleons, are packed together tightly in a nucleus. With a radius of about 10−15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10−10 meters. Nuclei are extremely dense compared to bulk matter, averaging $1.8 \times 10^{14}$ grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm3. If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately $6.4 \times 10^6$ meters, 30,000 times larger).
Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are nuclear reactions. To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction.
Nuclear Equations
A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways:
1. The sum of the mass numbers of the reactants equals the sum of the mass numbers of the products.
2. The sum of the charges of the reactants equals the sum of the charges of the products.
If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that $\ce{^{17}_8O}$ is a product of the nuclear reaction of $\ce{^{14}_7N}$ and $\ce{^4_2He}$ if we knew that a proton, $\ce{^1_1H}$, was one of the two products. Example $1$ shows how we can identify a nuclide by balancing the nuclear reaction.
Example $1$: Balancing Equations for Nuclear Reactions
The reaction of an α particle with magnesium-25 $(\ce{^{25}_{12}Mg})$ produces a proton and a nuclide of another element. Identify the new nuclide produced.
Solution
The nuclear reaction can be written as:
$\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber$
where
• $\ce A$ is the mass number and
• $\ce Z$ is the atomic number of the new nuclide, $\ce X$.
Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products:
$\mathrm{25+4=A+1} \nonumber$
so
$\mathrm{A=28} \nonumber$
Similarly, the charges must balance, so:
$\mathrm{12+2=Z+1} \nonumber$
so
$\mathrm{Z=13} \nonumber$
Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is $\ce{^{28}_{13}Al}$.
Exercise $1$
The nuclide $\ce{^{125}_{53}I}$ combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction?
Answer
$\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber$
The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions. In a nuclear decay reaction, also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a nuclear transmutation reaction, a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is more massive than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit.
Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced.
Nuclear Decay Reactions
Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions.
To describe nuclear decay reactions, chemists have extended the $^A _Z \textrm{X}$ notation for nuclides to include radioactive emissions. Table $1$ lists the name and symbol for each type of emitted radiation. The most notable addition is the positron, a particle that has the same mass as an electron but a positive charge rather than a negative charge.
Table $1$: Nuclear Decay Emissions and Their Symbols
Identity Symbol Charge Mass (amu)
helium nucleus $^4_2\alpha$ +2 4.001506
electron $^0_{-1}\beta$ or $\beta ^-$ −1 0.000549
photon $_0^0\gamma$
neutron $^1_0\textrm n$ 0 1.008665
proton $^1_1\textrm p$ +1 1.007276
positron $^0_{+1}\beta$ or $\beta ^+$ +1 0.000549
Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, A = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus.
Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, Z = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so Z = 0. In the case of an electron, Z = −1, and for a positron, Z = +1. Because γ rays are high-energy photons, both A and Z are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol $^0_{-1}\textrm e$, which is usually simplified to e, represents a free electron or an electron associated with an atom, whereas the symbol $^0_{-1}\beta$, which is often simplified to β, denotes an electron that originates from within the nucleus, which is a β particle. Similarly, $^4_{2}\textrm{He}^{2+}$ refers to the nucleus of a helium atom, and $^4_{2}\alpha$ denotes an identical particle that has been ejected from a heavier nucleus.
There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in Figure $1$. The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions.
Alpha $\alpha$ Decay
Many nuclei with mass numbers greater than 200 undergo alpha (α) decay, which results in the emission of a helium-4 nucleus as an alpha (α) particle, $^4_{2}\alpha$. The general reaction is as follows:
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\ \textrm{particle}}{^4_2 \alpha}\label{Eq1}$
The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number A − 4 and a nuclear charge Z − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222:
$^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{Eq2}$
Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced.
Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction.
Beta $\beta^-$ Decay
Nuclei that contain too many neutrons often undergo beta (β) decay, in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle:
$\underset{\textrm{unstable} \ \textrm{neutron in} \ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \ \textrm{retained} \ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \ \textrm{emitted by} \ \textrm{nucleus}}{^0_{-1} \beta}\label{Eq3}$
The general reaction for beta decay is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z+1} \textrm X'}+\underset{\textrm{beta particle}}{^0_{-1} \beta}\label{Eq4}$
Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14:
$^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta \nonumber$
Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent.
Positron $\beta^+$ Emission
Because a positron has the same mass as an electron but opposite charge, positron emission is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron:
$^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{Eq6}$
The general reaction for positron emission is therefore
$\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\underset{\textrm{positron}}{^0_{+1} \beta^+} \nonumber$
Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11:
$^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \nonumber$
Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide.
Electron Capture
A neutron-poor nucleus can decay by either positron emission or electron capture (EC), in which an electron in an inner shell reacts with a proton to produce a neutron:
$^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{Eq9}$
When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus
$\underset{\textrm{parent}}{^A_Z \textrm X}+\underset{\textrm{electron}}{^0_{-1} \textrm e}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\textrm{x-ray} \nonumber$
Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows:
$^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
The atomic numbers of the parent and daughter nuclides differ in Equation 20.2.11, although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation:
$^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber$
Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different.
Gamma $\gamma$ Emission
Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. Gamma ($\gamma$) emission can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state:
$^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \ \textrm{nuclear} \ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th} + \ce{^0_0\gamma} \nonumber$
If we disregard the decay event that created the excited nucleus, then
$^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th} + ^{0}_{0}\gamma \nonumber$
or more generally,
$^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X} + ^{0}_{0}\gamma \nonumber$
Gamma emission can also occur after a significant delay. For example, technetium-99m has a half-life of about 6 hours before emitting a $γ$ ray to form technetium-99 (the m is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction.
Spontaneous Fission
Only very massive nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with Z ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of $^{254}_{98}\textrm{Cf}$, which gives a distribution of fission products; one possible set of products is shown in the following equation:
$^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n} \nonumber$
Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide.
Example $2$
Write a balanced nuclear equation to describe each reaction.
1. the beta decay of $^{35}_{16}\textrm{S}$
2. the decay of $^{201}_{80}\textrm{Hg}$ by electron capture
3. the decay of $^{30}_{15}\textrm{P}$ by positron emission
Given: radioactive nuclide and mode of decay
Asked for: balanced nuclear equation
Strategy:
A Identify the reactants and the products from the information given.
B Use the values of A and Z to identify any missing components needed to balance the equation.
Solution
a.
A We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as $^{A}_{Z}\textrm{X}$: $^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber$
B Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of A = 35 − 0 = 35 and an atomic number of Z = 16 − (−1) = 17. The element with Z = 17 is chlorine, so the balanced nuclear equation is as follows: $^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber$
b.
A We know the identities of both reactants: $^{201}_{80}\textrm{Hg}$ and an inner electron, $^{0}_{-1}\textrm{e}$. The reaction is as follows: $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber$
B Both protons and neutrons are conserved, so the mass number of the product must be A = 201 + 0 = 201, and the atomic number of the product must be Z = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus $^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au} \nonumber$
c.
A As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore $^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta \nonumber$
B The mass number of the second product is A = 30 − 0 = 30, and its atomic number is Z = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: $^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber$
Exercise $2$
Write a balanced nuclear equation to describe each reaction.
1. $^{11}_{6}\textrm{C}$ by positron emission
2. the beta decay of molybdenum-99
3. the emission of an α particle followed by gamma emission from $^{185}_{74}\textrm{W}$
Answer a
$^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta$
Answer d
$^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta$
Answer c
$^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma$
Example $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{45}_{22}\textrm{Ti}$
2. $^{242}_{94}\textrm{Pu}$
3. $^{12}_{5}\textrm{B}$
4. $^{256}_{100}\textrm{Fm}$
Given: nuclide
Asked for: type of nuclear decay
Strategy:
Based on the neutron-to-proton ratio and the value of Z, predict the type of nuclear decay reaction that will produce a more stable nuclide.
Solution
1. This nuclide has a neutron-to-proton ratio of only 1.05, which is much less than the requirement for stability for an element with an atomic number in this range. Nuclei that have low neutron-to-proton ratios decay by converting a proton to a neutron. The two possibilities are positron emission, which converts a proton to a neutron and a positron, and electron capture, which converts a proton and a core electron to a neutron. In this case, both are observed, with positron emission occurring about 86% of the time and electron capture about 14% of the time.
2. Nuclei with Z > 83 are too heavy to be stable and usually undergo alpha decay, which decreases both the mass number and the atomic number. Thus $^{242}_{94}\textrm{Pu}$ is expected to decay by alpha emission.
3. This nuclide has a neutron-to-proton ratio of 1.4, which is very high for a light element. Nuclei with high neutron-to-proton ratios decay by converting a neutron to a proton and an electron. The electron is emitted as a β particle, and the proton remains in the nucleus, causing an increase in the atomic number with no change in the mass number. We therefore predict that $^{12}_{5}\textrm{B}$ will undergo beta decay.
4. This is a massive nuclide, with an atomic number of 100 and a mass number much greater than 200. Nuclides with A ≥ 200 tend to decay by alpha emission, and even heavier nuclei tend to undergo spontaneous fission. We therefore predict that $^{256}_{100}\textrm{Fm}$ will decay by either or both of these two processes. In fact, it decays by both spontaneous fission and alpha emission, in a 97:3 ratio.
Exercise $3$
Predict the kind of nuclear change each unstable nuclide undergoes when it decays.
1. $^{32}_{14}\textrm{Si}$
2. $^{43}_{21}\textrm{Sc}$
3. $^{231}_{91}\textrm{Pa}$
Answer a
beta decay
Answer d
positron emission or electron capture
Answer c
alpha decay
Radioactive Decay Series
The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases Z by only 2, and positron emission or electron capture decreases Z by only 1, it is impossible for any nuclide with Z > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with Z > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a radioactive decay series. The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions (Figure $2$). Although a radioactive decay series can be written for almost any isotope with Z > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic.
Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin.
Induced Nuclear Reactions
The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction.
The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process:
$^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{Eq17}$
Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows:
$^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{Eq18}$
Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus (Figure $3$). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays:
$^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{Eq19}$
Example $4$
In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each 27Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction.
Given: reactants in a nuclear transmutation reaction
Asked for: product nuclide and balanced nuclear equation
Strategy:
A Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of Z and A of the product nuclide and thus its identity.
B Write the balanced nuclear equation for the reaction.
Solution
A Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum (Z = 13) will be converted to phosphorus (Z = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, $^{30}_{15}\textrm{P}$.
B The balanced nuclear equation for the reaction is as follows:
$^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n} \nonumber$
Exercise $4$
Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei $(^{2}_{1}\textrm{H})$ produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction.
Answer
neutron, $^{1}_{0}\textrm{n}$ ; $^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}$ :
We noted earlier in this section that very heavy nuclides, corresponding to Z ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of Z, such as the isotopes of uranium (Z = 92) and plutonium (Z = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as $^{235}_{92}\textrm{U}$ and $^{239}_{94}\textrm{Pu}$ undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in Figure $4$:
$^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{Eq20}$
Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a fissile isotope.
During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in Figure $4$. Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in Figure $5$, a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2.
Synthesis of Transuranium Elements
Uranium (Z = 92) is the heaviest naturally occurring element. Consequently, all the elements with Z > 92, the transuranium elements, are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium (Z = 93), which was synthesized in 1940 by bombarding a 238U target with neutrons. As shown in Equation 20.21, this reaction occurs in two steps. Initially, a neutron combines with a 238U nucleus to form 239U, which is unstable and undergoes beta decay to produce 239Np:
$^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{Eq21}$
Subsequent beta decay of 239Np produces the second transuranium element, plutonium (Z = 94):
$^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{Eq22}$
Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus (Table $2$). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability."
Table $2$: Some Reactions Used to Synthesize Transuranium Elements
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{242}_{96}\textrm{Cm}+\,^{1}_{0}\textrm{n}$
$^{239}_{94}\textrm{Pu}+\,^{4}_{2}\alpha \rightarrow \,^{241}_{95}\textrm{Am}+\,^{1}_{1}\textrm{p}+\,^{1}_{0}\textrm{n}$
$^{242}_{96}\textrm{Cm}+\,^{4}_{2}\alpha \rightarrow \,^{243}_{97}\textrm{Bk}+\,^{1}_{1}\textrm{p}+2^{1}_{0}\textrm{n}$
$^{253}_{99}\textrm{Es}+\,^{4}_{2}\alpha \rightarrow \,^{256}_{101}\textrm{Md}+\,^{1}_{0}\textrm{n}$
$^{238}_{92}\textrm{U}+\,^{12}_{6}\textrm{C} \rightarrow \,^{246}_{98}\textrm{Cf}+4^{1}_{0}\textrm{n}$
$^{252}_{98}\textrm{Cf}+\,^{10}_{5}\textrm{B} \rightarrow \,^{256}_{103}\textrm{Lr}+6^{1}_{0}\textrm{n}$
A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator (Figure $6$), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long.
To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target.
The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate.
Summary and Key Takeaway
• Nuclear decay reactions occur spontaneously under all conditions and produce more stable daughter nuclei, whereas nuclear transmutation reactions are induced and form a product nucleus that is more massive than the starting material.
In nuclear decay reactions (or radioactive decay), the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with A ≥ 200 and Z > 83) are unstable and tend to decay by emitting an α particle. When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. Alpha decay results in the emission of an α particle, $^4 _2 \alpha$, and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. Beta decay converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. Positron emission is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In electron capture (EC), an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In gamma emission, a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo spontaneous fission, in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a radioactive decay series—a succession of some combination of alpha- and beta-decay reactions. In nuclear transmutation reactions, a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All transuranium elements—elements with Z > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons.
Key Equations
alpha decay
$^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha \nonumber$
beta decay
$^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta \nonumber$
positron emission
$^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta \nonumber$
electron capture
$^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray} \nonumber$
gamma emission
$^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma \nonumber$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.03%3A_Types_of_Radioactivity.txt |
Learning Objectives
• To understand the factors that affect nuclear stability.
Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus.
The Atomic Nucleus
Each element can be represented by the notation \(^A_Z \textrm X\), where A, the mass number, is the sum of the number of protons and the number of neutrons, and Z, the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleons, and an atom with a particular number of protons and neutrons is called a nuclide. Nuclides with the same number of protons but different numbers of neutrons are called isotopes. Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways:
stable isotopes of oxygen represented in different ways
\(^A_Z \textrm X\) \(\ce{^{16}_8 O}\) \(\ce{^{17}_8 O}\) \(\ce{^{18}_8 O}\)
\(^A \textrm X\) \(\ce{^{16} O}\) \(\ce{^{17} O}\) \(\ce{^{18} O}\)
\(\textrm{element-A:}\) \(\textrm{oxygen-16}\) \(\textrm{oxygen-17}\) \(\textrm{oxygen-18}\)
Because the number of neutrons is equal to AZ, we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s natural abundance. For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17. Any nucleus that is unstable and decays spontaneously is said to be radioactive, emitting subatomic particles and electromagnetic radiation. The emissions are collectively called radioactivity and can be measured. Isotopes that emit radiation are called radioisotopes.
Nuclear Stability
The nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear force, an extremely powerful but very short-range attractive force between nucleons (Figure \(1\)). All stable nuclei except the hydrogen-1 nucleus (1H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability.
The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in Figure \(2\). The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1H and 3He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., \(^4_2 \textrm{He}\), \(^{10}_5 \textrm{B}\), and \(^{40}_{20} \textrm{Ca}\)). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive.
As shown in Figure \(3\), more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin (Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium (Z = 49) and antimony (Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are \(^4_2 \textrm{He}\), with 2 protons and 2 neutrons, and \(^{208}_{82} \textrm{Pb}\), with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element.
Most stable nuclei contain even numbers of both neutrons and protons
The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in Figure \(2\), the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei.
Origin of the Magic Numbers
Multiple models have been formulated to explain the origin of the magic numbers and two popular ones are the Nuclear Shell Model and the Liquid Drop Model. Unfortuneatly, both require advanced quantum mechanics to fully understand and are beyond the scope of this text.
Example \(1\)
Classify each nuclide as stable or radioactive.
1. \(\ce{_{15}^{30} P}\)
2. \(\ce{_{43}^{98} Tc}\)
3. tin-118
4. \(\ce{_{94}^{239} Pu}\)
Given: mass number and atomic number
Asked for: predicted nuclear stability
Strategy:
Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide.
Solution:
a. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in Figure \(2\), its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, \(_{15}^{30} \textrm P\) is predicted to be radioactive, and it is.
b. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places \(_{43}^{98} \textrm{Tc}\) near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that \(_{43}^{98} \textrm{Tc}\) might be stable. However, \(_{43}^{98} \textrm{Tc}\) has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, \(_{43}^{98} \textrm{Tc}\) is predicted to be radioactive, and it is.
c. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus \(_{50}^{118} \textrm{Sn}\)should be particularly stable.
d. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, \(_{94}^{239} \textrm{Pu}\) must be radioactive.
Exercise \(1\)
Classify each nuclide as stable or radioactive.
1. \(\ce{_{90}^{232} Th}\)
2. \(\ce{_{20}^{40} Ca}\)
3. \(\ce{_8^{15} O}\)
4. \(\ce{_{57}^{139} La}\)
Answer a
radioactive
Answer b
stable
Answer c
radioactive
Answer d
stable
Superheavy Elements
In addition to the “peninsula of stability” there is a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elements, with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232Th. With an estimated half-life greater than 108 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements.
Summary
Subatomic particles of the nucleus (protons and neutrons) are called nucleons. A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive, and its emissions are collectively called radioactivity. Isotopes that emit radiation are called radioisotopes. Each nucleon is attracted to other nucleons by the strong nuclear force. Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements, with atomic numbers near 126, may even be stable enough to exist in nature.
21.05: Detecting Radioactivity
Learning Objectives
• Understand how the Geiger counter can be used to quantify the rate of ionization radiation.
When alpha, beta or gamma particles collide with a target, some of the energy in the particle is transferred to the target, typically resulting in the promotion of an electron to an “excited state”. In many “targets”, especially gasses, this results in ionization. Alpha, beta and gamma radiation are broadly referred to as ionizing radiation. A Geiger counter (or Geiger-Müller counter) takes advantage of this in order to detect these particles. In a Geiger tube, the electron produced by ionization of a captive gas travels to the anode and the change in voltage is detected by the attached circuitry. Most counters of this type are designed to emit an audible “click” in response to the change in voltage, and to also show it on a digital or analog meter. A simple schematic of a Geiger counter is shown in Figure \(1\).
Although scientists were not aware at the time of the Geiger counter's invention, all of us are subjected to a certain amount of radiation every day. This radiation is called background radiation and comes from a variety of natural and artificial radiation sources. Approximately 82% of background radiation comes from natural sources. These natural sources include:
1. Sources in the earth—including naturally occurring radioactive elements—which are incorporated in building materials, and also in the human body.
2. Sources from space in the form of cosmic rays.
3. Sources in the atmosphere, such as radioactive radon gas released from the earth; and radioactive atoms like carbon-14, produced in the atmosphere by bombardment from high-energy cosmic rays. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.04%3A_The_Valley_of_Stability-_Predicting_the_Type_of_Radioactivity.txt |
Learning Objectives
• To know how to use half-lives to describe the rates of first-order reactions
Another approach to describing reaction rates is based on the time required for the concentration of a reactant to decrease to one-half its initial value. This period of time is called the half-life of the reaction, written as t1/2. Thus the half-life of a reaction is the time required for the reactant concentration to decrease from [A]0 to [A]0/2. If two reactions have the same order, the faster reaction will have a shorter half-life, and the slower reaction will have a longer half-life.
The half-life of a first-order reaction under a given set of reaction conditions is a constant. This is not true for zeroth- and second-order reactions. The half-life of a first-order reaction is independent of the concentration of the reactants. This becomes evident when we rearrange the integrated rate law for a first-order reaction to produce the following equation:
$\ln\dfrac{[\textrm A]_0}{[\textrm A]}=kt \label{21.4.1}$
Substituting [A]0/2 for [A] and t1/2 for t (to indicate a half-life) into Equation $\ref{21.4.1}$ gives
$\ln\dfrac{[\textrm A]_0}{[\textrm A]_0/2}=\ln 2=kt_{1/2}$
Substituting $\ln{2} \approx 0.693$ into the equation results in the expression for the half-life of a first-order reaction:
$t_{1/2}=\dfrac{0.693}{k} \label{21.4.2}$
Thus, for a first-order reaction, each successive half-life is the same length of time, as shown in Figure $1$, and is independent of [A].
If we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.
Number of Half-Lives Percentage of Reactant Remaining
1 $\dfrac{100\%}{2}=50\%$ $\dfrac{1}{2}(100\%)=50\%$
2 $\dfrac{50\%}{2}=25\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right)(100\%)=25\%$
3 $\dfrac{25\%}{2}=12.5\%$ $\dfrac{1}{2}\left(\dfrac{1}{2}\right )\left (\dfrac{1}{2}\right)(100\%)=12.5\%$
n $\dfrac{100\%}{2^n}$ $\left(\dfrac{1}{2}\right)^n(100\%)=\left(\dfrac{1}{2}\right)^n\%$
As you can see from this table, the amount of reactant left after n half-lives of a first-order reaction is (1/2)n times the initial concentration.
For a first-order reaction, the concentration of the reactant decreases by a constant with each half-life and is independent of [A].
Example $1$
The anticancer drug cis-platin hydrolyzes in water with a rate constant of 1.5 × 10−3 min−1 at pH 7.0 and 25°C. Calculate the half-life for the hydrolysis reaction under these conditions. If a freshly prepared solution of cis-platin has a concentration of 0.053 M, what will be the concentration of cis-platin after 5 half-lives? after 10 half-lives? What is the percent completion of the reaction after 5 half-lives? after 10 half-lives?
Given: rate constant, initial concentration, and number of half-lives
Asked for: half-life, final concentrations, and percent completion
Strategy:
1. Use Equation $\ref{21.4.2}$ to calculate the half-life of the reaction.
2. Multiply the initial concentration by 1/2 to the power corresponding to the number of half-lives to obtain the remaining concentrations after those half-lives.
3. Subtract the remaining concentration from the initial concentration. Then divide by the initial concentration, multiplying the fraction by 100 to obtain the percent completion.
Solution
A We can calculate the half-life of the reaction using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}=\dfrac{0.693}{1.5\times10^{-3}\textrm{ min}^{-1}}=4.6\times10^2\textrm{ min}$
Thus it takes almost 8 h for half of the cis-platin to hydrolyze.
B After 5 half-lives (about 38 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^5}=\dfrac{0.053\textrm{ M}}{32}=0.0017\textrm{ M}$
After 10 half-lives (77 h), the remaining concentration of cis-platin will be as follows:
$\dfrac{0.053\textrm{ M}}{2^{10}}=\dfrac{0.053\textrm{ M}}{1024}=5.2\times10^{-5}\textrm{ M}$
C The percent completion after 5 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-0.0017\textrm{ M})(100)}{0.053}=97\%$
The percent completion after 10 half-lives will be as follows:
$\textrm{percent completion}=\dfrac{(0.053\textrm{ M}-5.2\times10^{-5}\textrm{ M})(100)}{0.053\textrm{ M}}=100\%$
Thus a first-order chemical reaction is 97% complete after 5 half-lives and 100% complete after 10 half-lives.
Exercise $1$
Ethyl chloride decomposes to ethylene and HCl in a first-order reaction that has a rate constant of 1.6 × 10−6 s−1 at 650°C.
1. What is the half-life for the reaction under these conditions?
2. If a flask that originally contains 0.077 M ethyl chloride is heated at 650°C, what is the concentration of ethyl chloride after 4 half-lives?
Answer a
4.3 × 105 s = 120 h = 5.0 days;
Answer b
4.8 × 10−3 M
Radioactive Decay Rates
Radioactivity, or radioactive decay, is the emission of a particle or a photon that results from the spontaneous decomposition of the unstable nucleus of an atom. The rate of radioactive decay is an intrinsic property of each radioactive isotope that is independent of the chemical and physical form of the radioactive isotope. The rate is also independent of temperature. In this section, we will describe radioactive decay rates and how half-lives can be used to monitor radioactive decay processes.
In any sample of a given radioactive substance, the number of atoms of the radioactive isotope must decrease with time as their nuclei decay to nuclei of a more stable isotope. Using N to represent the number of atoms of the radioactive isotope, we can define the rate of decay of the sample, which is also called its activity (A) as the decrease in the number of the radioisotope’s nuclei per unit time:
$A=-\dfrac{\Delta N}{\Delta t} \label{21.4.3}$
Activity is usually measured in disintegrations per second (dps) or disintegrations per minute (dpm).
The activity of a sample is directly proportional to the number of atoms of the radioactive isotope in the sample:
$A = kN \label{21.4.4}$
Here, the symbol k is the radioactive decay constant, which has units of inverse time (e.g., s−1, yr−1) and a characteristic value for each radioactive isotope. If we combine Equation $\ref{21.4.3}$ and Equation $\ref{21.4.4}$, we obtain the relationship between the number of decays per unit time and the number of atoms of the isotope in a sample:
$-\dfrac{\Delta N}{\Delta t}=kN \label{21.4.5}$
Equation $\ref{21.4.5}$ is the same as the equation for the reaction rate of a first-order reaction, except that it uses numbers of atoms instead of concentrations. In fact, radioactive decay is a first-order process and can be described in terms of either the differential rate law (Equation $\ref{21.4.5}$) or the integrated rate law:
$N = N_0e^{−kt}$
$\ln \dfrac{N}{N_0}=-kt \label{21.4.6}$
Because radioactive decay is a first-order process, the time required for half of the nuclei in any sample of a radioactive isotope to decay is a constant, called the half-life of the isotope. The half-life tells us how radioactive an isotope is (the number of decays per unit time); thus it is the most commonly cited property of any radioisotope. For a given number of atoms, isotopes with shorter half-lives decay more rapidly, undergoing a greater number of radioactive decays per unit time than do isotopes with longer half-lives. The half-lives of several isotopes are listed in Table 14.6, along with some of their applications.
Table $2$: Half-Lives and Applications of Some Radioactive Isotopes
Radioactive Isotope Half-Life Typical Uses
*The m denotes metastable, where an excited state nucleus decays to the ground state of the same isotope.
hydrogen-3 (tritium) 12.32 yr biochemical tracer
carbon-11 20.33 min positron emission tomography (biomedical imaging)
carbon-14 5.70 × 103 yr dating of artifacts
sodium-24 14.951 h cardiovascular system tracer
phosphorus-32 14.26 days biochemical tracer
potassium-40 1.248 × 109 yr dating of rocks
iron-59 44.495 days red blood cell lifetime tracer
cobalt-60 5.2712 yr radiation therapy for cancer
technetium-99m* 6.006 h biomedical imaging
iodine-131 8.0207 days thyroid studies tracer
radium-226 1.600 × 103 yr radiation therapy for cancer
uranium-238 4.468 × 109 yr dating of rocks and Earth’s crust
americium-241 432.2 yr smoke detectors
Note
Radioactive decay is a first-order process.
Radioisotope Dating Techniques
In our earlier discussion, we used the half-life of a first-order reaction to calculate how long the reaction had been occurring. Because nuclear decay reactions follow first-order kinetics and have a rate constant that is independent of temperature and the chemical or physical environment, we can perform similar calculations using the half-lives of isotopes to estimate the ages of geological and archaeological artifacts. The techniques that have been developed for this application are known as radioisotope dating techniques.
The most common method for measuring the age of ancient objects is carbon-14 dating. The carbon-14 isotope, created continuously in the upper regions of Earth’s atmosphere, reacts with atmospheric oxygen or ozone to form 14CO2. As a result, the CO2 that plants use as a carbon source for synthesizing organic compounds always includes a certain proportion of 14CO2 molecules as well as nonradioactive 12CO2 and 13CO2. Any animal that eats a plant ingests a mixture of organic compounds that contains approximately the same proportions of carbon isotopes as those in the atmosphere. When the animal or plant dies, the carbon-14 nuclei in its tissues decay to nitrogen-14 nuclei by a radioactive process known as beta decay, which releases low-energy electrons (β particles) that can be detected and measured:
$\ce{^{14}C \rightarrow ^{14}N + \beta^{−}} \label{21.4.7}$
The half-life for this reaction is 5700 ± 30 yr.
The 14C/12C ratio in living organisms is 1.3 × 10−12, with a decay rate of 15 dpm/g of carbon (Figure $2$). Comparing the disintegrations per minute per gram of carbon from an archaeological sample with those from a recently living sample enables scientists to estimate the age of the artifact, as illustrated in Example 11.Using this method implicitly assumes that the 14CO2/12CO2 ratio in the atmosphere is constant, which is not strictly correct. Other methods, such as tree-ring dating, have been used to calibrate the dates obtained by radiocarbon dating, and all radiocarbon dates reported are now corrected for minor changes in the 14CO2/12CO2 ratio over time.
Example $2$
In 1990, the remains of an apparently prehistoric man were found in a melting glacier in the Italian Alps. Analysis of the 14C content of samples of wood from his tools gave a decay rate of 8.0 dpm/g carbon. How long ago did the man die?
Given: isotope and final activity
Asked for: elapsed time
Strategy:
A Use Equation $\ref{21.4.4}$ to calculate N0/N. Then substitute the value for the half-life of 14C into Equation $\ref{21.4.2}$ to find the rate constant for the reaction.
B Using the values obtained for N0/N and the rate constant, solve Equation $\ref{21.4.6}$ to obtain the elapsed time.
Solution
We know the initial activity from the isotope’s identity (15 dpm/g), the final activity (8.0 dpm/g), and the half-life, so we can use the integrated rate law for a first-order nuclear reaction (Equation $\ref{21.4.6}$) to calculate the elapsed time (the amount of time elapsed since the wood for the tools was cut and began to decay).
\begin{align}\ln\dfrac{N}{N_0}&=-kt \ \dfrac{\ln(N/N_0)}{k}&=t\end{align}
A From Equation $\ref{21.4.4}$, we know that A = kN. We can therefore use the initial and final activities (A0 = 15 dpm and A = 8.0 dpm) to calculate N0/N:
$\dfrac{A_0}{A}=\dfrac{kN_0}{kN}=\dfrac{N_0}{N}=\dfrac{15}{8.0}$
Now we need only calculate the rate constant for the reaction from its half-life (5730 yr) using Equation $\ref{21.4.2}$:
$t_{1/2}=\dfrac{0.693}{k}$
This equation can be rearranged as follows:
$k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730\textrm{ yr}}=1.22\times10^{-4}\textrm{ yr}^{-1}$
B Substituting into the equation for t,
$t=\dfrac{\ln(N_0/N)}{k}=\dfrac{\ln(15/8.0)}{1.22\times10^{-4}\textrm{ yr}^{-1}}=5.2\times10^3\textrm{ yr}$
From our calculations, the man died 5200 yr ago.
Exercise $2$
It is believed that humans first arrived in the Western Hemisphere during the last Ice Age, presumably by traveling over an exposed land bridge between Siberia and Alaska. Archaeologists have estimated that this occurred about 11,000 yr ago, but some argue that recent discoveries in several sites in North and South America suggest a much earlier arrival. Analysis of a sample of charcoal from a fire in one such site gave a 14C decay rate of 0.4 dpm/g of carbon. What is the approximate age of the sample?
Answer
30,000 yr
Summary
• The half-life of a first-order reaction is independent of the concentration of the reactants.
• The half-lives of radioactive isotopes can be used to date objects.
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions. The rate of decay, or activity, of a sample of a radioactive substance is the decrease in the number of radioactive nuclei per unit time. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.06%3A_The_Kinetics_of_Radioactive_Decay_and_Radiometric_Dating.txt |
Learning Objectives
• Explain nuclear fission
• Relate the concepts of critical mass and nuclear chain reactions
• Summarize basic requirements for nuclear fission
Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56. Sometimes neutrons are also produced. This decomposition is called fission, the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(1\).
Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure \(2\). Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium.
A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 1010 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal.
As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear chain reaction (Figure \(3\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur.
Material that can sustain a nuclear fission chain reaction is said to be fissile or fissionable. (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a critical mass. An amount of fissionable material that cannot sustain a chain reaction is a subcritical mass. An amount of material in which there is an increasing rate of fission is known as a supercritical mass. The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(4\)).
An atomic bomb (Figure \(5\)) contains several pounds of fissionable material, \(\ce{^{235}_{92}U}\) or \(\ce{^{239}_{94}Pu}\), a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result.
Fission Reactors
Chain reactions of fissionable materials can be controlled and sustained without an explosion in a nuclear reactor (Figure \(6\)). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity.
Nuclear Fuels
Nuclear fuel consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U3O8; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation.
In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF6 (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF6 to pass through. The slightly lighter 235UF6 molecules diffuse through the barrier slightly faster than the heavier 238UF6 molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of 235UF6 to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF6 gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO2) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil.
Nuclear Moderators
Neutrons produced by nuclear reactions move too fast to cause fission (Figure \(4\)). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A nuclear moderator is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water \(\ce{( ^2_1H2O)}\) or light water (ordinary H2O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite.
Reactor Coolants
A nuclear reactor coolant is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts.
Control Rods
Nuclear reactors use control rods (Figure \(8\)) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles:
\[\ce{^{10}_5B + ^1_0n⟶ ^7_3Li + ^4_2He}\]
When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods.
Shield and Containment System
During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a containment system (or shield) that consists of three parts:
1. The reactor vessel, a steel shell that is 3–20-centimeters thick and, with the moderator, absorbs much of the radiation produced by the reactor
2. A main shield of 1–3 meters of high-density concrete
3. A personnel shield of lighter materials that protects operators from γ rays and X-rays
In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident.
Video \(1\): Click here to watch a 3-minute video from the Nuclear Energy Institute on how nuclear reactors work.
Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents.
Nuclear Accidents
The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima).
In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen:
\[\ce{Zr}(s)+\ce{2H2O}(g)⟶\ce{ZrO2}(s)+\ce{2H2}(g)\]
The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process.
Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure \(8\)).
Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland.
In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events.
An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure \(1\)0).
The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate.
Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.07%3A_The_Discovery_of_Fission-_The_Atomic_Bomb_and_Nuclear_Power.txt |
Learning Objectives
• To calculate a mass-energy balance and a nuclear binding energy.
• To understand the differences between nuclear fission and fusion.
Nuclear reactions, like chemical reactions, are accompanied by changes in energy. The energy changes in nuclear reactions, however, are enormous compared with those of even the most energetic chemical reactions. In fact, the energy changes in a typical nuclear reaction are so large that they result in a measurable change of mass. In this section, we describe the relationship between mass and energy in nuclear reactions and show how the seemingly small changes in mass that accompany nuclear reactions result in the release of enormous amounts of energy.
Mass–Energy Balance
The relationship between mass (m) and energy (E) is expressed in the following equation:
$E = mc^2 \label{Eq1}$
where
• $c$ is the speed of light ($2.998 \times 10^8\; m/s$), and
• $E$ and $m$ are expressed in units of joules and kilograms, respectively.
Albert Einstein first derived this relationship in 1905 as part of his special theory of relativity: the mass of a particle is directly proportional to its energy. Thus according to Equation $\ref{Eq1}$, every mass has an associated energy, and similarly, any reaction that involves a change in energy must be accompanied by a change in mass. This implies that all exothermic reactions should be accompanied by a decrease in mass, and all endothermic reactions should be accompanied by an increase in mass. Given the law of conservation of mass, how can this be true? The solution to this apparent contradiction is that chemical reactions are indeed accompanied by changes in mass, but these changes are simply too small to be detected. As you may recall, all particles exhibit wavelike behavior, but the wavelength is inversely proportional to the mass of the particle (actually, to its momentum, the product of its mass and velocity). Consequently, wavelike behavior is detectable only for particles with very small masses, such as electrons. For example, the chemical equation for the combustion of graphite to produce carbon dioxide is as follows:
$\textrm{C(graphite)} + \frac{1}{2}\textrm O_2(\textrm g)\rightarrow \mathrm{CO_2}(\textrm g)\hspace{5mm}\Delta H^\circ=-393.5\textrm{ kJ/mol} \label{Eq2}$
Combustion reactions are typically carried out at constant pressure, and under these conditions, the heat released or absorbed is equal to ΔH. When a reaction is carried out at constant volume, the heat released or absorbed is equal to ΔE. For most chemical reactions, however, ΔE ≈ ΔH. If we rewrite Einstein’s equation as
$\Delta{E}=(\Delta m)c^2 \label{Eq3}$
we can rearrange the equation to obtain the following relationship between the change in mass and the change in energy:
$\Delta m=\dfrac{\Delta E}{c^2} \label{Eq4}$
Because 1 J = 1 (kg•m2)/s2, the change in mass is as follows:
$\Delta m=\dfrac{-393.5\textrm{ kJ/mol}}{(2.998\times10^8\textrm{ m/s})^2}=\dfrac{-3.935\times10^5(\mathrm{kg\cdot m^2})/(\mathrm{s^2\cdot mol})}{(2.998\times10^8\textrm{ m/s})^2}=-4.38\times10^{-12}\textrm{ kg/mol} \label{Eq5}$
This is a mass change of about 3.6 × 10−10 g/g carbon that is burned, or about 100-millionths of the mass of an electron per atom of carbon. In practice, this mass change is much too small to be measured experimentally and is negligible.
In contrast, for a typical nuclear reaction, such as the radioactive decay of 14C to 14N and an electron (a β particle), there is a much larger change in mass:
$^{14}\textrm C\rightarrow \,^{14}\textrm N+\,^0_{-1}\beta \label{Eq6}$
We can use the experimentally measured masses of subatomic particles and common isotopes given in Table 20.1 to calculate the change in mass directly. The reaction involves the conversion of a neutral 14C atom to a positively charged 14N ion (with six, not seven, electrons) and a negatively charged β particle (an electron), so the mass of the products is identical to the mass of a neutral 14N atom. The total change in mass during the reaction is therefore the difference between the mass of a neutral 14N atom (14.003074 amu) and the mass of a 14C atom (14.003242 amu):
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}} \&=14.003074\textrm{ amu} - 14.003242\textrm{ amu} = - 0.000168\textrm{ amu}\end{align} \label{Eq7}
The difference in mass, which has been released as energy, corresponds to almost one-third of an electron. The change in mass for the decay of 1 mol of 14C is −0.000168 g = −1.68 × 10−4 g = −1.68 × 10−7 kg. Although a mass change of this magnitude may seem small, it is about 1000 times larger than the mass change for the combustion of graphite. The energy change is as follows:
\begin{align}\Delta E &=(\Delta m)c^2=(-1.68\times10^{-7}\textrm{ kg})(2.998\times10^8\textrm{ m/s})^2 \ &=-1.51\times10^{10}(\mathrm{kg\cdot m^2})/\textrm s^2=-1.51\times10^{10}\textrm{ J}=-1.51\times10^7\textrm{ kJ}\end{align} \label{Eq8}
The energy released in this nuclear reaction is more than 100,000 times greater than that of a typical chemical reaction, even though the decay of 14C is a relatively low-energy nuclear reaction.
Because the energy changes in nuclear reactions are so large, they are often expressed in kiloelectronvolts (1 keV = 103 eV), megaelectronvolts (1 MeV = 106 eV), and even gigaelectronvolts (1 GeV = 109 eV) per atom or particle. The change in energy that accompanies a nuclear reaction can be calculated from the change in mass using the relationship 1 amu = 931 MeV. The energy released by the decay of one atom of 14C is thus
$\mathrm{(-1.68\times10^{-4}\, amu) \left(\dfrac{931\, MeV}{amu}\right) = -0.156\, MeV = -156\, keV}\label{Eq9}$
Example $1$
Calculate the changes in mass (in atomic mass units) and energy (in joules per mole and electronvolts per atom) that accompany the radioactive decay of 238U to 234Th and an α particle. The α particle absorbs two electrons from the surrounding matter to form a helium atom.
Given: nuclear decay reaction
Asked for: changes in mass and energy
Strategy:
A Use the mass values in Table 20.1 to calculate the change in mass for the decay reaction in atomic mass units.
B Use Equation $\ref{Eq4}$ to calculate the change in energy in joules per mole.
C Use the relationship between atomic mass units and megaelectronvolts to calculate the change in energy in electronvolts per atom.
Solution
A Using particle and isotope masses from Table 20.1, we can calculate the change in mass as follows:
\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}}=(\mathrm{mass \;^{234}Th+mass\;^4_2He})-\mathrm{mass\;^{238}U} \&=(234.043601\textrm{ amu}+4.002603\textrm{ amu}) - 238.050788\textrm{ amu} = - 0.004584\textrm{ amu}\end{align}
B Thus the change in mass for 1 mol of 238U is −0.004584 g or −4.584 × 10−6 kg. The change in energy in joules per mole is as follows:
ΔE = (Δm)c2 = (−4.584 × 10−6 kg)(2.998 × 108 m/s)2 = −4.120 × 1011 J/mol
C The change in energy in electronvolts per atom is as follows:
$\Delta E = -4.584\times10^{-3}\textrm{ amu}\times\dfrac{\textrm{931 MeV}}{\textrm{amu}}\times\dfrac{1\times10^6\textrm{ eV}}{\textrm{1 MeV}}=-4.27\times10^6\textrm{ eV/atom}$
Exercise $1$
Calculate the changes in mass (in atomic mass units) and energy (in kilojoules per mole and kiloelectronvolts per atom) that accompany the radioactive decay of tritium (3H) to 3He and a β particle.
Answer
Δm = −2.0 × 10−5 amu; ΔE = −1.9 × 106 kJ/mol = −19 keV/atom
Nuclear Binding Energies
We have seen that energy changes in both chemical and nuclear reactions are accompanied by changes in mass. Einstein’s equation, which allows us to interconvert mass and energy, has another interesting consequence: The mass of an atom is always less than the sum of the masses of its component particles. The only exception to this rule is hydrogen-1 (1H), whose measured mass of 1.007825 amu is identical to the sum of the masses of a proton and an electron. In contrast, the experimentally measured mass of an atom of deuterium (2H) is 2.014102 amu, although its calculated mass is 2.016490 amu:
\begin{align}m_{^2\textrm H}&=m_{\textrm{neutron}}+m_{\textrm{proton}}+m_{\textrm{electron}} \&=1.008665\textrm{ amu}+1.007276\textrm{ amu}+0.000549\textrm{ amu}=2.016490\textrm{ amu} \end{align}\label{Eq10}
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defect of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to Equation $\ref{Eq4}$, this release of energy must be accompanied by a decrease in the mass of the nucleus.
The amount of energy released when a nucleus forms from its component nucleons is the nuclear binding energy (Figure $1$). In the case of deuterium, the mass defect is 0.002388 amu, which corresponds to a nuclear binding energy of 2.22 MeV for the deuterium nucleus. Because the magnitude of the mass defect is proportional to the nuclear binding energy, both values indicate the stability of the nucleus.
Just as a molecule is more stable (lower in energy) than its isolated atoms, a nucleus is more stable than its isolated components.
Not all nuclei are equally stable. Chemists describe the relative stability of different nuclei by comparing the binding energy per nucleon, which is obtained by dividing the nuclear binding energy by the mass number (A) of the nucleus. As shown in Figure $2$, the binding energy per nucleon increases rapidly with increasing atomic number until about Z = 26, where it levels off to about 8–9 MeV per nucleon and then decreases slowly. The initial increase in binding energy is not a smooth curve but exhibits sharp peaks corresponding to the light nuclei that have equal numbers of protons and neutrons (e.g., 4He, 12C, and 16O). As mentioned earlier, these are particularly stable combinations.
Because the maximum binding energy per nucleon is reached at 56Fe, all other nuclei are thermodynamically unstable with regard to the formation of 56Fe. Consequently, heavier nuclei (toward the right in Figure $2$) should spontaneously undergo reactions such as alpha decay, which result in a decrease in atomic number. Conversely, lighter elements (on the left in Figure $2$) should spontaneously undergo reactions that result in an increase in atomic number. This is indeed the observed pattern.
Heavier nuclei spontaneously undergo nuclear reactions that decrease their atomic number. Lighter nuclei spontaneously undergo nuclear reactions that increase their atomic number.
Example $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 56Fe. The experimental mass of the nuclide is given in Table A4.
Given: nuclide and mass
Asked for: nuclear binding energy and binding energy per nucleon
Strategy:
A Sum the masses of the protons, electrons, and neutrons or, alternatively, use the mass of the appropriate number of 1H atoms (because its mass is the same as the mass of one electron and one proton).
B Calculate the mass defect by subtracting the experimental mass from the calculated mass.
C Determine the nuclear binding energy by multiplying the mass defect by the change in energy in electronvolts per atom. Divide this value by the number of nucleons to obtain the binding energy per nucleon.
Solution
A An iron-56 atom has 26 protons, 26 electrons, and 30 neutrons. We could add the masses of these three sets of particles; however, noting that 26 protons and 26 electrons are equivalent to 26 1H atoms, we can calculate the sum of the masses more quickly as follows:
\begin{align*}\textrm{calculated mass}&=26(\textrm{mass }^1_1\textrm H)+30(\textrm{mass }^1_0 \textrm n)\[4pt] &=26(1.007825)\textrm{amu}+30(1.008665)\textrm{amu}=56.463400\textrm{ amu}\ \textrm{experimental mass} &=55.934938 \end{align*} \nonumber
B We subtract to find the mass defect:
\begin{align*}\textrm{mass defect}&=\textrm{calculated mass}-\textrm{experimental mass} \&=56.463400\textrm{ amu}-55.934938\textrm{ amu}=0.528462\textrm{ amu}\end{align*} \nonumber
C The nuclear binding energy is thus 0.528462 amu × 931 MeV/amu = 492 MeV. The binding energy per nucleon is 492 MeV/56 nucleons = 8.79 MeV/nucleon.
Exercise $2$
Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for 238U.
Answer
1800 MeV/238U; 7.57 MeV/nucleon
Summary
Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. Nuclear reactions are accompanied by large changes in energy, which result in detectable changes in mass. The change in mass is related to the change in energy according to Einstein’s equation: ΔE = (Δm)c2. Large changes in energy are usually reported in kiloelectronvolts or megaelectronvolts (thousands or millions of electronvolts). With the exception of 1H, the experimentally determined mass of an atom is always less than the sum of the masses of the component particles (protons, neutrons, and electrons) by an amount called the mass defect of the nucleus. The energy corresponding to the mass defect is the nuclear binding energy, the amount of energy released when a nucleus forms from its component particles. In nuclear fission, nuclei split into lighter nuclei with an accompanying release of multiple neutrons and large amounts of energy. The critical mass is the minimum mass required to support a self-sustaining nuclear chain reaction. Nuclear fusion is a process in which two light nuclei combine to produce a heavier nucleus plus a great deal of energy. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.08%3A_Converting_Mass_to_Energy-_Mass_Defect_and_Nuclear_Binding_Energy.txt |
Learning Objectives
• Describe the nuclear reactions in a nuclear fusion reaction
• Quantify the energy released or absorbed in a fusion reaction
The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called fusion. The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events:
$\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}n}$
A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 1011 kJ of energy per mole of $\ce{^4_2He}$ produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 1010 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ).
It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, $^2_1H$ and a triton, $^3_1H$, undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron:
$\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n}$
This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 109 kilojoules per mole of $\ce{^4_2He}$ formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide.
The most important fusion process in nature is the one that powers stars. In the 20th century, it was realized that the energy released from nuclear fusion reactions accounted for the longevity of the Sun and other stars as a source of heat and light. The fusion of nuclei in a star, starting from its initial hydrogen and helium abundance, provides that energy and synthesizes new nuclei as a byproduct of that fusion process. The prime energy producer in the Sun is the fusion of hydrogen to form helium, which occurs at a solar-core temperature of 14 million kelvin. The net result is the fusion of four protons into one alpha particle, with the release of two positrons, two neutrinos (which changes two of the protons into neutrons), and energy (Figure $2$).
Example $1$
Calculate the energy released in each of the following hypothetical processes.
1. $\ce{3 ^4_2He \rightarrow ^{12}_6C}$
2. $\ce{6 ^1_1H + 6 ^1_0n \rightarrow ^{12}_6C}$
3. $\ce{6 ^2_1D \rightarrow ^{12}_6C}$
Solution
1. $Q_a = 3 \times 4.0026 - 12.000) \,amu \times (1.4924\times 10^{-10} \,J/amu) = 1.17 \times 10^{-12} \,J$
2. $Q_b = (6 \times (1.007825 + 1.008665) - 12.00000)\, amu \times (1.4924\times 10^{1-0} J/amu) = 1.476\times 10^{-11} \,J$
3. $Q_c = 6 \times 2.014102 - 12.00000 \, amu \times (1.4924\times 10^{-10} \, J/amu) = 1.263\times 10^{-11}\, J$
Fusion of $\ce{He}$ to give $\ce{C}$ releases the least amount of energy, because the fusion to produce He has released a large amount. The difference between the second and the third is the binding energy of deuterium. The conservation of mass-and-energy is well illustrated in these calculations. On the other hand, the calculation is based on the conservation of mass-and-energy.
Nuclear Reactors
Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur.
Another much more beneficial way to create fusion reactions is in a fusion reactor, a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure $3$). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods.Contributors | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.09%3A_Nuclear_Fusion_-_The_Power_of_the_Sun.txt |
Learning Objectives
• Describe the synthesis of transuranium nuclides
After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War II, combined to usher in the Nuclear Age in the mid-twentieth century. Science learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society’s needs during peace.
Synthesis of Nuclides
Nuclear transmutation is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford’s laboratory in 1919 by a transmutation reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed α particles from a natural radioactive isotope of radium and observed protons resulting from the reaction:
$\ce{^{14}_7N + ^4_2He ⟶ ^{17}_8O + ^1_1H} \nonumber$
The $\ce{^{17}_8O}$ and $\ce{^1_1H}$ nuclei that are produced are stable, so no further (nuclear) changes occur.
To reach the kinetic energies necessary to produce transmutation reactions, devices called particle accelerators are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors.
CERN Particle Accelerator
Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure $1$). In the LHC, particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers.
Figure $1$: A small section of the LHC is shown with workers traveling along it. (credit: Christophe Delaere)
In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2103 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously.
Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are:
\begin{align*} \ce{^{238}_{92}U + ^1_0n} &⟶ \ce{^{239}_{92}U} && \ \ce{^{239}_{92}U} &⟶ \ce{^{239}_{93}Np + ^0_{−1}e} &&\textrm{half-life}=\mathrm{23.5\: min} \ \ce{^{239}_{93}Np } &⟶ \ce{^{239}_{94}Pu + ^0_{−1}e} &&\textrm{half-life}=\mathrm{2.36\: days} \end{align*}
Plutonium is now mostly formed in nuclear reactors as a byproduct during the decay of uranium. Some of the neutrons that are released during U-235 decay combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. It is possible to summarize these equations as:
$\mathrm{\ce{^{238}_{92}U} + {^1_0n}⟶ \ce{^{239}_{92}U} \xrightarrow{β^-} \ce{^{239}_{93}Np} \xrightarrow{β^-} \ce{^{239}_{94}Pu}}$
Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years.
Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses.
The elements beyond element 92 (uranium) are called transuranium elements. As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table $1$.
Table $1$: Preparation of Some of the Transuranium Elements
Name Symbol Atomic Number Reaction
americium Am 95 $\ce{^{239}_{94}Pu + ^1_0n ⟶ ^{240}_{95}Am + ^0_{−1}e}$
curium Cm 96 $\ce{^{239}_{94}Pu + ^4_2He ⟶ ^{242}_{96}Cm + ^1_0n}$
californium Cf 98 $\ce{^{242}_{96}Cm + ^4_2He⟶ ^{243}_{97}Bk + 2^1_0n}$
einsteinium Es 99 $\ce{^{238}_{92}U + 15^1_0n⟶ ^{253}_{99}Es + 7^0_{−1}e}$
mendelevium Md 101 $\ce{^{253}_{99}Es + ^4_2He ⟶ ^{256}_{101}Md + ^1_0n}$
nobelium No 102 $\ce{^{246}_{96}Cm + ^{12}_6C ⟶ ^{254}_{102}No + 4 ^1_0n}$
rutherfordium Rf 104 $\ce{^{249}_{98}Cf + ^{12}_6C⟶ ^{257}_{104}Rf + 4 ^1_0n}$
seaborgium
Sg
106
$\ce{^{206}_{82}Pb + ^{54}_{24}Cr ⟶ ^{257}_{106}Sg + 3 ^1_0n}$
$\ce{^{249}_{98}Cf + ^{18}_8O ⟶ ^{263}_{106}Sg + 4 ^1_0n}$
meitnerium Mt 107 $\ce{^{209}_{83}Bi + ^{58}_{26}Fe ⟶ ^{266}_{109}Mt + ^1_0n}$ | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.10%3A_Nuclear_Transmutation_and_Transuranium_Elements.txt |
Learning Objectives
• To know the differences between ionizing and nonionizing radiation and their effects on matter.
• To identify natural and artificial sources of radiation.
Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He2+) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation.
Ionizing versus Nonionizing Radiation
The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiation is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling.
In contrast, ionizing radiation is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions:
$\mathrm{atom + ionizing\: radiation \rightarrow ion^+ + \, {e^-}\label{Eq1}}$
Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials (Figure $1$). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle:
$\text{1 MeV/particle} = \text{96 billion J/mol}. \nonumber$
The Effects of Ionizing Radiation on Matter
The effects of ionizing radiation depend on four factors:
1. The type of radiation, which dictates how far it can penetrate into matter
2. The energy of the individual particles or photons
3. The number of particles or photons that strike a given area per unit time
4. The chemical nature of the substance exposed to the radiation
The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in Figure $2$. Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal.
Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source outside the body. Alpha particles, however, are the most damaging if their source is inside the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in Table $1$.
Table $1$: Some Properties of Ionizing Radiation
Type Energy Range (MeV) Penetration Distance in Water* Penetration Distance in Air*
*Distance at which half of the radiation has been absorbed.
α particles 3–9 < 0.05 mm < 10 cm
β particles ≤ 3 < 4 mm 1 m
x-rays <10−2 < 1 cm < 3 m
γ rays 10−2–101 < 20 cm > 3 m
There are many different ways to measure radiation exposure, or the dose. The roentgen (R), which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure.Named after the German physicist Wilhelm Röntgen (1845–1923; Nobel Prize in Physics, 1901), who discovered x-rays. The roentgen is actually defined as the amount of radiation needed to produce an electrical charge of 2.58 × 10−4 C in 1 kg of dry air. Damage to biological tissues, however, is proportional to the amount of energy absorbed by tissues, not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose); the SI equivalent is the gray (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram:
$\mathrm{1\: rad = 0.010\: J/kg \hspace{25 pt} 1\: Gy = 1\: J/kg \label{Eq2}}$
Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10−4 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10−5 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle.
Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in man) was devised to describe the actual amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the RBE (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in millirems (1 mrem = 10−3 rem).
Wilhelm Röntgen
Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics.
Natural Sources of Radiation
We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr (Figure $3$). One component of background radiation is cosmic rays, high-energy particles and $\gamma$ rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure.
A second component of background radiation is cosmogenic radiation, produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as $\ce{^{14}C}$:
$\ce{^{14}_7 N + ^1_0 n \rightarrow ^{14}_6 C + ^1_1p }\label{Eq3}$
The carbon atoms react with oxygen atoms to form CO2, which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 1012 of the carbon atoms in our bodies is radioactive 14C, which decays by beta emission. About 5000 14C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium (3H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to 14C is estimated to be 1 mrem/yr, while that due to 3H is about 1000 times less.
The third major component of background radiation is terrestrial radiation, which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as $\ce{^{232}Th}$ and $\ce{^{238}U}$ as well as radioactive daughter isotopes, such as$\ce{^{226}Ra}$. The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the $K^+$ ion. Naturally occurring potassium contains 0.0117% $\ce{^{40}K}$, which decays by emitting both a β particle and a (\gamma\) ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 $\ce{^{40}K}$ nuclei disintegrated in your body.
By far the most important source of background radiation is radon, the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of238U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, 222Rn undergoes two successive alpha-decay events to give 214Pb:
$\ce{^{222}_{86} Rn \rightarrow ^4_2\alpha + ^{218}_{84} Po + ^4_2\alpha + ^{214}_{82} Pb } \label{Eq4}$
Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of $\ce{^{218}Po}$ releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The $\ce{^{218}Po}$ isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States.
Artificial Sources of Radiation
In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium (3H) and promethium (147Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight).
Example $1$
Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring 40K in his or her body, which contains about 140 g of potassium (as the K+ ion). The natural abundance of 40K is 0.0117%. Each 1.00 mol of 40K undergoes 1.05 × 107 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle.
Given: mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle
Asked for: annual radiation dose in rads
Strategy:
1. Calculate the number of moles of 40K present using its mass, molar mass, and natural abundance.
2. Determine the number of decays per year for this amount of 40K.
3. Multiply the number of decays per year by the energy associated with each decay event. To obtain the annual radiation dose, use the mass of the student to convert this value to rads.
Solution
A The number of moles of 40K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as 40K divided by the atomic mass of 40K:
$\textrm{moles }^{40}\textrm K= 140\textrm{ g K} \times \dfrac{0.0117\textrm{ mol }^{40}\textrm K}{100\textrm{ mol K}}\times\dfrac{1\textrm{ mol K}}{40.0\textrm{ g K}}=4.10\times10^{-4}\mathrm{\,mol\,^{40}K} \nonumber$
B We are given the number of atoms of 40K that decay per second in 1.00 mol of 40K, so the number of decays per year is as follows:
$\dfrac{\textrm{decays}}{\textrm{year}}=4.10\times10^{-4}\mathrm{\,mol^{40}\,K}\times\dfrac{1.05\times10^7\textrm{ decays/s}}{\mathrm{1.00\,mol\,^{40}K}}\times\dfrac{60\textrm{ s}}{1\textrm{ min}}\times\dfrac{60\textrm{ min}}{1\textrm{ h}}\times\dfrac{24\textrm{ h}}{1\textrm{ day}}\times\dfrac{365\textrm{ days}}{1\textrm{ yr}}$
C The total energy the body receives per year from the decay of 40K is equal to the total number of decays per year multiplied by the energy associated with each decay event:
\begin{align*}\textrm{total energy per year}&=\dfrac{1.36\times10^{11}\textrm{ decays}}{\textrm{yr}}\times\dfrac{1.32\textrm{ MeV}}{\textrm{decays}}\times\dfrac{10^6\textrm{ eV}}{\textrm{MeV}}\times\dfrac{1.602\times10^{-19}\textrm{ J}}{\textrm{eV}}\&=2.87\times10^{-2}\textrm{ J/yr}\end{align*} \nonumber
We use the definition of the rad (1 rad = 10−2 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows:
\begin{align*}\textrm{radiation dose per year}&=\dfrac{2.87\times10^{-2}\textrm{ J/yr}}{\textrm{70.0 kg}}\times\dfrac{1\textrm{ rad}}{1\times10^{-2}\textrm{ J/kg}}\&=4.10\times10^{-2}\textrm{ rad/yr}=41\textrm{ mrad/yr}\end{align*} \nonumber
This corresponds to almost half of the normal background radiation most people experience.
Exercise $1$
Because strontium is chemically similar to calcium, small amounts of the Sr2+ ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca2+. Consequently, the radioactive strontium (90Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of 90Sr undergoes 4.55 × 1014 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were 90Sr?
Answer
5.7 × 103 rad/yr (which is 10 times the fatal dose)
Assessing the Impact of Radiation Exposure
One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in Table $2$. Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid.
Table $2$: The Effects of a Single Radiation Dose on a 70 kg Human
Dose (rem) Symptoms/Effects
< 5 no observable effect
5–20 possible chromosomal damage
20–100 temporary reduction in white blood cell count
50–100 temporary sterility in men (up to a year)
100–200 mild radiation sickness, vomiting, diarrhea, fatigue; immune system suppressed; bone growth in children retarded
> 300 permanent sterility in women
> 500 fatal to 50% within 30 days; destruction of bone marrow and intestine
> 3000 fatal within hours
Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess.
The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in Figure $4$, but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that all exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure.
Summary
Nonionizing radiation is relatively low in energy and can be used as a heat source, whereas ionizing radiation, which is higher in energy, can penetrate biological tissues and is highly reactive. The effects of radiation on matter depend on the energy of the radiation. Nonionizing radiation is relatively low in energy, and the energy is transferred to matter in the form of heat. Ionizing radiation is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the roentgen (R), the amount of energy absorbed by dry air, and the rad (radiation absorbed dose), the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The rem (roentgen equivalent in man) measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include cosmic radiation, consisting of high-energy particles and γ rays emitted by the sun and other stars; cosmogenic radiation, which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and terrestrial radiation, from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.11%3A_The_Effects_of_Radiation_on_Life.txt |
Learning Objectives
• List common applications of radioactive isotopes
Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a radioactive tracer (or radioactive label). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more.
Radioisotopes have revolutionized medical practice, where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 $\ce{(^{99}_{43}Tc)}$, thallium-201 $\ce{(^{201}_{81}Tl)}$, iodine-131 $\ce{(^{131}_{53}I)}$, and sodium-24 $\ce{(^{24}_{11}Na)}$. Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure $1$) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood.
Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure $2$). The parent nuclide Mo-99 is part of a molybdate ion, $\ce{MoO4^2-}$; when it decays, it forms the pertechnetate ion, $\ce{TcO4-}$. These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests.
Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. Radiation therapy is the use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing (Figure $3$). A cancer patient may receive external beam radiation therapy delivered by a machine outside the body, or internal radiation therapy (brachytherapy) from a radioactive substance that has been introduced into the body. Note that chemotherapy is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells.
Cobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is:
$\ce{^{59}_{27}Co + ^1_0n⟶ ^{60}_{27}Co⟶ ^{60}_{28}Ni + ^0_{−1}β + 2^0_0γ} \nonumber$
The overall decay scheme for this is shown graphically in Figure $4$.
Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants.
For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is:
$\ce{6CO2}(g)+\ce{6H2O}(l)⟶\ce{C6H12O6}(s)+\ce{6O2}(g), \nonumber$
but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO2 containing a high concentration of $\ce{^{14}_6C}$. At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction.
Commercial applications of radioactive materials are equally diverse (Figure $5$). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil.
Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure $6$). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm.
Summary
Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally.
Glossary
chemotherapy
similar to internal radiation therapy, but chemical rather than radioactive substances are introduced into the body to kill cancer cells
external beam radiation therapy
radiation delivered by a machine outside the body
internal radiation therapy
(also, brachytherapy) radiation from a radioactive substance introduced into the body to kill cancer cells
radiation therapy
use of high-energy radiation to damage the DNA of cancer cells, which kills them or keeps them from dividing
radioactive tracer
(also, radioactive label) radioisotope used to track or follow a substance by monitoring its radioactive emissions | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/21%3A_Radioactivity_and_Nuclear_Chemistry/21.12%3A_Radioactivity_in_Medicine_and_Other_Applications.txt |
• 22.1: Fragrances and Odor
• 22.2: Hydrocarbons- Compounds Containing Only Carbon and Hydrocarbon
The simplest organic compounds are hydrocarbons and are composed of carbon and hydrogen. Hydrocarbons can be aliphatic or aromatic; aliphatic hydrocarbons are divided into alkanes, alkenes, and alkynes. The combustion of hydrocarbons is a primary source of energy for our society.
• 22.3: Polymers
• 22.4: Alkanes- Saturated Hydrocarbons
Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit.
• 22.5: Alkenes and Alkynes
As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
• 22.6: Hydrocarbon Reactions
The alkanes and cycloalkanes, with the exception of cyclopropane, are probably the least chemically reactive class of organic compounds. Alkanes contain strong carbon-carbon single bonds and strong carbon-hydrogen bonds. The carbon-hydrogen bonds are only very slightly polar. Alkanes can be burned, alkanes can react with some of the halogens, breaking carbon-hydrogen bonds, and alkanes can crack by breaking the carbon-carbon bonds.
• 22.7: Aromatic Hydrocarbons
Aromatic hydrocarbons contain ring structures with delocalized π electron systems.
• 22.8: Functional Groups
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups
• 22.9: Alcohols
In the IUPAC system, alcohols are named by changing the ending of the parent alkane name to -ol. Alcohols are classified according to the number of carbon atoms attached to the carbon atom that is attached to the OH group. Alcohols have higher boiling points than do ethers and alkanes of similar molar masses because the OH group allows alcohol molecules to engage in hydrogen bonding. Alcohols of four or fewer carbon atoms are soluble in water because the alcohol molecules engage in hydrogen bond
• 22.10: Aldehydes and Ketones
The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The polar carbon-to-oxygen double bond causes aldehydes and ketones to have higher boiling points than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols that engage in intermolecular hydrogen bonding. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation.
• 22.11: Carboxylic Acids and Esters
Aldehydes and ketones are characterized by the presence of a carbonyl group (C=O), and their reactivity can generally be understood by recognizing that the carbonyl carbon contains a partial positive charge ( δ+ ) and the carbonyl oxygen contains a partial negative charge ( δ− ). Aldehydes are typically more reactive than ketones.
• 22.12: Ethers
To give ethers common names, simply name the groups attached to the oxygen atom, followed by the generic name ether. If both groups are the same, the group name should be preceded by the prefix di-. Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points for a given molar mass. Ether molecules have an oxygen atom and can engage in hydrogen bonding with water molecules. An ether molecule has about the same solubility in water as
• 22.13: Amines
An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom. Primary and secondary amines have higher boiling points than those of alkanes or ethers of similar molar mass because they can engage in intermolecular hydrogen bonding. Amines are bases; they react with acids to form salts.
22: Organic Chemistry
Learning Objectives
• Identify alkanes, alkenes, alkynes, and aromatic compounds.
• List some properties of hydrocarbons.
The simplest organic compounds are those composed of only two elements: carbon and hydrogen. These compounds are called hydrocarbons. Hydrocarbons themselves are separated into two types: aliphatic hydrocarbons and aromatic hydrocarbons.
Aliphatic hydrocarbons are hydrocarbons based on chains of C atoms. There are three types of aliphatic hydrocarbons. Alkanes are aliphatic hydrocarbons with only single covalent bonds. Alkenes are hydrocarbons that contain at least one C–C double bond, and Alkynes are hydrocarbons that contain a C–C triple bond. Occasionally, we find an aliphatic hydrocarbon with a ring of C atoms; these hydrocarbons are called cycloalkanes (or cycloalkenes or cycloalkynes).
Aromatic hydrocarbons have a special six-carbon ring called a benzene ring. Electrons in the benzene ring have special energetic properties that give benzene physical and chemical properties that are markedly different from alkanes. Originally, the term aromatic was used to describe this class of compounds because they were particularly fragrant. However, in modern chemistry the term aromatic denotes the presence of a six-membered ring that imparts different and unique properties to a molecule.
The simplest alkanes have their C atoms bonded in a straight chain; these are called normal alkanes. They are named according to the number of C atoms in the chain. The smallest alkane is methane:
The next-largest alkane has two C atoms that are covalently bonded to each other. For each C atom to make four covalent bonds, each C atom must be bonded to three H atoms. The resulting molecule, whose formula is C2H6, is ethane:
Propane has a backbone of three C atoms surrounded by H atoms. You should be able to verify that the molecular formula for propane is C3H8:
The diagrams representing alkanes are called structural formulas because they show the structure of the molecule. As molecules get larger, structural formulas become more and more complex. One way around this is to use a condensed structural formula, which lists the formula of each C atom in the backbone of the molecule. For example, the condensed structural formula for ethane is CH3CH3, while for propane it is CH3CH2CH3. Table $1$ - The First 10 Alkanes, gives the molecular formulas, the condensed structural formulas, and the names of the first 10 alkanes.
Table $1$ The First 10 Alkanes
Molecular Formula Condensed Structural Formula Name
CH4 CH4 methane
C2H6 CH3CH3 ethane
C3H8 CH3CH2CH3 propane
C4H10 CH3CH2CH2CH3 butane
C5H12 CH3CH2CH2CH2CH3 pentane
C6H14 CH3(CH2)4CH3 hexane
C7H16 CH3(CH2)5CH3 heptane
C8H18 CH3(CH2)6CH3 octane
C9H20 CH3(CH2)7CH3 nonane
C10H22 CH3(CH2)8CH3 decane
Because alkanes have the maximum number of H atoms possible according to the rules of covalent bonds, alkanes are also referred to as saturated hydrocarbons.
Alkenes have a C–C double bond. Because they have less than the maximum number of H atoms possible, they are unsaturated hydrocarbons. The smallest alkene—ethene—has two C atoms and is also known by its common name ethylene:
The next largest alkene—propene—has three C atoms with a C–C double bond between two of the C atoms. It is also known as propylene:
What do you notice about the names of alkanes and alkenes? The names of alkenes are the same as their corresponding alkanes except that the ending is -ene, rather than -ane. Using a stem to indicate the number of C atoms in a molecule and an ending to represent the type of organic compound is common in organic chemistry, as we shall see.
With the introduction of the next alkene, butene, we begin to see a major issue with organic molecules: choices. With four C atoms, the C–C double bond can go between the first and second C atoms or between the second and third C atoms:
2 structural formulas for butene, with the first butene having the double bond on the first and second carbon from the left and the latter having its double bond on the second and third carbon from the left.
(A double bond between the third and fourth C atoms is the same as having it between the first and second C atoms, only flipped over.) The rules of naming in organic chemistry require that these two substances have different names. The first molecule is named 1-butene, while the second molecule is named 2-butene. The number at the beginning of the name indicates where the double bond originates. The lowest possible number is used to number a feature in a molecule; hence, calling the second molecule 3-butene would be incorrect. Numbers are common parts of organic chemical names because they indicate which C atom in a chain contains a distinguishing feature.
The compounds 1-butene and 2-butene have different physical and chemical properties, even though they have the same molecular formula—C4H8. Different molecules with the same molecular formula are called isomers. Isomers are common in organic chemistry and contribute to its complexity.
Example $1$
Based on the names for the butene molecules, propose a name for this molecule.
Solution
With five C atoms, we will use the pent- stem, and with a C–C double bond, this is an alkene, so this molecule is a pentene. In numbering the C atoms, we use the number 2 because it is the lower possible label. So this molecule is named 2-pentene.
Exercise $1$
Based on the names for the butene molecules, propose a name for this molecule.
A structural formula of a six carbon molecule with a double bond on the third and fourth carbon from the left. There are twelve hydrogen atoms in total.
Answer
3-hexene
Alkynes, with a C–C triple bond, are named similarly to alkenes except their names end in -yne. The smallest alkyne is ethyne, which is also known as acetylene:
Propyne has the structure
Structural formula showing three carbon molecules with a triple bond present between the first and second carbon atom. The appropriate number of hydrogen atoms is attached to each carbon atom.
With butyne, we need to start numbering the position of the triple bond, just as we did with alkenes:
Two structural formula of butyne. One butyne has a triple bond between the first and second carbon atom, while two butyne has the triple bond between the second and third carbon atom.
Aromatic compounds contain the benzene unit. Benzene itself is composed of six C atoms in a ring, with alternating single and double C–C bonds:
The six carbons are arranged in a hexagon pattern with one hydrogen atom emerging outwards from each carbon atom. The presence of a double bond is alternated between every other carbon atom.
The alternating single and double C–C bonds give the benzene ring a special stability, and it does not react like an alkene as might be suspected. Benzene has the molecular formula C6H6; in larger aromatic compounds, a different atom replaces one or more of the H atoms.
As fundamental as hydrocarbons are to organic chemistry, their properties and chemical reactions are rather mundane. Most hydrocarbons are nonpolar because of the close electronegativities of the C and H atoms. As such, they dissolve only sparingly in H2O and other polar solvents. Small hydrocarbons, such as methane and ethane, are gases at room temperature, while larger hydrocarbons, such as hexane and octane, are liquids. Even larger hydrocarbons are solids at room temperature and have a soft, waxy consistency.
Hydrocarbons are rather unreactive, but they do participate in some classic chemical reactions. One common reaction is substitution with a halogen atom by combining a hydrocarbon with an elemental halogen. Light is sometimes used to promote the reaction, such as this one between methane and chlorine:
$CH_{4}+Cl_{2}\overset{light}{\rightarrow} CH_{3}Cl+HCl\nonumber$
Halogens can also react with alkenes and alkynes, but the reaction is different. In these cases, the halogen reacts with the C–C double or triple bond and inserts itself onto each C atom involved in the multiple bonds. This reaction is called an addition reaction. One example is
The reaction conditions are usually mild; in many cases, the halogen reacts spontaneously with an alkene or an alkyne.
Hydrogen can also be added across a multiple bond; this reaction is called a hydrogenation reaction. In this case, however, the reaction conditions may not be mild; high pressures of H2 gas may be necessary. A platinum or palladium catalyst is usually employed to get the reaction to proceed at a reasonable pace:
$CH_{2}=CH_{2}+H_{2}\overset{metal\: catalyst}{\rightarrow} CH_{3}CH_{3}\nonumber$
By far the most common reaction of hydrocarbons is combustion, which is the combination of a hydrocarbon with O2 to make CO2 and H2O. The combustion of hydrocarbons is accompanied by a release of energy and is a primary source of energy production in our society (Figure $2$ - Combustion). The combustion reaction for gasoline, for example, which can be represented by C8H18, is as follows:
$2C^{8}H_{18}+25O_{2}\rightarrow 16CO_{2}+18H_{2}O+\sim 5060kJ\nonumber$
Key Takeaways
• The simplest organic compounds are hydrocarbons and are composed of carbon and hydrogen.
• Hydrocarbons can be aliphatic or aromatic; aliphatic hydrocarbons are divided into alkanes, alkenes, and alkynes.
• The combustion of hydrocarbons is a primary source of energy for our society.
Exercise $2$
1. Define hydrocarbon. What are the two general types of hydrocarbons?
2. What are the three different types of aliphatic hydrocarbons? How are they defined?
3. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
4. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
5. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
6. Indicate whether each molecule is an aliphatic or an aromatic hydrocarbon; if aliphatic, identify the molecule as an alkane, an alkene, or an alkyne.
7. Name and draw the structural formulas for the four smallest alkanes.
8. Name and draw the structural formulas for the four smallest alkenes.
9. What does the term aromatic imply about an organic molecule?
10. What does the term normal imply when used for alkanes?
11. Explain why the name 1-propene is incorrect. What is the proper name for this molecule?
12. Explain why the name 3-butene is incorrect. What is the proper name for this molecule?
13. Name and draw the structural formula of each isomer of pentene.
14. Name and draw the structural formula of each isomer of hexyne.
15. Write a chemical equation for the reaction between methane and bromine.
16. Write a chemical equation for the reaction between ethane and chlorine.
17. Draw the structure of the product of the reaction of bromine with propene.
18. Draw the structure of the product of the reaction of chlorine with 2-butene.
19. Draw the structure of the product of the reaction of hydrogen with 1-butene.
20. Draw the structure of the product of the reaction of hydrogen with 1-butene.
21. Write the balanced chemical equation for the combustion of heptane.
22. Write the balanced chemical equation for the combustion of nonane.
Answers
1. an organic compound composed of only carbon and hydrogen; aliphatic hydrocarbons and aromatic hydrocarbons
1. aliphatic; alkane
2. aromatic
3. aliphatic; alkene
1. aliphatic; alkane
2. aliphatic; alkene
3. aromatic
2. Aromatic means that the molecule has a benzene ring.
3. The 1 is not necessary. The name of the compound is simply propene.
4. CH4 + Br2 → CH3Br + HBr
5. C7H16 + 11O2 → 7CO2 + 8H2O | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.02%3A_Hydrocarbons-_Compounds_Containing_Only_Carbon_and_Hydrocarbon.txt |
Learning Objective
• To understand polymer synthesis.
• To understand the differences between synthetic and biological polymers.
Most of the solids discussed so far have been molecules or ions with low molecular masses, ranging from tens to hundreds of atomic mass units. Many of the molecular materials in consumer goods today, however, have very high molecular masses, ranging from thousands to millions of atomic mass units, and are formed from a carefully controlled series of reactions that produce giant molecules called polymersA giant molecule that consists of many basic structural units (monomers) connected in a chain or network by covalent bonds. (from the Greek poly and meros, meaning “many parts”). Polymers are used in corrective eye lenses, plastic containers, clothing and textiles, and medical implant devices, among many other uses. They consist of basic structural units called monomersThe basic structural unit of a polymer., which are repeated many times in each molecule. As shown schematically in Figure 8.8.1 , polymerizationA process by which monomers are connected into chains or networks by covalent bonds. is the process by which monomers are connected into chains or networks by covalent bonds. Polymers can form via a condensation reaction, in which two monomer molecules are joined by a new covalent bond and a small molecule such as water is eliminated, or by an addition reaction, a variant of a condensation reaction in which the components of a species AB are added to adjacent atoms of a multiple bond. Many people confuse the terms plastics and polymers. PlasticThe property of a material that allows it to be molded into almost any shape. is the property of a material that allows it to be molded into almost any shape. Although many plastics are polymers, many polymers are not plastics. In this section, we introduce the reactions that produce naturally occurring and synthetic polymers.
Figure 8.8.1 Polymer formation during a polymerization reaction, a large number of monomers become connected by covalent bonds to form a single long molecule, a polymer.
Note the Pattern
Polymers are formed via condensation or addition reactions.
Naturally Occurring Polymers: Peptides and Proteins
Polymers that occur naturally are crucial components of all organisms and form the fabric of our lives. Hair, silk, skin, feathers, muscle, and connective tissue are all primarily composed of proteins, the most familiar kind of naturally occurring, or biological, polymer. The monomers of many biological polymers are the amino acids each called an amino acid residue. The residues are linked together by amide bonds, also called peptide bonds, via a condensation reaction where H2O is eliminated:
In the above equation, R represents an alkyl or aryl group, or hydrogen, depending on the amino acid. We write the structural formula of the product with the free amino group on the left (the N-terminus) and the free carboxylate group on the right (the C-terminus). For example, the structural formula for the product formed from the amino acids glycine and valine (glycyl-valine) is as follows:
The most important difference between synthetic and naturally occurring polymers is that the former usually contain very few different monomers, whereas biological polymers can have as many as 20 different kinds of amino acid residues arranged in many different orders. Chains with less than about 50 amino acid residues are called peptidesBiological polymers with less than about 50 amino acid residues., whereas those with more than about 50 amino acid residues are called proteinsBiological polymers with more than 50 amino acid residues linked together by amide bonds.. Many proteins are enzymesCatalysts that occur naturally in living organisms and that catalyze biological reactions., which are catalysts that increase the rate of a biological reaction.
Note the Pattern
Synthetic polymers usually contain only a few different monomers, whereas biological polymers can have many kinds of monomers, such as amino acids arranged in different orders.
Many small peptides have potent physiological activities. The endorphins, for example, are powerful, naturally occurring painkillers found in the brain. Other important peptides are the hormones vasopressin and oxytocin. Although their structures and amino acid sequences are similar, vasopressin is a blood pressure regulator, whereas oxytocin induces labor in pregnant women and milk production in nursing mothers. Oxytocin was the first biologically active peptide to be prepared in the laboratory by Vincent du Vigneaud (1901–1978), who was awarded the Nobel Prize in Chemistry in 1955.
Synthetic Polymers
Many of the synthetic polymers we use, such as plastics and rubbers, have commercial advantages over naturally occurring polymers because they can be produced inexpensively. Moreover, many synthetic polymers are actually more desirable than their natural counterparts because scientists can select monomer units to tailor the physical properties of the resulting polymer for particular purposes. For example, in many applications, wood has been replaced by plastics that are more durable, lighter, and easier to shape and maintain. Polymers are also increasingly used in engineering applications where weight reduction and corrosion resistance are required. Steel rods used to support concrete structures, for example, are often coated with a polymeric material when the structures are near ocean environments where steel is vulnerable to corrosion (For more information on corrosion, see Section 17.6.) In fact, the use of polymers in engineering applications is a very active area of research.
Probably the best-known example of a synthetic polymer is nylon (Figure 8.8.2). Its monomers are linked by amide bonds (which are called peptide bonds in biological polymers), so its physical properties are similar to those of some proteins because of their common structural unit—the amide group. Nylon is easily drawn into silky fibersA particle of a synthetic polymer that is more than 100 times longer than it is wide. that are more than a hundred times longer than they are wide and can be woven into fabrics. Nylon fibers are so light and strong that during World War II, all available nylon was commandeered for use in parachutes, ropes, and other military items. With polymer chains that are fully extended and run parallel to the fiber axis, nylon fibers resist stretching, just like naturally occurring silk fibers, although the structures of nylon and silk are otherwise different. Replacing the flexible –CH2– units in nylon by aromatic rings produces a stiffer and stronger polymer, such as the very strong polymer known as Kevlar. Kevlar fibers are so strong and rigid that they are used in lightweight army helmets, bulletproof vests, and even sailboat and canoe hulls, all of which contain multiple layers of Kevlar fabric.
Figure 8.8.2 The Synthesis of Nylon Nylon is a synthetic condensation polymer created by the reaction of a dicarboxylic acid and a diamine to form amide bonds and water.
Figure 8.8.3 Synthesis of Nylon: A video showing the synthesis of nylon 6,10 by Mabakken
Not all synthetic polymers are linked by amide bonds—for example, polyesters contain monomers that are linked by ester bonds. Polyesters are sold under trade names such as Dacron, Kodel, and Fortrel, which are used in clothing, and Mylar, which is used in magnetic tape, helium-filled balloons, and high-tech sails for sailboats. Although the fibers are flexible, properly prepared Mylar films are almost as strong as steel.
Polymers based on skeletons with only carbon are all synthetic. Most of these are formed from ethylene (CH2=CH2), a two-carbon building block, and its derivatives. The relative lengths of the chains and any branches control the properties of polyethylene. For example, higher numbers of branches produce a softer, more flexible, lower-melting-point polymer called low-density polyethylene (LDPE), whereas high-density polyethylene (HDPE) contains few branches. Substances such as glass that melt at relatively low temperatures can also be formed into fibers, producing fiberglass.
Figure 8.8.4 Commercial Polyethene production: A video discussing the commercial production of polyethene from the Royal Society of Chemistry
Because most synthetic fibers are neither soluble nor low melting, multistep processes are required to manufacture them and form them into objects. Graphite fibers are formed by heating a precursor polymer at high temperatures to decompose it, a process called pyrolysisA high-temperature decomposition reaction that can be used to form fibers of synthetic polymers.. The usual precursor for graphite is polyacrylonitrile, better known by its trade name—Orlon. A similar approach is used to prepare fibers of silicon carbide using an organosilicon precursor such as polydimethylsilane {[–(CH3)2Si–]n}. A new type of fiber consisting of carbon nanotubes, hollow cylinders of carbon just one atom thick, is lightweight, strong, and impact resistant. Its performance has been compared to that of Kevlar, and it is being considered for use in body armor, flexible solar panels, and bombproof trash bins, among other uses.
Because there are no good polymer precursors for elemental boron or boron nitride, these fibers have to be prepared by time-consuming and costly indirect methods. Even though boron fibers are about eight times stronger than metallic aluminum and 10% lighter, they are significantly more expensive. Consequently, unless an application requires boron’s greater resistance to oxidation, these fibers cannot compete with less costly graphite fibers.
Example 8.8.1
Polyethylene is used in a wide variety of products, including beach balls and the hard plastic bottles used to store solutions in a chemistry laboratory. Which of these products is formed from the more highly branched polyethylene?
Given: type of polymer
Asked for: application
Strategy:
Determine whether the polymer is LDPE, which is used in applications that require flexibility, or HDPE, which is used for its strength and rigidity.
Solution:
A highly branched polymer is less dense and less rigid than a relatively unbranched polymer. Thus hard, strong polyethylene objects such as bottles are made of HDPE with relatively few branches. In contrast, a beach ball must be flexible so it can be inflated. It is therefore made of highly branched LDPE.
Exercise
Which products are manufactured from LDPE and which from HPDE?
1. lawn chair frames
2. rope
3. disposable syringes
4. automobile protective covers
Answer
1. HDPE
2. LDPE
3. HDPE
4. LDPE
Summary
Polymers are giant molecules that consist of long chains of units called monomers connected by covalent bonds. Polymerization is the process of linking monomers together to form a polymer. Plastic is the property of a material that allows it to be molded. Biological polymers formed from amino acid residues are called peptides or proteins, depending on their size. Enzymes are proteins that catalyze a biological reaction. A particle that is more than a hundred times longer than it is wide is a fiber, which can be formed by a high-temperature decomposition reaction called pyrolysis.
Key Takeaway
• Polymers are giant molecules formed from addition or condensation reactions and can be classified as either biological or synthetic polymers.
Conceptual Problems
1. How are amino acids and proteins related to monomers and polymers? Draw the general structure of an amide bond linking two amino acid residues.
2. Although proteins and synthetic polymers (such as nylon) both contain amide bonds, different terms are used to describe the two types of polymer. Compare and contrast the terminology used for the
1. smallest repeating unit.
2. covalent bond connecting the units.
Contributors
• Anonymous
Modified by Joshua Halpern, Scott Sinex and Scott Johnson
Nylon synthesis from MA Bakken @ YouTube
Polyethylene production from Royal Society of Chemistry @ YouTube | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.03%3A_Polymers.txt |
Learning Objectives
• To identify and name simple (straight-chain) alkanes given formulas and write formulas for straight-chain alkanes given their names.
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). Saturated, in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules.
The word saturated has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C).
We previously introduced the three simplest alkanes—methane (CH4), ethane (C2H6), and propane (C3H8) and they are shown again in Figure $1$.
The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure $2$).
Methane (CH4), ethane (C2H6), and propane (C3H8) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH2 unit. The first 10 members of this series are given in Table $1$.
Table $1$: The First 10 Straight-Chain Alkanes
Name Molecular Formula (CnH2n + 2) Condensed Structural Formula Number of Possible Isomers
methane CH4 CH4
ethane C2H6 CH3CH3
propane C3H8 CH3CH2CH3
butane C4H10 CH3CH2CH2CH3 2
pentane C5H12 CH3CH2CH2CH2CH3 3
hexane C6H14 CH3CH2CH2CH2CH2CH3 5
heptane C7H16 CH3CH2CH2CH2CH2CH2CH3 9
octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3 18
nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3 35
decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3 75
Consider the series in Figure $3$. The sequence starts with C3H8, and a CH2 unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH2 group) is called a homologous series. The members of such a series, called homologs, have properties that vary in a regular and predictable manner. The principle of homology gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series.
The principle of homology allows us to write a general formula for alkanes: CnH2n + 2. Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C8H(2 × 8) + 2 = C8H18.
Key Takeaway
• Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.04%3A_Alkanes-_Saturated_Hydrocarbons.txt |
Learning Objectives
• To name alkenes given formulas and write formulas for alkenes given names.
As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R2C=CR2) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas:
Some representative alkenes—their names, structures, and physical properties—are given in Table \(1\).
Table \(1\): Physical Properties of Some Selected Alkenes
IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C)
ethene C2H4 CH2=CH2 –169 –104
propene C3H6 CH2=CHCH3 –185 –47
1-butene C4H8 CH2=CHCH2CH3 –185 –6
1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30
1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63
1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94
1-octene C8H16 CH2=CH(CH2)5CH3 –102 121
We used only condensed structural formulas in Table \(1\). Thus, CH2=CH2 stands for
The double bond is shared by the two carbons and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C2H4, whereas that for ethane is C2H6.
The first two alkenes in Table \(1\), ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure \(1\)). Ethylene is a major commercial chemical. The US chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products.
Although there is only one alkene with the formula C2H4 (ethene) and only one with the formula C3H6 (propene), there are several alkenes with the formula C4H8.
Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC):
1. The longest chain of carbon atoms containing the double bond is considered the parent chain. It is named using the same stem as the alkane having the same number of carbon atoms but ends in -ene to identify it as an alkene. Thus the compound CH2=CHCH3 is propene.
2. If there are four or more carbon atoms in a chain, we must indicate the position of the double bond. The carbons atoms are numbered so that the first of the two that are doubly bonded is given the lower of the two possible numbers.The compound CH3CH=CHCH2CH3, for example, has the double bond between the second and third carbon atoms. Its name is 2-pentene (not 3-pentene).
3. Substituent groups are named as with alkanes, and their position is indicated by a number. Thus, the structure below is 5-methyl-2-hexene. Note that the numbering of the parent chain is always done in such a way as to give the double bond the lowest number, even if that causes a substituent to have a higher number. The double bond always has priority in numbering.
Example \(1\)
Name each compound.
Solution
1. The longest chain containing the double bond has five carbon atoms, so the compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the fourth carbon atom (rule 3), so the compound’s name is 4-methyl-2-pentene.
2. The longest chain containing the double bond has five carbon atoms, so the parent compound is a pentene (rule 1). To give the first carbon atom of the double bond the lowest number (rule 2), we number from the left, so the compound is a 2-pentene. There is a methyl group on the third carbon atom (rule 3), so the compound’s name is 3-methyl-2-pentene.
Exercise \(1\)
Name each compound.
1. CH3CH2CH2CH2CH2CH=CHCH3
Just as there are cycloalkanes, there are cycloalkenes. These compounds are named like alkenes, but with the prefix cyclo- attached to the beginning of the parent alkene name.
Example \(2\)
Draw the structure for each compound.
1. 3-methyl-2-pentene
2. cyclohexene
Solution
1. First write the parent chain of five carbon atoms: C–C–C–C–C. Then add the double bond between the second and third carbon atoms:
Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds.
• First, consider what each of the three parts of the name means. Cyclo means a ring compound, hex means 6 carbon atoms, and -ene means a double bond.
Exercise \(2\)
Draw the structure for each compound.
1. 2-ethyl-1-hexene
2. cyclopentene
Key Takeaway
• Alkenes are hydrocarbons with a carbon-to-carbon double bond. | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.05%3A_Alkenes_and_Alkynes.txt |
Learning Objectives
• To identify the main chemical properties of alkanes.
Alkane molecules are nonpolar and therefore generally do not react with ionic compounds such as most laboratory acids, bases, oxidizing agents, or reducing agents. Consider butane as an example:
Neither positive ions nor negative ions are attracted to a nonpolar molecule. In fact, the alkanes undergo so few reactions that they are sometimes called paraffins, from the Latin parum affinis, meaning “little affinity.”
Two important reactions that the alkanes do undergo are combustion and halogenation. Nothing happens when alkanes are merely mixed with oxygen ($O_2$) at room temperature, but when a flame or spark provides the activation energy, a highly exothermic combustion reaction proceeds vigorously. For methane (CH4), the reaction is as follows:
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O + \text{heat} \label{12.7.1}$
If the reactants are adequately mixed and there is sufficient oxygen, the only products are carbon dioxide ($CO_2$), water ($H_2O$), and heat—heat for cooking foods, heating homes, and drying clothes. Because conditions are rarely ideal, however, other products are frequently formed. When the oxygen supply is limited, carbon monoxide ($CO$) is a by-product:
$2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O\label{12.7.2}$
This reaction is responsible for dozens of deaths each year from unventilated or improperly adjusted gas heaters. (Similar reactions with similar results occur with kerosene heaters.)
Alkanes also react with the halogens chlorine ($Cl_2$) and bromine ($Br_2$) in the presence of ultraviolet light or at high temperatures to yield chlorinated and brominated alkanes. For example, chlorine reacts with excess methane ($CH_4$) to give methyl chloride ($CH_3Cl$).
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl\label{12.7.3}$
With more chlorine, a mixture of products is obtained: CH3Cl, CH2Cl2, CHCl3, and CCl4. Fluorine ($F_2$), the lightest halogen, combines explosively with most hydrocarbons. Iodine ($I_2$) is relatively unreactive. Fluorinated and iodinated alkanes are produced by indirect methods.
Key Takeaway
• Alkanes react with oxygen (combustion) and with halogens (halogenation).
22.07: Aromatic Hydrocarbons
Textbook, Hydrocarbons
Textbook, Hydrocarbons
Benzene, C6H6, is the simplest member of a large family of hydrocarbons, called aromatic hydrocarbons. These compounds contain ring structures and exhibit bonding that must be described using the resonance hybrid concept of valence bond theory or the delocalization concept of molecular orbital theory. (To review these concepts, refer to the earlier chapters on chemical bonding). The resonance structures for benzene, C6H6, are:
Valence bond theory describes the benzene molecule and other planar aromatic hydrocarbon molecules as hexagonal rings of sp2-hybridized carbon atoms with the unhybridized p orbital of each carbon atom perpendicular to the plane of the ring. Three valence electrons in the sp2 hybrid orbitals of each carbon atom and the valence electron of each hydrogen atom form the framework of σ bonds in the benzene molecule. The fourth valence electron of each carbon atom is shared with an adjacent carbon atom in their unhybridized p orbitals to yield the π bonds. Benzene does not, however, exhibit the characteristics typical of an alkene. Each of the six bonds between its carbon atoms is equivalent and exhibits properties that are intermediate between those of a C–C single bond and a $\mathrm{C=C}$ double bond. To represent this unique bonding, structural formulas for benzene and its derivatives are typically drawn with single bonds between the carbon atoms and a circle within the ring as shown in Figure $10$.
There are many derivatives of benzene. The hydrogen atoms can be replaced by many different substituents. Aromatic compounds more readily undergo substitution reactions than addition reactions; replacement of one of the hydrogen atoms with another substituent will leave the delocalized double bonds intact. The following are typical examples of substituted benzene derivatives:
Toluene and xylene are important solvents and raw materials in the chemical industry. Styrene is used to produce the polymer polystyrene.
Structure of Aromatic Hydrocarbons
One possible isomer created by a substitution reaction that replaces a hydrogen atom attached to the aromatic ring of toluene with a chlorine atom is shown here. Draw two other possible isomers in which the chlorine atom replaces a different hydrogen atom attached to the aromatic ring:
Solution
Since the six-carbon ring with alternating double bonds is necessary for the molecule to be classified as aromatic, appropriate isomers can be produced only by changing the positions of the chloro-substituent relative to the methyl-substituent:
Exercise $7$
Draw three isomers of a six-membered aromatic ring compound substituted with two bromines.
Answer | textbooks/chem/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/22%3A_Organic_Chemistry/22.06%3A_Hydrocarbon_Reactions.txt |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.