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Learning Objectives
• Identify when a reaction will evolve a gas.
Neutralization Reactions
Acids and bases react chemically with each other to form salts. A salt is a general chemical term for any ionic compound formed from an acid and a base. In reactions where the acid is a hydrogen-ion-containing compound and the base is a hydroxide-ion-containing compound, water is also a product. The general reaction is as follows:
$\text{acid + base} → \text{water + salt} \nonumber$
The reaction of acid and base to make water and a salt is called neutralization. Like any chemical equation, a neutralization chemical equation must be properly balanced. For example, the neutralization reaction between sodium hydroxide and hydrochloric acid is as follows:
$\ce{NaOH (aq) + HCl (aq) \rightarrow NaCl (aq) + H_2O (ℓ)} \label{Eq2}$
with coefficients all understood to be one. The neutralization reaction between sodium hydroxide and sulfuric acid is as follows:
$\ce{2NaOH (aq) + H_2SO_4 (aq) \rightarrow Na_2SO_4(aq) + 2H_2O (ℓ)} \label{Eq3}$
Example $1$: Neutralizing Nitric Acid
Nitric acid (HNO3(aq)) can be neutralized by calcium hydroxide (Ca(OH)2(aq)). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces.
Solution
Solutions to Example 7.8.1
Steps Explanation Equation
Write the unbalanced equation. This is a double displacement reaction, so the cations and anions swap to create new products. Ca(OH)2(aq) + HNO3(aq) → Ca(NO3)2(aq) + H2O(ℓ)
Balance the equation. Because there are two OH ions in the formula for Ca(OH)2, we need two moles of HNO3 to provide H+ ions Ca(OH)2(aq) + 2HNO3(aq) → Ca(NO3)2(aq) + 2H2O(ℓ)
Additional step: identify the salt. The salt formed is calcium nitrate.
Exercise $1$
Hydrocyanic acid ($\ce{HCN(aq)}$) can be neutralized by potassium hydroxide ($\ce{KOH(aq)}$). Write a balanced chemical equation for the reaction between these two compounds and identify the salt that it produces.
Answer
$\ce{KOH (aq) + HCN(aq) → KCN (aq) + H2O(ℓ)} \nonumber$
Gas Evolving Reactions
A gas evolution reaction is a chemical process that produces a gas, such as oxygen or carbon dioxide. In the following examples, an acid reacts with a carbonate, producing salt, carbon dioxide, and water, respectively. For example, nitric acid reacts with sodium carbonate to form sodium nitrate, carbon dioxide, and water (Table $1$):
$\ce{2HNO3(aq)+Na2CO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)} \nonumber$
Sulfuric acid reacts with calcium carbonate to form calcium sulfate, carbon dioxide, and water:
$\ce{H2SO4(aq) + CaCO3(aq) → CaSO4(aq) + CO2(g)+H2O(l)} \nonumber$
Hydrochloric acid reacts with calcium carbonate to form calcium chloride, carbon dioxide, and water:
$\ce{2HCl(aq) + CaCO3(aq) → CaCl2(aq) + CO2(g) + H2O(l)} \nonumber$
Figure $1$ demonstrates this type of reaction:
In this reaction setup, lime water, a dilute calcium hydroxide ($Ca(OH)_2$) solution, is poured into one of the test tubes and sealed with a stopper. A small amount of hydrochloric acid is carefully poured into the remaining test tube. A small amount of sodium carbonate is added to the acid, and the tube is sealed with a rubber stopper. The two tubes are connected. As a result of the acid-carbonate reaction, carbon dioxide is produced and the lime water turns milky.
Table $1$: Types of Compounds That Undergo Gas-Evolution Reactions
Reactant Type Intermediate Product Gas Evolved Example
sulfide none $\ce{H2S}$ $\ce{2HCl(aq) + K2S \rightarrow H2S (g) + 2KCl (aq)}$
carbonates and bicarbonates $\ce{H2CO3}$ $\ce{CO2}$ $\ce{2HCl(aq) + K2CO2 \rightarrow H2O (l) + CO2(g) + 2KCl (aq)}$
sulfites and bisulfites $\ce{H2SO3}$ $\ce{SO2}$ $\ce{2HCl(aq) + K2SO2 \rightarrow H2O (l) + SO2(g) + 2KCl (aq)}$
ammonia $\ce{NH4OH}$ $\ce{NH3}$ $\ce{NH4Cl(aq) + KOH \rightarrow H2O (l) + NH3(g) + 2KCl (aq)}$
The gas-evolving experiment lime water is illustrated in the following video:
Video $1$: Carbon Dioxide ($CO_2$) & Limewater (Chemical Reaction). As the reaction proceeds, the limewater on the turns from clear to milky; this is due to the $CO_2(g)$ reacting with the aqueous calcium hydroxide to form calcium carbonate, which is only slightly soluble in water.
When this experiment is repeated with nitric or sulfuric acid instead of $HCl$, it yields the same results: the clear limewater turns milky, indicating the production of carbon dioxide. Another method to chemically generate gas is the oxidation of metals in acidic solutions. This reaction will yield a metal salt and hydrogen gas.
$\ce{2HCl (aq) + Zn(s) \rightarrow ZnCl_2 (aq) + H_2 (g)} \nonumber$
Here, hydrochloric acid oxidizes zinc to produce an aqueous metal salt and hydrogen gas bubbles. Recall that oxidation refers to a loss of electrons, and reduction refers to the gain of electrons. In the above redox reaction, neutral zinc is oxidized to $Zn^{2+}$, and the acid, $H^+$, is reduced to $H_2(g)$. The oxidation of metals by strong acids is another common example of a gas evolution reaction. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.08%3A_AcidBase_and_Gas_Evolution_Reactions.txt |
Learning Objectives
• Define oxidation and reduction.
• Assign oxidation numbers to atoms in simple compounds.
• Recognize a reaction as an oxidation-reduction reaction.
In the course of a chemical reaction between a metal and a nonmetal, electrons are transferred from the metal atoms to the nonmetal atoms. For example, when zinc metal is mixed with sulfur and heated, the compound zinc sulfide is produced. Two valence electrons from each zinc atom are transferred to each sulfur atom.
Since the zinc is losing electrons in the reaction, it is being oxidized. The sulfur is gaining electrons and is thus being reduced. An oxidation-reduction reaction is a reaction that involves the full or partial transfer of electrons from one reactant to another. Oxidation is the full or partial loss of electrons or the gain of oxygen. Reduction is the full or partial gain of electrons or the loss of oxygen. A redox reaction is another term for an oxidation-reduction reaction.
Each of these processes can be shown in a separate equation called a half-reaction. A half-reaction is an equation that shows either the oxidation or the reduction reaction that occurs during a redox reaction.
$\underbrace{\ce{Zn→Zn^{2+}+2e^{−}}}_{\text{Oxidation}} \label{7.9.1}$
$\underbrace{\ce{S+ 2 e^{−} → S^{2−}}}_ {\text{Reduction}} \label{7.9.2}$
It is important to remember that the two half-reactions occur simultaneously. The resulting ions that are formed are then attracted to one another in an ionic bond.
Another example of an oxidation-reduction reaction involving electron transfer is the well-known combination of metallic sodium and chlorine gas to form sodium chloride:
$\ce{2Na+Cl_2→2NaCl} \label{7.9.3}$
The half reactions are as follows:
$\underbrace{\ce{2Na→2Na^{+} + 2e^{−}}}_{\text{Oxidation}} \label{7.941}$
$\underbrace{\ce{Cl_2 +2e^{−} → 2Cl^{−}}}_ {\text{Reduction}} \label{7.9.5}$
We will concern ourselves with the balancing of these equations at another time.
Oxidation Numbers
Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use oxidation numbers to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on four rules. Oxidation numbers are not necessarily equal to the charge on the atom (although sometimes they can be); we must keep the concepts of charge and oxidation numbers separate.
Assigning Oxidation Numbers
The rules for assigning oxidation numbers to atoms are as follows:
1. Atoms in their elemental state are assigned an oxidation number of 0.
In H2, both H atoms have an oxidation number of 0.
2. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, to differentiate them from charges.
In MgCl2, magnesium has an oxidation number of +2, while chlorine has an oxidation number of −1.
3. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number [except when it exists as the hydride ion (H−), in which case rule 2 prevails].
In H2O, the H atoms each have an oxidation number of +1, while the O atom has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound (rule 3). By contrast, by rule 3, each H atom in hydrogen peroxide (H2O2) has an oxidation number of +1, while each O atom has an oxidation number of −1.
4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral).
In SO2, each O atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the S atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it means only that the S atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound.
Example $1$: Assigning Oxidation States
Assign oxidation numbers to the atoms in each substance.
1. Cl2
2. GeO2
3. Ca(NO3)2
Solution
1. Cl2 is the elemental form of chlorine. Rule 1 states that each atom has an oxidation number of 0.
2. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (zero), the Ge atom is assigned an oxidation number of +4.
3. Ca(NO3)2 can be separated into two parts: the Ca2+ ion and the NO3 ion. Considering these separately, the Ca2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3 ion. Oxygen is assigned an oxidation number of −2, and there are three of them. According to rule 4, the sum of the oxidation numbers on all atoms must equal the charge on the species, so we have the simple algebraic equation $x + 3(−2) = −1 \nonumber$
where x is the oxidation number of the N atom and the −1 represents the charge on the species. Evaluating for x,
$x + (−6) = −1x = +5 \nonumber$
Thus the oxidation number on the N atom in the NO3 ion is +5.
Exercise $1$: Assigning Oxidation States
Assign oxidation numbers to the atoms in the following:
1. H3PO4
2. MgO
Answer a
H: +1; O: −2; P: +5
Answer b
Mg: +2, O: −2
All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Thus oxidation and reduction can also be defined in terms of increasing or decreasing oxidation numbers, respectively.
Example $2$: Formation of Sodium Bromide
Identify what is being oxidized and reduced in the following redox reaction.
$\ce{2Na + Br2 → 2NaBr} \nonumber$
Solution
Both reactants are the elemental forms of their atoms, so the Na and Br atoms have oxidation numbers of 0. In the ionic product, the Na+ ions have an oxidation number of +1, while the Br ions have an oxidation number of −1.
$2\underset{0}{Na}+\underset{0}{Br_{2}}\rightarrow 2\underset{+1 -1}{NaBr} \nonumber$
Sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; bromine is decreasing its oxidation number from 0 to −1, so it is being reduced:
Because oxidation numbers are changing, this is a redox reaction. The total number of electrons being lost by sodium (two, one lost from each Na atom) is gained by bromine (two, one gained for each Br atom).
Exercise $2$: Oxidation of Carbon
Identify what is being oxidized and reduced in this redox reaction.$C + O_2 → CO_2 \nonumber$
Answer
C is being oxidized from 0 to +4; O is being reduced from 0 to −2
Oxidation reactions can become quite complex, as attested by the following redox reaction:
$6H^{+}(aq)+2\underset{+7}{MnO_{4}^{-}}(aq)+5\underset{-1}{H_{2}O_{2}}(l)\rightarrow 2\underset{+2}{Mn^{2+}}(aq)+5\underset{0}{O_{2}}(g)+8H_{2}O(l) \nonumber$
To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation. Eventually, we will need to learn techniques for writing correct (i.e., balanced) redox reactions.
Combustion Reactions
A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve $\ce{O_2}$ as one reactant. The combustion of hydrogen gas produces water vapor.
$2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right) \label{water}$
Notice that this reaction also qualifies as a combination reaction.
The Hindenberg was a hydrogen-filled airship that suffered an accident upon its attempted landing in New Jersey in 1937. The hydrogen immediately combusted in a huge fireball, destroying the airship and killing 36 people. The chemical reaction was a simple one: hydrogen combining with oxygen to produce water (Equation \ref{water}).
Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amounts of heat energy. Propane $\left( \ce{C_3H_8} \right)$ is a gaseous hydrocarbon that is commonly used as the fuel source in gas grills.
$\ce{C_3H_8} \left( g \right) + 5 \ce{O_2} \left( g \right) \rightarrow 3 \ce{CO_2} \left( g \right) + 4 \ce{H_2O} \left( g \right) \nonumber$
Example $3$: Combustion of Ethanol
Ethanol can be used as a fuel source in an alcohol lamp. The formula for ethanol is $\ce{C_2H_5OH}$. Write the balanced equation for the combustion of ethanol.
Solution
Solutions to Example 7.9.3
Steps Example Solution
Write the unbalanced reaction.
$\ce{C_2H_5OH} \left( l \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( g \right)$
Ethanol and atmospheric oxygen are the reactants. As with a hydrocarbon, the products of the combustion of an alcohol are carbon dioxide and water.
Balance the equation. $\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( g \right)$
Think about your result. Combustion reactions must have oxygen as a reactant. Note that the water produced is in the gas state, rather than the liquid state, because of the high temperatures that accompany a combustion reaction.
Exercise $3$: Combustion of Hexane
Write the balanced equation for the combustion of hexane, C6H14
Answer
$\ce{2C6H14 (ℓ) + 19O2(g) → 12CO2(g) + 14H2O(ℓ)} \nonumber$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.09%3A_OxidationReduction_Reactions.txt |
Learning Objectives
• Classify a chemical reaction as a synthesis, decomposition, single replacement, double replacement, or a combustion reaction.
• Predict the products of simple reactions.
The chemical reactions we have described are only a tiny sampling of the infinite number of chemical reactions possible. How do chemists cope with this overwhelming diversity? How do they predict which compounds will react with one another and what products will be formed? The key to success is to find useful ways to categorize reactions. Familiarity with a few basic types of reactions will help you to predict the products that form when certain kinds of compounds or elements come in contact.
Most chemical reactions can be classified into one or more of five basic types: acid–base reactions, exchange reactions, condensation reactions (and the reverse, cleavage reactions), and oxidation–reduction reactions. The general forms of these five kinds of reactions are summarized in Table $1$, along with examples of each. It is important to note, however, that many reactions can be assigned to more than one classification, as you will see in our discussion.
Table $1$: Basic Types of Chemical Reactions
Name of Reaction General Form Examples
Oxidation–Reduction (redox) oxidant + reductant → reduced oxidant + oxidized reductant C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(g)
Acid–Base acid + base → salt NaOH(aq) + HNO3(aq) → NaNO3(aq) +H2O(l)
Exchange: Single Replacement AB + C → AC + B ZnCl2(aq)+ Mg(s) → MgCl2(aq)+ Zn(s)
Exchange: Double Replacement AB + CDAD + CB BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
Combination (Synthesis) A + B → AB
CO2(g) + H2O(l) → H2CO3(aq)
N2(g) + 2O2(g)→ 2NO2(g)
Decomposition AB → A + B CaCO3(s) → CaO(s) + CO2(g)
The classification scheme is only for convenience; the same reaction can be classified in different ways, depending on which of its characteristics is most important. Oxidation–reduction reactions, in which there is a net transfer of electrons from one atom to another, and condensation reactions are discussed in this section. Acid–base reactions are one kind of exchange reaction—the formation of an insoluble salt, such as barium sulfate, when solutions of two soluble salts are mixed together.
Combination Reactions
A combination reaction is a reaction in which two or more substances combine to form a single new substance. Combination reactions can also be called synthesis reactions. The general form of a combination reaction is:
$\ce{A} + \ce{B} \rightarrow \ce{AB} \nonumber$
One combination reaction is two elements combining to form a compound. Solid sodium metal reacts with chlorine gas to produce solid sodium chloride.
$2 \ce{Na} \left( s \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{NaCl} \left( s \right) \nonumber$
Notice that in order to write and balance the equation correctly, it is important to remember the seven elements that exist in nature as diatomic molecules ($\ce{H_2}$, $\ce{N_2}$, $\ce{O_2}$, $\ce{F_2}$, $\ce{Cl_2}$, $\ce{Br_2}$, and $\ce{I_2}$).
One type of combination reaction that occurs frequently is the reaction of an element with oxygen to form an oxide. Metals and nonmetals both react readily with oxygen under most conditions. Magnesium reacts rapidly and dramatically when ignited, combining with oxygen from the air to produce a fine powder of magnesium oxide:
$2 \ce{Mg} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{MgO} \left( s \right) \nonumber$
Sulfur reacts with oxygen to form sulfur dioxide:
$\ce{S} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{SO_2} \left( g \right) \nonumber$
When nonmetals react with one another, the product is a molecular compound. Often, the nonmetal reactants can combine in different ratios and produce different products. Sulfur can also combine with oxygen to form sulfur trioxide:
$2 \ce{S} \left( s \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_3} \left( g \right) \nonumber$
Transition metals are capable of adopting multiple positive charges within their ionic compounds. Therefore, most transition metals are capable of forming different products in a combination reaction. Iron reacts with oxygen to form both iron (II) oxide and iron (III) oxide:
$2 \ce{Fe} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{FeO} \left( s \right) \nonumber$
$4 \ce{Fe} \left( s \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{Fe_2O_3} \left( s \right) \nonumber$
Example $1$: Combustion of Solid Potassium
Potassium is a very reactive alkali metal that must be stored under oil in order to prevent it from reacting with air. Write the balanced chemical equation for the combination reaction of potassium with oxygen.
Solution
Solutions to Example 7.10.1
Steps Example Solution
Plan the problem. Make sure formulas of all reactants and products are correct before balancing the equation. Oxygen gas is a diatomic molecule. Potassium oxide is an ionic compound and so its formula is constructed by the crisscross method. Potassium as an ion becomes $\ce{K^+}$, while the oxide ion is $\ce{O^{2-}}$.
Solve.
The skeleton (unbalanced) equation:
$\ce{K} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{K_2O} \left( s \right) \nonumber$
The equation is then easily balanced with coefficients.
$4 \ce{K} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{K_2O} \left( s \right) \nonumber$
Think about your result. Formulas are correct and the resulting combination reaction is balanced.
Combination reactions can also take place when an element reacts with a compound to form a new compound composed of a larger number of atoms. Carbon monoxide reacts with oxygen to form carbon dioxide according to the equation:
$2 \ce{CO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) \nonumber$
Two compounds may also react to form a more complex compound. A very common example is the reactions of oxides with water. Calcium oxide reacts readily with water to produce an aqueous solution of calcium hydroxide:
$\ce{CaO} \left( s \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{Ca(OH)_2} \left( aq \right) \nonumber$
Sulfur trioxide gas reacts with water to form sulfuric acid. This is an unfortunately common reaction that occurs in the atmosphere in some places where oxides of sulfur are present as pollutants. The acid formed in the reaction falls to the ground as acid rain.
$\ce{SO_3} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_2SO_4} \left( aq \right) \nonumber$
Exercise $1$
1. Write the chemical equation for the synthesis of silver bromide, $\ce{AgBr}$.
2. Predict the products for the following reaction: $\ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)$
Answer a:
$2 \ce{Ag} + \ce{Br_2} \rightarrow 2 \ce{AgBr}$
Answer b:
$\ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_2CO_3}$
Decomposition Reactions
A decomposition reaction is a reaction in which a compound breaks down into two or more simpler substances. The general form of a decomposition reaction is:
$\ce{AB} \rightarrow \ce{A} + \ce{B} \nonumber$
Most decomposition reactions require an input of energy in the form of heat, light, or electricity.
Binary compounds are compounds composed of just two elements. The simplest kind of decomposition reaction is when a binary compound decomposes into its elements. Mercury (II) oxide, a red solid, decomposes when heated to produce mercury and oxygen gas:
$2 \ce{HgO} \left( s \right) \rightarrow 2 \ce{Hg} \left( l \right) + \ce{O_2} \left( g \right) \nonumber$
Video $2$: Mercury (II) oxide is a red solid. When it is heated, it decomposes into mercury metal and oxygen gas.
A reaction is also considered to be a decomposition reaction even when one or more of the products are still compounds. A metal carbonate decomposes into a metal oxide and carbon dioxide gas. For example, calcium carbonate decomposes into calcium oxide and carbon dioxide:
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \nonumber$
Metal hydroxides decompose on heating to yield metal oxides and water. Sodium hydroxide decomposes to produce sodium oxide and water:
$2 \ce{NaOH} \left( s \right) \rightarrow \ce{Na_2O} \left( s \right) + \ce{H_2O} \left( g \right) \nonumber$
Some unstable acids decompose to produce nonmetal oxides and water. Carbonic acid decomposes easily at room temperature into carbon dioxide and water:
$\ce{H_2CO_3} \left( aq \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber$
Example $2$: Electrolysis of Water
When an electric current is passed through pure water, it decomposes into its elements. Write a balanced equation for the decomposition of water.
Solution
Solutions to Example 7.10.2
Steps Example Solution
Plan the problem. Water is a binary compound composed of hydrogen and oxygen. The hydrogen and oxygen gases produced in the reaction are both diatomic molecules.
Solve.
The skeleton (unbalanced) equation:
$\ce{H_2O} \left( l \right) \overset{\text{elec}}{\rightarrow} \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \nonumber$
Note the abbreviation "$\text{elec}$" above the arrow to indicate the passage of an electric current to initiate the reaction. Balance the equation.
$2 \ce{H_2O} \left( l \right) \overset{\text{elec}}{\rightarrow} 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \nonumber$
Think about your result. The products are elements and the equation is balanced.
Exercise $2$
Write the chemical equation for the decomposition of:
1. $\ce{Al_2O_3}$
2. $\ce{Ag_2S}$
Answer a
$2 \ce{Al_2O_3} \rightarrow 4 \ce{Al} + 3 \ce{O_2}$
Answer b
$\ce{Ag_2S} \rightarrow 2 \ce{Ag} + \ce{S}$
Single Replacement Reactions
A third type of reaction is the single replacement reaction, in which one element replaces a similar element in a compound. The general form of a single-replacement (also called single-displacement) reaction is:
$\ce{A} + \ce{BC} \rightarrow \ce{AC} + \ce{B} \nonumber$
In this general reaction, element $\ce{A}$ is a metal and replaces element $\ce{B}$, also a metal, in the compound. When the element that is doing the replacing is a nonmetal, it must replace another nonmetal in a compound, and the general equation becomes:
$\ce{Y} + \ce{XZ} \rightarrow \ce{XY} + \ce{Z} \nonumber$
where $\ce{Y}$ is a nonmetal and replaces the nonmetal $\ce{Z}$ in the compound with $\ce{X}$.
Metal Replacement
Magnesium is a more reactive metal than copper. When a strip of magnesium metal is placed in an aqueous solution of copper (II) nitrate, it replaces the copper. The products of the reaction are aqueous magnesium nitrate and solid copper metal.
$\ce{Mg} \left( s \right) + \ce{Cu(NO_3)_2} \left( aq \right) \rightarrow \ce{Mg(NO_3)_2} \left( aq \right) + \ce{Cu} \left( s \right) \nonumber$
This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (copper).
Hydrogen Replacement
Many metals react easily with acids and when they do so, one of the products of the reaction is hydrogen gas. Zinc reacts with hydrochloric acid to produce aqueous zinc chloride and hydrogen (figure below).
$\ce{Zn} \left( s \right) + 2 \ce{HCl} \left( aq \right) \rightarrow \ce{ZnCl_2} \left( aq \right) + \ce{H_2} \left( g \right) \nonumber$
In a hydrogen replacement reaction, the hydrogen in the acid is replaced by an active metal. Some metals are so reactive that they are capable of replacing the hydrogen in water. The products of such a reaction are the metal hydroxide and hydrogen gas. All Group 1 metals undergo this type of reaction. Sodium reacts vigorously with water to produce aqueous sodium hydroxide and hydrogen (see figure below).
$2 \ce{Na} \left( s \right) + 2 \ce{H_2O} \left( l \right) \rightarrow 2 \ce{NaOH} \left( aq \right) + \ce{H_2} \left( g \right) \nonumber$
Halogen Replacement
The element chlorine reacts with an aqueous solution of sodium bromide to produce aqueous sodium chloride and elemental bromine:
$\ce{Cl_2} \left( g \right) + 2 \ce{NaBr} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{Br_2} \left( l \right) \nonumber$
The reactivity of the halogen group (group 17) decreases from top to bottom within the group. Fluorine is the most reactive halogen, while iodine is the least. Since chlorine is above bromine, it is more reactive than bromine and can replace it in a halogen replacement reaction.
Example $3$
What are the products of the reaction between solid aluminum ($\ce{Al}$) and iron (III) oxide ($\ce{Fe_2O_3}$)?
Solution
Solutions to Example 7.10.3
Steps Example Solution
Plan the problem. To predict the products, we need to know that aluminum will replace iron and form aluminum oxide (the metal will replace the metal ion in the compound). Aluminum has a charge of $+3$ and oxygen has a charge of $-2$. The compound formed between aluminum and oxygen, therefore, will be $\ce{Al_2O_3}$. Since iron is replaced in the compound by aluminum, the iron will now be the single element in the products.
Solve.
The unbalanced equation will be:
$\ce{Al} + \ce{Fe_2O_3} \rightarrow \ce{Al_2O_3} + \ce{Fe} \nonumber$
and the balanced equation will be:
$2 \ce{Al} + \ce{Fe_2O_3} \rightarrow \ce{Al_2O_3} + 2 \ce{Fe} \nonumber$
Think about your result. This is a single replacement reaction, and when balanced the coefficients accurately reflect that the iron and aluminum have the same charge in this reaction.
Exercise $3$
1. Write the chemical equation for the single replacement reaction between zinc solid and lead (II) nitrate solution to produce zinc nitrate solution and solid lead. (Note that zinc forms ions with a $+2$ charge.)
2. Predict the products for the following reaction: $\ce{Fe} + \ce{CuSO_4}$. (In this reaction, assume iron forms ions with a $+2$ charge.)
Answer a
$\ce{Zn} + \ce{Pb(NO_3)_2} \rightarrow \ce{Pb} + \ce{Zn(NO_3)_2}$
Answer b
$\ce{Fe} + \ce{CuSO_4} \rightarrow \ce{Cu} + \ce{FeSO_4}$
Double Replacement Reactions
A double-replacement reaction is a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds. The general form of a double-replacement (also called double-displacement) reaction is:
$\ce{AB} + \ce{CD} \rightarrow \ce{AD} + \ce{BC} \nonumber$
In this reaction, $\ce{A}$ and $\ce{C}$ are positively-charged cations, while $\ce{B}$ and $\ce{D}$ are negatively-charged anions. Double-replacement reactions generally occur between substances in aqueous solution. In order for a reaction to occur, one of the products is usually a solid precipitate, a gas, or a molecular compound such as water.
Formation of a Precipitate
A precipitate forms in a double-replacement reaction when the cations from one of the reactants combine with the anions from the other reactant to form an insoluble ionic compound. When aqueous solutions of potassium iodide and lead (II) nitrate are mixed, the following reaction occurs:
$2 \ce{KI} \left( aq \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{KNO_3} \left( aq \right) + \ce{PbI_2} \left( s \right) \label{eq10}$
There are very strong attractive forces that occur between $\ce{Pb^{2+}}$ and $\ce{I^-}$ ions and the result is a brilliant yellow precipitate (Figure $3$). The other product of the reaction, potassium nitrate, remains soluble.
Formation of a Gas
Some double-replacement reactions produce a gaseous product which then bubbles out of the solution and escapes into the air. When solutions of sodium sulfide and hydrochloric acid are mixed, the products of the reaction are aqueous sodium chloride and hydrogen sulfide gas:
$\ce{Na_2S} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{H_2S} \left( g \right) \nonumber$
Formation of a Molecular Compound
Another kind of double-replacement reaction is one that produces a molecular compound as one of its products. Many examples in this category are reactions that produce water. When aqueous hydrochloric acid is reacted with aqueous sodium hydroxide, the products are aqueous sodium chloride and water:
$\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{NaCl} \left( aq \right) + \ce{H_2O} \left( l \right) \nonumber$
Example $4$
Write a complete and balanced chemical equation for the double-replacement reaction $\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow$ (hydrogen cyanide gas is formed).
Solution
Solutions to Example 7.10.4
Steps Example Solution
Plan the problem. The production of a gas drives the reaction.
Solve.
The cations of both reactants are $+1$ charged ions, while the anions are $-1$ charged ions. After exchanging partners, the balanced equation is:
$\ce{NaCN} \left( aq \right) + \ce{HBr} \left( aq \right) \rightarrow \ce{NaBr} \left( aq \right) + \ce{HCN} \left( g \right) \nonumber$
Think about your result. This is a double replacement reaction. All formulas are correct and the equation is balanced.
Exercise $4$
Write a complete and balanced chemical equation for the double-replacement reaction $\ce{(NH_4)_2SO_4} \left( aq \right) + \ce{Ba(NO_3)_2} \left( aq \right) \rightarrow$ (a precipitate of barium sulfate forms).
Answer a:
$\ce{(NH_4)_2SO_4} \left( aq \right) + \ce{Ba(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{NH_4NO_3} \left( aq \right) + \ce{BaSO_4} \left( s \right) \nonumber$
Occasionally, a reaction will produce both a gas and a molecular compound. The reaction of a sodium carbonate solution with hydrochloric acid produces aqueous sodium chloride, carbon dioxide gas, and water:
$\ce{Na_2CO_3} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right) \nonumber$ | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.10%3A_Classifying_Chemical_Reactions.txt |
Learning Objectives
• Use the activity series to predict if a reaction will occur.
We see below two metals that can be exposed to water. The picture on the left is of sodium, which has a violent reaction when it comes in contact with water. The picture on the right is of silver, a metal so unreactive with water that it can be made into drinking vessels. Both metals have a single $s$ electron in their outer shell, so you would predict a similar reactivity from each. However, we have a tool that allows us to make better predictions about how certain elements will react with others.
The Activity Series
Single-replacement reactions only occur when the element that is doing the replacing is more reactive than the element that is being replaced. Therefore, it is useful to have a list of elements in order of their relative reactivity. The activity series is a list of elements in decreasing order of their reactivity. Since metals replace other metals, while nonmetals replace other nonmetals, they each have a separate activity series. The table $1$ below is an activity series of most common metals, and the table $2$ is an activity series of the halogens.
Table $1$: Activity Series of Metal Elements
Elements, from most to least reactive Reaction Occurring
$\ce{Li}$
$\ce{K}$
$\ce{Ba}$
$\ce{Sr}$
$\ce{Ca}$
$\ce{Na}$
React with cold water, replacing hydrogen.
$\ce{Mg}$
$\ce{Al}$
$\ce{Zn}$
$\ce{Cr}$
$\ce{Fe}$
$\ce{Cd}$
React with steam, but not cold water, replacing hydrogen.
$\ce{Co}$
$\ce{Ni}$
$\ce{Sn}$
$\ce{Pb}$
Do not react with water. React with acids, replacing hydrogen.
$\ce{H_2}$
$\ce{Cu}$
$\ce{Hg}$
$\ce{Ag}$
$\ce{Pt}$
$\ce{Au}$
Unreactive with water or acids.
Table $2$: Activity Series of Nonmetal elements
Elements, from most to least reactive
$\ce{F_2}$
$\ce{Cl_2}$
$\ce{Br_2}$
$\ce{I_2}$
For a single-replacement reaction, a given element is capable of replacing an element that is below it in the activity series. This can be used to predict if a reaction will occur. Suppose that small pieces of the metal nickel were placed into two separate aqueous solutions: one of iron (III) nitrate and one of lead (II) nitrate. Looking at the activity series, we see that nickel is below iron, but above lead. Therefore, the nickel metal will be capable of replacing the lead in a reaction, but will not be capable of replacing iron.
$\ce{Ni} \left( s \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow \ce{Ni(NO_3)_2} \left( aq \right) + \ce{Pb} \left( s \right) \nonumber$
$\ce{Ni} \left( s \right) + \ce{Fe(NO_3)_3} \left( aq \right) \rightarrow \text{NR (no reaction)} \nonumber$
In the descriptions that accompany the activity series of metals, a given metal is also capable of undergoing the reactions described below that section. For example, lithium will react with cold water, replacing hydrogen. It will also react with steam and with acids, since that requires a lower degree of reactivity.
Examples $1$
Use the activity series to predict if the following reactions will occur. If not, write $\text{NR}$. If the reaction does occur, write the products of the reaction and balance the equation.
1. $\ce{Al} \left( s \right) + \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow$
2. $\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow$
Solution
Solutions to Example 7.11.1
Steps
Example $\PageIndex{1A}$
$\ce{Al} \left( s \right) + \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow$
Example $\PageIndex{1B}$
$\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow$
Plan the problem. Compare the placements of aluminum and zinc on the activity series (Table $1$) Compare the placements of silver and hydrogen (Table $1$)
Solve.
Since aluminum is above zinc, it is capable of replacing it and a reaction will occur. The products of the reaction will be aqueous aluminum nitrate and solid zinc. Take care to write the correct formulas for the products before balancing the equation. Aluminum adopts a $+3$ charge in an ionic compound, so the formula for aluminum nitrate is $\ce{Al(NO_3)_3}$. The balanced equation is:
$2 \ce{Al} \left( s \right) + 3 \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{Al(NO_3)_3} \left( aq \right) + 3 \ce{Zn} \left( s \right)$
Since silver is below hydrogen, it is not capable of replacing hydrogen in a reaction with an acid.
$\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow \text{NR}$
Exercise $1$
Use the activity series to predict the products, if any, of each equation.
1. $\ce{FeCl2 + Zn →}$
2. $\ce{HNO3 + Au →}$
Answer a
The products are ZnCl2 + Fe.
Answer b
No reaction.
Summary
• Metals and halogens are ranked according to their ability to displace other metals or halogens below them in the activity series. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/07%3A_Chemical_Reactions/7.11%3A_The_Activity_Series-_Predicting_Spontaneous_Redox_Reactions.txt |
So far, we have talked about chemical reactions in terms of individual atoms and molecules. Although this works, most of the reactions occurring around us involve much larger amounts of chemicals. Even a tiny sample of a substance will contain millions, billions, or a hundred billion billions of atoms and molecules. How do we compare amounts of substances to each other, in chemical terms, when it is so difficult to count to a hundred billion billion? Actually, there are ways to do this, which we will explore in this chapter. In doing so, we will increase our understanding of stoichiometry, which is the study of the numerical relationships between the reactants and the products in a balanced chemical reaction.
• 8.1: Climate Change - Too Much Carbon Dioxide
Carbon dioxide (CO2) is the primary greenhouse gas emitted through human activities. In 2015, CO2 accounted for about 82.2% of all U.S. greenhouse gas emissions from human activities. Carbon dioxide is naturally present in the atmosphere as part of the Earth's carbon cycle (the natural circulation of carbon among the atmosphere, oceans, soil, plants, and animals).
• 8.2: Making Pancakes- Relationships Between Ingredients
• 8.3: Making Molecules- Mole-to-Mole Conversions
Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). In this section, we will extend the meaning of the coefficients in a chemical equation.
• 8.4: Making Molecules- Mass-to-Mass Conversions
We have used balanced equations to set up ratios, in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions—such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that relation to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles.
• 8.5: Stoichiometry
Chemical equations also provide us with the relative number of particles and moles that react to form products. In this section, you will explore the quantitative relationships that exist between the quantities of reactants and products in a balanced equation. This is known as stoichiometry. Stoichiometry, by definition, is the calculation of the quantities of reactants or products in a chemical reaction using the relationships found in the balanced chemical equation.
• 8.6: Limiting Reactant and Theoretical Yield
In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities, with none of the reactants left over at the end of the reaction. Often reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely, but will be left over when the reaction is completed.
• 8.7: Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Masses of Reactants
Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield. The limiting reagent is that reactant that produces the least amount of product. Mass-mass calculations can determine how much product is produced and how much of the other reactants remain.
• 8.8: Enthalpy Change is a Measure of the Heat Evolved or Absorbed
A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter q. The sign of q for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat is released by the system.
08: Quantities in Chemical Reactions
Carbon dioxide (CO2) is an important heat-trapping (greenhouse) gas, which is released through human activities such as deforestation and burning fossil fuels, as well as natural processes such as respiration and volcanic eruptions. Figure $1$ shows CO2 levels during the last three glacial cycles, as reconstructed from ice cores.
Carbon dioxide ($\ce{CO2}$) is the primary greenhouse gas emitted through human activities. In 2015, $\ce{CO2}$ accounted for about 82.2% of all U.S. greenhouse gas emissions from human activities. Carbon dioxide is naturally present in the atmosphere as part of the Earth's carbon cycle (the natural circulation of carbon among the atmosphere, oceans, soil, plants, and animals). Human activities are altering the carbon cycle, both by adding more $\ce{CO2}$ to the atmosphere and by influencing the ability of natural sinks, like forests, to remove $\ce{CO2}$ from the atmosphere. While $\ce{CO2}$ emissions come from a variety of natural sources, human-related emissions are responsible for the increase that has occurred in the atmosphere since the industrial revolution.
The main human activity that emits $\ce{CO2}$ is the combustion of fossil fuels (coal, natural gas, and oil) for energy and transportation, although certain industrial processes and land-use changes also emit $\ce{CO2}$. As an example of how $\ce{CO2}$ can be generated, consider the combustion of octane, a component of gasoline:
$\ce{2C8H18 (l) + 21O2 (g) \rightarrow 16CO2 (g) + 18 H2O (g)} \label{eq1}$
The balanced reaction in Equation \ref{eq1} demonstrates that for every two molecules of octane that are burned, 16 molecules of $\ce{CO2}$ are generated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/08%3A_Quantities_in_Chemical_Reactions/8.01%3A_Climate_Change_-_Too_Much_Carbon_Dioxide.txt |
/;Learning Objectives
• Use a balanced chemical equation to determine molar relationships between substances.
Previously, you learned to balance chemical equations by comparing the numbers of each type of atom in the reactants and products. The coefficients in front of the chemical formulas represent the numbers of molecules or formula units (depending on the type of substance). As follows, we will extend the meaning of the coefficients in a chemical equation.
Consider the simple chemical equation:
$2H_2 + O_2 → 2H_2O \nonumber$
The convention for writing balanced chemical equations is to use the lowest whole-number ratio for the coefficients. However, the equation is balanced as long as the coefficients are in a 2:1:2 ratio. For example, this equation is also balanced if we write it as
$4H_2 + 2O_2 → 4H_2O \nonumber$
The ratio of the coefficients is 4:2:4, which reduces to 2:1:2. The equation is also balanced if we were to write it as
$22H_2 + 11O_2 → 22H_2O \nonumber$
because 22:11:22 also reduces to 2:1:2.
Suppose we want to use larger numbers. Consider the following coefficients:
$12.044 \times 10^{23}\; \ce{H_2} + 6.022 \times 10^{23}\; \ce{O_2} → 12.044 \times 10^{23}\; \ce{H_2O} \nonumber$
These coefficients also have the ratio 2:1:2 (check it and see), so this equation is balanced. But 6.022 × 1023 is 1 mol, while 12.044 × 1023 is 2 mol (and the number is written that way to make this more obvious), so we can simplify this version of the equation by writing it as
$2 \;mol\; \ce{H_2} + 1\; mol\; \ce{O_2} → 2 \;mol\; \ce{H_2O} \nonumber$
We can leave out the word mol and not write the 1 coefficient (as is our habit), so the final form of the equation, still balanced, is
$\ce{2H_2 + O_2 → 2H_2O} \nonumber$
Now we interpret the coefficients as referring to molar amounts, not individual molecules. The lesson? Balanced chemical equations are balanced not only at the molecular level, but also in terms of molar amounts of reactants and products. Thus, we can read this reaction as “two moles of hydrogen react with one mole of oxygen to produce two moles of water.”
By the same token, the ratios we constructed to describe a molecular reaction can also be constructed in terms of moles rather than molecules. For the reaction in which hydrogen and oxygen combine to make water, for example, we can construct the following ratios:
$\mathrm{\dfrac{2\: mol\: H_2}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2}} \nonumber$
$\mathrm{\dfrac{2\: mol\: H_2O}{1\: mol\: O_2}\: or\: \dfrac{1\: mol\: O_2}{2\: mol\: H_2O}} \nonumber$
$\mathrm{\dfrac{2\: mol\: H_2}{2\: mol\: H_2O}\: or\: \dfrac{2\: mol\: H_2O}{2\: mol\: H_2}} \nonumber$
We can use these ratios to determine what amount of a substance, in moles, will react with or produce a given number of moles of a different substance. The study of the numerical relationships between the reactants and the products in balanced chemical reactions is called stoichiometry.
Example $1$
How many moles of oxygen react with hydrogen to produce 27.6 mol of $\ce{H2O}$?
Solution
Solutions to Example 8.3.1
Steps for Problem Solving How many moles of oxygen react with hydrogen to produce 27.6 mol of $\ce{H2O}$?
Find a balanced equation that describes the reaction.
Unbalanced: H2 + O2 → H2O
Balanced: 2H2 + O22H2O
Identify the "given" information and what the problem is asking you to "find." Given: moles H2O Find: moles oxygen
List other known quantities. 1 mol O2 = 2 mol H2O
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate.
$\mathrm{\cancel{27.6\: mol\: H_2O}\times\dfrac{1\: mol\: O_2}{\cancel{2\: mol\: H_2O}}=13.8\: mol\: O_2}$
To produce 27.6 mol of H2O, 13.8 mol of O2 react.
Think about your result. Since each mole of oxygen produces twice as many moles of water, it makes sense that the produced amount is greater than the reactant amount
Example $2$
How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?
Solution
Solutions to Example 8.3.2
Steps for Problem Solving How many moles of ammonia are produced if 4.20 moles of hydrogen are reacted with an excess of nitrogen?
Find a balanced equation that describes the reaction.
Unbalanced: N2 + H2 → NH3
Balanced: N2 + 3H2 2NH3
Identify the "given" information and what the problem is asking you to "find."
Given: $\ce{H_2} = 4.20 \: \text{mol}$
Find: $\text{mol}$ of $\ce{NH_3}$
List other known quantities. 3 mol H2 = 2 mol NH3
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate.
$\cancel{4.20 \: \text{mol} \: H_2} \times \dfrac{2 \: \text{mol} \: NH_3}{\cancel{3 \: \text{mol} \: H_2}} = 2.80 \: \text{mol} \: NH_3$
The reaction of $4.20 \: \text{mol}$ of hydrogen with excess nitrogen produces $2.80 \: \text{mol}$ of ammonia.
Think about your result. The result corresponds to the 3:2 ratio of hydrogen to ammonia from the balanced equation.
Exercise $3$
1. Given the following balanced chemical equation: $\ce{C5H12 + 8O2 → 5CO2 + 6H2O} \nonumber$, How many moles of $\ce{H2O}$ can be formed if 0.0652 mol of $\ce{C_{5}H_{12}}$ were to react?
2. Balance the following unbalanced equation and determine how many moles of $\ce{H2O}$ are produced when 1.65 mol of NH3 react: $\ce{NH3 + O2 → N2 + H2O} \nonumber$
Answer a
0.391 mol H2O
Answer b
4NH3 + 3O2 → 2N2 + 6H2O; 2.48 mol H2O
Summary
• The balanced chemical reaction can be used to determine molar relationships between substances. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/08%3A_Quantities_in_Chemical_Reactions/8.03%3A_Making_Molecules-_Mole-to-Mole_Conversions.txt |
Learning Objectives
• Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.
Mole to Mass Conversions
We have established that a balanced chemical equation is balanced in terms of moles, as well as atoms or molecules. We have used balanced equations to set up ratios, in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions—such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that relation to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:
As an example, consider the balanced chemical equation
$\ce{Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3} \label{Eq1}$
If we have 3.59 mol of $\ce{Fe2O3}$, how many grams of $\ce{SO3}$ can react with it? Using the mole-mass calculation sequence, we can determine the required mass of $\ce{SO3}$ in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of $\ce{SO3}$ needed. Then, using the molar mass of $\ce{SO3}$ as a conversion factor, we determine the mass that this number of moles of $\ce{SO3}$ has.
As usual, we start with the quantity we were given:
$\mathrm{3.59\: \cancel{ mol\: Fe_2O_3 } \times \left( \dfrac{3\: mol\: SO_3}{1\: \cancel{ mol\: Fe_2O_3}} \right) =10.77\: mol\: SO_3} \label{Eq2}$
The mol $\ce{Fe2O3}$ units cancel, leaving mol $\ce{SO3}$ unit. Now, we take this answer and convert it to grams of $\ce{SO3}$, using the molar mass of $\ce{SO3}$ as the conversion factor:
$\mathrm{10.77\: \bcancel{mol\: SO_3} \times \left( \dfrac{80.06\: g\: SO_3}{1\: \bcancel{ mol\: SO_3}} \right) =862\: g\: SO_3} \label{Eq3}$
Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of $\ce{SO3}$ will react with 3.59 mol of $\ce{Fe2O3}$. Many problems of this type can be answered in this manner.
The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:
$3.59 \cancel{\, mol \, Fe_2O_3} \times \underbrace{\left( \dfrac{ 3 \bcancel{ \, mol\, SO_3}}{ 1 \cancel{\, mol\, Fe_2O_3}} \right)}_{\text{converts to moles of SO}_3} \times \underbrace{ \left( \dfrac{ 80.06 {\, g \, SO_3}}{ 1 \, \bcancel{ mol\, SO_3}} \right)}_{\text{converts to grams of SO}_3} = 862\, g\, SO_3 \nonumber$
We get exactly the same answer when combining all math steps together.
Example $1$: Generation of Aluminum Oxide
How many moles of $\ce{HCl}$ will be produced when 249 g of $\ce{AlCl3}$ are reacted according to this chemical equation?
$\ce{2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g)} \nonumber$
Solution
Solutions to Example 8.5.1
Steps for Problem Solving Example $1$
Identify the "given" information and what the problem is asking you to "find." Given: 249 g AlCl3
Find: moles HCl
List other known quantities. 1 mol AlCl3 = 133.33 g AlCl3
6 mol of HCl to 2 mol AlCl3
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $249\, \cancel{g\, AlCl_{3}}\times \dfrac{1\, \cancel{mol\, AlCl_{3}}}{133.33\, \cancel{g\, AlCl_{3}}}\times \dfrac{6\, mol\, HCl}{2\, \cancel{mol\, AlCl_{3}}}=5.60\, mol\, HCl$
Think about your result. Since 249 g of AlCl3 is less than 266.66 g, the mass for 2 moles of AlCl3 and the relationship is 6 mol of HCl to 2 mol AlCl3 , the answer should be less than 6 moles of HCl.
Exercise $1$: Generation of Aluminum Oxide
How many moles of $\ce{Al2O3}$ will be produced when 23.9 g of $\ce{H2O}$ are reacted according to this chemical equation?
$\ce{2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g)} \nonumber$
Answer
0.442 mol Al2O3
Mass to Mass Conversions
It is a small step from mole-mass calculations to mass-mass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:
This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques. Flowchart of mole mass calculations: To convert from grams to moles of substance A, use molar mass conversion factor; To convert from moles of substance A to moles of substance B, use the mole ratio conversion factor, and to convert from moles to grams of substance B, use molar mass conversion factor
Example $2$: Decomposition of Ammonium Nitrate
Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.
$\ce{NH_4NO_3} \left( s \right) \rightarrow \ce{N_2O} \left( g \right) + 2 \ce{H_2O} \left( l \right) \nonumber$
In a certain experiment, $45.7 \: \text{g}$ of ammonium nitrate is decomposed. Find the mass of each of the products formed.
Solutions to Example 8.5.2
Steps for Problem Solving Example $2$
Identify the "given" information and what the problem is asking you to "find."
Given: $45.7 \: \text{g} \: \ce{NH_4NO_3}$
Find:
Mass $\ce{N_2O} = ? \: \text{g}$
Mass $\ce{H_2O} = ? \: \text{g}$
List other known quantities.
1 mol $\ce{NH_4NO_3} = 80.06 \: \text{g}$
1 mol $\ce{N_2O} = 44.02 \: \text{g}$
1 mol $\ce{H_2O} = 18.02 \: \text{g}$
1 mol NH4NO3 to 1 mol N2O to 2 mol H2O
Prepare two concept maps and use the proper conversion factor.
Cancel units and calculate.
$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \dfrac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \dfrac{1 \: \text{mol} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \dfrac{44.02 \: \text{g} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{N_2O}} = 25.1 \: \text{g} \: \ce{N_2O}$
$45.7 \: \text{g} \: \ce{NH_4NO_3} \times \dfrac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \dfrac{2 \: \text{mol} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \dfrac{18.02 \: \text{g} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} = 20.6 \: \text{g} \: \ce{H_2O}$
Think about your result. The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures.
Exercise $2$: Carbon Tetrachloride
Methane can react with elemental chlorine to make carbon tetrachloride ($\ce{CCl_4}$). The balanced chemical equation is as follows:
$\ce{CH4 (g) + 4 Cl2 (g) → CCl2 (l) + 4 HCl (l) } \nonumber$
How many grams of $\ce{HCl}$ are produced by the reaction of 100.0 g of $\ce{CH4}$?
Answer
908.7g HCl
Summary
• Calculations involving conversions between moles of a substance and the mass of that substance can be done using conversion factors.
• A balanced chemical reaction can be used to determine molar and mass relationships between substances. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Making_Molecules-_Mass-to-Mass_Conversions.txt |
Learning Objectives
• Explain the meaning of the term "stoichiometry".
• Determine the relative amounts of each substance in chemical equations.
You have learned that chemical equations provide us with information about the types of particles that react to form products. Chemical equations also provide us with the relative number of particles and moles that react to form products. In this section you will explore the quantitative relationships that exist between the quantities of reactants and products in a balanced equation. This is known as stoichiometry.
Stoichiometry, by definition, is the calculation of the quantities of reactants or products in a chemical reaction using the relationships found in the balanced chemical equation. The word stoichiometry is actually Greek from two words: $\sigma \tau \omicron \iota \kappa \eta \iota \omicron \nu$, which means "element", and $\mu \epsilon \tau \rho \omicron \nu), which means "measure". Interpreting Chemical Equations The mole, as you remember, is a quantitative measure that is equivalent to Avogadro's number of particles. So how does this relate to the chemical equation? Look at the chemical equation below. $2 \ce{CuSO_4} + 4 \ce{KI} \rightarrow 2 \ce{CuI} + 2 \ce{K_2SO_4} + \ce{I_2} \nonumber$ The coefficients used, as we have learned, tell us the relative amounts of each substance in the equation. So for every 2 units of copper (II) sulfate (\(\ce{CuSO_4}$) we have, we need to have 4 units of potassium iodide ($\ce{KI}$). For every two dozen copper (II) sulfates, we need 4 dozen potassium iodides. Because the unit "mole" is also a counting unit, we can interpret this equation in terms of moles, as well: For every two moles of copper (II) sulfate, we need 4 moles potassium iodide.
The production of ammonia $\left( \ce{NH_3} \right)$ from nitrogen and hydrogen gases is an important industrial reaction called the Haber process, after German chemist Fritz Haber.
$\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right) \nonumber$
The balanced equation can be analyzed in several ways, as shown in the figure below.
We see that 1 molecule of nitrogen reacts with 3 molecules of hydrogen to form 2 molecules of ammonia. This is the smallest possible relative amount of the reactants and products. To consider larger relative amounts, each coefficient can be multiplied by the same number. For example, 10 molecules of nitrogen would react with 30 molecules of hydrogen to produce 20 molecules of ammonia.
The most useful quantity for counting particles is the mole. So if each coefficient is multiplied by a mole, the balanced chemical equation tells us that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. This is the conventional way to interpret any balanced chemical equation.
Finally, if each mole quantity is converted to grams by using the molar mass, we can see that the law of conservation of mass is followed. $1 \: \ce{mol}$ of nitrogen has a mass of $28.02 \: \text{g}$, while $3 \: \text{mol}$ of hydrogen has a mass of $6.06 \: \text{g}$, and $2 \: \text{mol}$ of ammonia has a mass of $34.08 \: \text{g}$.
$28.02 \: \text{g} \: \ce{N_2} + 6.06 \: \text{g} \: \ce{H_2} \rightarrow 34.08 \: \text{g} \: \ce{NH_3} \nonumber$
Mass and the number of atoms must be conserved in any chemical reaction. The number of molecules is not necessarily conserved.
Example $1$
The equation for the combustion of ethane ($\ce{C_2H_6}$) is
$2 \ce{C_2H_6} + 7 \ce{O_2} \rightarrow 4 \ce{CO_2} + 6 \ce{H_2O} \nonumber$
1. Indicate the number of formula units or molecules in the balanced equation.
2. Indicate the number of moles present in the balanced equation.
Solution
1. Two molecules of $\ce{C_2H_6}$ plus seven molecules of $\ce{O_2}$ yields four molecules of $\ce{CO_2}$ plus six molecules of $\ce{H_2O}$.
2. Two moles of $\ce{C_2H_6}$ plus seven moles of $\ce{O_2}$ yields four moles of $\ce{CO_2}$ plus six moles of $\ce{H_2O}$.
Exercise $1$
For the following equation below, indicate the number of formula units or molecules, and the number of moles present in the balanced equation.
$\ce{KBrO_3} + 6 \ce{KI} + 6 \ce{HBr} \rightarrow 7 \ce{KBr} + 3 \ce{H_2O} \nonumber$
Answer
One molecules of $\ce{KBrO_3}$ plus six molecules of $\ce{KI}$ plus six molecules of $\ce{HBr}$ yields seven molecules of $\ce{KBr}$ plus three molecules of $\ce{I_2}$ and three molecules of $\ce{H_2O}$. One mole of $\ce{KBrO_3}$ plus six moles of $\ce{KI}$ plus six moles of $\ce{HBr}$ yields seven moles of $\ce{KBr}$ plus three moles of $\ce{I_2}$ plus three moles of $\ce{H_2O}$.
Summary
• Stoichiometry is the calculation of the quantities of reactants or products in a chemical reaction using the relationships found in a balanced chemical equation.
• The coefficients in a balanced chemical equation represent the reacting ratios of the substances in the reaction.
• The coefficients of a balanced equation can be used to determine the ratio of moles of all substances in the reaction.
Vocabulary
• Stoichiometry - The calculation of quantitative relationships of the reactants and products in a balanced chemical equation.
• Formula unit - The empirical formula of an ionic compound.
• Mole ratio - The ratio of the moles of one reactant or product to the moles of another reactant or product according to the coefficients in the balanced chemical equation. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/08%3A_Quantities_in_Chemical_Reactions/8.05%3A_Stoichiometry.txt |
Learning Objectives
• Identify the limiting reactant (limiting reagent) in a given chemical reaction.
• Calculate how much product will be produced from the limiting reactant.
• Calculate how much reactant(s) remains when the reaction is complete.
In all examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants were left over at the end of the reaction. This is often desirable—as in the case of a space shuttle—where excess oxygen or hydrogen is not only extra freight to be hauled into orbit, but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely, but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess.
Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is:
$1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{3.7.1}$
If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies.
PhET Simulation: Reactants, Products and Leftovers
View this interactive simulation illustrating the concepts of limiting and excess reactants.
Consider this concept now with regard to a chemical process, the reaction of hydrogen with chlorine to yield hydrogen chloride:
$\ce{ H2 + Cl2}(g)\rightarrow \ce{2HCl}(g) \nonumber$
The balanced equation shows that hydrogen and chlorine react in a 1:1 stoichiometric ratio. If these reactants are provided in any other amounts, one of the reactants will nearly always be entirely consumed, thus limiting the amount of product that may be generated. This substance is the limiting reactant, and the other substance is the excess reactant. Identifying the limiting and excess reactants for a given situation requires computing the molar amounts of each reactant provided and comparing them to the stoichiometric amounts represented in the balanced chemical equation. For example, imagine combining 3 moles of H2 and 2 moles of Cl2. This represents a 3:2 (or 1.5:1) ratio of hydrogen to chlorine present for reaction, which is greater than the stoichiometric ratio of 1:1. Hydrogen, therefore, is present in excess, and chlorine is the limiting reactant. Reaction of all the provided chlorine (2 mol) will consume 2 mol of the 3 mol of hydrogen provided, leaving 1 mol of hydrogen non-reacted.
An alternative approach to identifying the limiting reactant involves comparing the amount of product expected for the complete reaction of each reactant. Each reactant amount is used to separately calculate the amount of product that would be formed per the reaction’s stoichiometry. The reactant yielding the lesser amount of product is the limiting reactant. For the example, in the previous paragraph, complete reaction of the hydrogen would yield:
$\mathrm{mol\: HCl\: produced=3\: mol\:H_2\times \dfrac{2\: mol\: HCl}{1\: mol\:H_2}=6\: mol\: HCl} \nonumber$
Complete reaction of the provided chlorine would produce:
$\mathrm{mol\: HCl\: produced=2\: mol\:Cl_2\times \dfrac{2\: mol\: HCl}{1\: mol\:Cl_2}=4\: mol\: HCl} \nonumber$
The chlorine will be completely consumed once 4 moles of HCl have been produced. Since enough hydrogen was provided to yield 6 moles of HCl, there will be non-reacted hydrogen remaining once this reaction is complete. Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure $2$).
A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reactant; the other reactant or reactants are considered to be in excess. A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reactant and which is in excess.
How to Identify the Limiting Reactant (Limiting Reagent)
There are two ways to determine the limiting reactant. One method is to find and compare the mole ratio of the reactants used in the reaction (Approach 1). Another way is to calculate the grams of products produced from the given quantities of reactants; the reactant that produces the smallest amount of product is the limiting reactant (Approach 2). This section will focus more on the second method.
Approach 1 (The "Reactant Mole Ratio Method"): Find the limiting reactant by looking at the number of moles of each reactant.
1. Determine the balanced chemical equation for the chemical reaction.
2. Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
3. Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
4. Use the amount of limiting reactant to calculate the amount of product produced.
5. If necessary, calculate how much is left in excess of the non-limiting (excess) reactant.
Approach 2 (The "The Product Method"): Find the limiting reactant by calculating and comparing the amount of product that each reactant will produce.
1. Balance the chemical equation for the chemical reaction.
2. Convert the given information into moles.
3. Use stoichiometry for each individual reactant to find the mass of product produced.
4. The reactant that produces a lesser amount of product is the limiting reactant.
5. The reactant that produces a larger amount of product is the excess reactant.
6. To find the amount of remaining excess reactant, subtract the mass of excess reactant consumed from the total mass of excess reactant given.
The key to recognizing which reactant is the limiting reactant is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reactant. What we need to do is determine an amount of one product (either moles or mass) assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reactant. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful.
Example $1$: Identifying the Limiting Reactant
As an example, consider the balanced equation
$\ce{4 C2H3Br3 + 11 O2 \rightarrow 8 CO2 + 6 H2O + 6 Br2} \nonumber$
What is the limiting reactant if 76.4 grams of $\ce{C_2H_3Br_3}$ reacted with 49.1 grams of $\ce{O_2}$?
Solution
Using Approach 1:
Step 1: Balance the chemical equation.
The equation is already balanced with the relationship
4 mol $\ce{C2H3Br3}$ to 11 mol $\ce{O2}$ to 6 mol $\ce{H2O}$ to 6 mol $\ce{Br}$
Step 2: Convert all given information into moles.
$\mathrm{76.4\:\cancel{g \:C_2H_3Br_3} \times \dfrac{1\: mol \:C_2H_3Br_3}{266.72\:\cancel{g \:C_2H_3B_3}} = 0.286\: mol \: C_2H_3Br_3} \nonumber$
$\mathrm{49.1\: \cancel{g\: O_2} \times \dfrac{1\: mol\: O_2}{32.00\:\cancel{g\: O_2}} = 1.53\: mol\: O_2} \nonumber$
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
Assuming that all of the oxygen is used up,
$\mathrm{1.53 \: \cancel{mol O_2} \times \dfrac{4 \: mol C_2H_3Br_3 }{11 \: \cancel{mol O_2}}}$ = 0.556 mol C2H3Br3 are required.
Because 0.556 moles of C2H3Br3 required > 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reactant.
Using Approach 2:
Step 1: Balance the chemical equation.
The equation is already balanced with the relationship
4 mol C2H3Br3 to 11 mol O2 to 6 mol H2O to 6 mol Br2
Step 2 and Step 3: Convert mass to moles and stoichiometry.
$\mathrm{76.4\:\cancel{g\: C_2H_3Br_3} \times \dfrac{1\: \cancel{mol\: C_2H_3Br_3}}{266.72\:\cancel{g\: C_2H_3Br_3}} \times \dfrac{8\: \cancel{mol\: CO_2}}{4\: \cancel{mol\: C_2H_3Br_3}} \times \dfrac{44.01\:g\: CO_2}{1\: \cancel{mol\: CO_2}} = 25.2\:g\: CO_2} \nonumber$
$\mathrm{49.1\: \cancel{ g\: O_2} \times \dfrac{1\: \cancel{ mol\: O_2}}{32.00\: \cancel{ g\: O_2}} \times \dfrac{8\: \cancel{ mol\: CO_2}}{11\: \cancel{ mol\: O_2}} \times \dfrac{44.01\:g\: CO_2}{1\: \cancel{ mol\: CO_2}} = 49.1\:g\: CO_2} \nonumber$
Step 4: The reactant that produces a smaller amount of product is the limiting reactant.
Therefore, by either method, $\ce{C2H3Br3}$ is the limiting reactant.
Example $2$: Identifying the Limiting Reactant and the Mass of Excess Reactant
For example, in the reaction of magnesium metal and oxygen, calculate the mass of magnesium oxide that can be produced if 2.40 g $Mg$ reacts with 10.0 g $O_2$. Also determine the amount of excess reactant. $\ce{MgO}$ is the only product in the reaction.
Solution
Following Approach 1:
Step 1: Balance the chemical equation.
2 Mg (s) + O2 (g) 2 MgO (s)
The balanced equation provides the relationship of 2 mol Mg to 1 mol O2 to 2 mol MgO
Step 2 and Step 3: Convert mass to moles and stoichiometry.
$\mathrm{2.40\:\cancel{g\: Mg }\times \dfrac{1\: \cancel{mol\: Mg}}{24.31\:\cancel{g\: Mg}} \times \dfrac{2\: \cancel{mol\: MgO}}{2\: \cancel{mol\: Mg}} \times \dfrac{40.31\:g\: MgO}{1\: \cancel{mol\: MgO}} = 3.98\:g\: MgO} \nonumber$
$\mathrm{10.0\:\cancel{g\: O_2}\times \dfrac{1\: \cancel{mol\: O_2}}{32.00\:\cancel{g\: O_2}} \times \dfrac{2\: \cancel{mol\: MgO}}{1\:\cancel{ mol\: O_2}} \times \dfrac{40.31\:g\: MgO}{1\: \cancel{mol\: MgO}} = 25.2\: g\: MgO} \nonumber$
Step 4: The reactant that produces a smaller amount of product is the limiting reactant.
Mg produces less MgO than does O2 (3.98 g MgO vs. 25.2 g MgO), therefore Mg is the limiting reactant in this reaction.
Step 5: The reactant that produces a larger amount of product is the excess reactant.
O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O2 is the excess reactant in this reaction.
Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reactant consumed from the total mass of excess reactant given.
Mass of excess reactant calculated using the limiting reactant:
$\mathrm{2.40\: \cancel{ g\: Mg }\times \dfrac{1\: \cancel{ mol\: Mg}}{24.31\: \cancel{ g\: Mg}} \times \dfrac{1\: \cancel{ mol\: O_2}}{2\: \cancel{ mol\: Mg}} \times \dfrac{32.00\:g\: O_2}{1\: \cancel{ mol\: O_2}} = 1.58\:g\: O_2} \nonumber$
OR
Mass of excess reactant calculated using the mass of the product:
$\mathrm{3.98\: \cancel{ g\: MgO }\times \dfrac{1\: \cancel{ mol\: MgO}}{40.31\: \cancel{ g\: MgO}} \times \dfrac{1\: \cancel{ mol\: O_2}}{2\: \cancel{ mol\: MgO}} \times \dfrac{32.0\:g\: O_2}{1\: \cancel{ mol\: O_2}} = 1.58\:g\: O_2} \nonumber$
Mass of total excess reactant given – mass of excess reactant consumed in the reaction:
10.0g O2 - (available) 1.58g O2 (used) = 8.42g O2 (excess)
Example $3$: Limiting Reactant
What is the limiting reactant if 78.0 grams of Na2O2 were reacted with 29.4 grams of H2O? The unbalanced chemical equation is $\ce{Na2O2 (s) + H2O (l) → NaOH (aq) + H2O2 (l)} \nonumber$
Solution
Solutions to Example 8.4.3
Steps for Problem Solving- The Product Method Example $1$
Identify the "given" information and what the problem is asking you to "find."
Given: 78.0 grams of Na2O2
29.4 g H2O
Find: limiting reactant
List other known quantities.
1 mol Na2O2= 77.96 g/mol
1 mol H2O = 18.02 g/mol
Since the amount of product in grams is not required, only the molar mass of the reactants is needed.
Balance the equation.
Na2O2 (s) + 2H2O (l) 2NaOH (aq) + H2O2 (l)
The balanced equation provides the relationship of 1 mol Na2O2 to 2 mol H2O 2mol NaOH to 1 mol H2O2
Prepare a concept map and use the proper conversion factor.
Because the question only asks for the limiting reactant, we can perform two mass-mole calculations and determine which amount is less.
Cancel units and calculate.
$\mathrm{78.0\:g\: Na_2O_2 \times \dfrac{1\: mol\: Na_2O_2}{77.96\:g\: Na_2O_2} \times \dfrac{2\: mol\: NaOH}{1\: mol\: Na_2O_2} \times \dfrac{40\:g\: NaOH}{1\: mol\: NaOH} = 2.00\:mol\: NaOH} \nonumber$
$\mathrm{29.4\:g\: H_2O \times \dfrac{1\: mol\: H_2O}{18.02\:g\: H_2O} \times \dfrac{2\: mol\: NaOH}{2\: mol\: Na_2O_2} \times \dfrac{40\:g\: NaOH}{1\: mol\: NaOH} = 1.63\:mol\: NaOH} \nonumber$
Therefore, H2O is the limiting reactant.
Think about your result.
Example $4$: Limiting Reactant and Mass of Excess Reactant
A 5.00 g quantity of $\ce{Rb}$ is combined with 3.44 g of $\ce{MgCl2}$ according to this chemical reaction: $2R b(s) + MgCl_2(s) → Mg(s) + 2RbCl(s) \nonumber$
What mass of $\ce{Mg}$ is formed, and what mass of remaining reactant is left over?
Solution
Solutions to Example 8.4.10
Steps for Problem Solving- The Product Method Example $2$
Steps for Problem Solving
A 5.00 g quantity of Rb is combined with 3.44 g of MgCl2 according to this chemical reaction:
$2Rb(s) + MgCl_2(s) → Mg(s) + 2RbCl(s) \nonumber$
What mass of Mg is formed, and what mass of remaining reactant is left over?
Identify the "given" information and what the problem is asking you to "find."
Given: 5.00g Rb, 2.44g MgCl2
Find: mass of Mg formed, mass of remaining reactant
List other known quantities.
• molar mass: Rb = 85.47 g/mol
• molar mass: MgCl2 = 95.21 g/mol
• molar mass: Mg = 24.31 g/mol
Prepare concept maps and use the proper conversion factor.
Find mass Mg formed based on mass of Rb
Find mass of Mg formed based on mass of MgCl2
Use limiting reactant to determine amount of excess reactant consumed
Cancel units and calculate.
Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less.
$5.00\cancel{g\, Rb}\times \dfrac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \dfrac{1\cancel{mol\, Mg}}{2\cancel{mol\, Rb}}\times \dfrac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.711\, g\, Mg \nonumber$
$3.44\cancel{g\, MgCl_{2}}\times \dfrac{1\cancel{mol\, MgCl_{2}}}{95.21\cancel{g\, MgCl_{2}}}\times \dfrac{1\cancel{mol\, Mg}}{1\cancel{mol\, MgCl_{2}}}\times \dfrac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.878\, g\, Mg \nonumber$
The 0.711 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reactant. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl2 reacted with the 5.00 g of Rb, and then subtract the amount reacted from the original amount.
$5.00\cancel{g\, Rb}\times \dfrac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \dfrac{1\cancel{mol\, MgCl_{2}}}{2\cancel{mol\, Rb}}\times \dfrac{95.21\, g\, MgCl_{2}}{\cancel{1\, mol\, MgCl_{2}}}=2.78\, g\, MgCl_{2}\: \: reacted \nonumber$
Because we started with 3.44 g of MgCl2, we have
3.44 g MgCl2 − 2.78 g MgCl2 reacted = 0.66 g MgCl2 left
Think about your result. It usually is not possible to determine the limiting reactant using just the initial masses, as the reagents have different molar masses and coefficients.
Exercise $1$
Given the initial amounts listed, what is the limiting reactant, and what is the mass of the leftover reactant?
$\underbrace{22.7\, g}_{MgO(s)}+\underbrace{17.9\, g}_{H_2S}\rightarrow MgS(s)+H_{2}O(l) \nonumber$
Answer
H2S is the limiting reagent; 1.5 g of MgO are left over. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/08%3A_Quantities_in_Chemical_Reactions/8.06%3A_Limiting_Reactant_and_Theoretical_Yield.txt |
Learning Objectives
• Calculate percentage or actual yields from known amounts of reactants.
The world of pharmaceutical production is an expensive one. Many drugs have several steps in their synthesis and use costly chemicals. A great deal of research takes place to develop better ways to make drugs faster and more efficiently. Studying how much of a compound is produced in any given reaction is an important part of cost control.
Percent Yield
Chemical reactions in the real world don't always go exactly as planned on paper. In the course of an experiment, many things will contribute to the formation of less product than predicted. Besides spills and other experimental errors, there are usually losses due to an incomplete reaction, undesirable side reactions, etc. Chemists need a measurement that indicates how successful a reaction has been. This measurement is called the percent yield.
To compute the percent yield, it is first necessary to determine how much of the product should be formed based on stoichiometry. This is called the theoretical yield, the maximum amount of product that can be formed from the given amounts of reactants. The actual yield is the amount of product that is actually formed when the reaction is carried out in the laboratory. The percent yield is the ratio of the actual yield to the theoretical yield, expressed as a percentage.
$\text{Percent Yield} = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\% \nonumber$
Percent yield is very important in the manufacture of products. Much time and money is spent improving the percent yield for chemical production. When complex chemicals are synthesized by many different reactions, one step with a low percent yield can quickly cause a large waste of reactants and unnecessary expense.
Typically, percent yields are understandably less than $100\%$ because of the reasons indicated earlier. However, percent yields greater than $100\%$ are possible if the measured product of the reaction contains impurities that cause its mass to be greater than it actually would be if the product was pure. When a chemist synthesizes a desired chemical, he or she is always careful to purify the products of the reaction. Example $1$ illustrates the steps for determining percent yield.
Example $1$: Decomposition of Potassium Chlorate
Potassium chlorate decomposes upon slight heating in the presence of a catalyst, according to the reaction below:
$2 \ce{KClO_3} \left( s \right) \rightarrow 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right)\nonumber$
In a certain experiment, $40.0 \: \text{g} \: \ce{KClO_3}$ is heated until it completely decomposes. The experiment is performed and the oxygen gas is collected and its mass is found to be $14.9 \: \text{g}$.
1. What is the theoretical yield of oxygen gas?
2. What is the percent yield for the reaction?
Solution
a. Calculation of theoretical yield
First, we will calculate the theoretical yield based on the stoichiometry.
Step 1: Identify the "given" information and what the problem is asking you to "find".
Given: Mass of $\ce{KClO_3} = 40.0 \: \text{g}$
Mass of O2 collected = 14.9g
Find: Theoretical yield, g O2
Step 2: List other known quantities and plan the problem.
1 mol KClO3 = 122.55 g/mol
1 mol O2 = 32.00 g/mol
Step 4: Solve.
$40.0 \: \cancel{\text{g} \: \ce{KClO_3}} \times \dfrac{1 \: \cancel{\text{mol} \: \ce{KClO_3}}}{122.55 \: \cancel{\text{g} \: \ce{KClO_3}}} \times \dfrac{3 \: \cancel{\text{mol} \: \ce{O_2}}}{2 \: \cancel{\text{mol} \: \ce{KClO_3}}} \times \dfrac{32.00 \: \text{g} \: \ce{O_2}}{1 \: \cancel{\text{mol} \: \ce{O_2}}} = 15.7 \: \text{g} \: \ce{O_2}\nonumber$
The theoretical yield of $\ce{O_2}$ is $15.7 \: \text{g}$, 15.67 g unrounded.
Step 5: Think about your result.
The mass of oxygen gas must be less than the $40.0 \: \text{g}$ of potassium chlorate that was decomposed.
b. Calculation of percent yield
Now we will use the actual yield and the theoretical yield to calculate the percent yield.
Step 1: Identify the "given" information and what the problem is asking you to "find".
Given: Theoretical yield =15.67 g, use the un-rounded number for the calculation.
Actual yield = 14.9g
Find: Percent yield, % Yield
Step 3: Use the percent yield equation below.
$\text{Percent Yield} = \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$
Step 4: Solve.
$\text{Percent Yield} = \dfrac{14.9 \: \text{g}}{15.\underline{6}7 \: \text{g}} \times 100\% = 94.9\%$
Step 5: Think about your result.
Since the actual yield is slightly less than the theoretical yield, the percent yield is just under $100\%$.
Example $2$: Oxidation of Zinc
Upon reaction of 1.274 g of copper sulfate with excess zinc metal, 0.392 g copper metal was obtained according to the equation:
$\ce{CuSO4}(aq)+\ce{Zn}(s)\rightarrow \ce{Cu}(s)+\ce{ZnSO4}(aq) \nonumber$
What is the percent yield?
Solution
Solutions to Example 8.6.2
Steps for Problem Solving-The Product Method Example $1$
Identify the "given" information and what the problem is asking you to "find."
Given: 1.274 g CuSO4
Actual yield = 0.392 g Cu
Find: Percent yield
List other known quantities.
1 mol CuSO4= 159.62 g/mol
1 mol Cu = 63.55 g/mol
Since the amount of product in grams is not required, only the molar mass of the reactants is needed.
Balance the equation.
The chemical equation is already balanced.
The balanced equation provides the relationship of 1 mol CuSO4 to 1 mol Zn to 1 mol Cu to 1 mol ZnSO4.
Prepare a concept map and use the proper conversion factor.
The provided information identifies copper sulfate as the limiting reactant, and so the theoretical yield (g Cu) is found by performing mass-mass calculation based on the initial amount of CuSO4.
Cancel units and calculate.
$\mathrm{1.274\:\cancel{g\:Cu_SO_4}\times \dfrac{1\:\cancel{mol\:CuSO_4}}{159.62\:\cancel{g\:CuSO_4}}\times \dfrac{1\:\cancel{mol\: Cu}}{1\:\cancel{mol\:CuSO_4}}\times \dfrac{63.55\:g\: Cu}{1\:\cancel{mol\: Cu}}=0.5072\: g\: Cu}\nonumber$
Using this theoretical yield and the provided value for actual yield, the percent yield is calculated to be:
\begin{align*} \text{percent yield} &= \left( \dfrac{\text{actual yield} }{\text{theoretical yield}} \right) \times 100 \[4pt] &= \left(\dfrac{0.392\, g\, \ce{Cu}}{0.5072 \, g\, \ce{Cu}} \right) \times 100 \[4pt] &=77.3\% \end{align*}
Think about your result. Since the actual yield is slightly less than the theoretical yield, the percent yield is just under $100\%$.
Exercise $1$
What is the percent yield of a reaction that produces 12.5 g of the Freon CF2Cl2 from 32.9 g of CCl4 and excess HF?
$\ce{CCl4 + 2HF \rightarrow CF2Cl2 + 2HCl} \nonumber$
Answer
48.3%
Summary
Theoretical yield is calculated based on the stoichiometry of the chemical equation. The actual yield is experimentally determined. The percent yield is determined by calculating the ratio of actual yield to theoretical yield. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/08%3A_Quantities_in_Chemical_Reactions/8.07%3A_Limiting_Reactant_Theoretical_Yield_and_Percent_Yield_from_Initial_Masses_of_Reactants.txt |
When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The law of conservation of energy states that in any physical or chemical process, energy is neither created nor destroyed. In other words, the entire energy in the universe is conserved. In order to better understand the energy changes taking place during a reaction, we need to define two parts of the universe, called the system and the surroundings. The system is the specific portion of matter in a given space that is being studied during an experiment or an observation. The surroundings are everything in the universe that is not part of the system. In practical terms for a laboratory chemist, the system is the particular chemicals being reacted, while the surroundings is the immediate vicinity within the room. During most processes, energy is exchanged between the system and the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings.
A chemical reaction or physical change is endothermic if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter $q$. The sign of $q$ for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is exothermic if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of $q$ for an exothermic process is negative because the system is losing heat.
Enthalpy
Heat changes in chemical reactions are often measured in the laboratory under conditions in which the reacting system is open to the atmosphere. In that case, the system is at a constant pressure. Enthalpy $\left( H \right)$ is the heat content of a system at constant pressure. Chemists routinely measure changes in enthalpy of chemical systems as reactants are converted into products. The heat that is absorbed or released by a reaction at constant pressure is the same as the enthalpy change, and is given the symbol $\Delta H$. Unless otherwise specified, all reactions in this material are assumed to take place at constant pressure.
The change in enthalpy of a reaction is a measure of the differences in enthalpy of the reactants and products. The enthalpy of a system is determined by the energies needed to break chemical bonds and the energies needed to form chemical bonds. Energy needs to be put into the system in order to break chemical bonds, as they do not come apart spontaneously in most cases. Bond formation to produce products will involve release of energy. The change in enthalpy shows the trade-offs made in these two processes. Does it take more energy to break bonds than that needed to form bonds? If so, the reaction is endothermic and the enthalpy change is positive. If more energy is produced in bond formation than that needed for bond breaking, the reaction is exothermic and the enthalpy is negative.
Several factors influence the enthalpy of a system. Enthalpy is an extensive property, determined in part by the amount of material we work with. The state of reactants and products (solid, liquid, or gas) influences the enthalpy value for a system. The direction of the reaction affects the enthalpy value. A reaction that takes place in the opposite direction has the same numerical enthalpy value, but the opposite sign.
Thermochemical Equation
When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of $1 \: \text{mol}$ of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of the balanced equation:
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) + 890.4 \: \text{kJ}\nonumber$
The equation tells us that $1 \: \text{mol}$ of methane combines with $2 \: \text{mol}$ of oxygen to produce $1 \: \text{mol}$ of carbon dioxide and $2 \: \text{mol}$ of water. In the process, $890.4 \: \text{kJ}$ is released and so it is written as a product of the reaction. A thermochemical equation is a chemical equation that includes the enthalpy change of the reaction. The process in the above thermochemical equation can be shown visually in Figure $2$.
In the combustion of methane example, the enthalpy change is negative because heat is being released by the system. Therefore, the overall enthalpy of the system decreases. The heat of reaction is the enthalpy change for a chemical reaction. In the case above, the heat of reaction is $-890.4 \: \text{kJ}$. The thermochemical reaction can also be written in this way:
$\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -890.4 \: \text{kJ}\nonumber$
Heats of reaction are typically measured in kilojoules. It is important to include the physical states of the reactants and products in a thermochemical equation as the value of the $\Delta H$ depends on those states.
Endothermic reactions absorb energy from the surroundings as the reaction occurs. When $1 \: \text{mol}$ of calcium carbonate decomposes into $1 \: \text{mol}$ of calcium oxide and $1 \: \text{mol}$ of carbon dioxide, $177.8 \: \text{kJ}$ of heat is absorbed. The process is shown visually in Figure $\PageIndex{2B}$. The thermochemical reaction is shown below.
$\ce{CaCO_3} \left( s \right) + 177.8 \: \text{kJ} \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right)\nonumber$
Because the heat is absorbed by the system, the $177.8 \: \text{kJ}$ is written as a reactant. The heat of reaction is positive for an endothermic reaction.
$\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = 177.8 \: \text{kJ}\nonumber$
The way in which a reaction is written influences the value of the enthalpy change for the reaction. Many reactions are reversible, meaning that the product(s) of the reaction are capable of combining and reforming the reactant(s). If a reaction is written in the reverse direction, the sign of the $\Delta H$ changes. For example, we can write an equation for the reaction of calcium oxide with carbon dioxide to form calcium carbonate.
$\ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \rightarrow \ce{CaCO_3} \left( s \right) + 177.8 \: \text{kJ}\nonumber$
The reaction is exothermic and thus the sign of the enthalpy change is negative.
$\ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \rightarrow \ce{CaCO_3} \left( s \right) \: \: \: \: \: \Delta H = -177.8 \: \text{kJ}\nonumber$
Stoichiometric Calculations and Enthalpy Changes
Chemistry problems that involve enthalpy changes can be solved by techniques similar to stoichiometry problems. Refer again to the combustion reaction of methane. Since the reaction of $1 \: \text{mol}$ of methane released $890.4 \: \text{kJ}$, the reaction of $2 \: \text{mol}$ of methane would release $2 \times 890.4 \: \text{kJ} = 1781 \: \text{kJ}$. The reaction of $0.5 \: \text{mol}$ of methane would release $\dfrac{890,4 \: \text{kJ}}{2} = 445.2 \: \text{kJ}$. As with other stoichiometry problems, the moles of a reactant or product can be linked to mass or volume.
Example $1$
Sulfur dioxide gas reacts with oxygen to form sulfur trioxide in an exothermic reaction, according to the following thermochemical equation.
$2 \ce{SO_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_3} \left( g \right) + 198 \: \text{kJ} \nonumber \nonumber$
Calculate the enthalpy change that occurs when $58.0 \: \text{g}$ of sulfur dioxide is reacted with excess oxygen.
Step 1: List the known quantities and plan the problem.
• Mass $\ce{SO_2} = 58.0 \: \text{g}$
• Molar mass $\ce{SO_2} = 64.07 \: \text{g/mol}$
• $\Delta H = -198 \: \text{kJ}$ for the reaction of $2 \: \text{mol} \: \ce{SO_2}$
Unknown
• $\Delta H = ? \: \text{kJ}$
The calculation requires two steps. The mass of $\ce{SO_2}$ is converted to moles. Then the moles of $\ce{SO_2}$ is multiplied by the conversion factor of $\left( \dfrac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} \right)$.
Step 2: Solve.
$\Delta H = 58.0 \: \text{g} \: \ce{SO_2} \times \dfrac{1 \: \text{mol} \: \ce{SO_2}}{64.07 \: \text{g} \: \ce{SO_2}} \times \dfrac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} = 89.6 \: \text{kJ} \nonumber \nonumber$
Step 3: Think about your result.
The mass of sulfur dioxide is slightly less than $1 \: \text{mol}$. Since $198 \: \text{kJ}$ is released for every $2 \: \text{mol}$ of $\ce{SO_2}$ that reacts, the heat released when about $1 \: \text{mol}$ reacts is one half of 198. The $89.6 \: \text{kJ}$ is slightly less than half of 198. The sign of $\Delta H$ is negative because the reaction is exothermic. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/08%3A_Quantities_in_Chemical_Reactions/8.08%3A_Enthalpy_Change_is_a_Measure_of_the_Heat_Evolved_or_Absorbed.txt |
• 9.2: Light- Electromagnetic Radiation
Light acts like a wave, with a frequency and a wavelength. The frequency and wavelength of light are related by the speed of light, a constant. Light acts like a particle of energy, whose value is related to the frequency of light.
• 9.3: The Electromagnetic Spectrum
Electromagnetic waves have an extremely wide range of wavelengths, frequencies, and energies. The highest energy form of electromagnetic waves are gamma (γ) rays and the lowest energy form are radio waves.
• 9.4: The Bohr Model - Atoms with Orbits
Bohr's model suggests that each atom has a set of unchangeable energy levels, and electrons in the electron cloud of that atom must be in one of those energy levels. Bohr's model suggests that the atomic spectra of atoms is produced by electrons gaining energy from some source, jumping up to a higher energy level, then immediately dropping back to a lower energy level and emitting the energy difference between the two energy levels. The existence of the atomic spectra supports Bohr's model.
• 9.5: The Quantum-Mechanical Model- Atoms with Orbitals
Quantum mechanics involves the study of material at the atomic level. This field deals with probabilities since we cannot definitely locate a particle. Orbitals are mathematically derived regions of space with different probabilities of having an electron.
• 9.6: Quantum-Mechanical Orbitals and Electron Configurations
We look at the four quantum numbers for a given electron. Electron configuration notation simplifies the indication of where electrons are located in a specific atom. The Aufbau principle gives the order of electron filling in an atom. Hund's rule specifies the order of electron filling within a set of orbitals. Orbital filling diagrams are a way of indicating electron locations in orbitals.
• 9.7: Electron Configurations and the Periodic Table
The arrangement of electrons in atoms is responsible for the shape of the periodic table. Electron configurations can be predicted by the position of an atom on the periodic table.
• 9.8: The Explanatory Power of the Quantum-Mechanical Model
The chemical properties of elements are determined primarily by the number and distribution of valence electrons.
• 9.9: Periodic Trends - Atomic Size, Ionization Energy, and Metallic Character
Certain properties—notably atomic radius, ionization energy, electron affinity and metallic character—can be qualitatively understood by the positions of the elements on the periodic table.
• 9.E: Electrons in Atoms and the Periodic Table (Exercises)
09: Electrons in Atoms and the Periodic Table
Learning Objectives
• Define the terms wavelength and frequency with respect to wave-form energy.
• State the relationship between wavelength and frequency with respect to electromagnetic radiation.
During the summer, almost everyone enjoys going to the beach. Beach-goers can swim, have picnics, and work on their tans. But if a person gets too much sun, they can burn. A particular set of solar wavelengths are especially harmful to the skin. This portion of the solar spectrum is known as UV B, with wavelengths of $280$-$320 \: \text{nm}$. Sunscreens are effective in protecting skin against both the immediate skin damage and the long-term possibility of skin cancer.
Waves
Waves are characterized by their repetitive motion. Imagine a toy boat riding the waves in a wave pool. As the water wave passes under the boat, it moves up and down in a regular and repeated fashion. While the wave travels horizontally, the boat only travels vertically up and down. The figure below shows two examples of waves.
A wave cycle consists of one complete wave—starting at the zero point, going up to a wave crest, going back down to a wave trough, and back to the zero point again. The wavelength of a wave is the distance between any two corresponding points on adjacent waves. It is easiest to visualize the wavelength of a wave as the distance from one wave crest to the next. In an equation, wavelength is represented by the Greek letter lambda $\left( \lambda \right)$. Depending on the type of wave, wavelength can be measured in meters, centimeters, or nanometers $\left( 1 \: \text{m} = 10^9 \: \text{nm} \right)$. The frequency, represented by the Greek letter nu $\left( \nu \right)$, is the number of waves that pass a certain point in a specified amount of time. Typically, frequency is measured in units of cycles per second or waves per second. One wave per second is also called a Hertz $\left( \text{Hz} \right)$ and in SI units is a reciprocal second $\left( \text{s}^{-1} \right)$.
Figure B above shows an important relationship between the wavelength and frequency of a wave. The top wave clearly has a shorter wavelength than the second wave. However, if you picture yourself at a stationary point watching these waves pass by, more waves of the first kind would pass by in a given amount of time. Thus the frequency of the first wave is greater than that of the second wave. Wavelength and frequency are therefore inversely related. As the wavelength of a wave increases, its frequency decreases. The equation that relates the two is:
$c = \lambda \nu \nonumber$
The variable $c$ is the speed of light. For the relationship to hold mathematically, if the speed of light is used in $\text{m/s}$, the wavelength must be in meters and the frequency in Hertz.
Example $1$: Orange Light
The color orange within the visible light spectrum has a wavelength of about $620 \: \text{nm}$. What is the frequency of orange light?
Solution
Solutions to Example 9.2.1
Steps for Problem Solving Example $1$
Identify the "given" information and what the problem is asking you to "find."
Given :
• Wavelength $\left( \lambda \right) = 620 \: \text{nm}$
• Speed of light $\left( c \right) = 3.00 \times 10^8 \: \text{m/s}$
Find: Frequency (Hz)
List other known quantities. $1 \: \text{m} = 10^9 \: \text{nm}$
Identify steps to get the final answer.
1.Convert the wavelength to $\text{m}$.
2. Apply the equation $c = \lambda \nu$ and solve for frequency. Dividing both sides of the equation by $\lambda$ yields:
$\nu = \dfrac{c}{\lambda}$
Cancel units and calculate.
$620 \: \text{nm} \times \left( \dfrac{1 \: \text{m}}{10^9 \: \text{nm}} \right) = 6.20 \times 10^{-7} \: \text{m}$
$\nu = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8 \: \text{m/s}}{6.20 \times 10^{-7}} = 4.8 \times 10^{14} \: \text{Hz}$
Think about your result. The value for the frequency falls within the range for visible light.
Exercise $1$
What is the wavelength of light if its frequency is 1.55 × 1010 s−1?
Answer
0.0194 m, or 19.4 mm
Summary
All waves can be defined in terms of their frequency and intensity. $c = \lambda \nu$ expresses the relationship between wavelength and frequency. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.02%3A_Light-_Electromagnetic_Radiation.txt |
Learning Objectives
• Know the properties of different types of electromagnetic radiation.
Electromagnetic waves have an extremely wide range of wavelengths, frequencies, and energies. The highest energy form of electromagnetic waves are gamma (γ) rays and the lowest energy form are radio waves.
The figure below shows the electromagnetic spectrum, which is all forms of electromagnetic radiation. On the far left of Figure \(1\) are the highest energy electromagnetic waves. These are called gamma rays and can be quite dangerous, in large numbers, to living systems. The next lower energy form of electromagnetic waves are called x-rays. Most of you are familiar with the penetration abilities of these waves. They can also be dangerous to living systems. Humans are advised to limit as much as possible the number of medical x-rays they have per year. Next lower, in energy, are ultraviolet rays. These rays are part of sunlight and the upper end of the ultraviolet range can cause sunburn and perhaps skin cancer. The tiny section next in the spectrum is the visible range of light. The visible light spectrum has been greatly expanded in the bottom half of the figure so that it can be discussed in more detail. The visible range of electromagnetic radiation are the frequencies to which the human eye responds. Lower in the spectrum are infrared rays and radio waves.
The light energies that are in the visible range are electromagnetic waves that cause the human eye to respond when those frequencies enter the eye. The eye sends a signal to the brain and the individual "sees" various colors. The highest energy waves in the visible region cause the brain to see violet and as the energy decreases, the colors change to blue, green, yellow, orange, and red. When the energy of the wave is above or below the visible range, the eye does not respond to them. When the eye receives several different frequencies at the same time, the colors are blended by the brain. If all frequencies of light strike the eye together, the brain sees white. If there are no visible frequencies striking the eye, the brain sees black. The objects that you see around you are light absorbers—that is, the chemicals on the surface of the object will absorb certain frequencies and not others. Your eyes detect the frequencies that strike your eye. Therefore, if your friend is wearing a red shirt, it means the dye in that shirt absorbs every frequency except red and the red frequencies are reflected. If your only light source was one exact frequency of blue light and you shined it on a shirt that was red in sunlight, the shirt would appear black because no light would be reflected. The light from fluorescent types of lights do not contain all the frequencies of sunlight and so clothes inside a store may appear to be a slightly different color when you get them home.
Summary
• Electromagnetic radiation has a wide spectrum, including gamma rays, X-rays, UV rays, visible light, IR radiation, microwaves, and radio waves.
• The different colors of light differ in their frequencies (or wavelengths). | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.03%3A_The_Electromagnetic_Spectrum.txt |
Learning Objectives
• Define an energy level in terms of the Bohr model.
• Discuss how the Bohr model can be used to explain atomic spectra.
Electric light bulbs contain a very thin wire in them that emits light when heated. The wire is called a filament. The particular wire used in light bulbs is made of tungsten. A wire made of any metal would emit light under these circumstances, but tungsten was chosen because the light it emits contains virtually every frequency and therefore, the light emitted by tungsten appears white. A wire made of some other element would emit light of some color that was not convenient for our uses. Every element emits light when energized by heating or passing electric current through it. Elements in solid form begin to glow when they are heated sufficiently, and elements in gaseous form emit light when electricity passes through them. This is the source of light emitted by neon signs and is also the source of light in a fire.
Each Element Has a Unique Spectrum
The light frequencies emitted by atoms are mixed together by our eyes so that we see a blended color. Several physicists, including Angstrom in 1868 and Balmer in 1875, passed the light from energized atoms through glass prisms in such a way that the light was spread out so they could see the individual frequencies that made up the light. The emission spectrum (or atomic spectrum) of a chemical element is the unique pattern of light obtained when the element is subjected to heat or electricity.
When hydrogen gas is placed into a tube and electric current passed through it, the color of emitted light is pink. But when the color is spread out, we see that the hydrogen spectrum is composed of four individual frequencies. The pink color of the tube is the result of our eyes blending the four colors. Every atom has its own characteristic spectrum; no two atomic spectra are alike. The image below shows the emission spectrum of iron. Because each element has a unique emission spectrum, elements can be defined using them.
You may have heard or read about scientists discussing what elements are present in the sun or some more distant star, and after hearing that, wondered how scientists could know what elements were present in a place no one has ever been. Scientists determine what elements are present in distant stars by analyzing the light that comes from stars and finding the atomic spectrum of elements in that light. If the exact four lines that compose hydrogen's atomic spectrum are present in the light emitted from the star, that element contains hydrogen.
Bohr's Model of the Atom
By 1913, the concept of the atom had evolved from Dalton's indivisible spheres idea, to J. J. Thomson's plum pudding model, and then to Rutherford's nuclear atom theory. Rutherford, in addition to carrying out the brilliant experiment that demonstrated the presence of the atomic nucleus, also proposed that the electrons circled the nucleus in a planetary type motion. The solar system or planetary model of the atom was attractive to scientists because it was similar to something with which they were already familiar, namely the solar system.
Unfortunately, there was a serious flaw in the planetary model. It was already known that when a charged particle (such as an electron) moves in a curved path, it gives off some form of light and loses energy in doing so. This is, after all, how we produce TV signals. If the electron circling the nucleus in an atom loses energy, it would necessarily have to move closer to the nucleus as it loses energy, and would eventually crash into the nucleus. Furthermore, Rutherford's model was unable to describe how electrons give off light forming each element's unique atomic spectrum. These difficulties cast a shadow on the planetary model and indicated that, eventually, it would have to be replaced.
In 1913, the Danish physicist Niels Bohr proposed a model of the electron cloud of an atom in which electrons orbit the nucleus and were able to produce atomic spectra. Understanding Bohr's model requires some knowledge of electromagnetic radiation (or light).
Energy Levels
Bohr's key idea in his model of the atom is that electrons occupy definite orbitals that require the electron to have a specific amount of energy. In order for an electron to be in the electron cloud of an atom, it must be in one of the allowable orbitals and it must have the precise energy required for that orbit. Orbits closer to the nucleus would require smaller amounts of energy for an electron, and orbits farther from the nucleus would require the electron to have a greater amount of energy. The possible orbits are known as energy levels. One of the weaknesses of Bohr's model was that he could not offer a reason why only certain energy levels or orbits were allowed.
Bohr hypothesized that the only way electrons could gain or lose energy would be to move from one energy level to another, thus gaining or losing precise amounts of energy. The energy levels are quantized, meaning that only specific amounts are possible. It would be like a ladder that had rungs only at certain heights. The only way you can be on that ladder is to be on one of the rungs, and the only way you could move up or down would be to move to one of the other rungs. Suppose we had such a ladder with 10 rungs. Other rules for the ladder are that only one person can be on a rung in the normal state, and the ladder occupants must be on the lowest rung available. If the ladder had five people on it, they would be on the lowest five rungs. In this situation, no person could move down because all of the lower rungs are full. Bohr worked out rules for the maximum number of electrons that could be in each energy level in his model, and required that an atom in its normal state (ground state) had all electrons in the lowest energy levels available. Under these circumstances, no electron could lose energy because no electron could move down to a lower energy level. In this way, Bohr's model explained why electrons circling the nucleus did not emit energy and spiral into the nucleus.
Bohr's Model and Atomic Spectra
The evidence used to support Bohr's model came from the atomic spectra. He suggested that an atomic spectrum is made by the electrons in an atom moving energy levels. The electrons typically have the lowest energy possible, called the ground state. If the electrons are given energy (through heat, electricity, light, etc.) the electrons in an atom could absorb energy by jumping to a higher energy level, or excited state. The electrons then give off the energy in the form of a piece of light—called a photon—that they had absorbed, to fall back to a lower energy level. The energy emitted by electrons dropping back to lower energy levels will always be precise amounts of energy, because the differences in energy levels are precise. This explains why you see specific lines of light when looking at an atomic spectrum—each line of light matches a specific "step down" that an electron can take in that atom. This also explains why each element produces a different atomic spectrum. Because each element has different acceptable energy levels for its electrons, the possible steps each element's electrons can take differ from all other elements.
Summary
• Bohr's model suggests each atom has a set of unchangeable energy levels, and electrons in the electron cloud of that atom must be in one of those energy levels.
• Bohr's model suggests that the atomic spectra of atoms is produced by electrons gaining energy from some source, jumping up to a higher energy level, then immediately dropping back to a lower energy level and emitting the energy difference between the two energy levels.
• The existence of the atomic spectra is support for Bohr's model of the atom.
• Bohr's model was only successful in calculating energy levels for the hydrogen atom.
Vocabulary
• Emission spectrum (or atomic spectrum) - The unique pattern of light given off by an element when it is given energy.
• Energy levels - Possible orbits that an electron can have in the electron cloud of an atom.
• Ground state - To be in the lowest energy level possible.
• Excited state - To be in a higher energy level.
• Photon - A piece of electromagnetic radiation, or light, with a specific amount of energy.
• Quantized - To have a specific amount.
9.05: The Quantum-Mechanical Model- Atoms with Orbitals
Learning Objectives
• Define quantum mechanics
• Differentiate between an orbit and an orbital.
How do you study something that seemingly makes no sense? We talk about electrons being in orbits and it sounds like we can tell where that electron is at any moment. We can draw pictures of electrons in orbit, but the reality is that we don't know exactly where they are. We are going to take a look at an area of science that even leaves scientists puzzled. When asked about quantum mechanics, Niels Bohr (who proposed the Bohr model of the atom) said: "Anyone who is not shocked by quantum theory has not understood it". Richard Feynman (one of the founders of modern quantum theory) stated: "I think I can safely say that nobody understands quantum theory." So, let's take a short trip into a land that challenges our everyday world.
Quantum Mechanics
The study of motion of large objects such as baseballs is called mechanics, or more specifically, classical mechanics. Because of the quantum nature of the electron and other tiny particles moving at high speeds, classical mechanics is inadequate to accurately describe their motion. Quantum mechanics is the study of the motion of objects that are atomic or subatomic in size and thus demonstrate wave-particle duality. In classical mechanics, the size and mass of the objects involved effectively obscures any quantum effects, so that such objects appear to gain or lose energies in any amounts. Particles whose motion is described by quantum mechanics gain or lose energy in small pieces called quanta.
One of the fundamental (and hardest to understand) principles of quantum mechanics is that the electron is both a particle and a wave. In the everyday macroscopic world of things we can see, something cannot be both. But this duality can exist in the quantum world of the submicroscopic on the atomic scale.
At the heart of quantum mechanics is the idea that we cannot accurately specify the location of an electron. All we can say is that there is a probability that it exists within this certain volume of space. The scientist Erwin Schrödinger developed an equation that deals with these calculations, which we will not pursue at this time.
Erwin Schrödinger.
Recall that in the Bohr model, the exact path of the electron was restricted to very well-defined circular orbits around the nucleus. An orbital is the quantum mechanical refinement of Bohr’s orbit. In contrast to his concept of a simple circular orbit with a fixed radius, orbitals are mathematically derived regions of space with different probabilities of having an electron.
Summary
Quantum mechanics involves the study of material at the atomic level. This field deals with probabilities, since we cannot definitely locate a particle. Orbitals are mathematically derived regions of space with different probabilities of having an electron. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.04%3A_The_Bohr_Model_-_Atoms_with_Orbits.txt |
Learning Objectives
• Represent the organization of electrons by an electron configuration and orbital diagram.
The flight path of a commercial airliner is carefully regulated by the Federal Aviation Administration. Each airplane must maintain a distance of five miles from another plane flying at the same altitude and 2,000 feet above and below another aircraft (1,000 feet if the altitude is less than 29,000 feet). So, each aircraft only has certain positions it is allowed to maintain while it flies. As we explore quantum mechanics, we see that electrons have similar restrictions on their locations.
Orbitals
We can apply our knowledge of quantum numbers to describe the arrangement of electrons for a given atom. We do this with something called electron configurations. They are effectively a map of the electrons for a given atom. We look at the four quantum numbers for a given electron and then assign that electron to a specific orbital below.
s Orbitals
For any value of $n$, a value of $l=0$ places that electron in an $s$ orbital. This orbital is spherical in shape:
p Orbitals
For the table below, we see that we can have three possible orbitals when $l=1$. These are designated as $p$ orbitals and have dumbbell shapes. Each of the $p$ orbitals has a different orientation in three-dimensional space.
d Orbitals
When $l=2$, $m_l$ values can be $-2, \: -1, \: 0, \: +1, \: +2$ for a total of five $d$ orbitals. Note that all five of the orbitals have specific three-dimensional orientations.
f Orbitals
The most complex set of orbitals are the $f$ orbitals. When $l=3$, $m_l$ values can be $-3, \: -2, \: -1, \: 0, \: +1, \: +2, \: +3$ for a total of seven different orbital shapes. Again, note the specific orientations of the different $f$ orbitals.
Orbitals that have the same value of the principal quantum number form a shell. Orbitals within a shell are divided into subshells that have the same value of the angular quantum number. Some of the allowed combinations of quantum numbers are compared in Table $1$.
Table $1$: Electron Arrangement Within Energy Levels
Principal Quantum Number $\left( n \right)$ Allowable Sublevels Number of Orbitals per Sublevel Number of Orbitals per Principal Energy Level Number of Electrons per Sublevel Number of Electrons per Principal Energy Level
1 $s$ 1 1 2 2
2 $s$ 1 4 2 8
$p$ 3 6
3 $s$ 1 9 2 18
$p$ 3 6
$d$ 5 10
4 $s$ 1 16 2 32
$p$ 3 6
$d$ 5 10
$f$ 7 14
Electron Configurations
Can you name one thing that easily distinguishes you from the rest of the world? And we're not talking about DNA—that's a little expensive to sequence. For many people, it is their email address. Your email address allows people all over the world to contact you. It does not belong to anyone else, but serves to identify you. Electrons also have a unique set of identifiers in the quantum numbers that describe their location and spin. Chemists use an electronic configuration to represent the organization of electrons in shells and subshells in an atom. An electron configuration simply lists the shell and subshell labels, with a right superscript giving the number of electrons in that subshell. The shells and subshells are listed in the order of filling. Electrons are typically organized around an atom by starting at the lowest possible quantum numbers first, which are the shells-subshells with lower energies.
For example, an H atom has a single electron in the 1s subshell. Its electron configuration is
$\ce{H}:\, 1s^1 \nonumber$
He has two electrons in the 1s subshell. Its electron configuration is
$\ce{He}:\, 1s^2 \nonumber$
The three electrons for Li are arranged in the 1s subshell (two electrons) and the 2s subshell (one electron). The electron configuration of Li is
$\ce{Li}:\, 1s^22s^1 \nonumber$
Be has four electrons, two in the 1s subshell and two in the 2s subshell. Its electron configuration is
$\ce{Be}:\, 1s^22s^2 \nonumber$
Now that the 2s subshell is filled, electrons in larger atoms must go into the 2p subshell, which can hold a maximum of six electrons. The next six elements progressively fill up the 2p subshell:
• B: 1s22s22p1
• C: 1s22s22p2
• N: 1s22s22p3
• O: 1s22s22p4
• F: 1s22s22p5
• Ne: 1s22s22p6
Now that the 2p subshell is filled (all possible subshells in the n = 2 shell), the next electron for the next-larger atom must go into the n = 3 shell, s subshell.
Second Period Elements
Periods refer to the horizontal rows of the periodic table. Looking at a periodic table you will see that the first period contains only the elements hydrogen and helium. This is because the first principal energy level consists of only the $s$ sublevel and so only two electrons are required in order to fill the entire principal energy level. Each time a new principal energy level begins, as with the third element lithium, a new period is started on the periodic table. As one moves across the second period, electrons are successively added. With beryllium $\left( Z=4 \right)$, the $2s$ sublevel is complete and the $2p$ sublevel begins with boron $\left( Z=5 \right)$. Since there are three $2p$ orbitals and each orbital holds two electrons, the $2p$ sublevel is filled after six elements. Table $1$ shows the electron configurations of the elements in the second period.
Element Name Symbol Atomic Number Electron Configuration
Table $2$: Electron Configurations of Second-Period Elements
Lithium $\ce{Li}$ 3 $1s^2 2s^1$
Beryllium $\ce{Be}$ 4 $1s^2 2s^2$
Boron $\ce{B}$ 5 $1s^2 2s^2 2p^1$
Carbon $\ce{C}$ 6 $1s^2 2s^2 2p^2$
Nitrogen $\ce{N}$ 7 $1s^2 2s^2 2p^3$
Oxygen $\ce{O}$ 8 $1s^2 2s^2 2p^4$
Fluorine $\ce{F}$ 9 $1s^2 2s^2 2p^5$
Neon $\ce{Ne}$ 10 $1s^2 2s^2 2p^6$
Aufbau Principle
Construction of a building begins at the bottom. The foundation is laid and the building goes up step by step. You obviously cannot start with the roof since there is no place to hang it. The building goes from the lowest level to the highest level in a systematic way. In order to create ground state electron configurations for any element, it is necessary to know the way in which the atomic sublevels are organized in order of increasing energy. Figure $5$ shows the order of increasing energy of the sublevels.
The lowest energy sublevel is always the $1s$ sublevel, which consists of one orbital. The single electron of the hydrogen atom will occupy the $1s$ orbital when the atom is in its ground state. As we proceed with atoms with multiple electrons, those electrons are added to the next lowest sublevel: $2s$, $2p$, $3s$, and so on. The Aufbau principle states that an electron occupies orbitals in order from lowest energy to highest. The Aufbau (German: "building up, construction") principle is sometimes referred to as the "building up" principle. It is worth noting that in reality atoms are not built by adding protons and electrons one at a time, and that this method is merely an aid for us to understand the end result.
As seen in the figure above, the energies of the sublevels in different principal energy levels eventually begin to overlap. After the $3p$ sublevel, it would seem logical that the $3d$ sublevel should be the next lowest in energy. However, the $4s$ sublevel is slightly lower in energy than the $3d$ sublevel and thus fills first. Following the filling of the $3d$ sublevel is the $4p$, then the $5s$ and the $4d$. Note that the $4f$ sublevel does not fill until just after the $6s$ sublevel. Figure $6$ is a useful and simple aid for keeping track of the order of fill of the atomic sublevels.
Example $1$: Nitrogen Atoms
Nitrogen has 7 electrons. Write the electron configuration for nitrogen.
Solution:
Take a close look at Figure $5$, and use it to figure out how many electrons go into each sublevel, and also the order in which the different sublevels get filled.
1. Begin by filling up the 1s sublevel. This gives 1s2. Now all of the orbitals in the red n = 1 block are filled.
Since we used 2 electrons, there are 7 − 2 = 5 electrons left
2. Next, fill the 2s sublevel. This gives 1s22s2. Now all of the orbitals in the s sublevel of the orange n = 2 block are filled.
Since we used another 2 electrons, there are 5 − 2 = 3 electrons left
3. Notice that we haven't filled the entire n = 2 block yet… there are still the p orbitals!
The final 3 electrons go into the 2p sublevel. This gives 1s22s22p3
The overall electron configuration is: 1s22s22p3.
Example $2$: Potassium Atoms
Potassium has 19 electrons. Write the electron configuration code for potassium.
Solution
This time, take a close look at Figure $5$.
1. Begin by filling up the 1s sublevel. This gives 1s2. Now the n = 1 level is filled.
Since we used 2 electrons, there are 19 − 2 = 17 electrons left
2. Next, fill the 2s sublevel. This gives 1s22s2
Since we used another 2 electrons, there are 17 − 2 = 15 electrons left
3. Next, fill the 2p sublevel. This gives 1s22s22p6. Now the n = 2 level is filled.
Since we used another 6 electrons, there are 15 − 6 = 9 electrons left
4. Next, fill the 3s sublevel. This gives 1s22s22p63s2
Since we used another 2 electrons, there are 9 − 2 = 7 electrons left
5. Next, fill the 3p sublevel. This gives 1s22s22p63s23p6
Since we used another 6 electrons, there are 7 − 6 = 1 electron left
Here's where we have to be careful – right after 3p6!
Remember, 4s comes before 3d
6. The final electron goes into the 4s sublevel. This gives 1s22s22p63s23p64s1
The overall electron configuration is: 1s22s22p63s23p64s1
Exercise $1$: Magnesium and Sodium Atoms
What is the electron configuration for Mg and Na?
Answer Mg
Mg: 1s22s22p63s2
Answer Na
Na: 1s22s22p63s1
Pauli Exclusion Principle
When we look at the orbital possibilities for a given atom, we see that there are different arrangements of electrons for each different type of atom. Since each electron must maintain its unique identity, we intuitively sense that the four quantum numbers for any given electron must not match up exactly with the four quantum numbers for any other electron in that atom.
For the hydrogen atom, there is no problem since there is only one electron in the $\ce{H}$ atom. However, when we get to helium we see that the first three quantum numbers for the two electrons are the same: same energy level, same spherical shape. What differentiates the two helium electrons is their spin. One of the electrons has a $+\dfrac{1}{2}$ spin while the other electron has a $-\dfrac{1}{2}$ spin. So the two electrons in the $1s$ orbital are each unique and distinct from one another because their spins are different. This observation leads to the Pauli exclusion principle, which states that no two electrons in an atom can have the same set of four quantum numbers. The energy of the electron is specified by the principal, angular momentum, and magnetic quantum numbers. If those three numbers are identical for two electrons, the spin numbers must be different in order for the two electrons to be differentiated from one another. The two values of the spin quantum number allow each orbital to hold two electrons. Figure $7$ shows how the electrons are indicated in a diagram.
Hund's Rule
The last of the three rules for constructing electron arrangements requires electrons to be placed one at a time in a set of orbitals within the same sublevel. This minimizes the natural repulsive forces that one electron has for another. Hund's rule states that orbitals of equal energy are each occupied by one electron before any orbital is occupied by a second electron and that each of the single electrons must have the same spin. The figure below shows how a set of three $p$ orbitals is filled with one, two, three, and four electrons.
Orbital Filling Diagrams
An orbital filling diagram is the more visual way to represent the arrangement of all the electrons in a particular atom. In an orbital filling diagram, the individual orbitals are shown as circles (or squares) and orbitals within a sublevel are drawn next to each other horizontally. Each sublevel is labeled by its principal energy level and sublevel. Electrons are indicated by arrows inside of the circles. An arrow pointing upwards indicates one spin direction, while a downward pointing arrow indicates the other direction. The orbital filling diagrams for hydrogen, helium, and lithium are shown in the figure below.
According to the Aufbau process, sublevels and orbitals are filled with electrons in order of increasing energy. Since the $s$ sublevel consists of just one orbital, the second electron simply pairs up with the first electron as in helium. The next element is lithium and necessitates the use of the next available sublevel, the $2s$.
The filling diagram for carbon is shown in Figure $10$. There are two $2p$ electrons for carbon and each occupies its own $2p$ orbital.
Oxygen has four $2p$ electrons. After each $2p$ orbital has one electron in it, the fourth electron can be placed in the first $2p$ orbital with a spin opposite that of the other electron in that orbital.
If you keep your papers in manila folders, you can pick up a folder and see how much it weighs. If you want to know how many different papers (articles, bank records, or whatever else you keep in a folder), you have to take everything out and count. A computer directory, on the other hand, tells you exactly how much you have in each file. We can get the same information on atoms. If we use an orbital filling diagram, we have to count arrows. When we look at electron configuration data, we simply add up the numbers.
Example $3$: Carbon Atoms
Draw the orbital filling diagram for carbon and write its electron configuration.
Known
• Atomic number of carbon, Z=6
Use the order of fill diagram to draw an orbital filling diagram with a total of six electrons. Follow Hund's rule. Write the electron configuration.
Orbital filling diagram for carbon.
Electron configuration 1s22s22p2
Step 3: Think about your result.
Following the 2s sublevel is the 2p, and p sublevels always consist of three orbitals. All three orbitals need to be drawn even if one or more is unoccupied. According to Hund's rule, the sixth electron enters the second of those p orbitals and has the same spin as the fifth electron.
Exercise $2$: Electronic Configurations
Write the electron configurations and orbital diagrams for
1. Potassium atom: $\ce{K}$
2. Arsenic atom: $\ce{As}$
3. Phosphorus atom: $\ce{P}$
Answer a:
Potassium: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$
Answer b:
Arsenic: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3$
Answer c:
Phosphorus $1s^2 2s^2 2p^6 3s^2 3p^3$
Summary
There are four different classes of electron orbitals. These orbitals are determined by the value of the angular momentum quantum number . An orbital is a wave function for an electron defined by the three quantum numbers, n, and m. Orbitals define regions in space where you are likely to find electrons. s orbitals ( = 0) are spherical shaped. p orbitals ( = 1) are dumb-bell shaped. The three possible p orbitals are always perpendicular to each other.
Electron configuration notation simplifies the indication of where electrons are located in a specific atom. Superscripts are used to indicate the number of electrons in a given sublevel. The Aufbau principle gives the order of electron filling in an atom. It can be used to describe the locations and energy levels of every electron in a given atom. Hund's rule specifies the order of electron filling within a set of orbitals. Orbital filling diagrams are a way of indicating electron locations in orbitals. The Pauli exclusion principle specifies limits on how identical quantum numbers can be for two electrons in the same atom.
Vocabulary
principal quantum number (n)
Defines the energy level of the wave function for an electron, the size of the electron's standing wave, and the number of nodes in that wave.
quantum numbers
Integer numbers assigned to certain quantities in the electron wave function. Because electron standing waves must be continuous and must not "double over" on themselves, quantum numbers are restricted to integer values. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.06%3A_Quantum-Mechanical_Orbitals_and_Electron_Configurations.txt |
Learning Objectives
• Relate the electron configurations of the elements to the shape of the periodic table.
• Determine the expected electron configuration of an element by its place on the periodic table.
Previously, we introduced the periodic table as a tool for organizing the known chemical elements. A periodic table is shown in Figure \(1\). The elements are listed by atomic number (the number of protons in the nucleus), and elements with similar chemical properties are grouped together in columns.
Why does the periodic table have the structure it does? The answer is rather simple, if you understand electron configurations: the shape of the periodic table mimics the filling of the subshells with electrons.
The shape of the periodic table mimics the filling of the subshells with electrons.
Let us start with H and He. Their electron configurations are 1s1 and 1s2, respectively; with He, the n = 1 shell is filled. These two elements make up the first row of the periodic table (Figure \(2\))
The next two electrons, for Li and Be, would go into the 2s subshell. Figure \(3\) shows that these two elements are adjacent on the periodic table.
For the next six elements, the 2p subshell is being occupied with electrons. On the right side of the periodic table, these six elements (B through Ne) are grouped together (Figure \(4\)).
The next subshell to be filled is the 3s subshell. The elements when this subshell is being filled, Na and Mg, are back on the left side of the periodic table (Figure \(5\)).
Next, the 3p subshell is filled with the next six elements (Figure \(6\)).
Instead of filling the 3d subshell next, electrons go into the 4s subshell (Figure \(7\)).
After the 4s subshell is filled, the 3d subshell is filled with up to 10 electrons. This explains the section of 10 elements in the middle of the periodic table (Figure \(8\)).
...And so forth. As we go across the rows of the periodic table, the overall shape of the table outlines how the electrons are occupying the shells and subshells.
The first two columns on the left side of the periodic table are where the s subshells are being occupied. Because of this, the first two rows of the periodic table are labeled the s block. Similarly, the p block are the right-most six columns of the periodic table, the d block is the middle 10 columns of the periodic table, while the f block is the 14-column section that is normally depicted as detached from the main body of the periodic table. It could be part of the main body, but then the periodic table would be rather long and cumbersome. Figure \(9\) shows the blocks of the periodic table.
The electrons in the highest-numbered shell, plus any electrons in the last unfilled subshell, are called valence electrons; the highest-numbered shell is called the valence shell. (The inner electrons are called core electrons.) The valence electrons largely control the chemistry of an atom. If we look at just the valence shell’s electron configuration, we find that in each column, the valence shell’s electron configuration is the same. For example, take the elements in the first column of the periodic table: H, Li, Na, K, Rb, and Cs. Their electron configurations (abbreviated for the larger atoms) are as follows, with the valence shell electron configuration highlighted:
Electrons, electron configurations, and the valence shell electron configuration highlighted.
H: 1s1
Li: 1s22s1
Na: [Ne]3s1
K: [Ar]4s1
Rb: [Kr]5s1
Cs: [Xe]6s1
They all have a similar electron configuration in their valence shells: a single s electron. Because much of the chemistry of an element is influenced by valence electrons, we would expect that these elements would have similar chemistry—and they do. The organization of electrons in atoms explains not only the shape of the periodic table, but also the fact that elements in the same column of the periodic table have similar chemistry.
The same concept applies to the other columns of the periodic table. Elements in each column have the same valence shell electron configurations, and the elements have some similar chemical properties. This is strictly true for all elements in the s and p blocks. In the d and f blocks, because there are exceptions to the order of filling of subshells with electrons, similar valence shells are not absolute in these blocks. However, many similarities do exist in these blocks, so a similarity in chemical properties is expected.
Similarity of valence shell electron configuration implies that we can determine the electron configuration of an atom solely by its position on the periodic table. Consider Se, as shown in Figure \(10\). It is in the fourth column of the p block. This means that its electron configuration should end in a p4 electron configuration. Indeed, the electron configuration of Se is [Ar]4s23d104p4, as expected.
Example \(1\): Predicting Electron Configurations
From the element’s position on the periodic table, predict the valence shell electron configuration for each atom (Figure \(11\)).
1. Ca
2. Sn
Solution
1. Ca is located in the second column of the s block. We expect that its electron configuration should end with s2. Calcium’s electron configuration is [Ar]4s2.
2. Sn is located in the second column of the p block, so we expect that its electron configuration would end in p2. Tin’s electron configuration is [Kr]5s24d105p2.
Exercise \(1\)
From the element’s position on the periodic table, predict the valence shell electron configuration for each atom. Figure \(11\).
1. Ti
2. Cl
Answer a
[Ar]4s23d2
Answer b
[Ne]3s23p5
Summary
The arrangement of electrons in atoms is responsible for the shape of the periodic table. Electron configurations can be predicted by the position of an atom on the periodic table. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.07%3A_Electron_Configurations_and_the_Periodic_Table.txt |
Learning Objectives
• Give the name and location of specific groups on the periodic table, including alkali metals, alkaline earth metals, noble gases, halogens, and transition metals.
• Explain the relationship between the chemical behavior of families in the periodic table and their electron configurations.
• Identify elements that will have the most similar properties to a given element.
The chemical behavior of atoms is controlled by their electron configuration. Since the families of elements were organized by their chemical behavior, it is predictable that the individual members of each chemical family will have similar electron configurations.
Families of the Periodic Table
Remember that Mendeleev arranged the periodic table so that elements with the most similar properties were placed in the same group. A group is a vertical column of the periodic table. All of the 1A elements have one valence electron. This is what causes these elements to react in the same ways as the other members of the family. The elements in 1A are all very reactive and form compounds in the same ratios with similar properties with other elements. Because of their similarities in their chemical properties, Mendeleev put these elements into the same group. Group 1A is also known as the alkali metals. Although most metals tend to be very hard, these metals are actually soft and can be easily cut.
Group 2A is also called the alkaline earth metals. Once again, because of their similarities in electron configurations, these elements have similar properties to each other. The same pattern is true of other groups on the periodic table. Remember, Mendeleev arranged the table so that elements with the most similar properties were in the same group on the periodic table.
It is important to recognize a couple of other important groups on the periodic table by their group name. Group 7A (or 17) elements are also called halogens. This group contains very reactive nonmetal elements.
The noble gases are in group 8A. These elements also have similar properties to each other, the most significant property being that they are extremely unreactive, rarely forming compounds. The reason for this will be communicated later, when we discuss how compounds form. The elements in this group are also gases at room temperature.
An alternate numbering system numbers all of the \(s\), \(p\), and \(d\) block elements from 1-18. In this numbering system, group 1A is group 1; group 2A is group 2; the halogens (7A) are group 17; and the noble gases (8A) are group 18. You will come across periodic tables with both numbering systems. It is important to recognize which numbering system is being used, and to be able to find the number of valence electrons in the main block elements, regardless of which numbering system is being used.
Periods of the Periodic Table
If you can locate an element on the Periodic Table, you can use the element's position to figure out the energy level of the element's valence electrons. A period is a horizontal row of elements on the periodic table. For example, the elements sodium (\(\ce{Na}\)) and magnesium (\(\ce{Mg}\)) are both in period 3. The elements astatine (\(\ce{At}\)) and radon (\(\ce{Rn}\)) are both in period 6.
Summary
• The vertical columns on the periodic table are called groups or families because of their similar chemical behavior.
• All the members of a family of elements have the same number of valence electrons and similar chemical properties.
• The horizontal rows on the periodic table are called periods.
Vocabulary
• Group (family) - A vertical column of the periodic table.
• Alkali metals - Group 1A of the periodic table.
• Alkaline earth metals - Group 2A of the periodic table.
• Halogens - Group 7A of the periodic table.
• Noble gases - Group 8A of the periodic table.
• Transition elements - Groups 3 to 12 of the periodic table. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.08%3A_The_Explanatory_Power_of_the_Quantum-Mechanical_Model.txt |
Learning Objectives
• Be able to state how certain properties of atoms vary based on their relative position on the periodic table.
One of the reasons the periodic table is so useful is because its structure allows us to qualitatively determine how some properties of the elements vary versus their position on the periodic table. The variations of properties versus positions on the periodic table are called periodic trends. There is no other tool in science that allows us to judge relative properties of a class of objects like this, which makes the periodic table a very useful tool. Many periodic trends are general. There may be a few points where an opposite trend is seen, but there is an overall trend when considered across a whole row or down a whole column of the periodic table.
The first periodic trend we will consider is atomic radius. The atomic radius is an indication of the size of an atom. Although the concept of a definite radius of an atom is a bit fuzzy, atoms behave as if they have a certain radius. Such radii can be estimated from various experimental techniques, such as the x-ray crystallography of crystals.
As you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus. This trend can be summarized as follows:
$as\downarrow PT,atomic\; radius \uparrow \nonumber$
where PT stands for periodic table. Going across a row on the periodic table, left to right, the trend is different. Even though the valence shell maintains the same principal quantum number, the number of protons—and hence the nuclear charge—is increasing as you go across the row. The increasing positive charge casts a tighter grip on the valence electrons, so as you go across the periodic table, the atomic radii decrease. Again, we can summarize this trend as follows:
$as\rightarrow PT,atomic\; radius \downarrow \nonumber$
Figure $1$ shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius.
Example $1$: Atomic Radii
Referring only to a periodic table and not to Figure $1$, which atom is larger in each pair?
1. Si or S
2. S or Te
Solution
1. Si is to the left of S on the periodic table; it is larger because as you go across the row, the atoms get smaller.
2. S is above Te on the periodic table; Te is larger because as you go down the column, the atoms get larger.
Exercise $1$: Atomic Radii
Referring only to a periodic table and not to Figure $1$, which atom is smaller, Ca or Br?
Answer
Br
Ionization energy (IE) is the amount of energy required to remove an electron from an atom in the gas phase:
$A(g)\rightarrow A^{+}(g)+e^{-}\; \; \; \; \; \Delta H\equiv IE \nonumber$
IE is usually expressed in kJ/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus. Thus,
$as\downarrow PT,\; IE\downarrow \nonumber$
However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron; as a result, IE increases:
$as\rightarrow PT,\; IE\uparrow \nonumber$
Figure $2$ shows values of IE versus position on the periodic table. Again, the trend is not absolute, but the general trends going across and down the periodic table should be obvious.
IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with):
• First Ionization Energy (IE1):
$A(g) → A^+(g) + e^- \nonumber$
• Second Ionization Energy (IE2):
$A^{+}(g) → A^{2+}(g) + e^- \nonumber$
• Third Ionization Energy (IE3):
$A^{2+}(g) → A^{3+}(g) + e^- \nonumber$
and so forth.
Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration is 1s22s22p63s2:
• First Ionization Energy (IE1) = 738 kJ/mol:
$Mg(g) → Mg^{+}(g) + e^− \nonumber$
• Second Ionization Energy (IE2) = 1,450 kJ/mol:
$Mg^+(g) → Mg^{2+}(g) + e^− \nonumber$
• Third Ionization Energy (IE3) = 7,734 kJ/mol:
$Mg^{2+}(g) → Mg^{3+}(g) + e^− \nonumber$
The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over five times the previous one. Why is it so much larger? Because the first two electrons are removed from the 3s subshell, but the third electron has to be removed from the n = 2 shell (specifically, the 2p subshell, which is lower in energy than the n = 3 shell). Thus, it takes much more energy than just overcoming a larger ionic charge would suggest. It is trends like this that demonstrate that electrons within atoms are organized in groups.
Example $2$: Ionization Energies
Which atom in each pair has the larger first ionization energy?
1. Ca or Sr
2. K or K+
Solution
1. Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE.
2. Because K+ has a positive charge, it will be harder to remove another electron from it, so its IE is larger than that of K. Indeed, it will be significantly larger because the next electron in K+ to be removed comes from another shell.
Exercise $2$: Ionization Energies
Which atom has the lower ionization energy, C or F?
Answer
C
The opposite of IE is described by electron affinity (EA), which is the energy change when a gas-phase atom accepts an electron:
$A(g)+e^{-}\rightarrow A^{-}(g)\; \; \; \; \; \Delta H\equiv EA \nonumber$
EA is also usually expressed in kJ/mol. EA also demonstrates some periodic trends, although they are less obvious than the other periodic trends discussed previously. Generally, as you go across the periodic table, EA increases its magnitude:
$as\rightarrow PT,\; EA\uparrow \nonumber$
There is not a definitive trend as you go down the periodic table; sometimes EA increases, sometimes it decreases. Figure $3$ shows EA values versus position on the periodic table for the s- and p-block elements. The trend is not absolute, especially considering the large positive EA values for the second column. However, the general trend going across the periodic table should be obvious.
Example $3$: Electron Affinities
Predict which atom in each pair will have the highest magnitude of Electron Affinity.
1. C or F
2. Na or S
Solution
1. C and F are in the same row on the periodic table, but F is farther to the right. Therefore, F should have the larger magnitude of EA.
2. Na and S are in the same row on the periodic table, but S is farther to the right. Therefore, S should have the larger magnitude of EA.
Exercise $3$: Electron Affinities
Predict which atom will have the highest magnitude of Electron Affinity: As or Br.
Answer
Br
Metallic Character
The metallic character is used to define the chemical properties that metallic elements present. Generally, metals tend to lose electrons to form cations. Nonmetals tend to gain electrons to form anions. They also have a high oxidation potential—therefore they are easily oxidized and are strong reducing agents. Metals also form basic oxides; the more basic the oxide, the higher the metallic character.
As you move across the table from left to right, the metallic character decreases, because the elements easily accept electrons to fill their valance shells. Therefore, these elements take on the nonmetallic character of forming anions. As you move up the table, the metallic character decreases, due to the greater pull that the nucleus has on the outer electrons. This greater pull makes it harder for the atoms to lose electrons and form cations.
Uses of the Periodic Properties of Elements
1. Predict greater or smaller atomic size and radial distribution in neutral atoms and ions.
2. Measure and compare ionization energies.
3. Compare electron affinities and electronegativities.
• Predict redox potential.
• Compare metallic character with other elements; ability to form cations.
• Predict reactions that may or may not occur due to the trends.
• Determine greater cell potential (sum of oxidation and reduction potential) between reactions.
• Complete chemical reactions according to trends.
Summary
• Certain properties—notably atomic radius, ionization energies, and electron affinities—can be qualitatively understood by the positions of the elements on the periodic table. The major trends are summarized in the figure below.
• There are three factors that help in the prediction of the trends in the Periodic Table: number of protons in the nucleus, number of shells, and shielding effect. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.09%3A_Periodic_Trends_-_Atomic_Size_Ionization_Energy_and_Metallic_Character.txt |
9.3: The Electromagnetic Spectrum
1. Choose the correct word for the following statement. Blue light has a (longer or shorter) wavelength than red light.
2. Choose the correct word for the following statement. Yellow light has a (higher or lower) frequency than blue light.
3. Choose the correct word for the following statement. Green light has a (larger or smaller) energy than red light.
4. If "light A" has a longer wavelength than "light B", then "light A" has _______________ "light B".
(a) a lower frequency than
(b) a higher frequency than
(c) the same frequency as
5. If "light C" has a shorter wavelength than "light D", then "light C" has _______________ "light D".
(a) a larger energy than
(b) a smaller energy than
(c) the same energy as
6. If "light E" has a higher frequency than "light F", then "light E" has __________________ "light F".
(a) a longer wavelength than
(b) a shorter wavelength than
(c) the same wavelength as
7. If "light G" has a higher frequency than "light H", then "light G" has __________________ "light H".
(a) a larger energy than
(b) a smaller energy than
(c) the same energy as
8. If "light J" has larger energy than "light K", then "light J" has __________________ "light K".
(a) a shorter wavelength than
(b) a longer wavelength than
(c) the same wavelength as
9. Which of the following statements is true?
(a) The frequency of green light is higher than the frequency of blue light and the wavelength of green light is longer than the wavelength of blue light.
(b) The frequency of green light is higher than the frequency of blue light and the wavelength of green light is shorter than the wavelength of blue light.
(c) The frequency of green light is lower than the frequency of blue light and the wavelength of green light is shorter than the wavelength of blue light.
(d) The frequency of green light is lower than the frequency of blue light and the wavelength of green light is longer than the wavelength of blue light.
(e) The frequency of green light is the same as the frequency of blue light and the wavelength of green light is shorter than the wavelength of blue light.
10. As the wavelength of electromagnetic radiation increases:
(a) its energy increases.
(b) its frequency increases.
(c) its speed increases.
(d) more than one of the above statements is true.
(e) none of the above statements is true.
11. List three examples of electromagnetic waves.
12. Why do white objects appear white?
13. Name the colors present in white light in order of increasing frequency.
14. Why do objects appear black?
9.4: The Bohr Model: Atoms with Orbits
1. Decide whether each of the following statements is true or false:
(a) Niels Bohr suggested that the electrons in an atom were restricted to specific orbits and thus could only have certain energies.
(b) Bohr's model of the atom can be used to accurately predict the emission spectrum of hydrogen.
(c) Bohr's model of the atom can be used to accurately predict the emission spectrum of neon.
(d) According to the Bohr model, electrons have more or less energy depending on how far around an orbit they have traveled.
2. According to the Bohr model, electrons in an atom can only have certain, allowable energies. As a result, we say that the energies of these electrons are _______.
3. The Bohr model accurately predicts the emission spectra of atoms with…
(a) less than 1 electron.
(b) less than 2 electrons.
(c) less than 3 electrons.
(d) less than 4 electrons.
4. Consider an He+ atom. Like the hydrogen atom, the He+ atom only contains 1 electron, and thus can be described by the Bohr model. Fill in the blanks in the following statements.
(a) An electron falling from the n = 2 orbit of He+ to the n = 1 orbit of He+ releases ______ energy than an electron falling from the n = 3 orbit of He+ to the n = 1 orbit of He+.
(b) An electron falling from the n = 2 orbit of He+ to the n = 1 orbit of He+ produces light with a ______ wavelength than the light produced by an electron falling from the n = 3 orbit of He+ to the n = 1 orbit of He+.
(c) An electron falling from the n = 2 orbit of He+ to the n = 1 orbit of He+ produces light with a ______ frequency than the light produced by an electron falling from the n = 3 orbit of He+ to the n = 1 orbit of He+.
5. According to the Bohr model, higher energy orbits are located (closer to/further from) the atomic nucleus. This makes sense since negative electrons are (attracted to/repelled from) the positive protons in the nucleus, meaning it must take energy to move the electrons (away from/towards) the nucleus of the atom.
6. According to the Bohr model, what is the energy of an electron in the first Bohr orbit of hydrogen?
7. According to the Bohr model, what is the energy of an electron in the tenth Bohr orbit of hydrogen?
8. According to the Bohr model, what is the energy of an electron in the seventh Bohr orbit of hydrogen?
9. If an electron in a hydrogen atom has an energy of −6.06×10−20 J, which Bohr orbit is it in?
10. If an electron in a hydrogen atom has an energy of −2.69×10−20 J, which Bohr orbit is it in?
11. If an electron falls from the 5th Bohr orbital of hydrogen to the 3rd Bohr orbital of hydrogen, how much energy is released (you can give the energy as a positive number)?
12. If an electron falls from the 6th Bohr orbital of hydrogen to the 3rd Bohr orbital of hydrogen, what wavelength of light is emitted? Is this in the visible light range?
9.6: Quantum-Mechanical Orbitals and Electron Configurations
1. Match each quantum number with the property that they describe.
Match each quantum number with the property that they describe.
(a) n i. shape
(b) ii. orientation in space
(c) ml iii. number of nodes
2. A point in an electron wave where there is zero electron density is called a _________.
3. Choose the correct word in each of the following statements.
(a) The (higher/lower) the value of n, the more nodes there are in the electron standing wave.
(b) The (higher/lower) the value of n, the less energy the electron has.
(c) The (more/less) energy the electron has, the more nodes there are in its electron standing wave.
4. Fill in the blank. For lower values of n, the electron density is typically found ________ the nucleus of the atom, while for higher values of n, the electron density is typically found __________the nucleus of the atom.
5. Circle all of the statements that make sense: Schrödinger discovered that certain quantities in the electron wave equation had to be integers, because when they weren't, the wave equation described waves which…
(a) were discontinuous
(b) were too small
(c) were too long and narrow
(d) were too short and fat
(e) "doubled back" on themselves
6. What are the allowed values of for an electron standing wave with n = 4?
7. How many values of are possible for an electron standing wave with n = 9?
8. What are the allowed values of ml for an electron standing wave with = 3?
9. How many different orientations are possible for an electron standing wave with = 4?
10. What are the allowed values of ml for n = 2?
11. Fill in the blanks. When = 0, the electron orbital is _________ and when = 1, the electron orbital is _________ shaped.
12. The n = 1 s orbital has _____ nodes.
13. The n = 2 s orbital has _____ nodes.
14. The n = 2 p orbital has _____ nodes.
15. The n = 1 p orbital has _____ nodes.
16. There are ____ different p orbitals.
17. What energy level (or value of n) has s, p and d orbitals, but no f orbitals?
18. How many different d orbital orientations are there?
19. How many f orbital orientations are there?
20. How many different orbitals are there in the n = 3 energy level?
1. Write the electron configuration for beryllium. Beryllium has 4 electrons.
2. Write the electron configuration for silicon. Silicon has 14 electrons.
3. Write the electron configuration for nitrogen. Nitrogen has 7 electrons.
4. Write the electron configuration for chromium. Chromium has 24 electrons.
5. Write the electron configuration for silver. Silver has 47 electrons.
9.7: Electron Configurations and the Periodic Table
1. Use the Periodic Table to determine the energy level of the valence electrons in each of the following elements.
(a) B
(b) Ga
(c) Rb
(d) At
(e) He
2. Fill in the blanks:
(a) B is in the __ level block of the Periodic Table
(b) Sr is in the __ level block of the Periodic Table
(c) Fe is in the __ level block of the Periodic Table
(d) Cs is in the __ level block of the Periodic Table
(e) O is in the __ level block of the Periodic Table
3. Use the Periodic Table to determine the energy level and sublevel of the highest energy electrons in each of the following elements:
(a) N
(b) Ca
(c) Rb
(d) P
(e) In
4. Decide whether each of the following statements is true or false.
(a) Li has valence electrons in the n = 1 energy level.
(b) Si has valence electrons in the n = 3 energy level.
(c) Ga has valence electrons in the n = 3 energy level.
(d) Xe has valence electrons in the n = 5 energy level.
(e) P has valence electrons in the n = 2 energy level.
5. Match the element to the sublevel block it is found in:
Match the element to the sublevel block it is found in:
(a) C i. s sublevel block
(b) Cs ii. p sublevel block
(c) Ce iii. d sublevel block
(d) Cr iv. f sublevel block
6. The first row of the Periodic Table has:
(a) 1 element
(b) 2 elements
(c) 3 elements
(d) 4 elements
(e) 5 elements
7. Use the Periodic Table to determine which of the following elements has the highest energy valence electrons.
(a) Sr
(b) As
(c) H
(d) At
(e) Na
8. Use the Periodic Table to determine which of the following elements has the lowest energy valence electrons.
(a) Ga
(b) B
(c) Cs
(d) Bi
(e) Cl
9. Which energy level does the first row in the d sublevel block correspond to?
9.9: Periodic Trends: Atomic Size, Ionization Energy, and Metallic Character
1. Why is the atomic size considered to have "no definite boundary"?
2. How is atomic size measured?
(a) using a spectrophotomer
(b) using a tiny ruler (called a nano ruler)
(c) indirectly
(d) directly
3. Draw a visual representation of the atomic radii of an iodine molecule.
4. Which of the following would be smaller?
(a) In or Ga
(b) K or Cs
(c) Te or Po
5. Explain in your own words why Iodine is larger than Bromine.
6. What three factors determine the trend of atomic size going down a group?
7. What groups tend to show this trend?
8. Which of the following would have the largest atomic radii?
(a) Si
(b) C
(c) Sn
(d) Pb
9. Which of the following would have the smallest atomic radius?
(a) 2s2
(b) 4s24p3
(c) 2s22p4
(d) 4s2
10. Arrange the following in order of increasing atomic radii: Tl, B, Ga, Al, In.
11. Arrange the following in order of increasing atomic radii: Ge, Sn, C.
12. Which of the following would be larger?
(a) Rb or Sn
(b) Ca or As
13. Place the following in order of increasing atomic radii: Mg, Cl, S, Na.
14. Describe the atomic size trend for the rows in the Periodic Table.
15. Draw a visual representation of the periodic table describing the trend of atomic size.
16. Which of the following would have the largest atomic radii?
(a) Sr
(b) Sn
(c) Rb
(d) In
17. Which of the following would have the smallest atomic radii?
(a) K
(b) Kr
(c) Ga
(d) Ge
18. Place the following elements in order of increasing atomic radii: In, Ca, Mg, Sb, Xe.
19. Place the following elements in order of decreasing atomic radii: Al, Ge, Sr, Bi, Cs.
20. Knowing the trend for the rows, what would you predict to be the effect on the atomic radius if an atom were to gain an electron? Use an example in your explanation.
21. Knowing the trend for the rows, what would you predict to be the effect on the atomic radius if the atom were to lose an electron? Use an example in your explanation.
Ionization Energy
1. Define ionization energy and show an example ionization equation.
2. Draw a visual representation of the periodic table describing the trend of ionization energy.
3. Which of the following would have the largest ionization energy?
(a) Na
(b) Al
(c) H
(d) He
4. Which of the following would have the smallest ionization energy?
(a) K
(b) P
(c) S
(d) Ca
5. Place the following elements in order of increasing ionization energy: Na, O, Ca, Ne, K.
6. Place the following elements in order of decreasing ionization energy: N, Si, S, Mg, He.
7. Using experimental data, the first ionization energy for an element was found to be 600 kJ/mol. The second ionization energy for the ion formed was found to be 1,800 kJ/mol. The third ionization energy for the ion formed was found to be 2,700 kJ/mol. The fourth ionization energy for the ion formed was found to be 11,600 kJ/mol. And finally the fifth ionization energy was found to be 15,000 kJ/mol. Write the reactions for the data represented in this question. To which group does this element belong? Explain.
8. Using electron configurations and your understanding of ionization energy, which would you predict would have higher second ionization energy: Na or Mg?
9. Comparing the first ionization energy (IE1) of calcium, Ca, and magnesium, Mg, :
(a) Ca has a higher IE1 because its radius is smaller.
(b) Mg has a higher IE1 because its radius is smaller.
(c) Ca has a higher IE1 because its outer sub-shell is full.
(d) Mg has a higher IE1 because its outer sub-shell is full.
(e) they have the same IE1 because they have the same number of valence electrons.
10. Comparing the first ionization energy (IE1) of beryllium, Be, and boron, B:
(a) Be has a higher IE1 because its radius is smaller.
(b) B has a higher IE1 because its radius is smaller.
(c) Be has a higher IE1 because its s sub-shell is full.
(d) B has a higher IE1 because its s sub-shell is full.
(e) They have the same IE1 because B has only one more electron than Be.
Electron Affinity
1. Define electron affinity and show an example equation.
2. Choose the element in each pair that has the lower electron affinity:
(a) Li or N
(b) Na or Cl
(c) Ca or K
(d) Mg or F
3. Why is the electron affinity for calcium much higher than that of potassium?
4. Draw a visual representation of the periodic table describing the trend of electron affinity.
5. Which of the following would have the largest electron affinity?
(a) Se
(b) F
(c) Ne
(d) Br
6. Which of the following would have the smallest electron affinity?
(a) Na
(b) Ne
(c) Al
(d) Rb
7. Place the following elements in order of increasing electron affinity: Te, Br, S, K, Ar.
8. Place the following elements in order of decreasing electron affinity: S, Sn, Pb, F, Cs.
9. Describe the trend that would occur for electron affinities for elements in Period 3. Are there any anomalies? Explain.
10. Comparing the electron affinity (EA) of sulfur, S, and phosphorus, P:
(a) S has a higher EA because its radius is smaller.
(b) P has a higher EA because its radius is smaller.
(c) S has a higher EA because its p sub-shell is half full.
(d) P has a higher EA because its p sub-shell is half full.
(e) they have the same EA because they are next to each other in the Periodic Table. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.E%3A_Electrons_in_Atoms_and_the_Periodic_Table_%28Exercises%29.txt |
How do atoms make compounds? Typically they join together in such a way that they lose their identities as elements and adopt a new identity as a compound. These joins are called chemical bonds. But how do atoms join together? Ultimately, it all comes down to electrons. Before we discuss how electrons interact, we need to introduce a tool to simply illustrate electrons in an atom.
• 10.1: Bonding Models and AIDs Drugs
• 10.2: Representing Valence Electrons with Dots
The Lewis Structure of a molecule shows how the valence electrons are arranged among the atoms of the molecule. Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol. Lewis electron dot diagrams for ions have less (for cations) or more (for anions) dots than the corresponding atom. From experimentation, chemists have learned that when a stable compound forms, the atoms usually have a noble gas electron configuration—or eight valence electrons.
• 10.3: Lewis Structures of Ionic Compounds- Electrons Transferred
The tendency to form species that have eight electrons in the valence shell is called the octet rule. The attraction of oppositely charged ions caused by electron transfer is called an ionic bond. The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions.
• 10.4: Covalent Lewis Structures- Electrons Shared
Covalent bonds are formed when atoms share electrons. Lewis electron dot diagrams can be drawn to illustrate covalent bond formation. Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules.
• 10.5: Writing Lewis Structures for Covalent Compounds
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms.
• 10.6: Resonance - Equivalent Lewis Structures for the Same Molecule
Resonance structures are averages of different Lewis structure possibilities. Bond lengths are intermediate between covalent bonds and covalent double bonds.
• 10.7: Predicting the Shapes of Molecules
The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms.
• 10.8: Electronegativity and Polarity - Why Oil and Water Do not Mix
Covalent bonds can be nonpolar or polar, depending on the electronegativities of the atoms involved. Covalent bonds can be broken if energy is added to a molecule. The formation of covalent bonds is accompanied by energy given off. Covalent bond energies can be used to estimate the enthalpy changes of chemical reactions.
10: Chemical Bonding
Learning Objective
• Draw a Lewis electron dot diagram for an atom or a monatomic ion.
In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful.
A Lewis electron dot diagram (or electron dot diagram, or a Lewis diagram, or a Lewis structure) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. (The order in which the positions are used does not matter.) For example, the Lewis electron dot diagram for hydrogen is simply
$\mathbf{H}\mathbf{\cdot} \nonumber$
Because the side is not important, the Lewis electron dot diagram could also be drawn as follows:
$\mathbf{\dot{H}}\; \; or\; \mathbf{\cdot}\mathbf{H}\; \; \; or\; \; \; \mathbf{\underset{.}H} \nonumber$
The electron dot diagram for helium, with two valence electrons, is as follows:
$\mathbf{He}\mathbf{:} \nonumber$
By putting the two electrons together on the same side, we emphasize the fact that these two electrons are both in the 1s subshell; this is the common convention we will adopt, although there will be exceptions later. The next atom, lithium, has an electron configuration of 1s22s1, so it has only one electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used:
$\mathbf{Li}\mathbf{\cdot} \nonumber$
Beryllium has two valence electrons in its 2s shell, so its electron dot diagram is like that of helium:
$\mathbf{Be}\mathbf{:} \nonumber$
The next atom is boron. Its valence electron shell is 2s22p1, so it has three valence electrons. The third electron will go on another side of the symbol:
$\mathbf{\dot{B}}\mathbf{:} \nonumber$
Again, it does not matter on which sides of the symbol the electron dots are positioned.
For carbon, there are four valence electrons, two in the 2s subshell and two in the 2p subshell. As usual, we will draw two dots together on one side, to represent the 2s electrons. However, conventionally, we draw the dots for the two p electrons on different sides. As such, the electron dot diagram for carbon is as follows:
$\mathbf{\cdot \dot{C}}\mathbf{:} \nonumber$
With N, which has three p electrons, we put a single dot on each of the three remaining sides:
$\mathbf{\cdot}\mathbf{\dot{\underset{.}N}}\mathbf{:} \nonumber$
For oxygen, which has four p electrons, we now have to start doubling up on the dots on one other side of the symbol. When doubling up electrons, make sure that each side has no more than two electrons.
$\mathbf{\cdot}\mathbf{\ddot{\underset{.}O}}\mathbf{:} \nonumber$
Fluorine and neon have seven and eight dots, respectively:
$\mathbf{:}\mathbf{\ddot{\underset{.}F}}\mathbf{:} \nonumber$
$\mathbf{:}\mathbf{\ddot{\underset{.\: .}Ne}}\mathbf{:} \nonumber$
With the next element, sodium, the process starts over with a single electron because sodium has a single electron in its highest-numbered shell, the n = 3 shell. By going through the periodic table, we see that the Lewis electron dot diagrams of atoms will never have more than eight dots around the atomic symbol.
Example $1$: Lewis Dot Diagrams
What is the Lewis electron dot diagram for each element?
1. aluminum
2. selenium
Solution
1. The valence electron configuration for aluminum is 3s23p1. So it would have three dots around the symbol for aluminum, two of them paired to represent the 3s electrons:
$\dot{Al:} \nonumber$
1. The valence electron configuration for selenium is 4s24p4. In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows:
$\mathbf{\cdot }\mathbf{\dot{\underset{.\: .}Se}}\mathbf{:} \nonumber$
Exercise $1$
What is the Lewis electron dot diagram for each element?
1. phosphorus
2. argon
Answer a
$\mathbf{\cdot }\mathbf{\dot{\underset{.}P}}\mathbf{:} \nonumber$
Answer b
$\mathbf{:}\mathbf{\ddot{\underset{.\, .}Ar}}\mathbf{:} \nonumber$
Summary
• Lewis electron dot diagrams use dots to represent valence electrons around an atomic symbol.
• Lewis electron dot diagrams for ions have less (for cations) or more (for anions) dots than the corresponding atom. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/10%3A_Chemical_Bonding/10.02%3A_Representing_Valence_Electrons_with_Dots.txt |
Learning Objectives
• State the octet rule.
• Define ionic bond.
• Draw Lewis structures for ionic compounds.
In Section 4.7, we demonstrated that ions are formed by losing electrons to make cations, or by gaining electrons to form anions. The astute reader may have noticed something: many of the ions that form have eight electrons in their valence shell. Either atoms gain enough electrons to have eight electrons in the valence shell and become the appropriately charged anion, or they lose the electrons in their original valence shell; the lower shell, now the valence shell, has eight electrons in it, so the atom becomes positively charged. For whatever reason, having eight electrons in a valence shell is a particularly energetically stable arrangement of electrons. The octet rule explains the favorable trend of atoms having eight electrons in their valence shell. When atoms form compounds, the octet rule is not always satisfied for all atoms at all times, but it is a very good rule of thumb for understanding the kinds of bonding arrangements that atoms can make.
It is not impossible to violate the octet rule. Consider sodium: in its elemental form, it has one valence electron and is stable. It is rather reactive, however, and does not require a lot of energy to remove that electron to make the Na+ ion. We could remove another electron by adding even more energy to the ion, to make the Na2+ ion. However, that requires much more energy than is normally available in chemical reactions, so sodium stops at a 1+ charge after losing a single electron. It turns out that the Na+ ion has a complete octet in its new valence shell, the n = 2 shell, which satisfies the octet rule. The octet rule is a result of trends in energies and is useful in explaining why atoms form the ions that they do.
Now consider an Na atom in the presence of a Cl atom. The two atoms have these Lewis electron dot diagrams and electron configurations:
$\mathbf{Na\, \cdot }\; \; \; \; \; \; \; \; \; \; \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :} \nonumber$
$\left [ Ne \right ]3s^{1}\; \; \; \; \left [ Ne \right ]3s^{2}3p^{5} \nonumber$
For the Na atom to obtain an octet, it must lose an electron; for the Cl atom to gain an octet, it must gain an electron. An electron transfers from the Na atom to the Cl atom:
$\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :} \nonumber$
resulting in two ions—the Na+ ion and the Cl ion:
$\mathbf{Na}^{+}\; \; \; \; \; \; \; \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-} \nonumber$
$\left [ Ne \right ]\; \; \; \; \; \left [ Ne \right ]3s^{2}3p^{6} \nonumber$
Both species now have complete octets, and the electron shells are energetically stable. From basic physics, we know that opposite charges attract. This is what happens to the Na+ and Cl ions:
$\mathbf{Na}^{+}\; + \; \mathbf{:}\mathbf{\ddot{\underset{.\: .}Cl}}\mathbf{\: :}^{-}\rightarrow Na^{+}Cl^{-}\; \; or\; \; NaCl \nonumber$
where we have written the final formula (the formula for sodium chloride) as per the convention for ionic compounds, without listing the charges explicitly. The attraction between oppositely charged ions is called an ionic bond, and it is one of the main types of chemical bonds in chemistry. Ionic bonds are caused by electrons transferring from one atom to another.
In electron transfer, the number of electrons lost must equal the number of electrons gained. We saw this in the formation of NaCl. A similar process occurs between Mg atoms and O atoms, except in this case two electrons are transferred:
The two ions each have octets as their valence shell, and the two oppositely charged particles attract, making an ionic bond:
$\mathbf{Mg\,}^{2+}\; + \; \left[\mathbf{:}\mathbf{\ddot{\underset{.\: .}O}}\mathbf{\: :}\right]^{2-}\; \; \; \; \; Mg^{2+}O^{2-}\; or\; MgO \nonumber$
Remember, in the final formula for the ionic compound, we do not write the charges on the ions.
What about when an Na atom interacts with an O atom? The O atom needs two electrons to complete its valence octet, but the Na atom supplies only one electron:
$\mathbf{Na\, \cdot }\curvearrowright \mathbf{\cdot }\mathbf{\ddot{\underset{.}O}}\mathbf{\: :} \nonumber$
The O atom still does not have an octet of electrons. What we need is a second Na atom to donate a second electron to the O atom:
These three ions attract each other to give an overall neutral-charged ionic compound, which we write as Na2O. The need for the number of electrons lost being equal to the number of electrons gained explains why ionic compounds have the ratio of cations to anions that they do. This is required by the law of conservation of matter as well.
Example $1$: Synthesis of Calcium Chloride from Elements
With arrows, illustrate the transfer of electrons to form calcium chloride from $Ca$ atoms and $Cl$ atoms.
Solution
A $Ca$ atom has two valence electrons, while a $Cl$ atom has seven electrons. A $Cl$ atom needs only one more to complete its octet, while $Ca$ atoms have two electrons to lose. Thus we need two $Cl$ atoms to accept the two electrons from one $Ca$ atom. The transfer process looks as follows:
The oppositely charged ions attract each other to make CaCl2.
Exercise $1$
With arrows, illustrate the transfer of electrons to form potassium sulfide from $K$ atoms and $S$ atoms.
Answer
Summary
• The tendency to form species that have eight electrons in the valence shell is called the octet rule.
• The attraction of oppositely charged ions caused by electron transfer is called an ionic bond.
• The strength of ionic bonding depends on the magnitude of the charges and the sizes of the ions. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/10%3A_Chemical_Bonding/10.03%3A_Lewis_Structures_of_Ionic_Compounds-_Electrons_Transferred.txt |
Learning Objectives
• Define covalent bond.
• Illustrate covalent bond formation with Lewis electron dot diagrams.
Ionic bonding typically occurs when it is easy for one atom to lose one or more electrons and another atom to gain one or more electrons. However, some atoms won’t give up or gain electrons easily. Yet they still participate in compound formation. How? There is another mechanism for obtaining a complete valence shell: sharing electrons. When electrons are shared between two atoms, they make a bond called a covalent bond.
Let us illustrate a covalent bond by using H atoms, with the understanding that H atoms need only two electrons to fill the 1s subshell. Each H atom starts with a single electron in its valence shell:
$\mathbf{H\, \cdot }\; \; \; \; \; \mathbf{\cdot \: H} \nonumber$
The two H atoms can share their electrons:
$\mathbf{H}\: \mathbf{: H} \nonumber$
We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom’s valence shell:
Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. (This explains why hydrogen is one of the diatomic elements.) For simplicity’s sake, it is not unusual to represent the covalent bond with a dash, instead of with two dots:
H–H
Because two atoms are sharing one pair of electrons, this covalent bond is called a single bond. As another example, consider fluorine. F atoms have seven electrons in their valence shell:
These two atoms can do the same thing that the H atoms did; they share their unpaired electrons to make a covalent bond.
Note that each F atom has a complete octet around it now:
We can also write this using a dash to represent the shared electron pair:
There are two different types of electrons in the fluorine diatomic molecule. The bonding electron pair makes the covalent bond. Each F atom has three other pairs of electrons that do not participate in the bonding; they are called lone pair electrons. Each F atom has one bonding pair and three lone pairs of electrons.
Covalent bonds can be made between different elements as well. One example is HF. Each atom starts out with an odd number of electrons in its valence shell:
The two atoms can share their unpaired electrons to make a covalent bond:
We note that the H atom has a full valence shell with two electrons, while the F atom has a complete octet of electrons.
Example $1$:
Use Lewis electron dot diagrams to illustrate the covalent bond formation in HBr.
Solution
HBr is very similar to HF, except that it has Br instead of F. The atoms are as follows:
The two atoms can share their unpaired electron:
Exercise $1$
Use Lewis electron dot diagrams to illustrate the covalent bond formation in Cl2.
Answer
When working with covalent structures, it sometimes looks like you have leftover electrons. You apply the rules you learned so far, and there are still some electrons that remain unattached. You can't just leave them there. So where do you put them?
Multiple Covalent Bonds
Some molecules are not able to satisfy the octet rule by making only single covalent bonds between the atoms. Consider the compound ethene, which has a molecular formula of $\ce{C_2H_4}$. The carbon atoms are bonded together, with each carbon also bonded to two hydrogen atoms.
two $\ce{C}$ atoms $= 2 \times 4 = 8$ valence electrons
four $\ce{H}$ atoms $= 4 \times 1 = 4$ valence electrons
total of 12 valence electrons in the molecule
If the Lewis electron dot structure was drawn with a single bond between the carbon atoms and with the octet rule followed, it would look like this:
This Lewis structure is incorrect because it contains a total of 14 electrons. However, the Lewis structure can be changed by eliminating the lone pairs on the carbon atoms and having to share two pairs instead of only one pair.
A double covalent bond is a covalent bond formed by atoms that share two pairs of electrons. The double covalent bond that occurs between the two carbon atoms in ethane can also be represented by a structural formula and with a molecular model as shown in the figure below.
A triple covalent bond is a covalent bond formed by atoms that share three pairs of electrons. The element nitrogen is a gas that composes the majority of Earth's atmosphere. A nitrogen atom has five valence electrons, which can be shown as one pair and three single electrons. When combining with another nitrogen atom to form a diatomic molecule, the three single electrons on each atom combine to form three shared pairs of electrons.
Each nitrogen atom follows the octet rule with one lone pair of electrons, and six electrons that are shared between the atoms.
Summary
• Covalent bonds are formed when atoms share electrons.
• Lewis electron dot diagrams can be drawn to illustrate covalent bond formation.
• Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules.
Contributions & Attributions
• Anonymous by request | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/10%3A_Chemical_Bonding/10.04%3A_Covalent_Lewis_Structures-_Electrons_Shared.txt |
Learning Objectives
• Draw Lewis structures for covalent compounds.
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions.
How-to: Constructing Lewis electron structures
1. Determine the total number of valence electrons in the molecule or ion.
• Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.)
• If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion.
For CO32, for example, we add two electrons to the total because of the −2 charge.
2. Arrange the atoms to show specific connections.
• When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32, which both have C as the central atom), which is another clue to the compound’s structure.
• Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
• In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
• These electrons will usually be lone pairs.
5. If any electrons are left over, place them on the central atom.
• We will explain later that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet.
• This will not change the number of electrons on the terminal atoms.
7. Final check
• Always make sure all valence electrons are accounted for and that each atom has an octet of electrons, except for hydrogen (with two electrons).
• The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
Example $1$: Water
Write the Lewis Structure for H2O.
Solution
Solutions to Example 10.4.1
Steps for Writing Lewis Structures Example $1$
1. Determine the total number of valence electrons in the molecule or ion. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons.
2. Arrange the atoms to show specific connections.
H O H
Because H atoms are almost always terminal, the arrangement within the molecule must be HOH.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
Placing one bonding pair of electrons between the O atom and each H atom gives
H -O- H
with 4 electrons left over.
Each H atom has a full valence shell of 2 electrons.
5. If any electrons are left over, place them on the central atom.
Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure:
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. Not necessary.
7. Final check. The Lewis structure gives oxygen an octet and each hydrogen 2 electrons.
Example $2$
Write the Lewis structure for the $CH_2O$ molecule
Solution
Solutions to Example 10.4.2
Steps for Writing Lewis Structures Example $2$
1. Determine the total number of valence electrons in the molecule or ion. Each hydrogen atom (group 1) has 1 valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons.
2. Arrange the atoms to show specific connections.
Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
Placing a bonding pair of electrons between each pair of bonded atoms gives the following:
6 electrons are used, and 6 are left over.
4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following:
Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons.
5. If any electrons are left over, place them on the central atom.
Not necessary.
There are no electrons left to place on the central atom.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet.
To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond:
7. Final check Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid.
Exercise $1$
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Answer CO2
.
Answer SCl2
.
The United States Supreme Court has the unenviable task of deciding what the law is. This responsibility can be a major challenge when there is no clear principle involved or where there is a new situation not encountered before. Chemistry faces the same challenge in extending basic concepts to fit a new situation. Drawing of Lewis structures for polyatomic ions uses the same approach, but tweaks the process a little to fit a somewhat different set of circumstances.
Writing Lewis Structures for Polyatomic Ions (CK-12)
Recall that a polyatomic ion is a group of atoms that are covalently bonded together and which carry an overall electrical charge. The ammonium ion, $\ce{NH_4^+}$, is formed when a hydrogen ion $\left( \ce{H^+} \right)$ attaches to the lone pair of an ammonia $\left( \ce{NH_3} \right)$ molecule in a coordinate covalent bond.
When drawing the Lewis structure of a polyatomic ion, the charge of the ion is reflected in the number of total valence electrons in the structure. In the case of the ammonium ion:
$1 \: \ce{N}$ atom $= 5$ valence electrons
$4 \: \ce{H}$ atoms $= 4 \times 1 = 4$ valence electrons
subtract 1 electron for the $1+$ charge of the ion
total of 8 valence electrons in the ion
It is customary to put the Lewis structure of a polyatomic ion into a large set of brackets, with the charge of the ion as a superscript outside of the brackets.
Exercise $2$
Draw the Lewis electron dot structure for the sulfate ion.
Answer (CK12 License)
Exceptions to the Octet Rule (BC Campus)
As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations.
There are three violations to the octet rule. Odd-electron molecules represent the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are NO, NO2, and ClO2. The Lewis electron dot diagram for NO is as follows:
Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds.
Electron-deficient molecules represent the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell:
Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the B atom. A well-known example is BF3:
The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is PF5. The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds:
Formally, the P atom has 10 electrons in its valence shell.
Example $3$: Octet Violations
Identify each violation to the octet rule by drawing a Lewis electron dot diagram.
1. ClO
2. SF6
Solution
a. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows:
b. In SF6, the central S atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows:
Exercise $3$: Xenon Difluoride
Identify the violation to the octet rule in XeF2 by drawing a Lewis electron dot diagram.
Answer
The Xe atom has an expanded valence shell with more than eight electrons around it.
Summary
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms. Lewis structures for polyatomic ions follow the same rules as those for other covalent compounds. There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/10%3A_Chemical_Bonding/10.05%3A_Writing_Lewis_Structures_for_Covalent_Compounds.txt |
Learning Objectives
• Explain the concept of resonance and how it works with within molecules.
Resonance
There are some cases in which more than one viable Lewis structure can be drawn for a molecule. An example is the ozone $\left( \ce{O_3} \right)$ molecule in Figure $1$. There are a total of 18 electrons in the structure and so the following two structures are possible.
The structure on the left ($1$) can be converted to the structure on the right by a shifting of electrons without altering the positions of the atoms.
It was once thought that the structure of a molecule such as $\ce{O_3}$ consisted of one single bond and one double bond which then shifted back and forth as shown above. However, further studies showed that the two bonds are identical. Any double covalent bond between two given atoms is typically shorter than a single covalent bond. Studies of the $\ce{O_3}$ and other similar molecules showed that the bonds were identical in length. Interestingly, the length of the bond is in between the lengths expected for an $\ce{O-O}$ single bond and a double bond.
Resonance is the use of two or more Lewis structures to represent the covalent bonding in a molecule. One of the valid structures is referred to as a resonance structure. It is now understood that the true structure of a molecule which displays resonance is that of an average or a hybrid of all the resonance structures. In the case of the $\ce{O_3}$ molecule, each of the covalent bonds between $\ce{O}$ atoms are best thought of as being "one and a half" bonds, as opposed to either a pure single bond or a pure double bond. This "half-bond" can be shown as a dotted line in both the Lewis structure and the molecular model (Figure $2$).
Many polyatomic ions also display resonance. In some cases, the true structure may be an average of three valid resonance structures, as in the case of the nitrate ion, $\ce{NO_3^-}$ in Figure $3$.
The bond lengths between the central $\ce{N}$ atom and each $\ce{O}$ atom are identical and the bonds can be approximated as being equal to one and one-third bonds.
Summary
• Resonance structures are averages of different Lewis structure possibilities.
• Bond lengths are intermediate between covalent bonds and covalent double bonds.
10.07: Predicting the Shapes of Molecules
Learning Objective
• Determine the shape of simple molecules.
Molecules have shapes. There is an abundance of experimental evidence to that effect—from their physical properties to their chemical reactivity. Small molecules—molecules with a single central atom—have shapes that can be easily predicted. The basic idea in molecular shapes is called valence shell electron pair repulsion (VSEPR). VSEPR says that electron pairs, being composed of negatively charged particles, repel each other to get as far away from one another as possible. VSEPR makes a distinction between electron group geometry, which expresses how electron groups (bonds and nonbonding electron pairs) are arranged, and molecular geometry, which expresses how the atoms in a molecule are arranged. However, the two geometries are related.
There are two types of electron groups: any type of bond—single, double, or triple—and lone electron pairs. When applying VSEPR to simple molecules, the first thing to do is to count the number of electron groups around the central atom. Remember that a multiple bond counts as only one electron group.
Any molecule with only two atoms is linear. A molecule whose central atom contains only two electron groups orients those two groups as far apart from each other as possible—180° apart. When the two electron groups are 180° apart, the atoms attached to those electron groups are also 180° apart, so the overall molecular shape is linear. Examples include BeH2 and CO2:
The two molecules, shown in the figure below in a "ball and stick" model.
A molecule with three electron groups orients the three groups as far apart as possible. They adopt the positions of an equilateral triangle—120° apart and in a plane. The shape of such molecules is trigonal planar. An example is BF3:
Some substances have a trigonal planar electron group distribution but have atoms bonded to only two of the three electron groups. An example is GeF2:
From an electron group geometry perspective, GeF2 has a trigonal planar shape, but its real shape is dictated by the positions of the atoms. This shape is called bent or angular.
A molecule with four electron groups about the central atom orients the four groups in the direction of a tetrahedron, as shown in Figure \(1\) Tetrahedral Geometry. If there are four atoms attached to these electron groups, then the molecular shape is also tetrahedral. Methane (CH4) is an example.
This diagram of CH4 illustrates the standard convention of displaying a three-dimensional molecule on a two-dimensional surface. The straight lines are in the plane of the page, the solid wedged line is coming out of the plane toward the reader, and the dashed wedged line is going out of the plane away from the reader.
NH3 is an example of a molecule whose central atom has four electron groups, but only three of them are bonded to surrounding atoms.
Although the electron groups are oriented in the shape of a tetrahedron, from a molecular geometry perspective, the shape of NH3 is trigonal pyramidal.
H2O is an example of a molecule whose central atom has four electron groups, but only two of them are bonded to surrounding atoms.
Although the electron groups are oriented in the shape of a tetrahedron, the shape of the molecule is bent or angular. A molecule with four electron groups about the central atom, but only one electron group bonded to another atom, is linear because there are only two atoms in the molecule.
Double or triple bonds count as a single electron group. The Lewis electron dot diagram of formaldehyde (CH2O) is shown in Figure \(9\).
The central C atom has three electron groups around it because the double bond counts as one electron group. The three electron groups repel each other to adopt a trigonal planar shape.
(The lone electron pairs on the O atom are omitted for clarity.) The molecule will not be a perfect equilateral triangle because the C–O double bond is different from the two C–H bonds, but both planar and triangular describe the appropriate approximate shape of this molecule.
Table \(1\) summarizes the shapes of molecules based on the number of electron groups and surrounding atoms.
Table \(1\): Summary of Molecular Shapes
Number of Electron Groups on Central Atom Number of Bonding Groups Number of Lone Pairs Electron Geometry Molecular Shape
2 2 0 linear linear
3 3 0 trigonal planar trigonal planar
3 2 1 trigonal planar bent
4 4 0 tetrahedral tetrahedral
4 3 1 tetrahedral trigonal pyramidal
4 2 2 tetrahedral bent
Example \(1\)
What is the approximate shape of each molecule?
1. PCl3
2. NOF
Solution
The first step is to draw the Lewis structure of the molecule.
For \(\ce{PCl3}\), the electron dot diagram is as follows:
The lone electron pairs on the Cl atoms are omitted for clarity. The P atom has four electron groups with three of them bonded to surrounding atoms, so the molecular shape is trigonal pyramidal.
The electron dot diagram for \(\ce{NOF}\) is as follows:
The N atom has three electron groups on it, two of which are bonded to other atoms. The molecular shape is bent.
Exercise \(1\)
What is the approximate molecular shape of \(\ce{CH2Cl2}\)?
Answer
Tetrahedral
Exercise \(2\)
Ethylene (\(\ce{C2H4}\)) has two central atoms. Determine the geometry around each central atom and the shape of the overall molecule. (Hint: hydrogen is a terminal atom.)
Answer
Trigonal planar about both central C atoms.
Summary
The approximate shape of a molecule can be predicted from the number of electron groups and the number of surrounding atoms. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/10%3A_Chemical_Bonding/10.06%3A_Resonance_-_Equivalent_Lewis_Structures_for_the_Same_Molecule.txt |
Learning Objectives
• Explain how polar compounds differ from nonpolar compounds.
• Determine if a molecule is polar or nonpolar.
• Given a pair of compounds, predict which would have a higher melting or boiling point.
Bond Polarity
The ability of an atom in a molecule to attract shared electrons is called electronegativity. When two atoms combine, the difference between their electronegativities is an indication of the type of bond that will form. If the difference between the electronegativities of the two atoms is small, neither atom can take the shared electrons completely away from the other atom, and the bond will be covalent. If the difference between the electronegativities is large, the more electronegative atom will take the bonding electrons completely away from the other atom (electron transfer will occur), and the bond will be ionic. This is why metals (low electronegativities) bonded with nonmetals (high electronegativities) typically produce ionic compounds.
A bond may be so polar that an electron actually transfers from one atom to another, forming a true ionic bond. How do we judge the degree of polarity? Scientists have devised a scale called electronegativity, a scale for judging how much atoms of any element attract electrons. Electronegativity is a unitless number; the higher the number, the more an atom attracts electrons. A common scale for electronegativity is shown in Figure $1$.
The polarity of a covalent bond can be judged by determining the difference of the electronegativities of the two atoms involved in the covalent bond, as summarized in the following table:
difference of the electronegativities of the two atoms involved in the covalent bond
Electronegativity Difference Bond Type
0–0.4 pure covalent
0.5–2.0 polar covalent
>2.0 likely ionic
Nonpolar Covalent Bonds
A bond in which the electronegativity difference is less than 1.9 is considered to be mostly covalent in character. However, at this point we need to distinguish between two general types of covalent bonds. A nonpolar covalent bond is a covalent bond in which the bonding electrons are shared equally between the two atoms. In a nonpolar covalent bond, the distribution of electrical charge is balanced between the two atoms.
The two chlorine atoms share the pair of electrons in the single covalent bond equally, and the electron density surrounding the $\ce{Cl_2}$ molecule is symmetrical. Also note that molecules in which the electronegativity difference is very small (<0.5) are also considered nonpolar covalent. An example would be a bond between chlorine and bromine ($\Delta$EN $=3.0 - 2.8 = 0.2$).
Polar Covalent Bonds
A bond in which the electronegativity difference between the atoms is between 0.5 and 2.0 is called a polar covalent bond. A polar covalent bond is a covalent bond in which the atoms have an unequal attraction for electrons and so the sharing is unequal. In a polar covalent bond, sometimes simply called a polar bond, the distribution of electrons around the molecule is no longer symmetrical.
An easy way to illustrate the uneven electron distribution in a polar covalent bond is to use the Greek letter delta $\left( \delta \right)$.
The atom with the greater electronegativity acquires a partial negative charge, while the atom with the lesser electronegativity acquires a partial positive charge. The delta symbol is used to indicate that the quantity of charge is less than one. A crossed arrow can also be used to indicate the direction of greater electron density.
Electronegativity differences in bonding using Pauling scale. Differences in electronegativity classify bonds as covalent, polar covalent, or ionic.
Example $1$: Bond Polarity
What is the polarity of each bond?
1. C–H
2. O–H
Solution
Using Figure $1$, we can calculate the difference of the electronegativities of the atoms involved in the bond.
1. For the C–H bond, the difference in the electronegativities is 2.5 − 2.1 = 0.4. Thus we predict that this bond will be nonpolar covalent.
2. For the O–H bond, the difference in electronegativities is 3.5 − 2.1 = 1.4, so we predict that this bond will be polar covalent.
Exercise $1$
What is the polarity of each bond?
1. Rb–F
2. P–Cl
Answer a
likely ionic
Answer b
polar covalent
Molecular Polarity
To determine if a molecule is polar or nonpolar, it is generally useful to look at Lewis structures. Nonpolar compounds will be symmetric, meaning all of the sides around the central atom are identical—bonded to the same element with no unshared pairs of electrons. Polar molecules are asymmetric, either containing lone pairs of electrons on a central atom or having atoms with different electronegativities bonded. This works pretty well, as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming that you do, you can look at the structure of each one and decide if it is polar or not, whether or not you know the individual atom's electronegativity. This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in).
A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as $\ce{HF}$, is a polar molecule. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. A molecule with two poles is called a dipole (see figure below). Hydrogen fluoride is a dipole.
For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. The figure below shows a comparison between carbon dioxide and water. Carbon dioxide $\left( \ce{CO_2} \right)$ is a linear molecule. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the $\ce{C}$ atom to each $\ce{O}$ atom. However, since the dipoles are of equal strength and are oriented this way, they cancel out and the overall molecular polarity of $\ce{CO_2}$ is zero.
Water is a bent molecule because of the two lone pairs on the central oxygen atom. The individual dipoles point from the $\ce{H}$ atoms toward the $\ce{O}$ atom. Because of the shape, the dipoles do not cancel each other out and the water molecule is polar. In the figure below, the net dipole is shown in blue and points upward.
Some other molecules are shown in the figure below. Notice that a tetrahedral molecule such as $\ce{CH_4}$ is nonpolar. However, if one of the peripheral $\ce{H}$ atoms is replaced with another atom that has a different electronegativity, the molecule becomes polar. A trigonal planar molecule $\left( \ce{BF_3} \right)$ may be nonpolar if all three peripheral atoms are the same, but a trigonal pyramidal molecule $\left( \ce{NH_3} \right)$ is polar.
To summarize, to be polar, a molecule must:
1. Contain at least one polar covalent bond.
2. Have a molecular structure such that the sum of the vectors of each bond dipole moment do not cancel.
Steps to Identify Polar Molecules
1. Draw the Lewis structure.
2. Figure out the geometry (using VSEPR theory).
3. Visualize or draw the geometry.
4. Find the net dipole moment (you don't have to actually do calculations if you can visualize it).
5. If the net dipole moment is zero, it is non-polar. Otherwise, it is polar.
Properties of Polar Molecules
Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure $14$). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances.
While molecules can be described as "polar covalent" or "ionic", it must be noted that this is often a relative term, with one molecule simply being more polar or less polar than another. However, the following properties are typical of such molecules. Polar molecules tend to:
• have higher melting points than nonpolar molecules
• have higher boiling points than nonpolar molecules
• be more soluble in water (dissolve better) than nonpolar molecules
• have lower vapor pressures than nonpolar molecules
Example $2$:
Label each of the following as polar or nonpolar.
1. Water, H2O:
2. Methanol, CH3OH:
3. Hydrogen Cyanide, HCN:
4. Oxygen, O2:
5. Propane, C3H8:
Solution
1. Water is polar. Any molecule with lone pairs of electrons around the central atom is polar.
2. Methanol is polar. This is not a symmetric molecule. The $\ce{-OH}$ side is different from the other 3 $\ce{-H}$ sides.
3. Hydrogen cyanide is polar. The molecule is not symmetric. The nitrogen and hydrogen have different electronegativities, creating an uneven pull on the electrons.
4. Oxygen is nonpolar. The molecule is symmetric. The two oxygen atoms pull on the electrons by exactly the same amount.
5. Propane is nonpolar, because it is symmetric, with $\ce{H}$ atoms bonded to every side around the central atoms and no unshared pairs of electrons.
Exercise $2$
Label each of the following as polar or nonpolar.
1. $\ce{SO3}$
2. $\ce{NH3}$
Answer a
nonpolar
Answer b
polar | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/10%3A_Chemical_Bonding/10.08%3A_Electronegativity_and_Polarity_-_Why_Oil_and_Water_Do_not_Mix.txt |
Of the three basic phases of matter—solids, liquids, and gases—only one of them has predictable physical properties: gases. In fact, the study of the properties of gases was the beginning of the development of modern chemistry from its alchemical roots. The interesting thing about some of these properties is that they are independent of the identity of the gas. That is, it doesn’t matter if the gas is helium gas, oxygen gas, or sulfur vapors; some of their behavior is predictable and very similar. In this chapter, we will review some of the common behaviors of gases. Gases have no definite shape or volume; they tend to fill whatever container they are in. They can compress and expand, sometimes to a great extent. Gases have extremely low densities, a one-thousandth or less of the density of a liquid or solid. Combinations of gases tend to mix together spontaneously—that is, they form solutions. Air, for example, is a solution of mostly nitrogen and oxygen. Any understanding of the properties of gases must be able to explain these characteristics.
• 11.1: Extra-Long Straws
Straws work because sucking creates a pressure difference between the inside of the straw and the outside. If, when drinking orange soda, you formed a perfect vacuum within the straw, the pressure outside of the straw at sea level would be enough to push the orange soda (which is mostly water) to a total height of about 10.3 m.
• 11.2: Kinetic Molecular Theory- A Model for Gases
The physical behavior of gases is explained by the kinetic theory of gases. An ideal gas adheres exactly to the kinetic theory of gases.
• 11.3: Pressure - The Result of Constant Molecular Collisions
Pressure is a force exerted over an area. Pressure has several common units that can be converted.
• 11.4: Boyle’s Law - Pressure and Volume
Boyle’s Law relates the pressure and volume of a gas at constant temperature and amount.
• 11.5: Charles’s Law- Volume and Temperature
Charles’s Law relates the volume and temperature of a gas at constant pressure and amount. In gas laws, temperatures must always be expressed in kelvins.
• 11.6: The Combined Gas Law- Pressure, Volume, and Temperature
There are gas laws that relate any two physical properties of a gas. The Combined Gas Law relates pressure, volume, and temperature of a gas.
• 11.7: Avogadro’s Law- Volume and Moles
The original statement of Avogadro’s law states that equal volumes of different gases at the same temperature and pressure contain the same number of particles of gas. Because the number of particles is related to the number of moles, Avogadro’s law essentially states that equal volumes of different gases at the same temperature and pressure contain the same amount (moles, particles) of gas.
• 11.8: The Ideal Gas Law- Pressure, Volume, Temperature, and Moles
The Ideal Gas Law relates the four independent physical properties of a gas at any time. The Ideal Gas Law can be used in stoichiometry problems with chemical reactions involving gases. Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases. At STP, gases have a volume of 22.4 L per mole. The Ideal Gas Law can be used to determine densities of gases.
• 11.9: Mixtures of Gases - Why Deep-Sea Divers Breathe a Mixture of Helium and Oxygen
The pressure of a gas in a gas mixture is termed the partial pressure. Dalton’s Law of Partial Pressures states that the total pressure in a gas mixture is the sum of the individual partial pressures. Collecting gases over water requires that we take the vapor pressure of water into account. Mole fraction is another way to express the amounts of components in a mixture.
• 11.11: Gay-Lussac's Law- Temperature and Pressure
Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac's Law is very similar to Charles's Law, with the only difference being the type of container. Whereas the container in a Charles's Law experiment is flexible, it is rigid in a Gay-Lussac's Law experiment.
11: Gases
A drinking straw is a tube for transferring a beverage from its container to the mouth of the drinker and is typically a thin tube of plastic (such as polypropylene and polystyrene) or other material. Many people believe that when they drink a liquid they are sucking the liquid up, however the liquid is really being pushed up. A straw works because when you suck the air out of the straw, it creates a vacuum. This causes a decrease in air pressure on the inside of the straw. Since the atmospheric pressure is greater on the outside of the straw, liquid is forced into and up the straw and into your mouth (Figure \(1\)).
How Long of a Straw is Possible?
With the straw just sitting in the glass, the pressure on the surface of the tea is the same all over, including on the little bit of surface inside the straw. When you suck the air out of the straw, you decrease the pressure inside the straw, allowing the higher pressure on the rest of the surface to push the tea up the straw and into your mouth. Because it is really the atmosphere that is doing the pushing, the atmospheric pressure limits how high water will go up a straw.
If you formed a perfect vacuum within the straw, the pressure outside of the straw at sea level would be enough to push water to a total height of about 10.3 m. A 10.3-m column of water exerts the same pressure—101,325 N/m2 or 14.7 lb/in2 (psi)—as do the gas molecules in our atmosphere. At sea level, the air pressure is enough to support a column of water about thirty feet high. This means that even if you could suck all the air out of a forty-foot straw, the water would not rise more than thirty feet.
Contributions & Attributions
• Lisa Peck's Conceptual Physics Class | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/11%3A_Gases/11.01%3A_Extra-Long_Straws.txt |
Learning Objectives
• State the major concepts behind the kinetic theory of gases.
• Relate the general properties of gases to the kinetic theory.
Gases were among the first substances studied in terms of the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviors, suggesting that all gases could be described by one all-encompassing theory. Today, that theory is the kinetic theory of gases. It is based on the following statements:
1. Gases consist of tiny particles of matter that are in constant motion.
2. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic; that is, there is no net loss of energy from the collisions.
3. Gas particles are separated by large distances, with the size of a gas particle tiny compared to the distances that separate them.
4. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.
5. The average speed of gas particles is dependent on the temperature of the gas.
Figure \(1\) shows a representation of how we mentally picture the gas phase.
This model of gases explains some of the physical properties of gases. Because most of a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix, as the particles from the individual gases move and collide with each other.
An ideal gas is a gas that exactly follows the statements of the kinetic theory. Unfortunately, real gases are not ideal. Many gases deviate slightly from agreeing perfectly with the kinetic theory of gases. However, most gases adhere to the statements so well that the kinetic theory of gases is well accepted by the scientific community.
• The physical behavior of gases is explained by the kinetic theory of gases.
• An ideal gas adheres exactly to the kinetic theory of gases.
11.03: Pressure - The Result of Constant Molecular Collisions
Learning Objectives
• Define pressure.
• Learn the units of pressure and how to convert between them.
The kinetic theory of gases indicates that gas particles are always in motion and are colliding with other particles and the walls of the container holding them. Although collisions with container walls are elastic (i.e., there is no net energy gain or loss because of the collision), a gas particle does exert a force on the wall during the collision. The accumulation of all these forces distributed over the area of the walls of the container causes something we call pressure. Pressure ($P$) is defined as the force of all the gas particle/wall collisions divided by the area of the wall:
$\text{pressure}=\dfrac{\text{force}}{\text{area}} \nonumber$
All gases exert pressure; it is one of the fundamental measurable quantities of this phase of matter. Even our atmosphere exerts pressure—in this case, the gas is being “held in” by the earth’s gravity, rather than the gas being in a container. The pressure of the atmosphere is about 14.7 pounds of force for every square inch of surface area: 14.7 lb/in2.
Pressure has a variety of units. The formal, SI-approved unit of pressure is the pascal (Pa), which is defined as 1 N/m2 (one newton of force over an area of one square meter). However, this is usually too small in magnitude to be useful. A common unit of pressure is the atmosphere (atm), which was originally defined as the average atmospheric pressure at sea level.
However, “average atmospheric pressure at sea level” is difficult to pinpoint because of atmospheric pressure variations. A more reliable and common unit is millimeters of mercury (mmHg), which is the amount of pressure exerted by a column of mercury exactly 1 mm high. An equivalent unit is the torr, which equals 1 mmHg. (The torr is named after Evangelista Torricelli, a seventeenth-century Italian scientist who invented the mercury barometer.) With these definitions of pressure, the atmosphere unit is redefined: 1 atm is defined as exactly 760 mmHg, or 760 torr. We thus have the following equivalents:
1 atm=760 mmHg=760 torr
We can use these equivalents as with any equivalence—to perform conversions from one unit to another. Relating these to the formal SI unit of pressure, 1 atm = 101,325 Pa.
Example $1$: Pressure Conversion
How many atmospheres are there in 595 torr?
Solution
Solutions to Example 11.3.1
Steps for Problem Solving Unit Conversion
Identify the "given” information and what the problem is asking you to "find."
Given: 595 torr
Find: ? atm
List other known quantities. 1 atm = 760 torr
Prepare a concept map.
Cancel units and calculate. $595\, \cancel{torr}\times \dfrac{1\, atm}{760\, \cancel{torr}}=0.783\, atm$
Think about your result. 595 torr is less than 760 torr so the final answer should be less than 1 atm.
Exercise $1$
How many atmospheres are there in 1,022 torr?
Answer
1.345 atm
Example $2$: Mars
The atmosphere on Mars is largely CO2 at a pressure of 6.01 mmHg. What is this pressure in atmospheres?
Solution
Solutions to Example 11.3.2
Steps for Problem Solving Unit Conversion
Identify the "given” information and what the problem is asking you to "find."
Given: 6.01mmHg
Find: ? atm
List other known quantities. 1 atm = 760 mmHg
Prepare a concept map.
Cancel units and calculate. $6.01\, \cancel{mmHg}\times \dfrac{1\, atm}{760\, \cancel{mmHg}}=0.00791\, atm=7.91\times 10^{-3}atm$
Think about your result. 6.01 is a very small number relative to 760 mmHg, just like the value in atmospheres.
Exercise $2$
Atmospheric pressure is low in the eye of a hurricane. In a 1979 hurricane in the Pacific Ocean, a pressure of 0.859 atm was reported inside the eye. What is this pressure in torr?
Answer
652 torr
Summary
• Pressure is a force exerted over an area.
• Pressure has several common units that can be converted. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/11%3A_Gases/11.02%3A_Kinetic_Molecular_Theory-_A_Model_for_Gases.txt |
Learning Objectives
• Learn what is meant by the term gas laws.
• Learn and apply Boyle’s Law.
When seventeenth-century scientists began studying the physical properties of gases, they noticed some simple relationships between some of the measurable properties of the gas. Take pressure (P) and volume (V), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [n]), if the temperature (T) of the gas was kept constant, pressure and volume were related: as one increases, the other decreases. As one decreases, the other increases. This means that pressure and volume are inversely related.
There is more to it, however: pressure and volume of a given amount of gas at constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature:
$P × V = \text{ constant at constant n and T} \nonumber$
If either volume or pressure changes while amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labeled $P_1$ and $V_1$ and the new conditions are labeled $P_2$ and $V_2$, we have
$P_1V_1 = \text{constant} = P_2V_2 \nonumber$
where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply
$P_1V_1 = P_2V_2 \text{ at constant n and T} \nonumber$
This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behavior of a gas. This particular gas law is called Boyle's Law, after the English scientist Robert Boyle, who first announced it in 1662. Figure $1$ shows two representations of how Boyle’s Law works.
Boyle’s Law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle’s Law there are two pressure variables; they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won’t matter what the unit is, but the unit must be the same on both sides of the equation.
Example $1$
A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant?
Solution
Solutions to Example 11.8.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: P1 = 2.44 atm and V1 = 4.01 L
P2 = 1.93 atm
Find: V2 = ? L
List other known quantities. none
Plan the problem.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \dfrac{P_1 \times V_1}{P_2} \nonumber$
Cancel units and calculate.
Now substitute the known quantities into the equation and solve.
$V_2 = \dfrac{2.44 \: \cancel{\text{atm}} \times 4.01 \: \text{L}}{1.93 \: \cancel{atm}} = 5.07 \: \text{L} \nonumber$
Think about your result. We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle’s Law.
Exercise $1$
If P1 = 334 torr, V1 = 37.8 mL, and P2 = 102 torr, what is V2?
Answer
124 mL
As mentioned, you can use any units for pressure and volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.
Example $2$:
A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure?
Solution
Solutions to Example 11.8.2
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: P1 = 722 torr and V1 = 88.8 mL
V2 = 0.633 L
Find: P2 = ? torr
List other known quantities. 1 L = 1000 mL to have the same units for volume.
Plan the problem.
1. Perform the conversion of the second volume unit from L to mL.
2. Rearrange the equation algebraically to solve for $P_2$.
$P_2 = \dfrac{P_1 \times V_1}{V_2} \nonumber$
Cancel units and calculate.
1. $0.663\, \cancel{L}\times \dfrac{1000\, ml}{1\, \cancel{L}}=663\, ml \nonumber$
2. Substitute the known quantities into the equation and solve. $P_2 = \dfrac{722 \: \text{torr} \times 88.8 \: \cancel{\text{mL}}}{663 \: \cancel{\text{mL}}} = 96.7 \: \text{torr} \nonumber$
Think about your result. When the volume increased, the pressure decreased, which is as expected for Boyle’s Law.
Exercise $2$
If V1 = 456 mL, P1 = 308 torr, and P2 = 1.55 atm, what is V2?
Answer
119 mL
Summary
• The behavior of gases can be modeled with gas laws.
• Boyle’s Law relates the pressure and volume of a gas at constant temperature and amount. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/11%3A_Gases/11.04%3A_Boyles_Law_-_Pressure_and_Volume.txt |
Learning Objectives
• Learn and apply Charles's Law.
Everybody enjoys the smell and taste of freshly-baked bread. It is light and fluffy as a result of the action of yeast on sugar. The yeast converts the sugar to carbon dioxide, which at high temperatures causes the dough to expand. The end result is an enjoyable treat, especially when covered with melted butter.
Charles's Law
French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stop of molecular motion.
Mathematically, the direct relationship of Charles's Law can be represented by the following equation:
$\dfrac{V}{T} = k \nonumber$
As with Boyle's Law, $k$ is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.
Temperature $\left( \text{K} \right)$ Volume $\left( \text{mL} \right)$ $\dfrac{V}{T} = k$ $\left( \dfrac{\text{mL}}{\text{K}} \right)$
Table $1$: Temperature-Volume Data
50 20 0.40
100 40 0.40
150 60 0.40
200 80 0.40
300 120 0.40
500 200 0.40
1000 400 0.40
When this data is graphed, the result is a straight line, indicative of a direct relationship, shown in the figure below.
Notice that the line goes exactly toward the origin, meaning that as the absolute temperature of the gas approaches zero, its volume approaches zero. However, when a gas is brought to extremely cold temperatures, its molecules would eventually condense into the liquid state before reaching absolute zero. The temperature at which this change into the liquid state occurs varies for different gases.
Charles's Law can also be used to compare changing conditions for a gas. Now we use $V_1$ and $T_1$ to stand for the initial volume and temperature of a gas, while $V_2$ and $T_2$ stand for the final volume and temperature. The mathematical relationship of Charles's Law becomes:
$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \nonumber$
This equation can be used to calculate any one of the four quantities if the other three are known. The direct relationship will only hold if the temperatures are expressed in Kelvin. Temperatures in Celsius will not work. Recall the relationship that $\text{K} = \: ^\text{o} \text{C} + 273$.
Example $1$:
A balloon is filled to a volume of $2.20 \: \text{L}$ at a temperature of $22^\text{o} \text{C}$. The balloon is then heated to a temperature of $71^\text{o} \text{C}$. Find the new volume of the balloon.
Solution
Solutions to Example 11.5.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
$V_1 = 2.20 \: \text{L}$ and
$T_1 = 22^\text{o} \text{C} = 295 \: \text{K}$
$T_2 = 71^\text{o} \text{C} = 344 \: \text{K}$
Find: V2 = ? L
List other known quantities. The temperatures have first been converted to Kelvin.
Plan the problem.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \dfrac{V_1 \times T_2}{T_1} \nonumber$
Cancel units and calculate.
Now substitute the known quantities into the equation and solve.
$V_2 = \dfrac{2.20 \: \text{L} \times 344 \: \cancel{\text{K}}}{295 \: \cancel{\text{K}}} = 2.57 \: \text{L} \nonumber$
Think about your result. The volume increases as the temperature increases. The result has three significant figures.
Exercise $1$
If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K?
Answer
1.82 L
Example $2$:
A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 25.0 L?
Solution
Solutions to Example 11.5.2
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
Given:T1 = -27oC and V1 = 34.8 L
V2 = 25.0 L
Find: T2 = ? K
List other known quantities. K = -27oC + 273
Plan the problem.
1. Convert the initial temperature to Kelvin
2. Rearrange the equation algebraically to solve for $T_2$.
$T_2 = \dfrac{V_2 \times T_1}{V_1} \nonumber$
Cancel units and calculate.
1. −67°C + 273 = 206 K
2. Substitute the known quantities into the equation and solve.
$T_2 = \dfrac{25.0 \: \cancel{\text{L}} \times 206 \: \text{K}}{34.8 \: \cancel{\text{L}}} = 148 \: \text{K} \nonumber$
Think about your result. This is also equal to −125°C. As temperature decreases, volume decreases—which it does in this example.
Exercise $2$
If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2?
Answer
235 K, or −38°C
Summary
• Charles’s Law relates the volume and temperature of a gas at constant pressure and amount. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/11%3A_Gases/11.05%3A_Charless_Law-_Volume_and_Temperature.txt |
Learning Objectives
• Learn and apply the Combined Gas Law.
One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the Combined Gas Law, and its mathematical form is
$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}\; at\; constant\; n \nonumber$
This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature must be in Kelvin.
Example $1$:
A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?
Solution
Solutions to Example 11.4.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
V1 = 8.33 L, P1 = 1.82 atm, and T1 = 286 K
V2 = 5.72 L and T2 = 355 K
Find: P2 = ? atm
List other known quantities. none
Plan the problem.
First, rearrange the equation algebraically to solve for $V_2$.
$P_2 = \dfrac{P_1 V_1 T_2 }{T_1V_2}$
Calculate.
Now substitute the known quantities into the equation and solve.
$P_2 = \dfrac{(1.82\, atm)(8.33\, \cancel{L})(355\, \cancel{K})}{(286\, \cancel{K})(5.72\, \cancel{L})}=3.22 atm \nonumber$
Think about your result. Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing.
Exercise $1$
If P1 = 662 torr, V1 = 46.7 mL, T1 = 266 K, P2 = 409 torr, and T2 = 371 K, what is V2?
Answer
105 mL
As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.
Summary
• The Combined Gas Law relates pressure, volume, and temperature of a gas.
11.07: Avogadros Law- Volume and Moles
A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. How much air should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst.
Avogadro's Law
You have learned about Avogadro's hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles (or number of particles) of gas when the temperature and pressure are held constant. The mathematical expression of Avogadro's Law is:
$V = k \times n \nonumber$
or
$\dfrac{V_1}{n_1} = \dfrac{V_2}{n_2} \nonumber$
where $n$ is the number of moles of gas and $k$ is a constant. Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.
If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro's Law. Adding gas to a rigid container makes the pressure increase.
Example $1$
A balloon has been filled to a volume of $1.90 \: \text{L}$ with $0.0920 \: \text{mol}$ of helium gas. If $0.0210 \: \text{mol}$ of additional helium is added to the balloon while the temperature and pressure are held constant, what is the new volume of the balloon?
Solution
Solutions to Example 11.11.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
$V_1 = 1.90 \: \text{L}$
$n_1 = 0.0920 \: \text{mol}$
Find: $V_2 = ? \: \text{L}$
List other known quantities.
Note that the final number of moles has to be calculated by adding the original number of moles to the moles of added helium.
$n_2 = 0.0920 + 0.0210 = 0.1130 \: \text{mol}$
Plan the problem.
First, rearrange the equation algebraically to solve for $V_2$.
$V_2 = \dfrac{V_1 \times n_2}{n_1} \nonumber$
Calculate.
Now substitute the known quantities into the equation and solve.
$V_2 = \dfrac{1.90 \: \text{L} \times 0.1130 \: \cancel{\text{mol}}}{0.0920 \: \cancel{\text{mol}}} = 2.33 \: \text{L} \nonumber$
Think about your result. Since a relatively small amount of additional helium was added to the balloon, its volume increases slightly.
Exercise $1$
A 12.8 L volume of gas contains .000498 moles of oxygen gas. At constant temperature and pressure, what volume does .0000136 moles of the gas fill?
Answer
0.350 L
Summary
• Calculations for relationships between volume and number of moles of a gas can be performed using Avogadro's Law. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/11%3A_Gases/11.06%3A_The_Combined_Gas_Law-_Pressure_Volume_and_Temperature.txt |
Learning Objectives
• Explain the Ideal Gas Law.
There are a number of chemical reactions that require ammonia. In order to carry out the reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.
Ideal Gas Law
The Combined Gas Law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these together leaves us with the following equation:
$\dfrac{P_1 \times V_1}{T_1 \times n_1} = \dfrac{P_2 \times V_2}{T_2 \times n_2} \nonumber$
As with the other gas laws, we can also say that $\dfrac{\left( P \times V \right)}{\left( T \times n \right)}$ is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.
The Ideal Gas Law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable $R$ for the constant, the equation becomes:
$\dfrac{P \times V}{T \times n} = R \nonumber$
The Ideal Gas Law is conveniently rearranged to look this way, with the multiplication signs omitted:
$PV = nRT \nonumber$
The variable $R$ in the equation is called the ideal gas constant.
Evaluating the Ideal Gas Constant
The value of $R$, the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: $\text{kPa}$, $\text{atm}$, or $\text{mm} \: \ce{Hg}$. Therefore, $R$ can have three different values.
We will demonstrate how $R$ is calculated when the pressure is measured in $\text{kPa}$. The volume of $1.00 \: \text{mol}$ of any gas at STP (Standard temperature, 273.15 K and pressure, 1 atm) is measured to be $22.414 \: \text{L}$. We can substitute $101.325 \: \text{kPa}$ for pressure, $22.414 \: \text{L}$ for volume, and $273.15 \: \text{K}$ for temperature into the ideal gas equation and solve for $R$.
\begin{align*} R &= \dfrac{PV}{nT} \[4pt] &= \dfrac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} \[4pt] &= 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol} \end{align*} \nonumber
This is the value of $R$ that is to be used in the ideal gas equation when the pressure is given in $\text{kPa}$. The table below shows a summary of this and the other possible values of $R$. It is important to choose the correct value of $R$ to use for a given problem.
Unit of $P$ Unit of $V$ Unit of $n$ Unit of $T$ Value and Unit of $R$
Table $1$: Values of the Ideal Gas Constant
$\text{kPa}$ $\text{L}$ $\text{mol}$ $\text{K}$ $8.314 \: \text{J/K} \cdot \text{mol}$
$\text{atm}$ $\text{L}$ $\text{mol}$ $\text{K}$ $0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}$
$\text{mm} \: \ce{Hg}$ $\text{L}$ $\text{mol}$ $\text{K}$ $62.36 \: \text{L} \cdot \text{mm} \: \ce{Hg}/\text{K} \cdot \text{mol}$
Notice that the unit for $R$ when the pressure is in $\text{kPa}$ has been changed to $\text{J/K} \cdot \text{mol}$. A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule $\left( \text{J} \right)$.
Example $1$ Oxygen Gas
What volume is occupied by $3.76 \: \text{g}$ of oxygen gas at a pressure of $88.4 \: \text{kPa}$ and a temperature of $19^\text{o} \text{C}$? Assume the oxygen is ideal.
Solution
Solutions to Example 11.5.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
• $P = 88.4 \: \text{kPa}$
• $T = 19^\text{o} \text{C} = 292 \: \text{K}$
Mass $\ce{O_2} = 3.76 \: \text{g}$
Find: V = ? L
List other known quantities.
$\ce{O_2} = 32.00 \: \text{g/mol}$
$R = 8.314 \: \text{J/K} \cdot \text{mol}$
Plan the problem.
1. First, determine the number of moles of O2 from the given mass and the molar mass.
2. Then, rearrange the equation algebraically to solve for V
$V = \dfrac{nRT}{P} \nonumber$
Calculate.
1.
$3.76 \: \cancel{\text{g}} \times \dfrac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \cancel{\text{g}} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2} \nonumber$
2. Now substitute the known quantities into the equation and solve.
$V = \dfrac{nRT}{P} = \dfrac{0.1175 \: \cancel{\text{mol}} \times 8.314 \: \cancel{\text{J/K}} \cdot \cancel{\text{mol}} \times 292 \: \cancel{\text{K}}}{88.4 \: \cancel{\text{kPa}}} = 3.23 \: \text{L} \: \ce{O_2} \nonumber$
Think about your result. The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume $\left( 22.4 \: \text{L/mol} \right)$ since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for $T$ and $P$. Since a joule $\left( \text{J} \right) = \text{kPa} \cdot \text{L}$, the units cancel out correctly, leaving a volume in liters.
Example $2$: Argon Gas
A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?
Solution
Solutions to Example 11.5.2
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
n = 4.22 mol
P = 1.21 atm
T = 34°C
Find: V = ? L
List other known quantities. none
Plan the problem.
1. The first step is to convert temperature to Kelvin.
2. Then, rearrange the equation algebraically to solve for V
$V = \dfrac{nRT}{P} \nonumber$
Calculate.
1. 34 + 273 = 307 K
2. Now substitute the known quantities into the equation and solve.
\begin{align*} V&=\dfrac{(4.22\, \cancel{mol})(0.08205\dfrac{L.\cancel{atm}}{\cancel{mol.K}})(307\, \cancel{K)}}{1.21\cancel{atm}} \[4pt] &= 87.9 \,L \end{align*}
Think about your result. The number of moles of Ar is large so the expected volume should also be large.
Exercise $1$
A 0.0997 mol sample of O2 has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?
Answer
3.94 L
Exercise $2$
For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume?
Answer
8.21 L
Summary
• The Ideal Gas Law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/11%3A_Gases/11.08%3A_The_Ideal_Gas_Law-_Pressure_Volume_Temperature_and_Moles.txt |
Learning Objectives
• Explain Dalton's Law of Partial Pressures.
The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are $96.5\%$ carbon dioxide and $3\%$ nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus would contribute a pressure well over $2700 \: \text{mm} \: \ce{Hg}$. And there is no oxygen present, so we couldn't breathe there. Not that we would want to go to Venus, as the surface temperature is usually over $460^\text{o} \text{C}$.
Dalton's Law of Partial Pressures
Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about $78\%$ nitrogen and $21\%$ oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up $78\%$ of the gas particles in a given sample of air, it exerts $78\%$ of the pressure. If the overall atmospheric pressure is $1.00 \: \text{atm}$, then the pressure of just the nitrogen in the air is $0.78 \: \text{atm}$. The pressure of the oxygen in the air is $0.21 \: \text{atm}$.
The partial pressure of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of nitrogen is represented by $P_{N_2}$. Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton's Law can be expressed with the following equation:
$P_\text{total} = P_1 + P_2 + P_3 + \cdots \nonumber$
The figure below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure, $P_1$ and $P_2$, reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If $P_1 = 300 \: \text{mm} \: \ce{Hg}$ and $P_2 = 500 \: \text{mm} \: \ce{Hg}$, then $P_\text{total} = 800 \: \text{mm} \: \ce{Hg}$.
Collecting Gases Over Water
You need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple: you don't. All you need is the atmospheric pressure in the room. As the gas pushes out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside.
Gas Collection by Water Displacement
Gases that are produced in laboratory experiments are often collected by a technique called water displacement (Figure $2$). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing, which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid.
Because the gas is collected over water, it is not pure, but is mixed with vapor from the evaporation of the water. Dalton's Law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor.
$P_\text{Total} = P_g + P_{H_2O} \nonumber$
where $P_g$ is the pressure of the desired gas, which can be solved for:
$P_g = P_{Total} - P_{H_2O} \nonumber$
In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see table below). The sample problem illustrates the use of Dalton's Law when a gas is collected over water.
Table $1$: Vapor Pressure of Water $\left( \text{mm} \: \ce{Hg} \right)$ at Selected Temperatures $\left( ^\text{o} \text{C} \right)$
0 5 10 15 20 25 30 35 40 45 50 55 60
4.58 6.54 9.21 12.79 17.54 23.76 31.82 42.18 55.32 71.88 92.51 118.04 149.38
Example 14.14.1
A certain experiment generates $2.58 \: \text{L}$ of hydrogen gas, which is collected over water. The temperature is $20^\text{o} \text{C}$ and the atmospheric pressure is $98.60 \: \text{kPa}$. Find the volume that the dry hydrogen would occupy at STP.
Known
• $V_\text{Total} = 2.58 \: \text{L}$
• $T = 20^\text{o} \text{C} = 293 \: \text{K}$
• $P_\text{Total} = 98.60 \: \text{kPa} = 739.7 \: \text{mm} \: \ce{Hg}$
Unknown
• $V_{H_2}$ at STP $= ? \: \text{L}$
The atmospheric pressure is converted from $\text{kPa}$ to $\text{mm} \: \ce{Hg}$ in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law.
Step 2: Solve.
\begin{align*} P_{H_2} &= P_\text{Total} - P_{H_2O} \[4pt] &= 739,7 \: \text{mm} \: \ce{Hg} - 17.54 \: \text{mm} \: \ce{Hg} \[4pt] &= 722.2 \: \text{mm} \: \ce{Hg} \end{align*}
Now the combined gas law is used, solving for $V_2$, the volume of hydrogen at STP.
\begin{align*} V_2 &= \dfrac{P_1 \times V_1 \times T_2}{P_2 \times T_1} \[4pt] &= \dfrac{722.2 \: \text{mm} \: \ce{Hg} \times 2.58 \: \text{L} \times 273 \: \text{K}}{760 \: \text{mm} \: \ce{Hg} \times 293 \: \text{K}} \[4pt] &= 2.28 \: \text{L} \: \ce{H_2} \end{align*}
Step 3: Think about your result.
If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be $2.28 \: \text{L}$. This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes.
Summary
• Dalton's Law of Partial Pressures states that the total pressure in a system is equal to the sum of the partial pressures of the gases present.
• The vapor pressure due to water in a sample can be corrected for, in order to get the true value for the pressure of the gas.
11.11: Gay-Lussac's Law- Temperature and Pressure
Learning Objectives
• Explain Gay-Lussac's Law.
Propane tanks are widely used with barbeque grills. But it's not fun to find out halfway through grilling that you have run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out.
Gay-Lussac's Law
When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. Gay-Lussac's Law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac's Law is very similar to Charles's Law, with the only difference being the type of container. Whereas the container in a Charles's Law experiment is flexible, it is rigid in a Gay-Lussac's Law experiment.
The mathematical expressions for Gay-Lussac's Law are likewise similar to those of Charles's Law:
$\dfrac{P}{T} \: \: \: \text{and} \: \: \: \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \nonumber$
A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume, its pressure continually decreases until the gas condenses to a liquid.
Example $1$
The gas in an aerosol can is under a pressure of $3.00 \: \text{atm}$ at a temperature of $25^\text{o} \text{C}$. It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of $845^\text{o} \text{C}$?
Solution
Solutions to Example 11.10.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
$P_1 = 3.00 \: \text{atm}$
$T_1 = 25^\text{o} \text{C} = 298 \: \text{K}$
$T_2 = 845^\text{o} \text{C} = 1118 \: \text{K}$
Find: $P_2 = ? \: \text{atm}$
List other known quantities. The temperatures have first been converted to Kelvin.
Plan the problem.
First, rearrange the equation algebraically to solve for $P_2$.
$P_2 = \dfrac{P_1 \times T_2}{T_1} \nonumber$
Calculate.
Now substitute the known quantities into the equation and solve.
$P_2 = \dfrac{3.00 \: \text{atm} \times 1118 \: \cancel{\text{K}}}{298 \: \cancel{\text{K}}}= 11.3 \: \text{atm} \nonumber$
Think about your result. The pressure increases dramatically due to a large increase in temperature.
Summary
• Pressure and temperature at constant volume are directly proportional. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/11%3A_Gases/11.09%3A_Mixtures_of_Gases_-_Why_Deep-Sea_Divers_Breathe_a_Mixture_of_Helium_and_Oxygen.txt |
In Chapter 6, we discussed the properties of gases. In this chapter, we consider some properties of liquids and solids. As a review, the table below lists some general properties of the three phases of matter.
General properties of the three phases of matter.
Phase Shape Density Compressibility
Gas fills entire container low high
Liquid fills a container from bottom to top high low
Solid rigid high low
• 12.1: Interactions between Molecules
The melting point is the temperature at which a solid changes into a liquid. Intermolecular forces have a strong influence on melting point.
• 12.2: Properties of Liquids and Solids
All liquids evaporate. If volume is limited, evaporation eventually reaches a dynamic equilibrium, and a constant vapor pressure is maintained. All liquids experience surface tension, an imbalance of forces at the surface of the liquid. All liquids experience capillary action, demonstrating either capillary rise or capillary depression in the presence of other substances. Solids can be divided into amorphous solids and crystalline solids.
• 12.3: Intermolecular Forces in Action- Surface Tension and Viscosity
The surface tension of a liquid is a measure of the elastic force in the liquid's surface. Liquids with strong intermolecular forces have higher surface tensions than liquids with weaker forces.
• 12.4: Evaporation and Condensation
Evaporation is the conversion of a liquid to its vapor below the boiling temperature of the liquid. Condensation is the change of state from a gas to a liquid. As the temperature increases, the rate of evaporation increases.
• 12.5: Melting, Freezing, and Sublimation
Phase changes can occur between any two phases of matter. All phase changes occur with a simultaneous change in energy. All phase changes are isothermal.
• 12.6: Intermolecular Forces- Dispersion, Dipole–Dipole, Hydrogen Bonding, and Ion-Dipole
All substances experience dispersion forces between their particles. Substances that are polar experience dipole-dipole interactions. Substances with covalent bonds between an H atom and N, O, or F atoms experience hydrogen bonding. The preferred phase of a substance depends on the strength of the intermolecular force and the energy of the particles.
• 12.7: Types of Crystalline Solids
Crystalline substances can be described by the types of particles found within, and the types of chemical bonding that take place between the particles. There are four types of crystals: (1) ionic, (2) metallic, (3) covalent network, and (4) molecular.
• 12.8: Water - A Remarkable Molecule
Water has several properties that make it a unique substance among substances. It is an excellent solvent; it dissolves many other substances and allows those substances to react when in solution. In fact, water is sometimes called the universal solvent because of this ability. Water has unusually high melting and boiling points. Unlike most substances, the solid form of water is less dense than its liquid form, which allows ice to float on water.
12: Liquids Solids and Intermolecular Forces
In the winter, many people find the snow and ice beautiful; they enjoy getting out to ski or ice-skate. When the snow melts, however, the roads get very sloppy and messy. Some people look forward to spring, when the ice and snow are gone and the weather is warmer. All of these events and factors are dependent on the melting point of a solid and the freezing point of a liquid.
Melting Point
Solids are similar to liquids in that both are condensed states, with particles that are far closer together than those of a gas. However, while liquids are fluid, solids are not. The particles of most solids are packed tightly together in an orderly arrangement. The motion of individual atoms, ions, or molecules in a solid is restricted to vibrational motion about a fixed point. Solids are almost completely incompressible and are the densest of the three states of matter.
As a solid is heated, its particles vibrate more rapidly as it absorbs kinetic energy. Eventually, the organization of the particles within the solid structure begins to break down and the solid starts to melt. The melting point is the temperature at which a solid changes into a liquid. At its melting point, the disruptive vibrations of the particles of the solid overcome the attractive forces operating within the solid. As with boiling points, the melting point of a solid is dependent on the strength of those attractive forces. Sodium chloride $\left( \ce{NaCl} \right)$ is an ionic compound that consists of a multitude of strong ionic bonds. Sodium chloride melts at $801^\text{o} \text{C}$. Ice (solid $\ce{H_2O}$) is a molecular compound of molecules that are held together by hydrogen bonds. Though hydrogen bonds are the strongest of the intermolecular forces, the strength of hydrogen bonds is much less than that of ionic bonds. The melting point of ice is 0 °C.
The melting point of a solid is the same as the freezing point of the liquid. At that temperature, the solid and liquid states of the substance are in equilibrium. For water, this equilibrium occurs at $0^\text{o} \text{C}$.
$\ce{H_2O} \left( s \right) \rightleftharpoons \ce{H_2O} \left( l \right) \nonumber$
We tend to think of solids as those materials that are solid at room temperature. However, all materials have melting points of some sort. Gases become solids at extremely low temperatures, and liquids will also become solid if the temperature is low enough. The table below gives the melting points of some common materials.
Materials Melting Point (°C)
Table $1$: Melting Points of Common Materials
Hydrogen -259
Oxygen -219
Diethyl ether -116
Ethanol -114
Water 0
Pure silver 961
Pure gold 1063
Iron 1538
Summary
The melting point is the temperature at which a solid changes into a liquid. Intermolecular forces have a strong influence on melting point. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.01%3A_Interactions_between_Molecules.txt |
Learning Objectives
• Describe the solid and liquid phases.
Solids and liquids are collectively called condensed phases because their particles are in virtual contact. The two states share little else, however.
Solids
In the solid state, the individual particles of a substance are in fixed positions with respect to each other because there is not enough thermal energy to overcome the intermolecular interactions between the particles. As a result, solids have a definite shape and volume. Most solids are hard, but some (like waxes) are relatively soft. Many solids composed of ions can also be quite brittle.
Solids usually have their constituent particles arranged in a regular, three-dimensional array of alternating positive and negative ions called a crystal. The effect of this regular arrangement of particles is sometimes visible macroscopically, as shown in Figure \(1\). Some solids, especially those composed of large molecules, cannot easily organize their particles in such regular crystals and exist as amorphous (literally, “without form”) solids. Glass is one example of an amorphous solid.
Liquids
If the particles of a substance have enough energy to partially overcome intermolecular interactions, then the particles can move about each other while remaining in contact. This describes the liquid state. In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.
Gases
If the particles of a substance have enough energy to completely overcome intermolecular interactions, then the particles can separate from each other and move about randomly in space. Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either.
The change from solid to liquid usually does not significantly change the volume of a substance. However, the change from a liquid to a gas significantly increases the volume of a substance, by a factor of 1,000 or more. Figure \(3\) shows the differences among solids, liquids, and gases at the molecular level, while Table \(1\) lists the different characteristics of these states.
Table \(1\): Characteristics of the Three States of Matter
Characteristic Solid Liquid Gas
shape definite indefinite indefinite
volume definite definite indefinite
relative intermolecular interaction strength strong moderate weak
relative particle positions in contact and fixed in place in contact but not fixed not in contact, random positions
Example \(1\)
What state or states of matter does each statement describe?
1. This state has a definite volume.
2. This state has no definite shape.
3. This state allows the individual particles to move about while remaining in contact.
Solution
1. This statement describes either the liquid state or the solid state.
2. This statement describes either the liquid state or the gas state.
3. This statement describes the liquid state.
Exercise \(1\)
What state or states of matter does each statement describe?
1. This state has individual particles in a fixed position with regard to each other.
2. This state has individual particles far apart from each other in space.
3. This state has indefinite shape.
Answer a
solid
Answer b
gas
Answer c
liquid or gas
Looking Closer: Water, the Most Important Liquid
Earth is the only known body in our solar system that has liquid water existing freely on its surface; life on Earth would not be possible without the presence of liquid water.
Water has several properties that make it a unique substance among substances. It is an excellent solvent; it dissolves many other substances and allows those substances to react when in solution. In fact, water is sometimes called the universal solvent because of this ability. Water has unusually high melting and boiling points (0°C and 100°C, respectively) for such a small molecule. The boiling points for similar-sized molecules, such as methane (BP = −162°C) and ammonia (BP = −33°C), are more than 100° lower. Though a liquid at normal temperatures, water molecules experience a relatively strong intermolecular interaction that allows them to maintain the liquid phase at higher temperatures than expected.
Unlike most substances, the solid form of water is less dense than its liquid form, which allows ice to float on water. In colder weather, lakes and rivers freeze from the top, allowing animals and plants to continue to live underneath. Water also requires an unusually large amount of energy to change temperature. While 100 J of energy will change the temperature of 1 g of Fe by 230°C, this same amount of energy will change the temperature of 1 g of H2O by only 100°C. Thus, water changes its temperature slowly as heat is added or removed. This has a major impact on weather, as storm systems like hurricanes can be impacted by the amount of heat that ocean water can store.
Water’s influence on the world around us is affected by these properties. Isn’t it fascinating that such a small molecule can have such a big impact?
Key Takeaway
• Solids and liquids are phases that have their own unique properties.
12.03: Intermolecular Forces in Action- Surface Tension and Viscosity
Learning Objectives
• Explain the how the surface tension of a liquid relates to intermolecular forces.
The next time you are by a still body of water, take a close look at what is scooting along on the surface. You may see insects seemingly floating on top of the water. These creatures are known by a variety of names including water skaters, water striders, pond skaters, and other equally descriptive names. They take advantage of a property called surface tension to stay above the water and not sink. The force they exert downward is less than the forces exerted among the water molecules on the surface of the pond, so the insect does not penetrate beneath the surface of the water.
Surface Tension
Molecules within a liquid are pulled equally in all directions by intermolecular forces. However, molecules at the surface are pulled downwards and sideways by other liquid molecules, but not upwards away from the surface. The overall effect is that the surface molecules are pulled into the liquid, creating a surface that is tightened like a film (Figure \(\PageIndex{1A}\)). The surface tension of a liquid is a measure of the elastic force in the liquid's surface. Liquids that have strong intermolecular forces, like the hydrogen bonding in water, exhibit the greatest surface tension. Surface tension allows objects that are denser than water, such as the paper clip shown in B in the figure below, to nonetheless float on its surface. It is also responsible for the beading up of water droplets on a freshly waxed car, because there are no attractions between the polar water molecules and the nonpolar wax.
Other liquids, such as diethyl ether, do not demonstrate strong surface tension interactions. The intermolecular forces for the ether are the relatively weak dipole-dipole interactions that do not draw the molecules together as tightly as hydrogen bonds would.
Summary
• The surface tension of a liquid is a measure of the elastic force in the liquid's surface.
• Liquids with strong intermolecular forces have higher surface tensions than liquids with weaker forces. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.02%3A_Properties_of_Liquids_and_Solids.txt |
Learning Objectives
• Explain how intermolecular forces affect rates of vaporization, evaporation, and condensation.
On the roof of the house in the picture below is a device known as a "swamp cooler". This piece of equipment traces its origin back to the ancient Egyptians who hung wet blankets across the doors of their homes. As the warm air passed through the blankets, water would evaporate and cool the air. The royalty went one step further and had servants fan wet cloths over jugs of water to get more evaporation and cooling.
The origin of the term "swamp cooler" is not known - they certainly don't work in a swamp. Best conditions for cooling include a high temperature (over $80^\text{o} \text{F}$) and a low humidity (preferably less than $30\%$). These coolers work well in desert areas, but don't provide any cooling in the humid areas of the country.
Evaporation
A puddle of water left undisturbed eventually disappears. The liquid molecules escape into the gas phase, becoming water vapor. Vaporization is the process in which a liquid is converted to a gas. Evaporation is the conversion of a liquid to its vapor below the boiling temperature of the liquid. If the water is instead kept in a closed container, the water vapor molecules do not have a chance to escape into the surroundings and so the water level does not change. As some water molecules become vapor, an equal number of water vapor molecules condense back into the liquid state. Condensation is the change of state from a gas to a liquid.
In order for a liquid molecule to escape into the gas state, the molecule must have enough kinetic energy to overcome the intermolecular attractive forces in the liquid. Recall that a given liquid sample will have molecules with a wide range of kinetic energies. Liquid molecules that have this certain threshold kinetic energy escape the surface and become vapor. As a result, the liquid molecules that remain now have lower kinetic energy. As evaporation occurs, the temperature of the remaining liquid decreases. You have observed the effects of evaporative cooling. On a hot day, the water molecules in your perspiration absorb body heat and evaporate from the surface of your skin. The evaporating process leaves the remaining perspiration cooler, which in turn absorbs more heat from your body.
A given liquid will evaporate more quickly when it is heated. This is because the heating process results in a greater fraction of the liquid's molecules having the necessary kinetic energy to escape the surface of the liquid. The figure below shows the kinetic energy distribution of liquid molecules at two temperatures. The numbers of molecules that have the required kinetic energy to evaporate are shown in the shaded area under the curve at the right. The higher temperature liquid $\left( T_2 \right)$ has more molecules that are capable of escaping into the vapor phase than the lower temperature liquid $\left( T_1 \right)$.
At 29,029 feet $\left( 8848 \: \text{m} \right)$, Mount Everest in the Himalayan range on the border between China and Nepal is the highest point on the earth. Its altitude presents many practical problems to climbers. The oxygen content of the air is much lower than at sea level, making it necessary to bring oxygen tanks along (although a few climbers have reached the peak without oxygen). One other problem is that of boiling water for cooking food. Although water boils at $100^\text{o} \text{C}$ at sea level, the boiling point on top of Mount Everest is only about $70^\text{o} \text{C}$. This difference makes it very difficult to get a decent cup of tea (which definitely frustrated some of the British climbers).
Boiling
As a liquid is heated, the average kinetic energy of its particles increases. The rate of evaporation increases as more and more molecules are able to escape the liquid's surface into the vapor phase. Eventually a point is reached when the molecules all throughout the liquid have enough kinetic energy to vaporize. At this point the liquid begins to boil. The boiling point is the temperature at which the vapor pressure of a liquid is equal to the external pressure. The figure below illustrates the boiling of liquid.
In the picture on the left, the liquid is below its boiling point, yet some of the liquid evaporates. On the right, the temperature has been increased until bubbles begin to form in the body of the liquid. When the vapor pressure inside the bubble is equal to the external atmospheric pressure, the bubbles rise to the surface of the liquid and burst. The temperature at which this process occurs is the boiling point of the liquid.
The normal boiling point is the temperature at which the vapor pressure of the liquid is equal to standard pressure. Because atmospheric pressure can change based on location, the boiling point of a liquid changes with the external pressure. The normal boiling point is a constant because it is defined relative to the standard atmospheric pressure of $760 \: \text{mm} \: \ce{Hg}$ (or $1 \: \text{atm}$ or $101.3 \: \text{kPa}$).
Summary
• The boiling point is the temperature at which the vapor pressure of a liquid is equal to the external pressure.
• As the altitude increases, the boiling point decreases.
• Evaporation is the conversion of a liquid to its vapor below the boiling temperature of the liquid.
• Condensation is the change of state from a gas to a liquid.
• As the temperature increases, the rate of evaporation increases. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.04%3A_Evaporation_and_Condensation.txt |
Learning Objectives
• Define melting, freezing, and sublimation.
Depending on the surrounding conditions, normal matter usually exists as one of three phases: solid, liquid, or gas.
A phase change is a physical process in which a substance goes from one phase to another. Usually the change occurs when adding or removing heat at a particular temperature, known as the melting point or the boiling point of the substance. The melting point is the temperature at which the substance goes from a solid to a liquid (or from a liquid to a solid). The boiling point is the temperature at which a substance goes from a liquid to a gas (or from a gas to a liquid). The nature of the phase change depends on the direction of the heat transfer. Heat going into a substance changes it from a solid to a liquid, or a liquid to a gas. Removing heat from a substance changes a gas to a liquid, or a liquid to a solid.
Two key points are worth emphasizing. First, at a substance’s melting point or boiling point, two phases can exist simultaneously. Take water (H2O) as an example. On the Celsius scale, H2O has a melting point of 0°C and a boiling point of 100°C. At 0°C, both the solid and liquid phases of H2O can coexist. However, if heat is added, some of the solid H2O will melt and turn into liquid H2O. If heat is removed, the opposite happens: some of the liquid H2O turns into solid H2O. A similar process can occur at 100°C: adding heat increases the amount of gaseous H2O, while removing heat increases the amount of liquid H2O (Figure $1$).
Second, the temperature of a substance does not change as the substance goes from one phase to another. In other words, phase changes are isothermal (isothermal means “constant temperature”). Again, consider H2O as an example. Solid water (ice) can exist at 0°C. If heat is added to ice at 0°C, some of the solid changes phase to make liquid, which is also at 0°C. Remember, the solid and liquid phases of H2O can coexist at 0°C. Only after all of the solid has melted into liquid does the addition of heat change the temperature of the substance.
For each phase change of a substance, there is a characteristic quantity of heat needed to perform the phase change per gram (or per mole) of material. The heat of fusion (ΔHfus) is the amount of heat per gram (or per mole) required for a phase change that occurs at the melting point. The heat of vaporization (ΔHvap) is the amount of heat per gram (or per mole) required for a phase change that occurs at the boiling point. If you know the total number of grams or moles of material, you can use the ΔHfus or the ΔHvap to determine the total heat being transferred for melting or solidification using these expressions:
$\text{heat} = n \times ΔH_{fus} \label{Eq1a}$
where $n$ is the number of moles and $ΔH_{fus}$ is expressed in energy/mole or
$\text{heat} = m \times ΔH_{fus} \label{Eq1b}$
where $m$ is the mass in grams and $ΔH_{fus}$ is expressed in energy/gram.
For the boiling or condensation, use these expressions:
$\text{heat} = n \times ΔH_{vap} \label{Eq2a}$
where $n$ is the number of moles) and $ΔH_{vap}$ is expressed in energy/mole or
$\text{heat} = m \times ΔH_{vap} \label{Eq2b}$
where $m$ is the mass in grams and $ΔH_{vap}$ is expressed in energy/gram.
Remember that a phase change depends on the direction of the heat transfer. If heat transfers in, solids become liquids, and liquids become solids at the melting and boiling points, respectively. If heat transfers out, liquids solidify, and gases condense into liquids.
Example $1$
How much heat is necessary to melt 55.8 g of ice (solid H2O) at 0°C? The heat of fusion of H2O is 79.9 cal/g.
Solution
We can use the relationship between heat and the heat of fusion (Eq. $1$b) to determine how many joules of heat are needed to melt this ice:
\begin{align*} \text{heat} &= m \times ΔH_{fus} \[4pt] & = (55.8\, \cancel{g}) \left(\dfrac{79.9\, cal}{\cancel{g}}\right) \[4pt] &=4,460\, cal \end{align*}
Exercise $1$
How much heat is necessary to vaporize 685 g of H2O at 100°C? The heat of vaporization of H2O is 540 cal/g.
Table $1$ lists the heats of fusion and vaporization for some common substances. Note the units on these quantities; when you use these values in problem solving, make sure that the other variables in your calculation are expressed in units consistent with the units in the specific heats, or the heats of fusion and vaporization.
Table $1$: Heats of Fusion and Vaporization for Selected Substances
Substance ΔHfus (cal/g) ΔHvap (cal/g)
aluminum (Al) 94.0 2,602
gold (Au) 15.3 409
iron (Fe) 63.2 1,504
water (H2O) 79.9 540
sodium chloride (NaCl) 123.5 691
ethanol (C2H5OH) 45.2 200.3
benzene (C6H6) 30.4 94.1
Looking Closer: Sublimation
There is also a phase change where a solid goes directly to a gas:
$\text{solid} \rightarrow \text{gas} \label{Eq3}$
This phase change is called sublimation. Each substance has a characteristic heat of sublimation associated with this process. For example, the heat of sublimation (ΔHsub) of H2O is 620 cal/g.
We encounter sublimation in several ways. You may already be familiar with dry ice, which is simply solid carbon dioxide (CO2). At −78.5°C (−109°F), solid carbon dioxide sublimes, changing directly from the solid phase to the gas phase:
$\mathrm{CO_2(s) \xrightarrow{-78.5^\circ C} CO_2(g)} \label{Eq4}$
Solid carbon dioxide is called dry ice because it does not pass through the liquid phase. Instead, it goes directly to the gas phase. (Carbon dioxide can exist as liquid but only under high pressure.) Dry ice has many practical uses, including the long-term preservation of medical samples.
Even at temperatures below 0°C, solid H2O will slowly sublime. For example, a thin layer of snow or frost on the ground may slowly disappear as the solid H2O sublimes, even though the outside temperature may be below the freezing point of water. Similarly, ice cubes in a freezer may get smaller over time. Although frozen, the solid water slowly sublimes, redepositing on the colder cooling elements of the freezer, which necessitates periodic defrosting (frost-free freezers minimize this redeposition). Lowering the temperature in a freezer will reduce the need to defrost as often.
Under similar circumstances, water will also sublime from frozen foods (e.g., meats or vegetables), giving them an unattractive, mottled appearance called freezer burn. It is not really a “burn,” and the food has not necessarily gone bad, although it looks unappetizing. Freezer burn can be minimized by lowering a freezer’s temperature and by wrapping foods tightly so water does not have any space to sublime into.
Melting Point
Solids are similar to liquids in that both are condensed states, with particles that are far closer together than those of a gas. However, while liquids are fluid, solids are not. The particles of most solids are packed tightly together in an orderly arrangement. The motion of individual atoms, ions, or molecules in a solid is restricted to vibrational motion about a fixed point. Solids are almost completely incompressible and are the most dense of the three states of matter.
As a solid is heated, its particles vibrate more rapidly as the solid absorbs kinetic energy. Eventually, the organization of the particles within the solid structure begins to break down and the solid starts to melt. The melting point is the temperature at which a solid changes into a liquid. At its melting point, the disruptive vibrations of the particles of the solid overcome the attractive forces operating within the solid. As with boiling points, the melting point of a solid is dependent on the strength of those attractive forces. Sodium chloride $\left( \ce{NaCl} \right)$ is an ionic compound that consists of a multitude of strong ionic bonds. Sodium chloride melts at $801^\text{o} \text{C}$. Ice (solid $\ce{H_2O}$) is a molecular compound composed of molecules that are held together by hydrogen bonds. Though hydrogen bonds are the strongest of the intermolecular forces, the strength of hydrogen bonds is much less than that of ionic bonds. The melting point of ice is $0^\text{o} \text{C}$.
The melting point of a solid is the same as the freezing point of the liquid. At that temperature, the solid and liquid states of the substance are in equilibrium. For water, this equilibrium occurs at $0^\text{o} \text{C}$.
$\ce{H_2O} \left( s \right) \rightleftharpoons \ce{H_2O} \left( l \right) \nonumber$
We tend to think of solids as those materials that are solid at room temperature. However, all materials have melting points of some sort. Gases become solids at extremely low temperatures, and liquids will also become solid if the temperature is low enough. The table below gives the melting points of some common materials.
Materials Melting Point (ºC)
Table $2$: Melting Points of Common Materials
Hydrogen -259
Oxygen -219
Diethyl ether -116
Ethanol -114
Water 0
Pure silver 961
Pure gold 1063
Iron 1538
Exercise $2$
1. Explain what happens when heat flows into or out of a substance at its melting point or boiling point.
2. How does the amount of heat required for a phase change relate to the mass of the substance?
Answer a
The energy goes into changing the phase, not the temperature.
Answer b
The amount of heat is a constant per gram of substance.
Summary
• There is an energy change associated with any phase change.
• Sublimation is the change of state from a solid to a gas, without passing through the liquid state.
• Deposition is the change of state from a gas to a solid.
• Carbon dioxide is an example of a material that easily undergoes sublimation.
• The melting point is the temperature at which a solid changes into a liquid.
• Intermolecular forces have a strong influence on melting point. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.05%3A_Melting_Freezing_and_Sublimation.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/Introductory_Chemistry/Introductory_Chemistry/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.06%3A_Intermolecular_Forces-_Dispersion_DipoleDipole_Hydrogen_Bonding_and_Ion-Dipole.txt |
Learning Objectives
• Identify different types of solid substances.
As a society, we sometimes take things for granted. For example, it is often assumed that we will get electric power when we connect a plug to an electrical outlet. The wire that comprises that outlet is almost always copper, a material that conducts electricity well. The unique properties of the solid copper allow electrons to flow freely through the wire and into whatever device we connect it to. Then we can enjoy music, television, computer work, or whatever other activity we want to undertake. However, these activities—and the miracle of electricity itself—would not be possible without that copper wire!
Classes of Crystalline Solids
Crystalline substances can be described by the types of particles in them and the types of chemical bonding that take place between the particles. There are four types of crystals: (1) ionic, (2) metallic, (3) covalent network, and (4) molecular. Properties and several examples of each type are listed in the following table and are described in the table below.
Type of Crystalline Solid Examples (formulas) Melting Point (°C) Normal Boiling Point (°C)
Table $1$: Crystalline Solids: Melting and Boiling Points
Ionic $\ce{NaCl}$ 801 1413
$\ce{CaF_2}$ 1418 1533
Metallic $\ce{Hg}$ -39 630
$\ce{Na}$ 371 883
$\ce{Au}$ 1064 2856
$\ce{W}$ 3410 5660
Covalent Network $\ce{B}$ 2076 3927
$\ce{C}$ (diamond) 3500 3930
$\ce{SiO_2}$ 1600 2230
Molecular $\ce{H_2}$ -259 -253
$\ce{I_2}$ 114 184
$\ce{NH_3}$ -78 -33
$\ce{H_2O}$ 0 100
Ionic crystals - The ionic crystal structure consists of alternating positively-charged cations and negatively-charged anions (see figure below). The ions may either be monatomic or polyatomic. Generally, ionic crystals form from a combination of Group 1 or 2 metals and Group 16 or 17 nonmetals or nonmetallic polyatomic ions. Ionic crystals are hard and brittle and have high melting points. Ionic compounds do not conduct electricity as solids, but do conduct electricity when molten or in aqueous solution.
Metallic crystal - Metallic crystals consist of metal cations surrounded by a "sea" of mobile valence electrons (see figure below). These electrons, also referred to as delocalized electrons, do not belong to any one atom, but are capable of moving through the entire crystal. As a result, metals are good conductors of electricity. As seen in the table above, the melting points of metallic crystals span a wide range.
Covalent network crystals - A covalent network crystal consists of atoms at the lattice points of the crystal, with each atom being covalently bonded to its nearest neighbor atoms (see figure below). The covalently bonded network is three-dimensional and contains a very large number of atoms. Network solids include diamond, quartz, many metalloids, and oxides of transition metals and metalloids. Network solids are hard and brittle, with extremely high melting and boiling points. Being composed of atoms rather than ions, they do not conduct electricity in any state.
Molecular crystals - Molecular crystals typically consist of molecules at the lattice points of the crystal, held together by relatively weak intermolecular forces (see figure below). The intermolecular forces may be dispersion forces in the case of nonpolar crystals, or dipole-dipole forces in the case of polar crystals. Some molecular crystals, such as ice, have molecules held together by hydrogen bonds. When one of the noble gases is cooled and solidified, the lattice points are individual atoms rather than molecules. In all cases, the intermolecular forces holding the particles together are far weaker than either ionic or covalent bonds. As a result, the melting and boiling points of molecular crystals are much lower. Lacking ions or free electrons, molecular crystals are poor electrical conductors.
Some general properties of the four major classes of solids are summarized in Table $2$.
Table $2$: Properties of the Major Classes of Solids
Ionic Solids Molecular Solids Covalent Solids Metallic Solids
*Many exceptions exist. For example, graphite has a relatively high electrical conductivity within the carbon planes, and diamond has the highest thermal conductivity of any known substance.
poor conductors of heat and electricity poor conductors of heat and electricity poor conductors of heat and electricity* good conductors of heat and electricity
relatively high melting point low melting point high melting point melting points depend strongly on electron configuration
hard but brittle; shatter under stress soft very hard and brittle easily deformed under stress; ductile and malleable
relatively dense low density low density usually high density
dull surface dull surface dull surface lustrous
Example $1$
Classify $\ce{Ge}$, $\ce{RbI}$, $\ce{C6(CH3)6}$, and $\ce{Zn}$ as ionic, molecular, covalent, or metallic solids and arrange them in order of increasing melting points.
Given: compounds
Asked for: classification and order of melting points
Strategy:
1. Locate the component element(s) in the periodic table. Based on their positions, predict whether each solid is ionic, molecular, covalent, or metallic.
2. Arrange the solids in order of increasing melting points based on your classification, beginning with molecular solids.
Solution:
A. Germanium lies in the p block just under Si, along the diagonal line of semi-metallic elements, which suggests that elemental Ge is likely to have the same structure as Si (the diamond structure). Thus Ge is probably a covalent solid.
RbI contains a metal from group 1 and a nonmetal from group 17, so it is an ionic solid containing Rb+ and I ions.
The compound $\ce{C6(CH3)6}$ is a hydrocarbon (hexamethylbenzene), which consists of isolated molecules that stack to form a molecular solid with no covalent bonds between them.
Zn is a d-block element, so it is a metallic solid.
B. Arranging these substances in order of increasing melting points is straightforward, with one exception. We expect C6(CH3)6 to have the lowest melting point and Ge to have the highest melting point, with RbI somewhere in between. The melting points of metals, however, are difficult to predict based on the models presented thus far. Because Zn has a filled valence shell, it should not have a particularly high melting point, so a reasonable guess is
$\ce{C6(CH3)6 < Zn < RbI < Ge.} \nonumber$
The actual melting points are C6(CH3)6, 166°C; Zn, 419°C; RbI, 642°C; and Ge, 938°C. This agrees with our prediction.
Exercise $1$
Classify CO2, BaBr2, GaAs, and AgZn as ionic, covalent, molecular, or metallic solids and then arrange them in order of increasing melting points.
Answer
CO2 (molecular) < AgZn (metallic) ~ BaBr2 (ionic) < GaAs (covalent).
The actual melting points are: CO2, about -15.6°C; AgZn, about 700°C; BaBr2, 856°C; and GaAs, 1238°C.
Summary
• Ionic crystals are composed of alternating positive and negative ions.
• Metallic crystals consist of metal cations surrounded by a "sea" of mobile valence electrons.
• Covalent crystals are composed of atoms which are covalently bonded to one another.
• Molecular crystals are held together by weak intermolecular forces. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.07%3A_Types_of_Crystalline_Solids.txt |
Learning Objectives
• Interpret the unique properties of water in terms of a phase diagram.
Earth is the only known body in our solar system that has liquid water existing freely on its surface; life on Earth would not be possible without the presence of liquid water. Water has several properties that make it a unique substance among substances. It is an excellent solvent; it dissolves many other substances and allows those substances to react when in solution. In fact, water is sometimes called the universal solvent because of this ability. Water has unusually high melting and boiling points (0°C and 100°C, respectively) for such a small molecule. The boiling points for similar-sized molecules, such as methane (BP = −162°C) and ammonia (BP = −33°C), are more than 100° lower. Though a liquid at normal temperatures, water molecules experience a relatively strong intermolecular interaction that allows them to maintain the liquid phase at higher temperatures than expected.
Unlike most substances, the solid form of water is less dense than its liquid form, which allows ice to float on water. In colder weather, lakes and rivers freeze from the top, allowing animals and plants to continue to live underneath. Water also requires an unusually large amount of energy to change temperature. While 100 J of energy will change the temperature of 1 g of Fe by 230°C, this same amount of energy will change the temperature of 1 g of H2O by only 100°C. Thus, water changes its temperature slowly as heat is added or removed. This has a major impact on weather, as storm systems like hurricanes can be impacted by the amount of heat that ocean water can store. Water’s influence on the world around us is affected by these properties. Isn’t it fascinating that such a small molecule can have such a big impact?
Phase Diagram for Water
Water is a unique substance in many ways. One of these special properties is the fact that solid water (ice) is less dense than liquid water just above the freezing point. The phase diagram for water is shown in the figure below.
Notice one key difference between the general phase diagram and the phase diagram for water. In water's diagram, the slope of the line between the solid and liquid states is negative rather than positive. The reason for this is that water is an unusual substance, as its solid state is less dense than the liquid state. Ice floats in liquid water. Therefore, a pressure change has the opposite effect on those two phases. If ice is relatively near its melting point, it can be changed into liquid water by the application of pressure. The water molecules are actually closer together in the liquid phase than they are in the solid phase.
Refer again to water's phase diagram (figure above). Notice point $E$, labeled the critical point. What does that mean? At $373.99^\text{o} \text{C}$, particles of water in the gas phase are moving very, very rapidly. At any temperature higher than that, the gas phase cannot be made to liquefy, no matter how much pressure is applied to the gas. The critical pressure $\left( P_\text{C} \right)$ is the pressure that must be applied to the gas at the critical temperature in order to turn it into a liquid. For water, the critical pressure is very high, $217.75 \: \text{atm}$. The critical point is the intersection point of the critical temperature and the critical pressure.
Summary
• Solid water is less dense than liquid water just above the freezing point.
• The critical temperature $\left( T_\text{C} \right)$ of a substance is the highest temperature at which the substance can possibly exist as a liquid.
• The critical pressure $\left( P_\text{C} \right)$ is the pressure that must be applied to the gas at the critical temperature in order to turn it into a liquid.
• The critical point is the intersection point of the critical temperature and the critical pressure. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/12%3A_Liquids_Solids_and_Intermolecular_Forces/12.08%3A_Water_-_A_Remarkable_Molecule.txt |
Solutions play a very important role in many biological, laboratory, and industrial applications of chemistry. Of particular importance are solutions involving substances dissolved in water, or aqueous solutions. Solutions represent equilibrium systems, and the lessons learned in the last chapter will be of particular importance again. Quantitative measurements of solutions are another key component of this chapter. Solutions can involve all physical states—gases dissolved in gases (the air around us), solids dissolved in solids (metal alloys), and liquids dissolved in solids (amalgams—liquid mercury dissolved in another metal such as silver, tin or copper). This chapter is almost exclusively concerned with aqueous solutions, substances dissolved in water.
• 13.1: Tragedy in Cameroon
Lake Nyos is a crater lake in the Northwest Region of Cameroon, and is a deep lake high on the flank of an inactive volcano in the Oku volcanic plain along the Cameroon line of volcanic activity. A volcanic dam impounds the lake waters. A pocket of magma lies beneath the lake and leaks carbon dioxide into the water, changing it into carbonic acid. Nyos is one of only three known exploding lakes to be saturated with carbon dioxide in this way.
• 13.2: Solutions - Homogeneous Mixtures
There are two types of mixtures: mixtures in which the substances are evenly mixed together (called a homogenous mixture, or solution) and a mixture in which the substances are not evenly mixed (called a heterogeneous mixture). When a solution, or homogenous mixture, is said to have uniform properties throughout, the definition is referring to properties at the particle level.
• 13.3: Solutions of Solids Dissolved in Water- How to Make Rock Candy
Solutions can be formed in a variety of combinations using solids, liquids, and gases. We also know that solutions have constant composition and we can also vary this composition up to a point to maintain the homogeneous nature of the solution. Reasons for why solutions form will be explored in this section, along with a discussion of why water is used most frequently to dissolve substances of various types.
• 13.4: Solutions of Gases in Water
Several factors affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases. This is one of the major impacts resulting from the thermal pollution of natural bodies of water.
• 13.5: Specifying Solution Concentration- Mass Percent
To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as "dilute" or "concentrated" are used to describe solutions that have a little or a lot of dissolved solute, respectively. However "dilute" and "concentrated" are relative terms, and have meanings dependent on various factors. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100.
• 13.6: Specifying Solution Concentration- Molarity
Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution. The symbol for molarity is MM or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance.
• 13.7: Solution Dilution
We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms—dilute and concentrated.
• 13.8: Solution Stoichiometry
Double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and hence their amounts (i.e. volume of solutions or mass of precipitates).
• 13.9: Freezing Point Depression and Boiling Point Elevation
Freezing point depression and boiling point elevation are "colligative properties" that depend on the concentration of solute in a solvent, but not on the type of solute. What this means for the example above is that people in colder climates don't necessarily need salt to get the same effect on the roads—any solute will work. The higher the concentration of solute, the more these colligative properties will change.
• 13.10: Osmosis
Osmotic pressure is caused by concentration differences between solutions separated by a semipermeable membrane and is an important biological issue.
13: Solutions
Lake Nyos is a deep crater lake in the Northwest region of Cameroon, high on the flank of an inactive volcano in the Oku volcanic plain along the Cameroon line of volcanic activity. A volcanic dam impounds the lake waters. A pocket of magma lies beneath the lake and leaks carbon dioxide (\(\ce{CO_2}\)) into the water, changing it into carbonic acid. Nyos is one of only three known exploding lakes to be saturated with carbon dioxide in this way.
In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure \(1\)), a deep lake in a volcanic crater. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO2 were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.
Following the Lake Nyos tragedy, scientists investigated other African lakes to see if a similar phenomenon could happen elsewhere. Lake Kivu in Democratic Republic of Congo, 2,000 times larger than Lake Nyos, was also found to be supersaturated, and geologists found evidence for out-gassing events around the lake about every one thousand years. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.01%3A_Tragedy_in_Cameroon.txt |
Learning Objectives
• Learn terminology involving solutions.
• Explain the significance of the statement "like dissolves like."
• Explain why certain substances dissolve in other substances.
The major component of a solution is called the solvent. The minor component of a solution is called the solute. By major and minor we mean whichever component has the greater or lesser presence by mass or by moles. Sometimes this becomes confusing, especially with substances with very different molar masses. However, here we will confine the discussion to solutions for which the major component and the minor component are obvious.
Solutions exist for every possible phase of the solute and the solvent. Salt water, for example, is a solution of solid \(\ce{NaCl}\) in liquid water, while air is a solution of a gaseous solute (\(\ce{O2}\)) in a gaseous solvent (\(\ce{N2}\)). In all cases, however, the overall phase of the solution is the same phase as the solvent. Table \(1\) lists some common types of solutions, with examples of each.
Table \(1\): Types of Solutions
Solvent Phase Solute Phase Example
gas gas air
liquid gas carbonated beverages
liquid liquid ethanol (C2H5OH) in H2O (alcoholic beverages)
liquid solid salt water
solid gas H2 gas absorbed by Pd metal
solid liquid Hg(ℓ) in dental fillings
solid solid steel alloys
Example \(1\): Sugar and Water
A solution is made by dissolving 1.00 g of sucrose (\(\ce{C12H22O11}\)) in 100.0 g of liquid water. Identify the solvent and solute in the resulting solution.
Solution
Either by mass or by moles, the obvious minor component is sucrose, so it is the solute. Water—the majority component—is the solvent. The fact that the resulting solution is the same phase as water also suggests that water is the solvent.
Exercise \(1\)
A solution is made by dissolving 3.33 g of \(\ce{HCl(g)}\) in 40.0 g of liquid methyl alcohol (\(\ce{CH3OH}\)). Identify the solvent and solute in the resulting solution.
Answer
solute: \(\ce{HCl(g)}\)
solvent: \(\ce{CH3OH}\)
Like Dissolves Like
A simple way to predict which compounds will dissolve in other compounds is the phrase "like dissolves like". What this means is that polar compounds dissolve polar compounds, nonpolar compounds dissolve nonpolar compounds, but polar and nonpolar do not dissolve in each other.
Even some nonpolar substances dissolve in water but only to a limited degree. Have you ever wondered why fish are able to breathe? Oxygen gas, a nonpolar molecule, does dissolve in water—it is this oxygen that the fish take in through their gills. The reason we can enjoy carbonated sodas is also due to a nonpolar compound that dissolves in water. Pepsi-cola and all the other sodas have carbon dioxide gas, \(\ce{CO_2}\), a nonpolar compound, dissolved in a sugar-water solution. In this case, to keep as much gas in solution as possible, the sodas are kept under pressure.
This general trend of "like dissolves like" is summarized in the following table:
Table \(2\): Summary of Solubilities
Solute (Polarity of Compound) Solvent (Polarity of Compound) Dominant Intermolecular Force Is Solution Formed?
Polar Polar Dipole-Dipole Force and/or Hydrogen Bond yes
Non-polar Non-polar Dispersion Force yes
Polar Non-polar no
Non-polar Polar no
Ionic Polar Ion-Dipole yes
Ionic Non-polar no
Note that every time charged particles (ionic compounds or polar substances) are mixed, a solution is formed. When particles with no charges (nonpolar compounds) are mixed, they will form a solution. However, if substances with charges are mixed with other substances without charges, a solution does not form. When an ionic compound is considered "insoluble", it doesn't necessarily mean the compound is completely untouched by water. All ionic compounds dissolve to some extent. An insoluble compound just doesn't dissolve in any noticeable or appreciable amount.
What is it that makes a solute soluble in some solvents but not others?
The answer is intermolecular interactions. The intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding (as described in Chapter 10). From experimental studies, it has been determined that if molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl—a very polar substance because it is composed of ions—dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane, but not in polar water.
Example \(2\): Polar and Nonpolar Solvents
Would \(\ce{I2}\) be more soluble in \(\ce{CCl4}\) or \(\ce{H2O}\)? Explain your answer.
Solution
\(\ce{I2}\) is nonpolar. Of the two solvents, \(\ce{CCl4}\) is nonpolar and \(\ce{H2O}\) is polar, so \(\ce{I2}\) would be expected to be more soluble in \(\ce{CCl4}\).
Exercise \(2\)
Would \(\ce{C3H7OH}\) be more soluble in \(\ce{CCl4}\) or \(\ce{H2O}\)? Explain your answer.
Answer
\(\ce{H2O}\), because both experience hydrogen bonding.
Example \(3\)
Water is considered a polar solvent. Which substances should dissolve in water?
1. methanol (\(\ce{CH3OH}\))
2. sodium sulfate (\ce{Na2SO4}\))
3. octane (\(\ce{C8H18}\))
Solution
Because water is polar, substances that are polar or ionic will dissolve in it.
1. Because of the OH group in methanol, we expect its molecules to be polar. Thus, we expect it to be soluble in water. As both water and methanol are liquids, the word miscible can be used in place of soluble.
2. Sodium sulfate is an ionic compound, so we expect it to be soluble in water.
3. Like other hydrocarbons, octane is nonpolar, so we expect that it would not be soluble in water.
Exercise \(3\): Toluene
Toluene (\(\ce{C6H5CH3}\)) is widely used in industry as a nonpolar solvent. Which substances should dissolve in toluene?
1. water (H2O)
2. sodium sulfate (Na2SO4)
3. octane (C8H18)
Answer
Octane (\(\ce{C8H18}\)) will dissolve. It is also non-polar.
Summary
• Solutions are composed of a solvent (major component) and a solute (minor component).
• “Like dissolves like” is a useful rule for deciding if a solute will be soluble in a solvent. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.02%3A_Solutions_-_Homogeneous_Mixtures.txt |
Learning Objectives
• Define electrolytes and non electrolytes
• Explain why solutions form.
• Discuss the idea of water as the "universal solvent".
• Explain how water molecules attract ionic solids when they dissolve in water.
We have learned that solutions can be formed in a variety of combinations using solids, liquids, and gases. We also know that solutions have constant composition, and that this composition can be varied up to a point to maintain the homogeneous nature of the solution. But how exactly do solutions form? Why is it that oil and water will not form a solution, and yet vinegar and water will? Why could we dissolve table salt in water, but not in vegetable oil? The reasons why solutions will form will be explored in this section, along with a discussion of why water is used most frequently to dissolve substances of various types.
Solubility and Saturation
Table salt $\left( \ce{NaCl} \right)$ readily dissolves in water. In most cases, only a certain maximum amount of solute can be dissolved in a given amount of solvent. This maximum amount is specified as the solubility of the solute. It is usually expressed in terms of the amount of solute that can dissolve in 100 g of the solvent at a given temperature. Table $1$ lists the solubilities of some simple ionic compounds. These solubilities vary widely. NaCl can dissolve up to 31.6 g per 100 g of H2O, while AgCl can dissolve only 0.00019 g per 100 g of H2O.
Table $1$: Solubilities of Some Ionic Compounds
Solute Solubility (g per 100 g of H2O at 25°C)
AgCl 0.00019
CaCO3 0.0006
KBr 70.7
NaCl 36.1
NaNO3 94.6
When the maximum amount of solute has been dissolved in a given amount of solvent, we say that the solution is saturated with solute. When less than the maximum amount of solute is dissolved in a given amount of solute, the solution is unsaturated. These terms are also qualitative terms because each solute has its own solubility. A solution of 0.00019 g of AgCl per 100 g of H2O may be saturated, but with so little solute dissolved, it is also rather dilute. A solution of 36.1 g of NaCl in 100 g of H2O is also saturated, but rather concentrated. In some circumstances, it is possible to dissolve more than the maximum amount of a solute in a solution. Usually, this happens by heating the solvent, dissolving more solute than would normally dissolve at regular temperatures, and letting the solution cool down slowly and carefully. Such solutions are called supersaturated solutions and are not stable; given an opportunity (such as dropping a crystal of solute in the solution), the excess solute will precipitate from the solution.The figure below illustrates the above process and shows the distinction between unsaturated and saturated.
How can you tell if a solution is saturated or unsaturated? If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium, but which has extra undissolved solute at the bottom of the container, must be saturated.
Electrolyte Solutions: Dissolved Ionic Solids
When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte (good conductor). If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, the substance is a weak electrolyte (does not conduct electricity as well).
Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit (Figure $1$).
Water and other polar molecules are attracted to ions, as shown in Figure $2$. The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes.
Example $1$: Identifying Ionic Compounds
Which compound(s) will dissolve in solution to separate into ions?
1. $\ce{LiF}$
2. $\ce{P_2F_5}$
3. $\ce{C_2H_5OH}$
Solution
$\ce{LiF}$ will separate into ions when dissolved in solution, because it is an ionic compound. $\ce{P_2F_5}$ and $\ce{C_2H_5OH}$ are both covalent and will stay as molecules in a solution.
Exercise $1$
Which compounds will dissolve in solution to separate into ions?
1. C6H12O11, glucose
2. CCl4
3. CaCl2
4. AgNO3
Answer
c & d
How Temperature Influences Solubility
The solubility of a substance is the amount of that substance that is required to form a saturated solution in a given amount of solvent at a specified temperature. Solubility is often measured as the grams of solute per $100 \: \text{g}$ of solvent. The solubility of sodium chloride in water is $36.0 \: \text{g}$ per $100 \: \text{g}$ water at $20^\text{o} \text{C}$. The temperature must be specified because solubility varies with temperature. For gases, the pressure must also be specified. Solubility is specific for a particular solvent. We will consider solubility of material in water as solvent.
The solubility of the majority of solid substances increases as the temperature increases. However, the effect is difficult to predict and varies widely from one solute to another. The temperature dependence of solubility can be visualized with the help of a solubility curve, a graph of the solubility vs. temperature (Figure $4$).
Notice how the temperature dependence of $\ce{NaCl}$ is fairly flat, meaning that an increase in temperature has relatively little effect on the solubility of $\ce{NaCl}$. The curve for $\ce{KNO_3}$, on the other hand, is very steep and so an increase in temperature dramatically increases the solubility of $\ce{KNO_3}$.
Several substances—$\ce{HCl}$, $\ce{NH_3}$, and $\ce{SO_2}$—have solubility that decreases as temperature increases. They are all gases at standard pressure. When a solvent with a gas dissolved in it is heated, the kinetic energy of both the solvent and solute increase. As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases.
Solubility curves can be used to determine if a given solution is saturated or unsaturated. Suppose that $80 \: \text{g}$ of $\ce{KNO_3}$ is added to $100 \: \text{g}$ of water at $30^\text{o} \text{C}$. According to the solubility curve, approximately $48 \: \text{g}$ of $\ce{KNO_3}$ will dissolve at $30^\text{o} \text{C}$. This means that the solution will be saturated since $48 \: \text{g}$ is less than $80 \: \text{g}$. We can also determine that there will be $80 - 48 = 32 \: \text{g}$ of undissolved $\ce{KNO_3}$ remaining at the bottom of the container. Now suppose that this saturated solution is heated to $60^\text{o} \text{C}$. According to the curve, the solubility of $\ce{KNO_3}$ at $60^\text{o} \text{C}$ is about $107 \: \text{g}$. Now the solution is unsaturated since it contains only the original $80 \: \text{g}$ of dissolved solute. Now suppose the solution is cooled all the way down to $0^\text{o} \text{C}$. The solubility at $0^\text{o} \text{C}$ is about $14 \: \text{g}$, meaning that $80 - 14 = 66 \: \text{g}$ of the $\ce{KNO_3}$ will re-crystallize.
Summary
• Solubility is the specific amount of solute that can dissolve in a given amount of solvent.
• Saturated and unsaturated solutions are defined.
• Ionic compounds dissolve in polar solvents, especially water. This occurs when the positive cation from the ionic solid is attracted to the negative end of the water molecule (oxygen) and the negative anion of the ionic solid is attracted to the positive end of the water molecule (hydrogen).
• Water is considered the universal solvent since it can dissolve both ionic and polar solutes, as well as some nonpolar solutes (in very limited amounts).
• The solubility of a solid in water increases with an increase in temperature.
Vocabulary
• Miscible - Liquids that have the ability to dissolve in each other.
• Immiscible - Liquids that do not have the ability to dissolve in each other.
• Electrostatic attraction - The attraction of oppositely charged particles. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.03%3A_Solutions_of_Solids_Dissolved_in_Water-_How_to_Make_Rock_Candy.txt |
Learning Objectives
• Explain how temperature and pressure affect the solubility of gases.
In an earlier module of this chapter, the effect of intermolecular attractive forces on solution formation was discussed. The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl3. Considering the role of the solvent’s chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C6H14, is approximately 20 times greater than it is in water.
Other factors also affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases (Figure $1$). This is one of the major impacts resulting from the thermal pollution of natural bodies of water.
When the temperature of a river, lake, or stream is raised abnormally high, usually due to the discharge of hot water from some industrial process, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills (Figure $2$).
The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO2 at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure $3$). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.”
"Fizz"
The dissolution in a liquid, also known as fizz, usually involves carbon dioxide under high pressure. When the pressure is reduced, the carbon dioxide is released from the solution as small bubbles, which causes the solution to become effervescent, or fizzy. A common example is the dissolving of carbon dioxide in water, resulting in carbonated water.
Carbon dioxide is weakly soluble in water, therefore it separates into a gas when the pressure is released. This process is generally represented by the following reaction, where a pressurized dilute solution of carbonic acid in water releases gaseous carbon dioxide at decompression:
$\ce{H2CO3(aq) → H2O(l) + CO2(g)} \nonumber$
In simple terms, it is the result of the chemical reaction occurring in the liquid which produces a gaseous product.
For many gaseous solutes, the relation between solubility, $C_g$, and partial pressure, $P_g$, is a proportional one:
$C_\ce{g}=kP_\ce{g} \nonumber$
where k is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry’s law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.
Example $1$: Application of Henry’s Law
At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10−3 mol L−1. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth’s atmosphere.
Solution
According to Henry’s law, for an ideal solution the solubility, $C_g$, of a gas (1.38 × 10−3 mol L−1, in this case) is directly proportional to the pressure, $P_g$, of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both $C_g$ and $P_g$, we can rearrange this expression to solve for k.
\begin{align*} C_\ce{g}&=kP_\ce{g}\ k&=\dfrac{C_\ce{g}}{P_\ce{g}}\ &=\mathrm{\dfrac{1.38×10^{−3}\:mol\:L^{−1}}{101.3\:kPa}}\ &=\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}}\ &\hspace{15px}\mathrm{(1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1})} \end{align*} \nonumber
Now we can use k to find the solubility at the lower pressure.
$C_\ce{g}=kP_\ce{g} \nonumber$
$\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}×20.7\:kPa\ (or\:1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1}×155\:torr)\ =2.82×10^{−4}\:mol\:L^{−1}} \nonumber$
Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.
Exercise $1$
A 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10−3 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr).
Answer
7.25 × 10−3 g
Case Study: Decompression Sickness (“The Bends”)
Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law.
As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventative measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure $4$).
Deviations from Henry’s law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water does not increase as rapidly with increasing pressure as predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions.
This reaction diagram shows three H atoms bonded to an N atom above, below, and two the left of the N. A single pair of dots is present on the right side of the N. This is followed by a plus, then two H atoms bonded to an O atom to the left and below the O. Two pairs of dots are present on the O, one above and the other to the right of the O. A double arrow, with a top arrow pointing right and a bottom arrow pointing left follows. To the right of the double arrow, four H atoms are shown bonded to a central N atom. These 5 atoms are enclosed in brackets with a superscript plus outside. A plus follows, then an O atom linked by a bond to an H atom on its right. The O atom has pairs of dots above, to the left, and below the atom. The linked O and H are enclosed in brackets with superscript minus outside.
Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.04%3A_Solutions_of_Gases_in_Water.txt |
Learning Objectives
• Express the amount of solute in a solution in various concentration units.
To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms with meanings that depend on various factors.
Introduction
Concentration is the measure of how much of a given substance is mixed with another substance. Solutions are said to be either dilute or concentrated. When we say that vinegar is $5\%$ acetic acid in water, we are giving the concentration. If we said the mixture was $10\%$ acetic acid, this would be more concentrated than the vinegar solution.
A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. To make juice, you have to mix the frozen juice concentrate from inside these containers with three or four times the container size full of water. Therefore, you are diluting the concentrated juice. In terms of solute and solvent, the concentrated solution has a lot of solute versus the dilute solution that would have a smaller amount of solute.
The terms "concentrated" and "dilute" provide qualitative methods of describing concentration. Although qualitative observations are necessary and have their place in every part of science, including chemistry, we have seen throughout our study of science that there is a definite need for quantitative measurements in science. This is particularly true in solution chemistry. In this section, we will explore some quantitative methods of expressing solution concentration.
Mass Percent
There are several ways of expressing the concentration of a solution by using a percentage. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100:
$\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber$
mass of solution = mass of solute + mass solvent
If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration.
Suppose that a solution was prepared by dissolving $25.0 \: \text{g}$ of sugar into $100.0 \: \text{g}$ of water.
The mass of the solution is
mass of solution = 25.0g sugar + 100.0g water = 125.0 g
The percent by mass would be calculated by:
$\text{Percent by mass} = \dfrac{25.0 \: \text{g sugar}}{125.0 \: \text{g solution}} \times 100\% = 20.0\% \: \text{sugar} \nonumber$
Example $1$
A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?
Solution
We can substitute the quantities given in the equation for mass/mass percent:
$\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}$
Exercise $1$
A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution?
Answer
7.90 %
Using Mass Percent in Calculations
Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use. Using mass percent as a conversion can be useful in this type of problem. The mass percent can be expressed as a conversion factor in the form $\dfrac{g \; \rm{solute}}{100 \; \rm{g solution}}$ or $\dfrac{100 \; \rm g solution}{g\; \rm{solute}}$
For example, if you need to make $3000.0 \: \text{g}$ of a $5.00\%$ solution of sodium chloride, the mass of solute needs to be determined.
Solution
Given: 3000.0 g NaCl solution
5.00% NaCl solution
Find: mass of solute = ? g NaCl
Other known quantities: 5.00 g NaCl is to 100 g solution
The appropriate conversion factor (based on the given mass percent) can be used follows:
To solve for the mass of NaCl, the given mass of solution is multiplied by the conversion factor.
$g NaCl = 3,000.0 \cancel{g \: NaCl \:solution} \times \dfrac{5.00 \:g \: NaCl}{100\cancel{g \: NaCl \: solution}} = 150.0g \: NaCl \nonumber$
You would need to weigh out $150 \: \text{g}$ of $\ce{NaCl}$ and add it to $2850 \: \text{g}$ of water. Notice that it was necessary to subtract the mass of the $\ce{NaCl}$ $\left( 150 \: \text{g} \right)$ from the mass of solution $\left( 3000 \: \text{g} \right)$ to calculate the mass of the water that would need to be added.
Exercise $1$
What is the amount (in g) of hydrogen peroxide (H2O2) needed to make a 6.00 kg, 3.00 % (by mass) H2O2 solution?
180 g H2O2 | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.05%3A_Specifying_Solution_Concentration-_Mass_Percent.txt |
Learning Objectives
• Use molarity to determine quantities in chemical reactions.
• Use molarity as a conversion factor in calculations.
Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution.
$\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \label{defMolarity}$
The symbol for molarity is $\text{M}$ or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance. For example, the expression $\left[ \ce{Ag^+} \right]$ refers to the molarity of the silver ion in solution. Solution concentrations expressed in molarity are the easiest to perform calculations with, but the most difficult to make in the lab. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.
It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is
$\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber$
Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl (aq).” This is read as "a 3.00 molar sodium chloride solution," meaning that there are 3.00 moles of NaOH dissolved per one liter of solution.
Be sure to note that molarity is calculated as the total volume of the entire solution, not just volume of solvent! The solute contributes to total volume.
If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L?
Step 1: First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol):
$22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.5\cancel{gHCl}}=0.614\, mol\; HCl \nonumber$
Step 2: Now we can use the definition of molarity to determine a concentration:
$M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber$
Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.
Example $1$
A solution is prepared by dissolving $42.23 \: \text{g}$ of $\ce{NH_4Cl}$ into enough water to make $500.0 \: \text{mL}$ of solution. Calculate its molarity.
Solution
Solutions to Example 13.6.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
Mass $= 42.23 \: \text{g} \: \ce{NH_4Cl}$
Volume solution $= 500.0 \: \text{mL} = 0.5000 \: \text{L}$
Find: Molarity = ? M
List other known quantities. Molar mass $\ce{NH_4Cl} = 53.50 \: \text{g/mol}$
Plan the problem.
1. The mass of the ammonium chloride is first converted to moles.
2. Then the molarity is calculated by dividing by liters. Note the given volume has been converted to liters.
$\mathrm{M=\dfrac{mol\: NH_4Cl}{L\: solution}}$
Cancel units and calculate.
Now substitute the known quantities into the equation and solve.
\begin{align*} 42.23 \ \cancel{\text{g} \: \ce{NH_4Cl}} \times \dfrac{1 \: \text{mol} \: \ce{NH_4Cl}}{53.50 \: \cancel{\text{g} \: \ce{NH_4Cl}}} &= 0.7893 \: \text{mol} \: \ce{NH_4Cl} \ \dfrac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L solution}} &= 1.579 \: \text{M} \end{align*}
Think about your result. The molarity is $1.579 \: \text{M}$, meaning that a liter of the solution would contain $1.579 \: \text{mol} \: \ce{NH_4Cl}$. Four significant figures are appropriate.
Exercise $\PageIndex{1A}$
What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution?
Answer
1.57 M C6H12O6
Exercise $\PageIndex{1B}$
What is the concentration, in $\text{mol/L}$, where $137 \: \text{g}$ of $\ce{NaCl}$ has been dissolved in enough water to make $500 \: \text{mL}$ of solution?
Answer
4.69 M NaCl
Using Molarity in Calculations
Concentration can be a conversion factor between the amount of solute and the amount of solution or solvent (depending on the definition of the concentration unit). As such, concentrations can be useful in a variety of stoichiometry problems. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor.
A simple example of using a concentration unit as a conversion factor is one in which we use the definition of the concentration unit and rearrange; we can do the calculation again as a unit conversion, rather than as a definition.
Example $2$: Determining Moles of Solute, Given the Concentration and Volume of a Solution
For example, suppose we ask how many moles of solute are present in 0.108 L of a 0.887 M NaCl solution. Because 0.887 M means 0.887 mol/L, we can use this second expression for the concentration as a conversion factor:
Solution
$0.108\cancel{L\, solution}\times \dfrac{0.887\, mol\, NaCl}{\cancel{1L\, solution}}=0.0958\, mol\, NaCl \nonumber$
If we used the definition approach, we get the same answer, but now we are using conversion factor skills.
Like any other conversion factor that relates two different types of units, the reciprocal of the concentration can be also used as a conversion factor.
Example $3$: Determining Volume of a Solution, Given the Concentration and Moles of Solute
Using concentration as a conversion factor, how many liters of 2.35 M CuSO4 are needed to obtain 4.88 mol of CuSO4?
Solution
This is a one-step conversion, but the concentration must be written as the reciprocal for the units to work out:
$4.88\cancel{mol\, CuSO_{4}}\times \dfrac{1\, L\, solution}{2.35\cancel{mol\, CuSO_{4}}}=2.08\, L\, of \, solution \nonumber$
In a laboratory situation, a chemist must frequently prepare a given volume of solutions of a known molarity. The task is to calculate the mass of the solute that is necessary. The molarity equation can be rearranged to solve for moles, which can then be converted to grams. The following example illustrates this.
Example $4$
A chemist needs to prepare $3.00 \: \text{L}$ of a $0.250 \: \text{M}$ solution of potassium permanganate $\left( \ce{KMnO_4} \right)$. What mass of $\ce{KMnO_4}$ does she need to make the solution?
Solution
Solutions to Example 13.6.4
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
Molarity $= 0.250 \: \text{M}$
Volume $= 3.00 \: \text{L}$
Find: Mass $\ce{KMnO_4} = ? \: \text{g}$
List other known quantities.
Molar mass $\ce{KMnO_4} = 158.04 \: \text{g/mol}$
0.250 mol KMnO4 to 1 L of KMnO4 solution
Plan the problem.
Cancel units and calculate.
Now substitute the known quantities into the equation and solve.
\begin{align*} \text{mol} \: \ce{KMnO_4} = 0.250 \: \text{M} \: \ce{KMnO_4} \times 3.00 \: \text{L} &= 0.750 \: \text{mol} \: \ce{KMnO_4} \ \cancel{3.00 \: L \: solution} \times \dfrac{0.250 \: \cancel{\text{mol} \: \ce{KMnO_4}}}{1\cancel{L \: solution}} \times \dfrac{158.04 \: \text{g} \: \ce{KMnO_4}}{1 \: \cancel{\text{mol} \: \ce{KMnO_4}}} &= 119 \: \text{g} \: \ce{KMnO_4} \end{align*}
Think about your result. When $119 \: \text{g}$ of potassium permanganate is dissolved into water to make $3.00 \: \text{L}$ of solution, the molarity is $0.250 \: \text{M}$.
Exercise $\PageIndex{4A}$
Using concentration as a conversion factor, how many liters of 0.0444 M CH2O are needed to obtain 0.0773 mol of CH2O?
Answer
1.74 L
Exercise $\PageIndex{4B}$
Answer the problems below using concentration as a conversion factor.
1. What mass of solute is present in 1.08 L of 0.0578 M H2SO4?
2. What volume of 1.50 M HCl solution contains 10.0 g of hydrogen chloride?
Answer a
6.12 g
Answer b
183 mL or 0.183L
How to Indicate Concentration
• Square brackets are often used to represent concentration, e.g., [NaOH] = 0.50 M.
• Use the capital letter M for molarity, not a lower case m (this is a different concentration unit called molality).
Watch as the Flinn Scientific Tech Staff demonstrates "How To Prepare Solutions."
It is important to note that there are many different ways you can set up and solve your chemistry equations. Some students prefer to answer multi-step calculations in one long step, while others prefer to work out each step individually. Neither method is necessarily better or worse than the other method—whichever makes the most sense to you is the one that you should use. In this text, we will typically use unit analysis (also called dimension analysis or factor analysis).
Contributors and Attributions
• Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.06%3A_Specifying_Solution_Concentration-_Molarity.txt |
Learning Objectives
• Explain how concentrations can be changed in the lab.
• Understand how stock solutions are used in the laboratory.
We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms: dilute and concentrated.
• A dilute solution is one in which there is a relatively small amount of solute dissolved in the solution.
• A concentrated solution contains a relatively large amount of solute.
These two terms do not provide any quantitative information (actual numbers), but they are often useful in comparing solutions in a more general sense. These terms also do not tell us whether or not the solution is saturated or unsaturated, or whether the solution is "strong" or "weak". These last two terms will have special meanings when we discuss acids and bases, so be careful not to confuse them.
Stock Solutions
It is often necessary to have a solution with a concentration that is very precisely known. Solutions containing a precise mass of solute in a precise volume of solution are called stock (or standard) solutions. To prepare a standard solution, a piece of lab equipment called a volumetric flask should be used. These flasks range in size from 10 mL to 2000 mL and are carefully calibrated to a single volume. On the narrow stem is a calibration mark. The precise mass of solute is dissolved in a bit of the solvent, and this is added to the flask. Then, enough solvent is added to the flask until the level reaches the calibration mark.
Often, it is convenient to prepare a series of solutions of known concentrations by first preparing a single stock solution, as described in the previous section. Aliquots (carefully measured volumes) of the stock solution can then be diluted to any desired volume. In other cases, it may be inconvenient to weigh a small mass of sample accurately enough to prepare a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration.
Dilutions of Stock (or Standard) Solutions
Imagine we have a salt water solution with a certain concentration. That means we have a certain amount of salt (a certain mass or a certain number of moles) dissolved in a certain volume of solution. Next, we will dilute this solution. This is done by adding more water, not more salt:
$\rightarrow$
Before Dilution and After Dilution
The molarity of solution 1 is
$M_1 = \dfrac{\text{moles}_1}{\text{liter}_1} \nonumber$
and the molarity of solution 2 is
$M_2 = \dfrac{\text{moles}_2}{\text{liter}_2} \nonumber$
rearrange the equations to find moles:
$\text{moles}_1 = M_1 \text{liter}_1 \nonumber$
and
$\text{moles}_2 = M_2 \text{liter}_2 \nonumber$
What stayed the same and what changed between the two solutions? By adding more water, we changed the volume of the solution. Doing so also changed its concentration. However, the number of moles of solute did not change. So,
$moles_1 = moles_2 \nonumber$
Therefore
$\boxed{M_1V_1= M_2V_2 } \label{diluteEq}$
where
• $M_1$ and $M_2$ are the concentrations of the original and diluted solutions
• $V_1$ and $V_2$ are the volumes of the two solutions
Preparing dilutions is a common activity in the chemistry lab and elsewhere. Once you understand the above relationship, the calculations are simple.
Suppose that you have $100. \: \text{mL}$ of a $2.0 \: \text{M}$ solution of $\ce{HCl}$. You dilute the solution by adding enough water to make the solution volume $500. \: \text{mL}$. The new molarity can easily be calculated by using the above equation and solving for $M_2$.
$M_2 = \dfrac{M_1 \times V_1}{V_2} = \dfrac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl} \nonumber$
The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value.
Another common dilution problem involves calculating what amount of a highly concentrated solution is required to make a desired quantity of solution of lesser concentration. The highly concentrated solution is typically referred to as the stock solution.
Example $1$: Diluting Nitric Acid
Nitric acid $\left( \ce{HNO_3} \right)$ is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is $16 \: \text{M}$. How much of the stock solution of nitric acid needs to be used to make $8.00 \: \text{L}$ of a $0.50 \: \text{M}$ solution?
Solution
Solutions to Example13.7.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
M1, Stock $\ce{HNO_3} = 16 \: \text{M}$
$V_2 = 8.00 \: \text{L}$
$M_2 = 0.50 \: \text{M}$
Find: Volume stock $\ce{HNO_3} \left( V_1 \right) = ? \: \text{L}$
List other known quantities. none
Plan the problem.
First, rearrange the equation algebraically to solve for $V_1$.
$V_1 = \dfrac{M_2 \times V_2}{M_1} \nonumber$
Calculate and cancel units.
Now substitute the known quantities into the equation and solve.
$V_1 = \dfrac{0.50 \: \text{M} \times 8.00 \: \text{L}}{16 \: \text{M}} = 0.25 \: \text{L}$
Think about your result. $0.25 \: \text{L} \: (250 \: \text{mL})$ of the stock $\ce{HNO_3}$ needs to be diluted with water to a final volume of $8.00 \: \text{L}$. The dilution is by a factor of 32 to go from $16 \: \text{M}$ to $0.5 \: \text{M}$.
Exercise $1$
A 0.885 M solution of KBr with an initial volume of 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution?
Answer
135.4 mL
Note that the calculated volume will have the same dimensions as the input volume, and dimensional analysis tells us that in this case we don't need to convert to liters, since L cancels when we divide M (mol/L) by M (mol/L).
Diluting and Mixing Solutions
Diluting and Mixing Solutions
How to Dilute a Solution by CarolinaBiological | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.07%3A_Solution_Dilution.txt |
Learning Objectives
• Determine amounts of reactants or products in aqueous solutions.
As we learned previously, double replacement reactions involve the reaction between ionic compounds in solution and, in the course of the reaction, the ions in the two reacting compounds are “switched” (they replace each other). Because these reactions occur in aqueous solution, we can use the concept of molarity to directly calculate the number of moles of reactants or products that will be formed, and therefore their amounts (i.e. volume of solutions or mass of precipitates).
As an example, lead (II) nitrate and sodium chloride react to form sodium nitrate and the insoluble compound, lead (II) chloride.
$\ce{Pb(NO_3)_2 (aq) + 2 NaCl (aq) → PbCl_2 (s) + 2 NaNO3 (aq)} \label{EQ1}$
In the reaction shown above, if we mixed 0.123 L of a 1.00 M solution of $\ce{NaCl}$ with 1.50 M solution of $\ce{Pb(NO3)2}$, we could calculate the volume of $\ce{Pb(NO3)2}$ solution needed to completely precipitate the $\ce{Pb^{2+}}$ ions.
The molar concentration can also be expressed as the following:
$1.00\,M \,\ce{NaCl} = \dfrac{1.00 \; mol \; \ce{NaCl}}{1 \; L \; \ce{NaCl} \; \text{solution}} \nonumber$
and
$1.50\,M\, \ce{Pb(NO3)2} = \dfrac{1.50 \; mol \; \ce{Pb(NO_3)_2}}{1 \; L \; \ce{Pb(NO_3)_2} \text{solution}} \nonumber$
First, we must examine the reaction stoichiometry in the balanced reaction (Equation \ref{EQ1}). In this reaction, one mole of $\ce{Pb(NO3)2}$ reacts with two moles of $\ce{NaCl}$ to give one mole of $\ce{PbCl2}$ precipitate. Thus, the concept map utilizing the stoichiometric ratios is:
so the volume of lead (II) nitrate that reacted is calculated as:
$0.123 \; L \; \ce{NaCl} \, \text{solution} \times \dfrac{1.00 \; mol \; \ce{NaCl}}{1 \; L \; \ce{NaCl} \; \text{solution}} \times \dfrac{1 \; mol \; \ce{Pb(NO_3)_2}}{2 \; mol \; \ce{NaCl}} \times \dfrac{1 \; L \; \ce{Pb(NO_3)_2} \; \text{solution}}{1.5 \; mol \; \ce{Pb(NO_3)_2}} = 0.041 \; \ce{Pb(NO_3)_2}\,L \; \text{solution} \nonumber$
This volume makes intuitive sense for two reasons: (1) the number of moles of $\ce{Pb(NO3)2}$ required is half of the number of moles of $\ce{NaCl}$, based off of the stoichiometry in the balanced reaction (Equation \ref{EQ1}); (2) the concentration of $\ce{Pb(NO3)2}$ solution is 50% greater than the $\ce{NaCl}$ solution, so less volume is needed.
Example $1$
What volume (in L) of 0.500 M sodium sulfate will react with 275 mL of 0.250 M barium chloride to completely precipitate all $\ce{Ba^{2+}}$ in the solution?
Solution
Solutions to Example 13.8.1
Steps for Problem Solving Example $1$
Identify the "given" information and what the problem is asking you to "find."
Given: 275 mL BaCl2
0.250 M $\ce{BaCl2}$ or $\displaystyle \dfrac{0.250\; mol BaCl_2}{1\; L\; BaCl_2\; solution}$
0.500 M $\ce{Na2SO4}$ or $\displaystyle \dfrac{0.500\; mol Na_2SO_4}{1\; L\; Na_2SO_4\; solution}$
Find: Volume $\ce{Na2SO4}$ solution.
Set up and balance the chemical equation.
$\ce{Na2SO4(aq) + BaCl2(aq) -> BaSO4(s)} + \underline{2} \ce{NaCl (aq)}$
An insoluble product is formed after the reaction.
List other known quantities.
1 mol of Na2SO4 to 1 mol BaCl2
1000 mL = 1 L
Prepare a concept map and use the proper conversion factor.
Cancel units and calculate. $\displaystyle 275\cancel{mL \; BaCl_2 \; solution}\times \dfrac{1\cancel{L}}{1000\cancel{mL}}\times \dfrac{0.250 \cancel{mol \; BaCl_2}}{1 \cancel{L\;BaCl_2 \; solution}}\times \dfrac{1 \cancel{mol \; Na_2SO_4}}{1 \cancel{mol \; BaCl_2}}\times \dfrac{1\; L \; Na_2SO_4 \; solution}{0.500 \cancel{mol Na_2SO_4}}$ = 0.1375 L sodium sulfate
Think about your result. The lesser amount (almost half) of sodium sulfate is to be expected as it is more concentrated than barium chloride. Also, the units are correct.
Exercise $1$
What volume of 0.250 M lithium hydroxide will completely react with 0.500 L of 0.250 M of sulfuric acid solution?
Answer
0.250 L $\ce{LiOH}$ solution | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.08%3A_Solution_Stoichiometry.txt |
Learning Objectives
• Explain what the term "colligative" means, and list the colligative properties.
• Indicate what happens to the boiling point and the freezing point of a solvent when a solute is added to it.
• Calculate boiling point elevations and freezing point depressions for a solution.
People who live in colder climates have seen trucks put salt on the roads when snow or ice is forecast. Why is this done? As a result of the information you explore in this section, you will understand why these events occur. You will also learn to calculate exactly how much of an effect a specific solute can have on the boiling point or freezing point of a solution.
The example given in the introduction is an example of a colligative property. Colligative properties are properties that differ based on the concentration of solute in a solvent, but not on the type of solute. What this means for the example above is that people in colder climates do not necessarily need salt to get the same effect on the roads—any solute will work. However, the higher the concentration of solute, the more these properties will change.
Boiling Point Elevation
Water boils at $100^\text{o} \text{C}$ at $1 \: \text{atm}$ of pressure, but a solution of saltwater does not . When table salt is added to water, the resulting solution has a higher boiling point than the water did by itself. The ions form an attraction with the solvent particles that prevents the water molecules from going into the gas phase. Therefore, the saltwater solution will not boil at $100^\text{o} \text{C}$. In order for the saltwater solution to boil, the temperature must be raised about $100^\text{o} \text{C}$. This is true for any solute added to a solvent; the boiling point will be higher than the boiling point of the pure solvent (without the solute). In other words, when anything is dissolved in water, the solution will boil at a higher temperature than pure water would.
The boiling point elevation due to the presence of a solute is also a colligative property. That is, the amount of change in the boiling point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A $0.20 \: \text{m}$ solution of table salt and a $0.20 \: \text{m}$ solution of hydrochloric acid would have the same effect on the boiling point.
Freezing Point Depression
The effect of adding a solute to a solvent has the opposite effect on the freezing point of a solution as it does on the boiling point. A solution will have a lower freezing point than a pure solvent. The freezing point is the temperature at which the liquid changes to a solid. At a given temperature, if a substance is added to a solvent (such as water), the solute-solvent interactions prevent the solvent from going into the solid phase. The solute-solvent interactions require the temperature to decrease further in order to solidify the solution. A common example is found when salt is used on icy roadways. Salt is put on roads so that the water on the roads will not freeze at the normal $0^\text{o} \text{C}$ but at a lower temperature, as low as $-9^\text{o} \text{C}$. The de-icing of planes is another common example of freezing point depression in action. A number of solutions are used, but commonly a solution such as ethylene glycol, or a less toxic monopropylene glycol, is used to de-ice an aircraft. The aircrafts are sprayed with the solution when the temperature is predicted to drop below the freezing point. The freezing point depression is the difference in the freezing points of the solution from the pure solvent. This is true for any solute added to a solvent; the freezing point of the solution will be lower than the freezing point of the pure solvent (without the solute). Thus, when anything is dissolved in water, the solution will freeze at a lower temperature than pure water would.
The freezing point depression due to the presence of a solute is also a colligative property. That is, the amount of change in the freezing point is related to the number of particles of solute in a solution and is not related to the chemical composition of the solute. A $0.20 \: \text{m}$ solution of table salt and a $0.20 \: \text{m}$ solution of hydrochloric acid would have the same effect on the freezing point.
Comparing the Freezing and Boiling Point of Solutions
Recall that covalent and ionic compounds do not dissolve in the same way. Ionic compounds break up into cations and anions when they dissolve. Covalent compounds typically do not break up. For example a sugar/water solution stays as sugar and water, with the sugar molecules staying as molecules. Remember that colligative properties are due to the number of solute particles in the solution. Adding 10 molecules of sugar to a solvent will produce 10 solute particles in the solution. When the solute is ionic, such as $\ce{NaCl}$ however, adding 10 formulas of solute to the solution will produce 20 ions (solute particles) in the solution. Therefore, adding enough $\ce{NaCl}$ solute to a solvent to produce a $0.20 \: \text{m}$ solution will have twice the effect of adding enough sugar to a solvent to produce a $0.20 \: \text{m}$ solution. Colligative properties depend on the number of solute particles in the solution.
"$i$" is the number of particles that the solute will dissociate into upon mixing with the solvent. For example, sodium chloride, $\ce{NaCl}$, will dissociate into two ions so for $\ce{NaCl}$, $i = 2$; for lithium nitrate, $\ce{LiNO_3}$, $i = 2$; and for calcium chloride, $\ce{CaCl_2}$, $i = 3$. For covalent compounds, $i$ is always equal to 1.
By knowing the molality of a solution and the number of particles a compound will dissolve to form, it is possible to predict which solution in a group will have the lowest freezing point. To compare the boiling or freezing points of solutions, follow these general steps:
1. Label each solute as ionic or covalent.
2. If the solute is ionic, determine the number of ions in the formula. Be careful to look for polyatomic ions.
3. Multiply the original molality ($\text{m}$) of the solution by the number of particles formed when the solution dissolves. This will give you the total concentration of particles dissolved.
4. Compare these values. The higher total concentration will result in a higher boiling point and a lower freezing point.
Example $1$
Rank the following solutions in water in order of increasing (lowest to highest) freezing point:
• $0.1 \: \text{m} \: \ce{NaCl}$
• $0.1 \: \text{m} \: \ce{C_6H_{12}O_6}$
• $0.1 \: \text{m} \: \ce{CaI_2}$
Solution
To compare freezing points, we need to know the total concentration of all particles when the solute has been dissolved.
• $0.1 \: \text{m} \: \ce{NaCl}$: This compound is ionic (metal with nonmetal), and will dissolve into 2 parts. The total final concentration is: $\left( 0.1 \: \text{m} \right) \left( 2 \right) = 0.2 \: \text{m}$
• $0.1 \: \text{m} \: \ce{C_6H_{12}O_6}$: This compound is covalent (nonmetal with nonmetal), and will stay as 1 part. The total final concentration is: $\left( 0.1 \: \text{m} \right) \left(1 \right) = 0.1 \: \text{m}$
• $0.1 \: \text{m} \: \ce{CaI_2}$: This compound is ionic (metal with nonmetal), and will dissolve into 3 parts. The total final concentration is: $\left( 0.1 \: \text{m} \right) \left( 3 \right) = 0.3 \: \text{m}$
Remember, the greater the concentration of particles, the lower the freezing point will be. $0.1 \: \text{m} \: \ce{CaI_2}$ will have the lowest freezing point, followed by $0.1 \: \text{m} \: \ce{NaCl}$, and the highest of the three solutions will be $0.1 \: \text{m} \: \ce{C_6H_{12}O_6}$, but all three of them will have a lower freezing point than pure water.
The boiling point of a solution is higher than the boiling point of a pure solvent, and the freezing point of a solution is lower than the freezing point of a pure solvent. However, the amount to which the boiling point increases or the freezing point decreases depends on the amount of solute that is added to the solvent. A mathematical equation is used to calculate the boiling point elevation or the freezing point depression.
The boiling point elevation is the amount that the boiling point temperature increases compared to the original solvent. For example, the boiling point of pure water at $1.0 \: \text{atm}$ is $100^\text{o} \text{C}$ while the boiling point of a $2\%$ saltwater solution is about $102^\text{o} \text{C}$. Therefore, the boiling point elevation would be $2^\text{o} \text{C}$. The freezing point depression is the amount that the freezing temperature decreases.
Both the boiling point elevation and the freezing point depression are related to the molality of the solution. Looking at the formula for the boiling point elevation and freezing point depression, we see similarities between the two. The equation used to calculate the increase in the boiling point is:
$\Delta T_b = k_b \cdot \text{m} \cdot i \label{BP}$
Where:
• $\Delta T_b =$ the amount the boiling point increases.
• $k_b =$ the boiling point elevation constant which depends on the solvent (for water, this number is $0.515^\text{o} \text{C/m}$).
• $\text{m} =$ the molality of the solution.
• $i =$ the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1).
The following equation is used to calculate the decrease in the freezing point:
$\Delta T_f = k_f \cdot \text{m} \cdot i \label{FP}$
Where:
• $\Delta T_f =$ the amount the freezing temperature decreases.
• $k_f =$ the freezing point depression constant which depends on the solvent (for water, this number is $1.86^\text{o} \text{C/m}$).
• $\text{m} =$ the molality of the solution.
• $i =$ the number of particles formed when that compound dissolves (for covalent compounds, this number is always 1).
Example $2$: Adding Antifreeze to Protein Engines
Antifreeze is used in automobile radiators to keep the coolant from freezing. In geographical areas where winter temperatures go below the freezing point of water, using pure water as the coolant could allow the water to freeze. Since water expands when it freezes, freezing coolant could crack engine blocks, radiators, and coolant lines. The main component in antifreeze is ethylene glycol, $\ce{C_2H_4(OH)_2}$. What is the concentration of ethylene glycol in a solution of water, in molality, if the freezing point dropped by $2.64^\text{o} \text{C}$? The freezing point constant, $k_f$, for water is $1.86^\text{o} \text{C/m}$.
Solution
Use the equation for freezing point depression of solution (Equation $\ref{FP}$):
$\Delta T_f = k_f \cdot \text{m} \cdot i \nonumber$
Substituting in the appropriate values we get:
$2.64^\text{o} \text{C} = \left( 1.86^\text{o} \text{C/m} \right) \left( \text{m} \right) \left( 1 \right) \nonumber$
Solve for $\text{m}$ by dividing both sides by $1.86^\text{o} \text{C/m}$.
$\text{m} = 1.42 \nonumber$
Example $3$: Adding Salt to Elevate Boiling Temperature
A solution of $10.0 \: \text{g}$ of sodium chloride is added to $100.0 \: \text{g}$ of water in an attempt to elevate the boiling point. What is the boiling point of the solution? $k_b$ for water is $0.52^\text{o} \text{C/m}$.
Solution
Use the equation for boiling point elevation of solution (Equation $\ref{BP}$):
$\Delta T_b = k_b \cdot \text{m} \cdot i \nonumber$
We need to be able to substitute each variable into this equation.
• $k_b = 0.52^\text{o} \text{C/m}$
• $\text{m}$: We must solve for this using stoichiometry. Given: $10.0 \: \text{g} \: \ce{NaCl}$ and $100.0 \: \text{g} \: \ce{H_2O}$ Find: $\text{mol} \: \ce{NaCl}/\text{kg} \: \ce{H_2O}$. Ratios: molar mass of $\ce{NaCl}$, $1000 \: \text{g} = 1 \: \text{kg}$
$\dfrac{10.0 \: \cancel{\text{g} \: \ce{NaCl}}}{100.0 \: \cancel{\text{g} \: \ce{H_2O}}} \cdot \dfrac{1 \: \text{mol} \: \ce{NaCl}}{58.45 \: \cancel{\text{g} \: \ce{NaCl}}} \cdot \dfrac{1000 \: \cancel{\text{g} \: \ce{H_2O}}}{1 \: \text{kg} \: \ce{H_2O}} = 1.71 \: \text{m} \nonumber$
• For $\ce{NaCl}$, $i = 2$
Substitute these values into the equation $\Delta T_b = k_b \cdot \text{m} \cdot i$. We get:
$\Delta T_b = \left( 0.52 \dfrac{^\text{o} \text{C}}{\cancel{\text{m}}} \right) \left( 1.71 \: \cancel{\text{m}} \right) \left( 2 \right) = 1.78^\text{o} \text{C} \nonumber$
Water normally boils at $100^\text{o} \text{C}$, but our calculation shows that the boiling point increased by $1.78^\text{o} \text{C}$. Our new boiling point is $101.78^\text{o} \text{C}$.
Note: Since sea water contains roughly 28.0 g of NaCl per liter, this saltwater solution is approximately four times more concentrated than sea water (all for a 2° C rise of boiling temperature).
Summary
• Colligative properties are properties that are due only to the number of particles in solution, and are not related to the chemical properties of the solute.
• Boiling points of solutions are higher than the boiling points of the pure solvents.
• Freezing points of solutions are lower than the freezing points of the pure solvents.
• Ionic compounds split into ions when they dissolve, forming more particles. Covalent compounds stay as complete molecules when they dissolve.
Vocabulary
• Colligative property - A property that is due only to the number of particles in solution, and not the type of the solute.
• Boiling point elevation - The amount that the boiling point of a solution increases from the boiling point of the pure solvent.
• Freezing point depression - The amount that the freezing point of a solution decreases from the freezing point of the pure solvent. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.09%3A_Freezing_Point_Depression_and_Boiling_Point_Elevation.txt |
Learning Objectives
• Explain the following laws within the Ideal Gas Law
Before we introduce the final colligative property, we need to present a new concept. A semipermeable membrane is a thin membrane that will pass certain small molecules, but not others. A thin sheet of cellophane, for example, acts as a semipermeable membrane. Consider the system in Figure $1$.
1. A semipermeable membrane separates two solutions having the different concentrations marked. Curiously, this situation is not stable; there is a tendency for water molecules to move from the dilute side (on the left) to the concentrated side (on the right) until the concentrations are equalized, as in Figure $\PageIndex{1b}$.
2. This tendency is called osmosis. In osmosis, the solute remains in its original side of the system; only solvent molecules move through the semipermeable membrane. In the end, the two sides of the system will have different volumes. Because a column of liquid exerts a pressure, there is a pressure difference (Π) on the two sides of the system that is proportional to the height of the taller column. This pressure difference is called the osmotic pressure, which is a colligative property.
The osmotic pressure of a solution is easy to calculate:
$\Pi = MRT \nonumber$
where $Π$ is the osmotic pressure of a solution, M is the molarity of the solution, R is the ideal gas law constant, and T is the absolute temperature. This equation is reminiscent of the ideal gas law we considered in Chapter 6.
Example $5$: Osmotic Pressure
What is the osmotic pressure of a 0.333 M solution of C6H12O6 at 25°C?
Solution
First we need to convert our temperature to kelvins:
T = 25 + 273 = 298 K
Now we can substitute into the equation for osmotic pressure, recalling the value for R:
$Π =(0.333M)\left (0.08205\dfrac{L.atm}{mol.K} \right )(298K) \nonumber$
The units may not make sense until we realize that molarity is defined as moles per liter:
$Π =\left ( 0.333\dfrac{mol}{L} \right )\left (0.08205\dfrac{L.atm}{mol.K} \right )(298K) \nonumber$
Now we see that the moles, liters, and kelvins cancel, leaving atmospheres, which is a unit of pressure. Solving,
$Π = 8.14 \,atm \nonumber$
This is a substantial pressure! It is the equivalent of a column of water 84 m tall.
Exercise $5$
What is the osmotic pressure of a 0.0522 M solution of C12H22O11 at 55°C?
Answer
1.40 atm
Osmotic pressure is important in biological systems because cell walls are semipermeable membranes. In particular, when a person is receiving intravenous (IV) fluids, the osmotic pressure of the fluid needs to be approximately the same as blood serum to avoid any negative consequences. Figure $3$ shows three red blood cells:
• A healthy red blood cell.
• A red blood cell that has been exposed to a lower concentration than normal blood serum (a hypotonic solution); the cell has plumped up as solvent moves into the cell to dilute the solutes inside.
• A red blood cell exposed to a higher concentration than normal blood serum (hypertonic); water leaves the red blood cell, so it collapses onto itself. Only when the solutions inside and outside the cell are the same (isotonic) will the red blood cell be able to do its job.
Osmotic pressure is also the reason you should not drink seawater if you’re stranded on a lifeboat in the ocean; seawater has a higher osmotic pressure than most of the fluids in your body. You can drink the water, but ingesting it will pull water out of your cells as osmosis works to dilute the seawater. Ironically, your cells will die of thirst, and you will also die. (It is okay to drink the water if you are stranded on a body of freshwater, at least from an osmotic pressure perspective.) Osmotic pressure is also thought to be important—in addition to capillary action—in getting water to the tops of tall trees.
Summary
• Osmotic pressure is caused by concentration differences between solutions separated by a semipermeable membrane, and is an important biological consideration. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/13%3A_Solutions/13.10%3A_Osmosis.txt |
Acids and bases are common substances found in many every day items, from fruit juices and soft drinks to soap. In this chapter, we will examine the properties of acids and bases, and learn about the chemical nature of these important compounds. We will cover pH, and how to calculate the pH of a solution.
• 14.1: Sour Patch Kids and International Spy Movies
Sour Patch Kids are a soft candy with a coating of invert sugar and sour sugar (a combination of citric acid, tartaric acid and sugar). The candy's slogan, "Sour. Sweet. Gone.", refers to its sour-to-sweet taste.
• 14.2: Acids- Properties and Examples
Acids are very common in some of the foods that we eat. Citrus fruits such as oranges and lemons contain citric acid and ascorbic acid, which is better known as vitamin C. Carbonated sodas contain phosphoric acid. Vinegar contains acetic acid. Your own stomach utilizes hydrochloric acid to digest food. Acids are a distinct class of compounds because of the properties of their aqueous solutions.
• 14.3: Bases- Properties and Examples
A base is thought of as a substance which can accept protons, or any chemical compound that yields hydroxide ions (OH-) in solution. It is also commonly referred to as any substance that can react with an acid to decrease or neutralize its acidic properties, change the color of indicators (e.g. turn red litmus paper blue), feel slippery to the touch when in solution, taste bitter, react with acids to form salts, and promote certain chemical reactions (e.g. base catalysis).
• 14.4: Molecular Definitions of Acids and Bases
Although the properties of acids and bases had been recognized for a long time, it was Svante Arrhenius in the 1880s who determined that the properties of acids were due to the presence of hydrogen ions, and that the properties of bases were due to the presence of hydroxide ions.
• 14.5: Reactions of Acids and Bases
When an acid and a base are combined, water and a salt are the products. Salts are ionic compounds containing a positive ion other than H+ and a negative ion other than the hydroxide ion, OH-. Double displacement reactions of this type are called neutralization reactions. Salt solutions do not always have a pH of 7, however. Through a process known as hydrolysis, the ions produced when an acid and base combine may react with the water to produce slightly acidic or basic solutions.
• 14.6: Acid–Base Titration
Acid-base titrations are lab procedures used to determine the concentration of a solution. One of the standard laboratory exercises in General Chemistry is an acid-base titration. During an acid-base titration, an acid with a known concentration (a standard solution) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base. The indicator will signal, by color change, when the base has been neutralized (when [H+] = [OH-]).
• 14.7: Strong and Weak Acids and Bases
Acids are classified as either strong or weak, based on their ionization in water. A strong acid is an acid which is completely ionized in an aqueous solution. A weak acid is an acid that ionizes only slightly in an aqueous solution. Acetic acid (found in vinegar) is a very common weak acid.
• 14.8: Water - Acid and Base in One
Water is an interesting compound in many respects. Here, we will consider its ability to behave as an acid or a base. In some circumstances, a water molecule will accept a proton and thus act as a Brønsted-Lowry base.
• 14.9: The pH and pOH Scales - Ways to Express Acidity and Basicity
pH and pOH are defined as the negative log of hydrogen ion concentration and hydroxide concentration, respectively. Knowledge of either can be used to calculate either [H+] of [OH-]. pOH is related to pH and can be easily calculated from pH.
• 14.10: Buffers- Solutions that Resist pH Change
A buffer is a solution that resists dramatic changes in pH. Buffers do so by being composed of certain pairs of solutes: either a weak acid plus a salt derived from that weak acid, or a weak base plus a salt of that weak base.
• 14.11: Prelude - Sour Patch Kids
Sour Patch Kids are a soft candy with a coating of invert sugar and sour sugar (a combination of citric acid, tartaric acid and sugar). The candy's slogan, "Sour. Sweet. Gone.", refers to its sour-to-sweet taste.
14: Acids and Bases
Sour Patch Kids are a soft candy with a coating of invert sugar and sour sugar. The candy's slogan, "Sour. Sweet. Gone.", refers to its sour-to-sweet taste.
Sour sugar is a food ingredient that is used to impart a sour flavor, made from citric or tartaric acid and sugar. It is used to coat sour candies like Sour Patch Kids. Eating large amounts of sour sugar can cause irritation of the tongue because of the acid. It can also cause irreversible dental erosion. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.01%3A_Sour_Patch_Kids_and_International_Spy_Movies.txt |
Learning Objectives
• Examine properties of acids.
Many people enjoy drinking coffee. A cup first thing in the morning helps start the day. But keeping the coffee maker clean can be a problem. Lime deposits build up after a while and slow down the brewing process. The best cure for this is to put vinegar (dilute acetic acid) in the pot and run it through the brewing cycle. The vinegar dissolves the deposits and cleans the maker, which will speed up the brewing process back to its original rate. Just be sure to run water through the brewing process after the vinegar, or you will get some really horrible coffee.
Acids
Acids are very common in some of the foods that we eat. Citrus fruits such as oranges and lemons contain citric acid and ascorbic acid, which is better known as vitamin C. Carbonated sodas contain phosphoric acid. Vinegar contains acetic acid. Your own stomach utilizes hydrochloric acid to digest food. Acids are a distinct class of compounds because of the properties of their aqueous solutions as outlined below:
1. Aqueous solutions of acids are electrolytes, meaning that they conduct electrical current. Some acids are strong electrolytes because they ionize completely in water, yielding a great many ions. Other acids are weak electrolytes that exist primarily in a non-ionized form when dissolved in water.
2. Acids have a sour taste. Lemons, vinegar, and sour candies all contain acids.
3. Acids change the color of certain acid-base indicates. Two common indicators are litmus and phenolphthalein. Blue litmus turns red in the presence of an acid, while phenolphthalein turns colorless.
4. Acids react with active metals to yield hydrogen gas. Recall that an activity series is a list of metals in descending order of reactivity. Metals that are above hydrogen in the activity series will replace the hydrogen from an acid in a single-replacement reaction, as shown below:
$\ce{Zn} \left( s \right) + \ce{H_2SO_4} \left( aq \right) \rightarrow \ce{ZnSO_4} \left( aq \right) + \ce{H_2} \left( g \right) \label{eq1}$
5. Acids react with bases to produce a salt compound and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The products of this reaction are an ionic compound, which is labeled as a salt, and water.
It should not be hard for you to name several common acids (but you might find that listing bases is a little more difficult). Below is a partial list of some common acids, along with some chemical formulas:
Table $1$: Common Acids and Their Uses
Chemist Name
Common Name Uses
hydrochloric acid, HCl muriatic acid (used in pools) and stomach acid is HCl Used in cleaning (refining) metals, in maintenance of swimming pools, and for household cleaning.
sulfuric acid, H2SO4 Used in car batteries, and in the manufacture of fertilizers.
nitric acid, HNO3 Used in the manufacture of fertilizers, explosives and in extraction of gold.
acetic acid, HC2H3O2 vinegar Main ingredient in vinegar.
carbonic acid, H2CO3 responsible for the "fizz" in carbonated drinks As an ingredient in carbonated drinks.
citric acid, C6H8O7 Used in food and dietary supplements. Also added as an acidulant in creams, gels, liquids, and lotions.
acetylsalicylic acid, C6H4(OCOCH3)CO2H aspirin The active ingredient in aspirin.
What exactly makes an acid an acid, and what makes a base act as a base? Take a look at the formulas given in the above table and take a guess.
Hydrochloric Acid
Hydrochloric acid is a corrosive, strong mineral acid with many industrial uses. A colorless, highly pungent solution of hydrogen chloride (HCl) in water. Hydrochloric acid is usually prepared by treating $\ce{HCl}$ with water.
$\ce{\displaystyle HCl (g) + H2O (l) \longrightarrow H_3O^{+}(aq) + Cl^{-} (aq) } \nonumber$
Hydrochloric acid can therefore be used to prepare chloride salts. Hydrochloric acid is a strong acid, since it is completely dissociated in water. Hydrochloric acid is the preferred acid in titration for determining the amount of bases.
Sulfuric Acid
Sulfuric acid is a highly corrosive strong mineral acid with the molecular formula $\ce{H2SO4}$. Sulfuric acid is a diprotic acid and has a wide range of applications including use in domestic acidic drain cleaners,[as an electrolyte in lead-acid batteries, and in various cleaning agents. It is also a central substance in the chemical industry.
Because the hydration of sulfuric acid is thermodynamically favorable (and is highly exothermic) and the affinity of it for water is sufficiently strong, sulfuric acid is an excellent dehydrating agent. Concentrated sulfuric acid has a very powerful dehydrating property, removing water ($\ce{H2O}$) from other compounds including sugar and other carbohydrates and producing carbon, heat, steam. Sulfuric acid behaves as a typical acid in its reaction with most metals by generating hydrogen gas (Equation \ref{Eq1}).
$\ce{M + H2SO4 → M(SO4) + H2 } \label{Eq1}$
Nitric Acid
Nitric acid ($\ce{HNO3}$) is a highly corrosive mineral acid and is also commonly used as a strong oxidizing agent. Nitric acid is normally considered to be a strong acid at ambient temperatures. Nitric acid can be made by reacting nitrogen dioxide ($\ce{NO_2(g)}$) with water.
$\ce{3 NO2(g) + H2O (l)→ 2 HNO3 (ag) + NO(g)} \nonumber$
Nitric acid reacts with most metals, but the details depend on the concentration of the acid and the nature of the metal. Dilute nitric acid behaves as a typical acid in its reaction with most metals (e.g., nitric acid with magnesium, manganese or zinc will liberate $\ce{H2}$ gas):
$\ce{Mg + 2 HNO3 → Mg(NO3)2 + H2 } \nonumber$
$\ce{Mn + 2 HNO3 → Mn(NO3)2 + H2 } \nonumber$
$\ce{Zn + 2 HNO3 → Zn(NO3)2 + H2 } \nonumber$
Nitric acid is a corrosive acid and a powerful oxidizing agent. The major hazard it poses is chemical burn, as it carries out acid hydrolysis with proteins (amide) and fats (ester) which consequently decomposes living tissue (Figure $2$). Concentrated nitric acid stains human skin yellow due to its reaction with the keratin
Carbonic Acid
Carbonic acid is a chemical compound with the chemical formula $\ce{H2CO3}$ and is also a name sometimes given to solutions of carbon dioxide in water (carbonated water), because such solutions contain small amounts of $\ce{H2CO3(aq)}$. Carbonic acid, which is a weak acid, forms two kinds of salts: the carbonates and the bicarbonates. In geology, carbonic acid causes limestone to dissolve, producing calcium bicarbonate—which leads to many limestone features such as stalactites and stalagmites. Carbonic acid is a polyprotic acid, specifically it is diprotic, meaning that it has two protons which may dissociate from the parent molecule.
When carbon dioxide dissolves in water, it exists in chemical equilibrium (discussed in Chapter 15), producing carbonic acid:
$\ce{CO2 + H2O <=> H2CO3} \nonumber$
The reaction can be pushed to favor the reactants to generate $\ce{CO2(g)}$ from solution, which is key to the bubbles observed in carbonated beverages (Figure $3$).
Formic Acid
Formic acid ($\ce{HCO2H}$) is the simplest carboxylic acid and is an important intermediate in chemical synthesis and occurs naturally, most notably in some ants. The word "formic" comes from the Latin word for ant, formica, referring to its early isolation by the distillation of ant bodies. Formic acid occurs widely in nature as its conjugate base formate.
Citric Acid
Citric acid ($\ce{C6H8O7}$) is a weak organic tricarboxylic acid that occurs naturally in citrus fruits. The citrate ion is an intermediate in the TCA cycle (Krebs cycle), a central metabolic pathway for animals, plants and bacteria. Because it is one of the stronger edible acids, the dominant use of citric acid is used as a flavoring and preservative in food and beverages, especially soft drinks.
Acetylsalicylic Acid
Acetylsalicylic acid (also known as aspirin) is a medication used to treat pain, fever, and inflammation. Aspirin, in the form of leaves from the willow tree, has been used for its health effects for at least 2,400 years.
Aspirin is a white, crystalline, weakly acidic substance.
Summary
A brief summary of key aspects of several acids commonly encountered by students was given. Acids are a distinct class of compounds because of the properties of their aqueous solutions.
Contributions & Attributions
• Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.02%3A_Acids-_Properties_and_Examples.txt |
Learning Objectives
• Examine properties of bases.
Perhaps you have eaten too much pizza and felt very uncomfortable hours later. This feeling is due to excess stomach acid being produced. The discomfort can be dealt with by taking an antacid. The base in the antacid will react with the $\ce{HCl}$ in the stomach and neutralize it, taking care of that unpleasant feeling.
Bases
Bases have properties that mostly contrast with those of acids.
1. Aqueous solutions of bases are also electrolytes. Bases can be either strong or weak, just as acids can.
2. Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.
3. Bases also change the color of indicators. Litmus turns blue in the presence of a base, while phenolphthalein turns pink.
4. Bases do not react with metals in the way that acids do.
5. Bases react with acids to produce a salt and water.
Warning!
Tasting chemicals and touching them are NOT good lab practices and should be avoided—in other words—don't do this at home.
Bases are less common as foods, but they are nonetheless present in many household products. Many cleaners contain ammonia, a base. Sodium hydroxide is found in drain cleaner. Antacids, which combat excess stomach acid, are comprised of bases such as magnesium hydroxide or sodium hydrogen carbonate. Various common bases and corresponding uses are given in Table $2$.
Table $1$: Common Bases and Corresponding Uses
Some Common Bases
Uses
sodium hydroxide, NaOH
(lye or caustic soda)
Used in the manufacture of soaps and detergents, and as the main ingredient in oven and drain cleaners.
potassium hydroxide, KOH
(lye or caustic potash)
Used in the production of liquid soaps and soft soaps. Used in alkaline batteries.
magnesium hydroxide, Mg(OH)2
(milk of magnesia)
Used as an ingredient in laxatives, antacids, and deodorants. Also used in the neutralization of acidic wastewater.
calcium hydroxide, Ca(OH)2
(slaked lime)
Used in the manufacture of cement and lime water. Also, added to neutralize acidic soil.
aluminum hydroxide Used in water purification and as an ingredient in antacids.
ammonia, NH3 Used as a building block for the synthesis of many pharmaceutical products and in many commercial cleaning products. Used in the manufacture of fertilizers.
Sodium Hydroxide
Sodium hydroxide, also known as lye and caustic soda, is an inorganic compound with formula $\ce{NaOH}$. It is a white solid ionic compound consisting of sodium cations $\ce{Na^{+}}$ and hydroxide anions $\ce{OH^{−}}$.
Dissolution of solid sodium hydroxide in water is a highly exothermic reaction:
$\ce{ NaOH (s) \rightarrow Na^{+} (aq) + OH^{-} (aq)} \nonumber$
The resulting solution is usually colorless and odorless and feels slippery when it comes in contact with skin.
Potassium Hydroxide
Potassium hydroxide is an inorganic compound with the formula $\ce{KOH}$, and is commonly called caustic potash. Along with sodium hydroxide (NaOH), this colorless solid is a prototypical strong base. It has many industrial and niche applications, most of which exploit its corrosive nature and its reactivity toward acids. Its dissolution in water is strongly exothermic.
$\ce{ KOH (s) \rightarrow K^{+} (aq) + OH^{-} (aq)} \nonumber$
Concentrated aqueous solutions are sometimes called potassium lyes.
Magnesium Hydroxide
Magnesium hydroxide is the inorganic compound with the chemical formula $\ce{Mg(OH)2}$. Magnesium hydroxide is a common component of antacids, such as milk of magnesia, as well as laxatives.
It is a white solid with low solubility in water. Combining a solution of many magnesium salts with basic water induces precipitation of solid $\ce{Mg(OH)2}$. However, a weak concentration of dissociated ions can be found in solution:
$\ce{Mg(OH)2 (s) <=> Mg^{2+} (aq) + 2 OH^{−}(aq) } \nonumber$
Calcium Hydroxide
Calcium hydroxide (traditionally called slaked lime) is an inorganic compound with the chemical formula $\ce{Ca(OH)2}$. It is a colorless crystal or white powder. It has many names including hydrated lime, caustic lime, builders' lime, slaked lime, cal, or pickling lime. Calcium hydroxide is used in many applications, including food preparation. Limewater is the common name for a saturated solution of calcium hydroxide.
Calcium hydroxide is relatively insoluble in water, but is large enough that its solutions are basic according to the following reaction:
$\ce{Ca(OH)2 (s) <=> Ca^{2+}(aq) + 2 OH^{−} (aq)} \nonumber$
Ammonia
Ammonia is a compound of nitrogen and hydrogen with the formula $\ce{NH3}$ and is a colorless gas with a characteristic pungent smell. It is the active product of “smelling salts,” and can quickly revive the faint of heart and light of head. Although common in nature and in wide use, ammonia is both caustic and hazardous in its concentrated form.
In aqueous solution, ammonia acts as a base, acquiring hydrogen ions from $\ce{H_2O}$ to yield ammonium and hydroxide ions:
$\ce{NH3 (g) + H2O (l) <=> NH4^{+} (aq) + OH^{-} (aq)} \nonumber$
Ammonia is also a building block for the synthesis of many pharmaceutical products and is used in many commercial cleaning products.
Summary
• A brief summary of properties of bases was given.
• The properties of bases mostly contrast those of acids.
• Bases have many, varied uses. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.03%3A_Bases-_Properties_and_Examples.txt |
Learning Objectives
• Identify an Arrhenius acid and an Arrhenius base.
• Identify a Brønsted-Lowry acid and a Brønsted-Lowry base.
• Identify conjugate acid-base pairs in an acid-base reaction.
There are three major classifications of substances known as acids or bases. The theory developed by Svante Arrhenius in 1883, the Arrhenius definition, states that an acid produces H+ in solution and a base produces OH-. Later, two more sophisticated and general theories were proposed. These theories are the Brønsted-Lowry and Lewis definitions of acids and bases. This section will cover the Arrhenius and Brønsted-Lowry theories; the Lewis theory is discussed elsewhere.
The Arrhenius Theory 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. An Arrhenius acid is a compound that increases the concentration of $\ce{H^{+}}$ ions that are present when added to water. These H+ ions form the hydronium ion (H3O+) when they combine with water molecules. This process is represented in a chemical equation by adding H2O to the reactants side.
$\ce{HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq) } \nonumber$
In this reaction, hydrochloric acid ($HCl$) dissociates completely into hydrogen (H+) and chlorine (Cl-) ions when dissolved in water, thereby releasing H+ ions into solution. Formation of the hydronium ion equation:
$\ce{ HCl(aq) + H_2O(l) \rightarrow H_3O^{+}(aq) + Cl^{-}(aq)} \nonumber$
An Arrhenius base is a compound that increases the concentration of $\ce{OH^{-}}$ ions that are present when added to water. The dissociation is represented by the following equation:
$\ce{ NaOH \; (aq) \rightarrow Na^{+} \; (aq) + OH^{-} \; (aq) } \nonumber$
In this reaction, sodium hydroxide (NaOH) disassociates into sodium ($\ce{Na^{+}}$) and hydroxide ($\ce{OH^{-}}$) ions when dissolved in water, thereby releasing OH- ions into solution.
Arrhenius acids are substances which produce hydrogen ions in solution and Arrhenius bases are substances which produce hydroxide ions in solution.
Limitations to the Arrhenius Theory
The Arrhenius theory has many more limitations than the other two theories. The theory does not explain the weak base ammonia (NH3), which in the presence of water, releases hydroxide ions into solution, but does not contain OH- itself. The Arrhenius definition of acid and base is also limited to aqueous (i.e., water) solutions.
The Brønsted-Lowry Theory of Acids and Bases
In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H+) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD), while a Brønsted-Lowry base is a proton acceptor (PA).
A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor.
Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products:
$\ce{NH3(aq) + H2O (ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1}$
What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows:
Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense.
Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion, we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H+ ion attaches itself to H2O to make H3O+, which is called the hydronium ion. For most purposes, H+ and H3O+ represent the same species, but writing H3O+ instead of H+ shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules.
The Hydronium Ion
A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like $\ce{H5O2^{+}}$ or $\ce{H9O4^{+}}$ rather than $\ce{H3O^{+}}$. It is simpler, however, to use $\ce{H3O^{+}}$ to represent the hydronium ion.
With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H2O:
$\ce{HCl(g) + H_2O (ℓ) \rightarrow H_3O^{+}(aq) + Cl^{−}(aq) }\label{Eq2}$
We can depict this process using Lewis electron dot diagrams:
Now we see that a hydrogen ion is transferred from the HCl molecule to the H2O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H2O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense, but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H2O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H2O a base in this circumstance.
• A Brønsted-Lowry acid is a proton (hydrogen ion) donor.
• A Brønsted-Lowry base is a proton (hydrogen ion) acceptor.
• All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases.
Example $1$
Aniline (C6H5NH2) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule, just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base.
Solution
C6H5NH2 and H2O are the reactants. When C6H5NH2 accepts a proton from H2O, it gains an extra H and a positive charge and leaves an OH ion behind. The reaction is as follows:
$\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber$
Because C6H5NH2 accepts a proton, it is the Brønsted-Lowry base. The H2O molecule, because it donates a proton, is the Brønsted-Lowry acid.
Exercise $1$
Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation.
$\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}} \nonumber$
Answer
Brønsted-Lowry acid: H2PO4-; Brønsted-Lowry base: H2O
Exercise $2$
Which of the following compounds is a Bronsted-Lowry base?
1. HCl
2. HPO42-
3. H3PO4
4. NH4+
5. CH3NH3+
Answer
A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H+. This eliminates $\ce{HCl}$, $\ce{H3PO4}$, $\ce{NH4^{+}}$ and $\ce{CH_3NH_3^{+}}$ because they are Bronsted-Lowry acids. They all give away protons. In the case of $\ce{HPO4^{2-}}$, consider the following equation:
$\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber$
Here, it is clear that HPO42- is the acid since it donates a proton to water to make H3O+ and PO43-. Now consider the following equation:
$\ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber$
In this case, HPO42- is the base since it accepts a proton from water to form H2PO4- and OH-. Thus, HPO42- is an acid and base together, making it amphoteric.
Since HPO42- is the only compound from the options that can act as a base, the answer is (b) HPO42-.
Conjugate Acid-Base Pair
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, $\ce{NH4+}$:
In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are $NH_4^+/NH_3$ and $H_2O/OH^−$.
Example $2$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{CH3CO2H + H2O <=> H3O^{+} + CH3CO2^{-}} \nonumber$
Solution
Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, $H_3O^+$ is the acid that donates a proton to the acetate ion, which acts as the base.
Once again, we have two conjugate acid-base pairs:
• the parent acid and its conjugate base ($CH_3CO_2H/CH_3CO_2^−$) and
• the parent base and its conjugate acid ($H_3O^+/H_2O$).
Example $3$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{(CH_{3})_{3}N + H_{2}O <=> (CH_{3})_{3}NH^{+} + OH^{-}} \nonumber$
Solution
One pair is H2O and OH, where H2O has one more H+ and is the conjugate acid, while OH has one less H+ and is the conjugate base.
The other pair consists of (CH3)3N and (CH3)3NH+, where (CH3)3NH+ is the conjugate acid (it has an additional proton) and (CH3)3N is the conjugate base.
Exercise $3$
Identify the conjugate acid-base pairs in this equilibrium.
$\ce{NH2^{-} + H2O\rightleftharpoons NH3 + OH^{-}} \nonumber$
Answer
H2O (acid) and OH (base); NH2 (base) and NH3 (acid) | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.04%3A_Molecular_Definitions_of_Acids_and_Bases.txt |
Learning Objectives
• Write acid-base neutralization reactions.
• Write reactions of acids with metals.
• Write reactions of bases with metals.
Neutralization Reactions
The reaction that happens when an acid, such as $\ce{HCl}$, is mixed with a base, such as $\ce{NaOH}$:
$\ce{HCl (aq) + NaOH (aq) → NaCl (aq) + H_2O (l)}\nonumber$
When an acid and a base are combined, water and a salt are the products. Salts are ionic compounds containing a positive ion other than $\ce{H^{+}}$ and a negative ion other than the hydroxide ion, $\ce{OH^{-}}$. Double displacement reactions of this type are called neutralization reactions. We can write an expanded version of this equation, with aqueous substances written in their longer form:
$\ce{H^{+} (aq) + Cl^{-} (aq) + Na^{+} (aq) + OH^{-} (aq) → Na^{+} (aq) + Cl^{-} (aq) + H_2O (l)}\nonumber$
After removing the spectator ions, we get the net ionic equation:
$\ce{H^{+} (aq) + OH^{-} (aq) → H_2O (l)}\nonumber$
When a strong acid and a strong base are combined in the proper amounts—when $[\ce{H^{+}}]$ equals $[\ce{OH^{-}}$]\)—a neutral solution results in which pH = 7. The acid and base have neutralized each other, and the acidic and basic properties are no longer present.
Salt solutions do not always have a pH of 7, however. Through a process known as hydrolysis, the ions produced when an acid and base combine may react with the water molecules to produce a solution that is slightly acidic or basic. As a general concept, if a strong acid is mixed with a weak base, the resulting solution will be slightly acidic. If a strong base is mixed with a weak acid, the solution will be slightly basic.
Example $1$: Propionic Acid + Calcium Hydroxide
Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH3CH2CO2H) with aqueous calcium hydroxide [Ca(OH)2].
Solution
Solutions to Example 14.5.1
Steps Reaction
Write the unbalanced equation.
This is a double displacement reaction, so the cations and anions swap to create the water and the salt.
CH3CH2CO2H(aq) + Ca(OH)2(aq)→(CH3CH2CO2)2Ca(aq) + H2O(l)
Balance the equation.
Because there are two OH ions in the formula for Ca(OH)2, we need two moles of propionic acid, CH3CH2CO2H, to provide H+ ions.
2CH3CH2CO2H(aq) + Ca(OH)2(aq)→(CH3CH2CO2)2Ca(aq) +2H2O(l)
Exercise $1$
Write a balanced chemical equation for the reaction of solid barium hydroxide with dilute acetic acid.
Answer
$\ce{Ba(OH)2(s) + 2CH3CO2H (aq)→Ba(CH3CO2)2 (aq) + 2H2O(l)} \nonumber \nonumber$
Acids and Bases React with Metals
Acids react with most metals to form a salt and hydrogen gas. As discussed previously, metals that are more active than acids can undergo a single displacement reaction. For example, zinc metal reacts with hydrochloric acid, producing zinc chloride and hydrogen gas.
$\ce{Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)}\nonumber$
Bases also react with certain metals, like zinc or aluminum, to produce hydrogen gas. For example, sodium hydroxide reacts with zinc and water to form sodium zincate and hydrogen gas.
$\ce{Zn(s) + 2NaOH (aq) + 2H2O(l) → Na2Zn(OH)4(aq) + H2 (g)}.\nonumber$
14.06: AcidBase Titration
Learning Objectives
• Understand the basics of acid-base titrations.
• Understand the use of indicators.
• Perform a titration calculation correctly.
The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because so many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a titration. A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions.
During an acid-base titration, an acid with a known concentration (a standard solution) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base. The indicator will signal, by color change, when the base has been neutralized (when [H+] = [OH-]). At that point—called the equivalence point, or end point—the titration is stopped. By knowing the volumes of acid and base used, and the concentration of the standard solution, calculations allow us to determine the concentration of the other solution.
It is important to accurately measure volumes when doing titrations. The instrument you would use is called a burette (or buret).
For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted:
# mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl
We also have the balanced chemical reaction between HCl and NaOH:
$\ce{HCl + NaOH → NaCl + H2O} \nonumber$
So we can construct a conversion factor to convert to number of moles of NaOH reacted:
$0.002766\cancel{mol\, HCl}\times \dfrac{1\, mol\, NaOH}{1\cancel{mol\, HCl}}=0.002766\, mol\, NaOH \nonumber$
Then we convert this amount to mass, using the molar mass of NaOH (40.00 g/mol):
$0.002766\cancel{mol\, HCl}\times \dfrac{40.00\, g\, NaOH}{1\cancel{mol\, HCl}}=0.1106\, g\, NaOH \nonumber$
This type of calculation is performed as part of a titration.
Example $1$: Equivalence Point
What mass of Ca(OH)2 is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO3? The balanced chemical equation is as follows:
$\ce{2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O} \nonumber$
Solution
In liters, the volume is 0.04402 L. We calculate the number of moles of titrant:
# moles HNO3 = (0.04402 L)(0.0885 M) = 0.00390 mol HNO3
Using the balanced chemical equation, we can determine the number of moles of Ca(OH)2 present in the analyte:
$0.00390\cancel{mol\, HNO_{3}}\times \dfrac{1\, mol\, Ca(OH)_{2}}{2\cancel{mol\, HNO_{3}}}=0.00195\, mol\, Ca(OH)_{2} \nonumber$
Then we convert this to a mass using the molar mass of Ca(OH)2:
$0.00195\cancel{mol\, Ca(OH)_{2}}\times \dfrac{74.1\, g\, Ca(OH)_{2}}{\cancel{mol\, Ca(OH)_{2}}}=0.144\, g\, Ca(OH)_{2} \nonumber$
Exercise $1$
What mass of H2C2O4 is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows:
$\ce{H2C2O4 + 2NaOH → Na2C2O4 + 2H2O} \nonumber$
Answer
0.182 g
Exercise $2$
If 25.00 mL of HCl solution with a concentration of 0.1234 M is neutralized by 23.45 mL of NaOH, what is the concentration of the base?
Answer
0.1316 M NaOH
Exercise $3$
A 20.0 mL solution of strontium hydroxide, Sr(OH)2, is placed in a flask and a drop of indicator is added. The solution turns color after 25.0 mL of a standard 0.0500 M HCl solution is added. What was the original concentration of the Sr(OH)2 solution?
Answer
$3.12 \times 10^{-2}\;M$ Sr(OH)2
Indicator Selection for Titrations
The indicator used depends on the type of titration performed. The indicator of choice should change color when enough of one substance (acid or base) has been added to exactly use up the other substance. Only when a strong acid and a strong base are produced will the resulting solution be neutral. The three main types of acid-base titrations, and suggested indicators, are:
The three main types of acid-base titrations, suggested indicators, and explanations
Titration between . . . Indicator Explanation
strong acid and strong base any
strong acid and weak base methyl orange changes color in the acidic range (3.2 - 4.4)
weak acid and strong base phenolphthalein changes color in the basic range (8.2 - 10.6)
Summary
A titration is the quantitative reaction of an acid and a base. Indicators are used to show that all the analyte has reacted with the titrant.
Contributions & Attributions
• Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.05%3A_Reactions_of_Acids_and_Bases.txt |
Learning Objectives
• Define a strong and a weak acid and base.
• Recognize an acid or a base as strong or weak.
• Determine if a salt produces an acidic or a basic solution.
Strong and Weak Acids
Except for their names and formulas, so far we have treated all acids as equals, especially in a chemical reaction. However, acids can be very different in a very important way. Consider HCl(aq). When HCl is dissolved in H2O, it completely dissociates into H+(aq) and Cl(aq) ions; all the HCl molecules become ions:
$HCl\overset{100\%}{\rightarrow}H^{+}(aq)+Cl^{-}(aq) \nonumber$
Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid. HC2H3O2 is an example of a weak acid:
$HC_{2}H_{3}O_{2}\overset{\sim 5\%}{\longrightarrow}H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq) \nonumber$
Because this reaction does not go 100% to completion, it is more appropriate to write it as a reversible reaction:
$HC_{2}H_{3}O_{2}\rightleftharpoons H^{+}(aq)+C_{2}H_{3}O_{2}^{-}(aq) \nonumber$
As it turns out, there are very few strong acids, which are given in Table $1$. If an acid is not listed here, it is a weak acid. It may be 1% ionized or 99% ionized, but it is still classified as a weak acid.
Any acid that dissociates 100% into ions is called a strong acid. If it does not dissociate 100%, it is a weak acid.
Table $1$: Strong Acids and Bases
Acids Bases
HCl LiOH
HBr NaOH
HI KOH
HNO3 RbOH
H2SO4 CsOH
HClO3 Mg(OH)2
HClO4 Ca(OH)2
Sr(OH)2
Ba(OH)2
Strong and Weak Bases
The issue is similar with bases: a strong base is a base that is 100% ionized in solution. If it is less than 100% ionized in solution, it is a weak base. There are very few strong bases (Table $1$); any base not listed is a weak base. All strong bases are OH compounds. So a base based on some other mechanism, such as NH3 (which does not contain OH ions as part of its formula), will be a weak base.
Example $1$: Identifying Strong and Weak Acids and Bases
Identify each acid or base as strong or weak.
1. HCl
2. Mg(OH)2
3. C5H5N
Solution
1. Because HCl is listed in Table $1$, it is a strong acid.
2. Because Mg(OH)2 is listed in Table $1$, it is a strong base.
3. The nitrogen in C5H5N would act as a proton acceptor and therefore can be considered a base, but because it does not contain an OH compound, it cannot be considered a strong base; it is a weak base.
Exercise $1$
Identify each acid or base as strong or weak.
1. $\ce{RbOH}$
2. $\ce{HNO_2}$
Answer a
strong base
Answer b
weak acid
Example $2$: Characterizing Base Ionization
Write the balanced chemical equation for the dissociation of Ca(OH)2 and indicate whether it proceeds 100% to products or not.
Solution
This is an ionic compound of Ca2+ ions and OH ions. When an ionic compound dissolves, it separates into its constituent ions:
$\ce{Ca(OH)2 → Ca^{2+}(aq) + 2OH^{−}(aq)} \nonumber$
Because Ca(OH)2 is listed in Table $1$, this reaction proceeds 100% to products.
Exercise $2$
Write the balanced chemical equation for the dissociation of hydrazoic acid (HN3) and indicate whether it proceeds 100% to products or not.
Answer a
The reaction is as follows:
$\ce{HN3 → H^{+}(aq) + N3^{−}(aq)} \nonumber$
It does not proceed 100% to products because hydrazoic acid is not a strong acid.
Key Takeaways
• Strong acids and bases are 100% ionized in aqueous solution.
• Weak acids and bases are less than 100% ionized in aqueous solution.
• Salts of weak acids or bases can affect the acidity or basicity of their aqueous solutions. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.07%3A_Strong_and_Weak_Acids_and_Bases.txt |
Learning Objectives
• Describe the autoionization of water.
• Calculate the concentrations of $\ce{H3O^{+}}$ and $\ce{OH^{−}}$ in aqueous solutions, knowing the other concentration.
We have already seen that $\ce{H2O}$ can act as an acid or a base:
$\color{blue}{\underbrace{\ce{NH3}}_{\text{base}}} + \color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} \color{black} \ce{<=> NH4^{+} + OH^{−}} \nonumber$
where $\ce{H2O}$ acts as an $\color{red}{\text{acid}}$ (in red).
$\color{red}{\underbrace{\ce{HCl}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{-> H3O^{+} + Cl^{−}} \nonumber$
where $\ce{H2O}$ acts as an $\color{blue}{\text{base}}$ (in blue).
It may not surprise you to learn, then, that within any given sample of water, some $\ce{H2O}$ molecules are acting as acids, and other $\ce{H2O}$ molecules are acting as bases. The chemical equation is as follows:
$\color{red}{\underbrace{\ce{H2O}}_{\text{acid}}} + \color{blue}{\underbrace{\ce{H2O}}_{\text{base}}} \color{black} \ce{<=> H3O^{+} + OH^{−}} \label{Auto}$
This occurs only to a very small degree: only about 6 in 108 $\ce{H2O}$ molecules are participating in this process, which is called the autoionization of water.
At this level, the concentration of both $\ce{H3O^{+}(aq)}$ and $\ce{OH^{−}(aq)}$ in a sample of pure $\ce{H2O}$ is about $1.0 \times 10^{−7}\, M$ (at room temperature). If we use square brackets—[ ]—around a dissolved species to imply the molar concentration of that species, we have
$\color{red}{\ce{[H3O^{+}]}} \color{black}{ = } \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-7} \label{eq5}$
for any sample of pure water because H2O can act as both an acid and a base. The product of these two concentrations is $1.0\times 10^{−14}$:
$\color{red}{\ce{[H3O^{+}]}} \color{black}{\times} \color{blue}{\ce{[OH^{-}]}} \color{black} = (1.0 \times 10^{-7})( 1.0 \times 10^{-7}) = 1.0 \times 10^{-14} \nonumber$
• For acids, the concentration of $\ce{H3O^{+}(aq)}$ (i.e., $\ce{[H3O^{+}]}$) is greater than $1.0 \times 10^{−7}\, M$.
• For bases the concentration of $\ce{OH^{−}(aq)}$ (i.e., $\ce{[OH^{−}]}$) is greater than $1.0 \times 10^{−7}\, M$.
However, the product of the two concentrations—$\ce{[H3O^{+}][OH^{−}]}$—is always equal to $1.0 \times 10^{−14}$, no matter whether the aqueous solution is an acid, a base, or neutral:
$\color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \nonumber$
This value of the product of concentrations is so important for aqueous solutions that it is called the autoionization constant of water and is denoted $K_w$:
$K_w = \color{red}{\ce{[H_3O^+]}} \color{blue}{\ce{[OH^{-}]}} \color{black} = 1.0 \times 10^{-14} \label{eq10}$
This means that if you know $\ce{[H3O^{+}]}$ for a solution, you can calculate what $\ce{[OH^{−}]}$) has to be for the product to equal $1.0 \times 10^{−14}$; or if you know $\ce{[OH^{−}]}$), you can calculate $\ce{[H3O^{+}]}$. This also implies that as one concentration goes up, the other must go down to compensate so that their product always equals the value of $K_w$.
Warning: Temperature Matters
The degree of autoionization of water (Equation \ref{Auto})—and hence the value of $K_w$—changes with temperature, so Equations \ref{eq5} - \ref{eq10} are accurate only at room temperature.
Example $1$: Hydroxide Concentration
What is $\ce{[OH^{−}]}$) of an aqueous solution if $\ce{[H3O^{+}]}$ is $1.0 \times 10^{−4} M$?
Solution
Solutions to Example 14.7.1
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: $\ce{[H3O^{+}]} =1.0 \times 10^{−4}\, M$
Find: [OH] = ? M
List other known quantities. none
Plan the problem.
Using the expression for $K_w$, (Equation \ref{eq10}), rearrange the equation algebraically to solve for [OH].
$\left [ \ce{OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ H_3O^+ \right ]} \nonumber$
Calculate.
Now substitute the known quantities into the equation and solve.
$\left [\ce{ OH^{-}} \right ]=\dfrac{1.0\times 10^{-14}}{1.0\times 10^{-4}}=1.0\times 10^{-10}M\nonumber$
It is assumed that the concentration unit is molarity, so $\ce{[OH^{−}]}$ is 1.0 × 10−10 M.
Think about your result. The concentration of the acid is high (> 1 x 10-7 M), so $\ce{[OH^{−}]}$ should be low.
Exercise $1$
What is $\ce{[OH^{−}]}$ in a 0.00032 M solution of H2SO4?
Hint
Assume both protons ionize from the molecule...although this is not the case.
Answer
$3.1 \times 10^{−11}\, M$
When you have a solution of a particular acid or base, you need to look at the formula of the acid or base to determine the number of H3O+ or OH ions in the formula unit because $\ce{[H_3O^{+}]}$ or $\ce{[OH^{−}]}$) may not be the same as the concentration of the acid or base itself.
Example $2$: Hydronium Concentration
What is $\ce{[H_3O^{+}]}$ in a 0.0044 M solution of $\ce{Ca(OH)_2}$?
Solution
Solutions to Example 14.7.2
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: $[\ce{Ca(OH)_2}] =0.0044 \,M$
Find: $\ce{[H_3O^{+}]}$ = ? M
List other known quantities.
We begin by determining $\ce{[OH^{−}]}$. The concentration of the solute is 0.0044 M, but because $\ce{Ca(OH)_2}$ is a strong base, there are two OH ions in solution for every formula unit dissolved, so the actual $\ce{[OH^{−}]}$ is two times this:
$\ce{[OH^{−}] = 2 \times 0.0044\, M = 0.0088 \,M.} \nonumber$
Plan the problem.
Use the expression for $K_w$ (Equation \ref{eq10}) and rearrange the equation algebraically to solve for $\ce{[H_3O^{+}]}$.
$\left [ H_3O^{+} \right ]=\dfrac{1.0\times 10^{-14}}{\left [ OH^{-} \right ]} \nonumber$
Calculate.
Now substitute the known quantities into the equation and solve.
$\left [ H_3O^{+} \right ]=\dfrac{1.0\times 10^{-14}}{(0.0088)}=1.1\times 10^{-12}M \nonumber$
$\ce{[H_3O^{+}]}$ has decreased significantly in this basic solution.
Think about your result. The concentration of the base is high (> 1 x 10-7 M) so $\ce{[H_3O^+}]}$ should be low.
Exercise $2$
What is $\ce{[H_3O^{+}]}$ of an aqueous solution if $\ce{[OH^{−}]}$ is $1.0 \times 10^{−9}\, M$?
Answer
1.0 × 10−5 M
In any aqueous solution, the product of $\ce{[H_3O^{+}]}$ and $\ce{[OH^{−}]}$ equals $1.0 \times 10^{−14}$ (at room temperature). | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.08%3A_Water_-_Acid_and_Base_in_One.txt |
Learning Objectives
• Define pH and pOH.
• Determine the pH of acidic and basic solutions.
• Determine the hydronium ion concentration and pOH from pH.
As we have seen, $[H_3O^+]$ and $[OH^−]$ values can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions.
$pH$ is a logarithmic function of $[H_3O^+]$:
$pH = −\log[H_3O^+] \label{pH}$
$pH$ is usually (but not always) between 0 and 14. Knowing the dependence of $pH$ on $[H_3O^+]$, we can summarize as follows:
• If pH < 7, then the solution is acidic.
• If pH = 7, then the solution is neutral.
• If pH > 7, then the solution is basic.
This is known as the $pH$ scale. The pH scale is the range of values from 0 to 14 that describes the acidity or basicity of a solution. You can use $pH$ to make a quick determination whether a given aqueous solution is acidic, basic, or neutral. Figure $1$ illustrates this relationship, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example $1.3 \times 10^{-3}\,M$), the log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of $[H_3O^+]$, which will give a positive value for pH.
Example $1$
Label each solution as acidic, basic, or neutral based only on the stated $pH$.
1. milk of magnesia, pH = 10.5
2. pure water, pH = 7
3. wine, pH = 3.0
Answer
1. With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH)2.)
2. Pure water, with a pH of 7, is neutral.
3. With a pH of less than 7, wine is acidic.
Exercise $1$
Identify each substance as acidic, basic, or neutral based only on the stated $pH$.
1. human blood with $pH$ = 7.4
2. household ammonia with $pH$ = 11.0
3. cherries with $pH$ = 3.6
Answer a
basic
Answer b
basic
Answer c
acidic
Calculating pH from Hydronium Concentration
The pH of solutions can be determined by using logarithms as illustrated in the next example for stomach acid. Stomach acid is a solution of $HCl$ with a hydronium ion concentration of $1.2 \times 10^{−3}\; M$, what is the $pH$ of the solution?
\begin{align} \mathrm{pH} &= \mathrm{-\log [H_3O^+]} \nonumber \ &=-\log(1.2 \times 10^{−3}) \nonumber \ &=−(−2.92)=2.92 \nonumber \end{align} \nonumber
Logarithms
To get the log value on your calculator, enter the number (in this case, the hydronium ion concentration) first, then press the LOG key.
If the number is 1.0 x 10-5 (for [H3O+] = 1.0 x 10-5 M) you should get an answer of "-5".
If you get a different answer, or an error, try pressing the LOG key before you enter the number.
Example $2$: Converting Ph to Hydronium Concentration
Find the pH, given the $[H_3O^+]$ of the following:
1. 1 ×10-3 M
2. 2.5 ×10-11 M
3. 4.7 ×10-9 M
Solution
Solutions to Example 14.9.2
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given:
1. [H3O+] =1 × 10−3 M
2. [H3O+] =2.5 ×10-11 M
3. [H3O+] = 4.7 ×10-9 M
Find: ? pH
Plan the problem.
Need to use the expression for pH (Equation \ref{pH}).
pH = - log [H3O+]
Calculate.
Now substitute the known quantity into the equation and solve.
1. pH = - log [1 × 10−3 ] = 3.0 (1 decimal places since 1 has 1 significant figure)
2. pH = - log [2.5 ×10-11] = 10.60 (2 decimal places since 2.5 has 2 significant figures)
3. pH = - log [4.7 ×10-9] = 8.30 (2 decimal places since 4.7 has 2 significant figures)
The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer:
Exercise $2$
Find the pH, given [H3O+] of the following:
1. 5.8 ×10-4 M
2. 1.0×10-7
Answer a
3.22
Answer b
7.00
Calculating Hydronium Concentration from pH
Sometimes you need to work "backwards"—you know the pH of a solution and need to find $[H_3O^+]$, or even the concentration of the acid solution. How do you do that? To convert pH into $[H_3O^+]$ we solve Equation \ref{pH} for $[H_3O^+]$. This involves taking the antilog (or inverse log) of the negative value of pH .
$[\ce{H3O^{+}}] = \text{antilog} (-pH) \nonumber$
or
$[\ce{H_3O^+}] = 10^{-pH} \label{ph1}$
As mentioned above, different calculators work slightly differently—make sure you can do the following calculations using your calculator.
Calculator Skills
We have a solution with a pH = 8.3. What is [H3O+] ?
With some calculators you will do things in the following order:
1. Enter 8.3 as a negative number (use the key with both the +/- signs, not the subtraction key).
2. Use your calculator's 2nd or Shift or INV (inverse) key to type in the symbol found above the LOG key. The shifted function should be 10x.
3. You should get the answer 5.0 × 10-9.
Other calculators require you to enter keys in the order they appear in the equation.
1. Use the Shift or second function to key in the 10x function.
2. Use the +/- key to type in a negative number, then type in 8.3.
3. You should get the answer 5.0 × 10-9.
If neither of these methods work, try rearranging the order in which you type in the keys. Don't give up—you must master your calculator!
Example $3$: Calculating Hydronium Concentration from pH
Find the hydronium ion concentration in a solution with a pH of 12.6. Is this solution an acid or a base? How do you know?
Solution
Solutions to Example 14.9.3
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: pH = 12.6
Find: [H3O+] = ? M
Plan the problem.
Need to use the expression for [H3O+] (Equation \ref{ph1}).
[H3O+] = antilog (-pH) or [H3O+] = 10-pH
Calculate.
Now substitute the known quantity into the equation and solve.
[H3O+] = antilog (12.60) = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places)
or
[H3O+] = 10-12.60 = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places)
The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer:
Exercise $3$
If moist soil has a pH of 7.84, what is [H3O+] of the soil solution?
Answer
1.5 x 10-8 M
The pOH scale
As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration.
$\text{pOH} = -\text{log} \left[ \ce{OH^-} \right] \nonumber$
The relation between the hydronium and hydroxide ion concentrations expressed as p-functions is easily derived from the $K_w$ expression:
$K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{$6$}$
$-\log K_\ce{w}=\mathrm{-\log([H_3O^+][OH^−])=-\log[H_3O^+] + -\log[OH^-]}\label{$7$}$
$\mathrm{p\mathit{K}_w=pH + pOH} \label{$8$}$
At 25 °C, the value of $K_w$ is $1.0 \times 10^{−14}$, and so:
$\mathrm{14.00=pH + pOH} \label{$9$}$
The hydronium ion molarity in pure water (or any neutral solution) is $1.0 \times 10^{-7}\; M$ at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore:
$\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{$1$0}$
$\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{$1$1}$
And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than $1.0 \times 10^{-7}\; M$ and hydroxide ion molarities less than $1.0 \times 10^{-7}\; M$ (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than $1.0 \times 10^{-7}\; M$ and hydroxide ion molarities greater than $1.0 \times 10^{-7}\; M$ (corresponding to pH values greater than 7.00 and pOH values less than 7.00).
Example $4$:
Find the pOH of a solution with a pH of 4.42.
Solution
Solutions to Example 14.9.4
Steps for Problem Solving
Identify the "given" information and what the problem is asking you to "find."
Given: pH =4.42
Find: ? pOH
Plan the problem.
Need to use the expression
pOH = 14 - pH
Calculate.
Now substitute the known quantity into the equation and solve.
pOH=14−4.42=9.58
Think about your result. The pH is that of an acidic solution, and the resulting pOH is the difference after subtracting from 14. The answer has two significant figures because the given pH has two decimal places.
Exercise $4$
The pH of a solution is 8.22. What is the pOH?
Answer
5.78
The diagram below shows all of the interrelationships between [H3O+][H3O+], [OH−][OH−], pH, and pOH. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.09%3A_The_pH_and_pOH_Scales_-_Ways_to_Express_Acidity_and_Basicity.txt |
Learning Objective
• Define buffer and describe how it reacts with an acid or a base.
Weak acids are relatively common, even in the foods we eat. But we occasionally come across a strong acid or base, such as stomach acid, that has a strongly acidic pH of 1–2. By definition, strong acids and bases can produce a relatively large amount of hydrogen or hydroxide ions and, as a consequence, have marked chemical activity. In addition, very small amounts of strong acids and bases can change the pH of a solution very quickly. If 1 mL of stomach acid [which we will approximate as 0.05 M HCl(aq)] is added to the bloodstream, and if no correcting mechanism is present, the pH of the blood would go from about 7.4 to about 4.9—a pH that is not conducive to life. Fortunately, the body has a mechanism for minimizing such dramatic pH changes.
This mechanism involves a buffer, a solution that resists dramatic changes in pH. Buffers do so by being composed of certain pairs of solutes: either a weak acid plus a salt derived from that weak acid, or a weak base plus a salt of that weak base. For example, a buffer can be composed of dissolved acetic acid (HC2H3O2, a weak acid) and sodium acetate (NaC2H3O2, a salt derived from that acid). Another example of a buffer is a solution containing ammonia (NH3, a weak base) and ammonium chloride (NH4Cl, a salt derived from that base).
Let us use an acetic acid–sodium acetate buffer to demonstrate how buffers work. If a strong base—a source of $\ce{OH^{-}(aq)}$ ions—is added to the buffer solution, those hydroxide ions will react with the acetic acid in an acid-base reaction:
$\ce{HC2H3O2(aq) + OH^{-}(aq) \rightarrow H2O(ℓ) + C2H3O^{-}2(aq)} \label{Eq1}$
Rather than changing the pH dramatically by making the solution basic, the added hydroxide ions react to make water, and the pH does not change much.
Many people are aware of the concept of buffers from buffered aspirin, which is aspirin that also has magnesium carbonate, calcium carbonate, magnesium oxide, or some other salt. The salt acts like a base, while aspirin is itself a weak acid.
If a strong acid—a source of H+ ions—is added to the buffer solution, the H+ ions will react with the anion from the salt. Because HC2H3O2 is a weak acid, it is not ionized much. This means that if lots of hydrogen ions and acetate ions (from sodium acetate) are present in the same solution, they will come together to make acetic acid:
$\ce{H^{+}(aq) + C2H3O^{−}2(aq) \rightarrow HC2H3O2(aq)} \label{Eq2}$
Rather than changing the pH dramatically and making the solution acidic, the added hydrogen ions react to make molecules of a weak acid. Figure $1$ illustrates both actions of a buffer.
Buffers made from weak bases and salts of weak bases act similarly. For example, in a buffer containing NH3 and NH4Cl, ammonia molecules can react with any excess hydrogen ions introduced by strong acids:
$\ce{NH3(aq) + H^{+}(aq) \rightarrow NH^{+}4(aq)} \label{Eq3}$
while the ammonium ion ($\ce{NH4^{+}(aq)}$) can react with any hydroxide ions introduced by strong bases:
$\ce{NH^{+}4(aq) + OH^{-}(aq) \rightarrow NH3(aq) + H2O(ℓ)} \label{Eq4}$
Example $1$: Making Buffer Solutions
Which solute combinations can make a buffer solution? Assume that all are aqueous solutions.
1. HCHO2 and NaCHO2
2. HCl and NaCl
3. CH3NH2 and CH3NH3Cl
4. NH3 and NaOH
Solution
1. Formic acid (HCHO2) is a weak acid, while NaCHO2 is the salt made from the anion of the weak acid—the formate ion (CHO2). The combination of these two solutes would make a buffer solution.
2. Hydrochloric acid (HCl) is a strong acid, not a weak acid, so the combination of these two solutes would not make a buffer solution.
3. Methylamine (CH3NH2) is like ammonia with one of its hydrogen atoms substituted with a CH3 (methyl) group. Because it is not on our list of strong bases, we can assume that it is a weak base. The compound CH3NH3Cl is a salt made from that weak base, so the combination of these two solutes would make a buffer solution.
4. Ammonia (NH3) is a weak base, but NaOH is a strong base. The combination of these two solutes would not make a buffer solution.
Exercise $1$
Which solute combinations can make a buffer solution? Assume that all are aqueous solutions.
1. NaHCO3 and NaCl
2. H3PO4 and NaH2PO4
3. NH3 and (NH4)3PO4
4. NaOH and NaCl
Answer a
Yes.
Answer b
No. Need a weak acid or base and a salt of its conjugate base or acid.
Answer c
Yes.
Answer d
No. Need a weak base or acid.
Buffers work well only for limited amounts of added strong acid or base. Once either solute is all reacted, the solution is no longer a buffer, and rapid changes in pH may occur. We say that a buffer has a certain capacity. Buffers that have more solute dissolved in them to start with have larger capacities, as might be expected.
Human blood has a buffering system to minimize extreme changes in pH. One buffer in blood is based on the presence of HCO3 and H2CO3 [H2CO3 is another way to write CO2(aq)]. With this buffer present, even if some stomach acid were to find its way directly into the bloodstream, the change in the pH of blood would be minimal. Inside many of the body’s cells, there is a buffering system based on phosphate ions.
Career Focus: Blood Bank Technology Specialist
At this point in this text, you should have the idea that the chemistry of blood is fairly complex. Because of this, people who work with blood must be specially trained to work with it properly.
A blood bank technology specialist is trained to perform routine and special tests on blood samples from blood banks or transfusion centers. This specialist measures the pH of blood, types it (according to the blood’s ABO+/− type, Rh factors, and other typing schemes), tests it for the presence or absence of various diseases, and uses the blood to determine if a patient has any of several medical problems, such as anemia. A blood bank technology specialist may also interview and prepare donors to give blood and may actually collect the blood donation.
Blood bank technology specialists are well trained. Typically, they require a college degree with at least a year of special training in blood biology and chemistry. In the United States, training must conform to standards established by the American Association of Blood Banks.
Key Takeaway
• A buffer is a solution that resists sudden changes in pH.
14.11: Prelude - Sour Patch Kids
Sour Patch Kids are a soft candy with a coating of invert sugar and sour sugar. The candy's slogan, "Sour. Sweet. Gone.", refers to its sour-to-sweet taste.
Sour sugar is a food ingredient that is used to impart a sour flavor, made from citric or tartaric acid and sugar. It is used to coat sour candies like Sour Patch Kids. Eating large amounts of sour sugar can cause irritation of the tongue because of the acid. It can also cause irreversible dental erosion.
• Wikipedia | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/14%3A_Acids_and_Bases/14.10%3A_Buffers-_Solutions_that_Resist_pH_Change.txt |
In previous science classes, you may have learned that one way to distinguish chemical changes from physical changes is that physical changes—such as the melting and freezing of water—are reversible, but that chemical changes are not. In this chapter, we will see that this simple answer is not necessarily what it seems.
15: Chemical Equilibrium
Learning Objectives
• Describe the conditions for successful collisions that cause reactions.
• Describe rate in terms of the conditions of successful collisions.
• Describe how changing the temperature, concentration of a reactant, or surface area of a reaction affects the rate of a reaction.
• Define a catalyst and how a catalyst affects the rate of a reaction.
We know that a chemical system can be made up of atoms ($\ce{H_2}$, $\ce{N_2}$, $\ce{K}$, etc.), ions ($\ce{NO_3^-}$, $\ce{Cl^-}$, $\ce{Na^+}$, etc.), or molecules ($\ce{H_2O}$, $\ce{C_{12}H_{22}O_{11}}$, etc.). We also know that in a chemical system, these particles are moving around in random motion. The collision theory explains why reactions occur at this particle level between these atoms, ions, and/or molecules. It also explains how it is possible to speed up or slow down reactions that are occurring.
Collision Theory
The collision theory provides us with the ability to predict what conditions are necessary for a successful reaction to take place. These conditions include:
1. The particles must collide with each other.
2. The particles must collide with sufficient energy to break the old bonds.
3. The particles must have proper orientation.
A chemical reaction involves breaking bonds in the reactants, rearranging the atoms into new groupings (the products), and forming new bonds in the products.
Therefore, a collision must not only occur between reactant particles, but the collision also has to have sufficient energy to break all the reactant bonds that need to be broken in order to form the products. Some reactions need less collision energy than others. The amount of energy that reactant particles must have in order to break the old bonds for a reaction to occur is called the activation energy, abbreviated $\text{E}_a$. Another way to think of this is to look at an energy diagram, as shown in the figure. Particles must be able to get over the "bump"—the activation energy—if they are going to react. If the reactant particles collide with less than the activation energy, the particles will rebound (bounce off of each other), and no reaction will occur.
Reaction Rate
Chemists use reactions to generate a product for which they have a use. For the most part, the reactions that produce some desired compound are only useful if the reaction occurs at a reasonable rate. For example, using a reaction to produce brake fluid would not be useful if the reaction required 8,000 years to complete the product. Such a reaction would also not be useful if the reaction was so fast that it was explosive. For these reasons, chemists wish to be able to control reaction rates. In some cases, chemists wish to speed up reactions that are too slow or slow down reactions that are too fast. In order to gain any control over reaction rates, we must know the factors that affect reaction rates. Chemists have identified many factors that affect the rate of a reaction.
The rate, or speed, at which a reaction occurs depends on the frequency of successful collisions. Remember, a successful collision occurs when two reactants collide with enough energy and with the right orientation. That means if there is an increase in the number of collisions, an increase in the number of particles that have enough energy to react, and/or an increase in the number of particles with the correct orientation, the rate of reaction will increase.
Effect of Temperature on Rate of Reaction
The rate of reaction was discussed in terms of three factors: collision frequency, the collision energy, and the geometric orientation. Remember that the collision frequency is the number of collisions per second. The collision frequency is dependent, among other factors, on the temperature of the reaction.
When the temperature is increased, the average velocity of the particles is increased. The average kinetic energy of these particles is also increased. The result is that the particles will collide more frequently, because the particles move around faster and will encounter more reactant particles. However, this is only a minor part of the reason why the rate is increased. Just because the particles are colliding more frequently does not mean that the reaction will definitely occur.
The major effect of increasing the temperature is that more of the particles that collide will have the amount of energy needed to have an effective collision. In other words, more particles will have the necessary activation energy.
At room temperature, the hydrogen and oxygen in the atmosphere do not have sufficient energy to attain the activation energy needed to produce water:
$\ce{O_2} \left( g \right) + \ce{H_2} \left( g \right) \rightarrow \text{No reaction} \nonumber$
At any one moment in the atmosphere, there are many collisions occurring between these two reactants. But what we find is that water is not formed from the oxygen and hydrogen molecules colliding in the atmosphere, because the activation energy barrier is just too high, and all the collisions are resulting in rebound. When we increase the temperature of the reactants or give them energy in some other way, the molecules have the necessary activation energy and are able to react to produce water:
$\ce{O_2} \left( g \right) + \ce{H_2} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) \nonumber$
There are times when the rate of a reaction needs to be slowed down. Lowering the temperature could also be used to decrease the number of collisions that would occur and lowering the temperature would also reduce the kinetic energy available for activation energy. If the particles have insufficient activation energy, the collisions will result in rebound rather than reaction. Using this idea, when the rate of a reaction needs to be lower, keeping the particles from having sufficient activation energy will definitely keep the reaction at a lower rate.
Society uses the effects of temperature on reaction rate every day. Food storage is a prime example of how the temperature effect on reaction rate is utilized by society. Consumers store food in freezers and refrigerators to slow down the processes that cause it to spoil. The decrease in temperature decreases the rate at which food will break down or be broken down by bacteria. In the early years of the 20$^\text{th}$ century, explorers were fascinated with being the first to reach the South Pole. In order to attempt such a difficult task at a time without most of the technology that we take for granted today, they devised a variety of ways of surviving. One method was to store their food in the snow to be used later during their advances to the pole. On some explorations, they buried so much food that they didn't need to use all of it, and some was left behind. Many years later, when this food was located and thawed, it was found to still be edible.
When milk, for example, is stored in the refrigerator, the molecules in the milk have less energy. This means that while molecules will still collide with other molecules, few of them will react (which means in this case "spoil") because the molecules do not have sufficient energy to overcome the activation energy barrier. The molecules do have energy and are colliding, however, and so, over time, even in the refrigerator, the milk will spoil. Eventually the higher energy molecules will gain the energy needed to react and when enough of these reactions occur, the milk becomes "soured".
However, if that same carton of milk was at room temperature, the milk would react (in other words, "spoil") much more quickly. Most of the molecules would have sufficient energy to overcome the energy barrier at room temperature, and many more collisions would occur. This allows for the milk to spoil in a fairly short amount of time. This is also the reason why most fruits and vegetables ripen in the summer when the temperature is much warmer. You may have experienced this first hand if you have ever bitten into an unripe banana—it was probably sour tasting and might even have felt like biting into a piece of wood! When a banana ripens, numerous reactions occur that produce all the compounds that we expect to taste in a banana. But this can only happen if the temperature is high enough to allow these reactions to make those products.
Effect of Concentration on Rate of Reaction
If you had an enclosed space, like a classroom, and there was one red ball and one green ball flying around the room in random motion, undergoing perfectly elastic collisions with the walls and with each other, in a given amount of time, the balls would collide with each other a certain number of times determined by probability. If you now put two red balls and one green ball in the room under the same conditions, the probability of a collision between a red ball and the green ball would exactly double. The green ball would have twice the chance of encountering a red ball in the same amount of time.
In terms of chemical reactions, a similar situation exists. Particles of two gaseous reactants or two reactants in solution have a certain probability of undergoing collisions with each other in a reaction vessel. If you double the concentration of either reactant, the probability of a collision doubles. The rate of reaction is proportional to the number of collisions per unit time. If one concentration is doubled, the number of collisions will also double. Assuming that the percent of collisions that are successful does not change, then having twice as many collisions will result in twice as many successful collisions. The rate of reaction is proportional to the number of collisions over time; increasing the concentration of either reactant increases the number of collisions, and therefore increases the number of successful collisions and the reaction rate.
For example, the chemical test used to identify a gas as oxygen, or not, relies on the fact that increasing the concentration of a reactant increases reaction rate. The reaction we call combustion refers to a reaction in which a flammable substance reacts with oxygen. If we light a wooden splint (a thin splinter of wood) on fire and then blow the fire out, the splint will continue to glow in air for a period of time. If we insert that glowing splint into any gas that does not contain oxygen, the splint will immediately cease to glow—that is, the reaction stops. Oxygen is the only gas that will support combustion, Air is approximately $20\%$ oxygen gas. If we take that glowing splint and insert it into pure oxygen gas, the reaction will increase its rate by a factor of five, since pure oxygen has 5 times the concentration of oxygen that is in the air. When the reaction occurring on the glowing splint increases its rate by a factor of five, the glowing splint will suddenly burst back into full flame. This test, of thrusting a glowing splint into a gas, is used to identify the gas as oxygen. Only a greater concentration of oxygen than that found in air will cause the glowing splint to burst into flame.
Effect of Surface Area on Rate of Reaction
The very first requirement for a reaction to occur between reactant particles is that the particles must collide with one another. The previous section pointed out how increasing the concentration of the reactants increases reaction rate because it increases the frequency of collisions between particles. It can be shown that the number of collisions that occur between reactant particles is also dependent on the surface area of solid reactants. Consider a reaction between reactant RED and reactant BLUE in which reactant blue is in the form of a single lump. Then compare this to the same reaction where reactant blue has been broken up into many smaller pieces.
In the diagram, only the blue particles on the outside surface of the lump are available for collision with reactant red. The blue particles on the interior of the lump are protected by the blue particles on the surface. In Figure A, if you count the number of blue particles available for collision, you will find that only 20 blue particles could be struck by a particle of reactant red. In Figure A, there are a number of blue particles on the interior of the lump that cannot be struck. In Figure B, however, the lump has been broken up into smaller pieces and all the interior blue particles are now on a surface and available for collision. In Figure B, more collisions between the blue and red will occur, and therefore, the reaction in Figure B will occur at a faster rate than the same reaction in Figure A. Increasing the surface area of a reactant increases the frequency of collisions and increases the reaction rate.
Several smaller particles have more surface area than one large particle. The more surface area that is available for particles to collide, the faster the reaction will occur. You can see an example of this in everyday life if you have ever tried to start a fire in the fireplace. If you hold a match up against a large log in an attempt to start the log burning, you will find it to be an unsuccessful effort. Holding a match against a large log will not cause enough reactions to occur in order to keep the fire going by providing sufficient activation energy for further reactions. In order to start a wood fire, it is common to break a log up into many small, thin sticks called kindling. These thinner sticks of wood provide many times the surface area of a single log. The match will successfully cause enough reactions in the kindling so that sufficient heat is given off to provide activation energy for further reactions.
There have been, unfortunately, cases where serious accidents were caused by the failure to understand the relationship between surface area and reaction rate. One such example occurred in flour mills. A grain of wheat is not very flammable. It takes significant effort to get a grain of wheat to burn. If the grain of wheat, however, is pulverized and scattered through the air, only a spark is necessary to cause an explosion. When the wheat is ground to make flour, it is pulverized into a fine powder and some of the powder gets scattered around in the air. A small spark then, is sufficient to start a very rapid reaction which can destroy the entire flour mill. In a 10-year period from 1988 to 1998, there were 129 grain dust explosions in mills in the United States. Efforts are now made in flour mills to have huge fans circulate the air in the mill through filters to remove the majority of the flour dust particles.
Another example is in the operation of coal mines. Coal will of course burn, but it takes an effort to get the coal started; once it is burning, it burns slowly because only the surface particles are available to collide with oxygen particles. The interior particles of coal have to wait until the outer surface of the coal lump burns off before they can collide with oxygen. In coal mines, huge blocks of coal must be broken up before the coal can be brought out of the mine. In the process of breaking up the huge blocks of coal, drills are used to drill into the walls of coal. This drilling produces fine coal dust that mixes into the air; then a spark from a tool can cause a massive explosion in the mine. There are explosions in coal mines for other reasons, but coal dust explosions have contributed to the death of many miners. In modern coal mines, lawn sprinklers are used to spray water through the air in the mine and this reduces the coal dust in the air, and eliminates coal dust explosions.
Effect of a Catalyst on Rate of Reaction
The final factor that affects the rate of the reaction is the effect of a catalyst. A catalyst is a substance that speeds up the rate of the reaction, without being consumed by the reaction itself.
In the reaction of potassium chlorate breaking down to potassium chloride and oxygen, a catalyst is available to make this reaction occur much faster than it would occur by itself under room conditions. The reaction is:
$2 \ce{KClO_3} \left( s \right) \overset{\ce{MnO_2} \left( s \right)}{\longrightarrow} 2 \ce{KCl} \left( s \right) + 3 \ce{O_2} \left( g \right) \nonumber$
The catalyst is manganese dioxide and its presence causes the reaction shown above to run many times faster than it occurs without the catalyst. When the reaction has reached completion, the $\ce{MnO_2}$ can be removed from the reaction vessel and its condition is exactly the same as it was before the reaction. This is part of the definition of a catalyst—that it is not consumed by the reaction. You should note that the catalyst is not written into the equation as a reactant or a product, but is noted above the yields arrow. This is standard notation for the use of a catalyst.
Some reactions occur very slowly without the presence of a catalyst. In other words, the activation energy for these reactions is very high. When the catalyst is added, the activation energy is lowered because the catalyst provides a new reaction pathway with lower activation energy.
In the figure on the right, the endothermic reaction shows the catalyst reaction in red with the lower activation energy, designated $\text{E}'_a$. The new reaction pathway has lower activation energy but has no effect on the energy of the reactants, the products, or the value of $\Delta H$. The same is true for the exothermic reaction. The activation energy of the catalyzed reaction is lower than that of the uncatalyzed reaction. The new reaction pathway provided by the catalyst affects the energy required for reactant bonds to break, and product bonds to form.
While many reactions in the laboratory can be increased by increasing the temperature, this is not possible for all of the reactions that occur in our bodies throughout our entire lives. In fact, the body needs to be maintained at a very specific temperature: $98.6^\text{o} \text{F}$ or $37^\text{o} \text{C}$. Of course there are times, such as when the body is fighting infection, when the body temperature may be increased. But generally, in a healthy person, the temperature is quite consistent. However, many of the reactions that a healthy body depends on could never occur at body temperature. The answer to this dilemma is catalysts—also referred to as enzymes. Many of these enzymes are made in human cells because human DNA carries the directions to make them. However, there are some enzymes required by the body that are not made by human cells. These catalysts must be supplied to our bodies in the food we eat and are called vitamins.
Reversible Reactions
Typically when we think of a chemical reaction, we think of the reactants getting totally used up so that none are left, and that we end up with only products. Also, we generally consider chemical reactions as one-way events. You may well have learned during earlier science classes that this is one way to distinguish chemical change from physical change—physical changes (such as the melting and freezing of ice) are easily reversed, but chemical changes cannot be reversed (pretty tough to un-fry an egg).
Throughout this chaper, we will see that this isn't always the case. We will see that many chemical reactions are, in fact, reversible under the right conditions. And because many reactions can be reversed, our idea of a reaction ending with no reactants left, only products, will need to be modified.
Here are some examples of reactions that can be reversed:
1.
Nitrogen dioxide, $\ce{NO_2}$, a reddish-brown gas, reacts to form colorless dinitrogen tetroxide, $\ce{N_2O_4}$ :
$\ce{2NO_2(g) \rightarrow N_2O_4(g)}$
But the reaction can also go the other waydinitrogen tetroxide also readily breaks down to form nitrogen dioxide:
$\ce{N_2O_4(g) \rightarrow 2NO_2(g)}$
We typically write a reaction that can go in both directions by using a double arrow (which will sometimes appear as ↔ in these online notes):
$\ce{2NO_2(g) \leftrightarrow N_2O_4(g)}$
Because the reaction continues in both directions at the same time, we never run out of either $\ce{NO_2}$ or $\ce{N_2O_4}$ . $\ce{NO_2}$ is continually being used up to form $\ce{N_2O_4}$, but at the same time $\ce{N_2O_4}$ is forming more $\ce{NO_2}$
2.
When hydrogen gas is passed over heated iron oxide, iron and steam are produced:
(1) $\ce{Fe_3O_4(s) + 4H_2 (g) \rightarrow 3Fe(s) + 4H_2O(g)}$
The reverse reaction can occur when steam is passed over red-hot iron:
(2) $\ce{3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)}$
We can write these two equations together as:
(3) $\ce{Fe_3O_4(s) + 4H_2(g) \leftrightarrow 3Fe(s) + 4H_2O(g)}$
When we have a reversible reaction written in this way, we need to be able to distinguish between which way the reaction is headed. As written above in Reaction (3), we would say that in the forward reaction, iron oxide and hydrogen gas, the reactants, produce the products iron and steam.
During the reverse reaction, iron reacts with steam to produce the products iron oxide and hydrogen gas.
It is important to understand the terminology, and to use the terms correctly.
Does it matter which way we write our reversible reaction? It can also be written as
$\ce{3Fe(s) + 4H_2O(g) \leftrightarrow Fe_3O_4(s) + 4H_2(g)}$
Now iron and steam are reactants of the forward direction, and iron oxide and hydrogen gas would be the reactants of the reverse direction.
Summary
• The collision theory explains why reactions occur between atoms, ions, and molecules.
• In order for a reaction to be effective, particles must collide with enough energy, and have the correct orientation.
• With an increase in temperature, there is an increase in energy that can be converted into activation energy in a collision, and that will increase the reaction rate. A decrease in temperature would have the opposite effect.
• With an increase in temperature, there is an increase in the number of collisions.
• Increasing the concentration of a reactant increases the frequency of collisions between reactants and will, therefore, increase the reaction rate.
• Increasing the surface area of a reactant (by breaking a solid reactant into smaller particles) increases the number of particles available for collision and will increase the number of collisions between reactants per unit time.
• A catalyst is a substance that speeds up the rate of the reaction without being consumed by the reaction itself. When a catalyst is added, the activation energy is lowered because the catalyst provides a new reaction pathway with lower activation energy.
Vocabulary
• Catalyst - A substance that speeds up the rate of the reaction without being consumed by the reaction itself.
• Surface area to volume ratio - The comparison of the volume inside a solid to the area exposed on the surface.
Further Reading/Supplemental Links
• Activation Energy: http://www.mhhe.com/physsci/chemistr...sh/activa2.swf
• learner.org/resources/series61.html The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos, but there is no charge. The website has one video that relates to this lesson called Molecules in Action.
• www.vitamins-guide.net
• en.Wikipedia.org/wiki
• Observing molecules during chemical reactions helps explain the role of catalysts. Dynamic equilibrium is also demonstrated. Molecules in Action (www.learner.org/vod/vod_window.html?pid=806)
• Surface science examines how surfaces react with each other at the molecular level. On the Surface (www.learner.org/vod/vod_window.html?pid=812) | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.02%3A_The_Rate_of_a_Chemical_Reaction.txt |
Learning Objectives
• Describe the three possibilities that exist when reactants come together.
• Describe what is occurring in a system at equilibrium.
Think for a minute about sitting down to a table to eat dinner. There are three possibilities that could happen when you eat dinner. You could (1) finish your entire dinner, (2) you could not want any of it and leave it all on your plate, or (3) you could eat some of it and leave some of it. Reactions have the same possibilities. Reactions do not always proceed all the way from start to finish. You may have reactions that (1) go to completion so that at the end, the reaction vessel contains all products and only products. Some reactions (2) may not start at all, so at the end the reaction vessel contains all reactants and only reactants. And some reactions (3) may start but not go to completion, that is, the reaction might start but not go completely to products. In this last case, the reaction vessel would contain some reactants and some products. In this section, we are going to take a closer look at the third type of reaction.
Reversible Reactions and Equilibrium
Consider the hypothetical reaction:
$\text{A} + \text{B} \rightarrow \text{C} + \text{D}. \nonumber$
If we looked at this reaction using what we have learned, this reaction will keep going, forming $\text{C}$ and $\text{D}$ until $\text{A}$ and $\text{B}$ run out. This is what we call an "irreversible reaction" or a "reaction that goes to completion".
Some reactions, however, are reversible, meaning the reaction can go backwards in which products react to form reactants, so that: $\text{A} + \text{B} \leftarrow \text{C} + \text{D}$. The direction of the arrow shows that $\text{C}$ and $\text{D}$ are reacting to form $\text{A}$ and $\text{B}$. What if the two reactions, the forward reaction and the reverse reaction, were occurring at the same time? What would this look like? If you could peer into the reaction, you would be able to find $\text{A}$, $\text{B}$, $\text{C}$, and $\text{D}$ particles. $\text{A}$ and $\text{B}$ would react to form $\text{C}$ and $\text{D}$ at the same time that $\text{C}$ and $\text{D}$ are reacting to form $\text{A}$ and $\text{B}$.
If the forward and reverse reactions are happening at the same rate, the reaction is said to be at equilibrium or dynamic equilibrium. At this point, the concentrations of $\text{A}$, $\text{B}$, $\text{C}$, and $\text{D}$ are not changing (or, are constant) and we would see no difference in our reaction container, but reactions are still occurring in both directions. It is important to point out that having constant amounts of reactants and products does NOT mean that the concentration of the reactants is equal to the concentration of the products. It means they are not changing. These reactions appear to have stopped before one of the reactants has run out.
Chemists use a double-headed arrow, $\rightleftharpoons$, to show that a reaction is at equilibrium. We would write the example reaction as:
$\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}. \nonumber$
The arrow indicates that both directions of the reaction are happening.
Another way to think about reversible and irreversible reactions is to compare them to two types of games of tag. Reversible reactions are in many ways like a traditional game of tag: the "it" person can become "not it" and somebody who is "not it" is tagged and becomes "it". In this way, it is a reversible change. It is also like a reaction at equilibrium, because overall no change is occurring. There is always a constant number of "it" people and "not it" people in the game. Also, having constant numbers of "it" and "not it" people in our game does not mean that the number of "it" people (reactants) is equal to the number of "not it" people (products). Furthermore, this is similar to equilibrium in that this game never truly ends (unless everybody gets tired of playing). The game could go on forever. We could write this as the following reversible reaction:
$\text{"It"} \rightleftharpoons \text{"Not it"} \nonumber$
Irreversible reactions (those that only go in one direction from reactants to products and cannot reach a state of equilibrium) are more like a game of sharks and minnows. In sharks and minnows, almost everybody starts out as a minnow. Once tagged, they become a shark. However, the difference here is that once you are a shark you are always a shark; there is no way to go back to becoming a minnow. The game continues until everybody has been tagged and becomes a shark. This is similar to irreversible reactions in that the reactants turn into products, but can't change back. Furthermore, the reaction will proceed until the reactants have been used up and there are not any more left. We could write the reaction as:
$\text{Minnow} \rightarrow \text{Shark} \nonumber$
Here's another example of a reversible reactiondissolving salt in a beaker of water, described by the following reaction:
$\ce{NaCl(s) \leftrightarrow NaCl(aq)} \nonumber$
If you keep adding more and more solid salt, eventually you'll reach the point where no more salt dissolves, and the excess sits at the bottom of the beaker. At this point we have a saturated solution. Has the dissolving reaction stopped? It would appear so, but that's not the case (wouldn't that be too easy?).
What happens in our saturated solution, which has reached the point of equilibrium, is that both the forward
$\ce{NaCl(s) \rightarrow NaCl(aq)} \nonumber$
and reverse
$\ce{NaCl(aq) \rightarrow NaCl(s)} \nonumber$
reactions are still going on, but at the same rate. This in effect cancels out any observable, or measurable, changes in our system. At the same rate that solid NaCl produces aqueous NaCl (dissolved salt), the dissolved salt is re-crystallizing to form more solid NaCl.
• Equilibrium is the state at which the rate of the forward reaction equals the rate of the reverse reaction.
• At the point of equilibrium, no more measurable or observable changes in the system can be noted.
It is important for you to understand that equilibrium means the rates of the forward and reverse reactions are equal; it does not mean that there are equal amounts of reactants and products present at equilibrium.
For example, the following reaction was allowed to come to the point of equilibrium, and concentrations of all reaction participants were measured at that time:
$\ce{H_2(g) + I_2(g) \leftrightarrow 2HI(g)}$
At equilibrium:
$\mathrm{[H_2]=0.022\: M}$
$\mathrm{[I_2]=0.022\: M}$
$\mathrm{[HI]=0.156\: M}$
For this particular reversible reaction, there is more HI at equilibrium (0.156 M) than there is of H2 and I2 (both at 0.022M). We say that the product side of the reaction is favored.
Equilibrium does not mean equal amounts at equilibrium!
Here is an example to help you understand how equilibrium works: imagine yourself on a escalator that is going down. You start at the top (reactants) and end up at the bottom (products). But when you are partway down, you start walking up the escalator as it continues going down. If you match your rate of walking up to the same rate that the escalator is going down, you make no progress and appear to be at a standstill. To an observer it would look as if you and the escalator had come to a stop, when actually both upward and downward movements continue.
Equilibrium is dynamic—both forward and reverse reactions continue, even though the reaction appears to have stopped. And this equilibrium does not need to occur right in the middle of two floors—you could be near the bottom, near the top, or anywhere in between when you carry out your reverse process.
In order for a reversible reaction to reach the point of equilibrium, the reaction must be carried out in a closed system—no additional reactants can be added or products removed. If, in our last example, the product HI was removed as it formed, the reaction would never reach the point of equilibrium; instead, H2 and I2 would continue to react to produce HI until one or both of the reactants was used up.
If reactants are constantly being added, and products removed as they form, the system would appear to be at equilibrium because to an outside observer it would appear that the reaction has stoppedbut that would not be the case. This situationwith new material constantly being added as products are removedis called a steady state system. A factory with an assembly line is a steady state systemnew raw materials are constantly being added; finished products are removed. A campfire with wood being added to the fire is another steady state system. Be careful not to confuse steady state with equilibrium.
How do the rates of the forward and reverse reactions change as the reaction heads towards equilibrium (before it reaches equilibrium)?
If we start our above reaction with H2 and I2, and with no HI, the two gases will react at a certain rate. But remember that the rate of a reaction slows down over time, as the reactants get used up (and lower their concentrations). Eventually, however, the amount of the product HI increases, and it will begin producing H2 and I2. Thus the rate of the reverse reaction starts out slowly (there is no HI present), but will speed up as the concentration of HI increases. Eventually both rates will level off (not always to the same level as shown by this example, however):
Chemists have found that there is a mathematical relationship that exists between the concentration of the reactants and products, once equilibrium has been reached, that is independent of the initial concentration of the participants. For any general reaction,
$\ce{aA + bB \rightleftharpoons cD + dD} \label{2.1.1}$
an equilibrium constant expression can be written as:
$\mathrm{K_{eq}=\dfrac{[C]^c \times [D]^d}{[A]^a \times [B]^b}} \label{2.1.2}$
This mathematical relationship exists for all equilibrium systems, and produces a constant ratio called the equilibrium constant, Keq .
Law of Mass-Action
Equation \ref{2.1.2} is sometimes called the Law of Mass-Action.
This relationship will be very important to us for the next few units, so it is important that you understand how to set this relationship up and what it tells us about an equilibrium system.
The products of the reaction (C and D) are placed in the numerator, and their concentrations are raised to the power of the coefficients from the balanced equation. The reactants (A and B) are placed in the denominator, with their concentrations raised to the power of their coefficients.
Example $1$
For the reaction between hydrogen and iodine gas to produce hydrogen iodide:
$\ce{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)} \label{2.1.3}$
the equilibrium constant expression will be:
$\mathrm{K_{eq}=\dfrac{[HI]^2}{[H_2] \times [I_2]}} \label{2.1.4}$
Using the example we examined in our last section, equilibrium concentrations for each substance were measured at equilibrium and found to be:
At equilibrium:
$\mathrm{[H_2]=0.022\: M}$
$\mathrm{[I_2]=0.022\: M}$
$\mathrm{[HI]=0.156\: M}$
We substitute these values into our equilibrium expression and solve for Keq:
$\mathrm{K_{eq}=\dfrac{[HI]^2}{[H_2] \times [I_2]}=\dfrac{(0.156)^2}{(0.022)(0.022)}=50.3} \nonumber$
The value of Keq, which has no units, is a constant for any particular reaction, and its value does not change unless the temperature of the system is changed. It does not depend on the initial concentrations used to reach the point of equilibrium.
For example, the following data were obtained for equilibrium concentrations of H2, I2 and HI, and the value of Keq was calculated for each trial:
Trial
$\ce{[HI]}$
$\ce{[ H2]}$
$\ce{[ I2]}$
$K_{eq}$
1
0.156
0.0220
0.0220
50.3
2
0.750
0.106
0.106
50.1
3
1.00
0.820
0.0242
50.4
4
1.00
0.0242
0.820
50.4
5
1.56
0.220
0.220
50.3
Aside from accounting for slight experimental variation between trials, the value for Keq is the same despite differences in equilibrium concentrations for the individual participants.
There is one other important point to make at this time.
Keq relates the concentrations of products to reactants at equilibrium.
For aqueous solutions, concentration is often measured as mol · L-1. For gases, concentration is often measured as partial pressure.
The concentrations of both aqueous solutions and gases change during the progress of a reaction. For reactions involving a solid or a liquid, while the amounts of the solid or liquid will change during a reaction, their concentrations (much like their densities) will not change during the reaction. Instead, their values will remain constant. Because they are constant, their values are not included in the equilibrium constant expression.
Example $2$
For example, consider the reaction showing the formation of solid calcium carbonate from solid calcium oxide and carbon dioxide gas:
$\ce{CaO(s) + CO_2(g) \leftrightarrow CaCO_3(s)} \nonumber$
The equilibrium constant for this reaction is (before modification):
$\mathrm{K_{eq}=\dfrac{[CaCO_3]}{[CaO] \times [CO_2]}} \nonumber$
But we remove those participants whose state is either a solid or a liquid, which leaves us with the following equilibrium constant expression:
$\mathrm{K_{eq}=\dfrac{1}{[CO_2]}} \nonumber$
Summary
• There are a few possible ways a reaction can go: it can go to completion ($\text{reactants} \rightarrow \text{products}$), and it can occur but not go to completion. Instead, it will reach chemical equilibrium ($\text{reactants} \rightleftharpoons \text{products}$).
• Chemical equilibrium occurs when the number of particles becoming products is equal to the number of particles becoming reactants.
• A dynamic equilibrium is a state where the rate of the forward reaction is equal to the rate of the reverse reaction.
Vocabulary
• Equilibrium - A state that occurs when the rate of the forward reaction is equal to the rate of the reverse reaction. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.03%3A_The_Idea_of_Dynamic_Chemical_Equilibrium.txt |
Learning Objectives
• Write equilibrium constant expressions.
• Use equilibrium constant expressions to solve for unknown concentrations.
• Use known concentrations to solve for the equilibrium constants.
• Explain what the value of $K$ means in terms of relative concentrations of reactants and products.
In the previous section, you learned about reactions that can reach a state of equilibrium, in which the concentration of reactants and products aren't changing. If these amounts are changing, we should be able to make a relationship between the amount of product and reactant when a reaction reaches equilibrium.
The Equilibrium Constant
Equilibrium reactions are those that do not go to completion, but are in a state where the reactants are reacting to yield products and the products are reacting to produce reactants. In a reaction at equilibrium, the equilibrium concentrations of all reactants and products can be measured. The equilibrium constant ($K$) is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants. Sometimes, subscripts are added to the equilibrium constant symbol $K$, such as $K_\text{eq}$, $K_\text{c}$, $K_\text{p}$, $K_\text{a}$, $K_\text{b}$, and $K_\text{sp}$. These are all equilibrium constants and are subscripted to indicate special types of equilibrium reactions.
There are some rules about writing equilibrium constant expressions that need to be learned:
1. Concentrations of products are multiplied on the top of the expression. Concentrations of reactants are multiplied together on the bottom.
2. Coefficients in the equation become exponents in the equilibrium constant expression.
3. Solids, liquids, and solvents are assigned a value of 1, so their concentrations do not affect the value of K.
Example $1$
Write the equilibrium constant expression for:
$\ce{CO} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons \ce{CH_4} \left( g \right) + \ce{H_2O} \left( g \right) \nonumber$
Solution
$K = \dfrac{\left[ \ce{CH_4} \right] \left[ \ce{H_2O} \right]}{\left[ \ce{CO} \right] \left[ \ce{H_2} \right]^3} \nonumber$
*Note that the coefficients become exponents. Also, note that the concentrations of products in the numerator are multiplied. The same is true of the reactants in the denominator.
Example $2$
Write the equilibrium constant expression for:
$2 \ce{TiCl_3} \left( s \right) + 2 \ce{HCl} \left( g \right) \rightleftharpoons 2 \ce{TiCl_4} \left( s \right) + \ce{H_2} \left( g \right) \nonumber$
Solution
$K = \dfrac{\left[ \ce{H_2} \right]}{\left[ \ce{HCl} \right]^2} \nonumber$
*Note that the solids have a value of 1, and multiplying or dividing by 1 does not change the value of K.
Example $2$
Write the equilibrium constant expression for:
$\ce{P_4} \left( s \right) + 6 \ce{Cl_2} \left( g \right) \rightleftharpoons 4 \ce{PCl_3} \left( s \right) \nonumber$
Solution
$K = \dfrac{1}{\left[ \ce{Cl_2} \right]^6} \nonumber$
*Note that the only product is a solid, which is defined to have a value of 1. That leaves just 1 on top in the numerator.
Example $3$
Write the equilibrium constant expression for:
$\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \nonumber$
Solution
$K = \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right] \nonumber$
*Note that the water is the solvent, and thus has a value of 1. Dividing by 1 does not change the value of K.
Equilibrium Constant Expressions
The equilibrium constant value is the ratio of the concentrations of the products over the reactants. This means that we can use the value of $K$ to predict whether there are more products or reactants at equilibrium for a given reaction. What can the value of Keq tell us about a reaction?
• If Keq is very large, the concentration of the products is much greater than the concentration of the reactants. The reaction essentially "goes to completion"; all, or most of, the reactants are used up to form the products.
• If Keq is very small, the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent—most of the reactants remain unchanged, and there are few products produced.
• When Keq is not very large or very small (close to a value of 1) then roughly equal amounts of reactants and products are present at equilibrium.
Here are some examples to consider:
Reaction Chemical Equations Equilibrium Constant
the decomposition of ozone, $\ce{O3}$ $\ce{2O_3(g) \rightleftharpoons 3O_2(g)}$ $\mathrm{K_{eq}=2.0 \times 10^{57}}$
$K_{eq}$ is very large, indicating that mostly $\ce{O2}$ is present in an equilibrium system, with very little $\ce{O3}$.
production of
nitrogen monoxide
$\ce{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)}$ $\mathrm{K_{eq}=1.0 \times 10^{-25}}$
Very little $\ce{NO}$ is produced by this reaction; $\ce{N2}$ and $\ce{O2}$ do not react readily to produce $\ce{NO}$ (this is lucky for us—otherwise we would have little oxygen to breath in our atmosphere!).
reaction of carbon monoxide and water $\ce{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}$ $\mathrm{K_{eq}=5.09} \,(\text{at 700 K})$
The concentrations of the reactants are very close to the concentrations of the products at equilibrium.
If the equilibrium constant is 1 or nearly 1, it indicates that the molarities of the reactants and products are about the same. If the equilibrium constant value is a large number, like 100, or a very large number, like $1 \times 10^{15}$, it indicates that the products (numerator) are a great deal larger than the reactants. This means that at equilibrium, the great majority of the material is in the form of products and it is said that the "products are strongly favored". If the equilibrium constant is small, like 0.10, or very small, like $1 \times 10^{-12}$, it indicates that the reactants are much larger than the products and the reactants are strongly favored. With large $K$ values, most of the material at equilibrium is in the form of products and with small $K$ values, most of the material at equilibrium is in the form of the reactants.
The equilibrium constant expression is an equation that we can use to solve for $K$ or for the concentration of a reactant or product.
Example $4$
Determine the value of $K$ for the reaction
$\ce{SO_2} \left( g \right) + \ce{NO_2} \left( g \right) \rightleftharpoons \ce{SO_3} \left( g \right) + \ce{NO} \left( g \right) \nonumber$
when the equilibrium concentrations are: $\left[ \ce{SO_2} \right] = 1.20 \: \text{M}$, $\left[ \ce{NO_2} \right] = 0.60 \: \text{M}$, $\left[ \ce{NO} \right] = 1.6 \: \text{M}$, and $\left[ \ce{SO_3} \right] = 2.2 \: \text{M}$.
Solution
Step 1: Write the equilibrium constant expression:
$K = \dfrac{\left[ \ce{SO_3} \right] \left[ \ce{NO} \right]}{\left[ \ce{SO_2} \right] \left[ \ce{NO_2} \right]} \nonumber$
Step 2: Substitute in given values and solve:
$K = \dfrac{\left( 2.2 \right) \left( 1.6 \right)}{\left( 1.20 \right) \left( 0.60 \right)} = 4.9 \nonumber$
Example $5$
Consider the following reaction:
$\ce{CO} \left( g \right) + \ce{H_2O} \left( g \right) \rightleftharpoons \ce{H_2} \left( g \right) + \ce{CO_2} \left( g \right) \nonumber$
with $K = 1.34$. If the $\left[ \ce{H_2O} \right] = 0.100 \: \text{M}$, $\left[ \ce{H_2} \right] = 0.100 \: \text{M}$, and $\left[ \ce{CO_2} \right] = 0.100 \: \text{M}$ at equilibrium, what is the equilibrium concentration of $\ce{CO}$?
Solution
Step 1: Write the equilibrium constant expression:
$K = \dfrac{\left[ \ce{H_2} \right] \left[ \ce{CO_2} \right]}{\left[ \ce{CO} \right] \left[ \ce{H_2O} \right]} \nonumber$
Step 2: Substitute in given values and solve:
$1.34 = \dfrac{\left( 0.100 \right) \left( 0.100 \right)}{\left[ \ce{CO} \right] \left( 0.100 \right)} \nonumber$
Solving for $\left[ \ce{CO} \right]$, we get: $\left[ \ce{CO} \right] = 0.0746 \: \text{M}$
Summary
• The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.
• If the value of $K$ is greater than 1, the products in the reaction are favored. If the value of $K$ is less than 1, the reactants in the reaction are favored. If $K$ is equal to 1, neither reactants nor products are favored.
Vocabulary
• Equilibrium constant ($K$) - A mathematical ratio that shows the concentrations of the products divided by the concentrations of the reactants.
Contributions & Attributions
• Modified by Tom Neils (Grand Rapids Community College) | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.04%3A_The_Equilibrium_Constant_-_A_Measure_of_How_Far_a_Reaction_Goes.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/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.06%3A_Calculating_and_Using_Equilibrium_Constants.txt |
Learning Objectives
• State Le Chatelier's Principle.
• Describe the effect of concentration on an equilibrium system.
• Describe the effect of temperature as a stress on an equilibrium system.
When a reaction has reached equilibrium with a given set of conditions, if the conditions are not changed, the reaction will remain at equilibrium forever. The forward and reverse reactions continue at the same equal and opposite rates and the macroscopic properties remain constant.
It is possible, however, to disturb that equilibrium by changing conditions. For example, you could increase the concentration of one of the products, or decrease the concentration of one of the reactants, or change the temperature. When a change of this type is made within a reaction at equilibrium, the reaction is no longer in equilibrium. When you alter something in a reaction at equilibrium, chemists say that you put stress on the equilibrium. When this occurs, the reaction will no longer be in equilibrium and the reaction itself will begin changing the concentrations of reactants and products until the reaction comes to a new position of equilibrium. How a reaction will change when a stress is applied can be explained and predicted. That is the topic of this section.
Le Chatelier's Principle
Think back to our escalator example, with you walking up a downward moving escalator. With the rate of the moving stairs and your walking evenly matched, you appear to be at a standstill. But what happens if the escalator begins moving just a little faster? If you want to maintain the same position you had, at some specific point between the bottom and the top of the stairs, you'll also need to make some adjustments. Chemical systems at equilibrium tend to make these adjustments as well.
In the late 1800's, a chemist by the name of Henry-Louis Le Chatelier was studying stresses that were applied to chemical equilibria. He formulated a principle from this research and, of course, the principle is called Le Chatelier's Principle. Le Chatelier's Principle states that when a stress is applied to a system at equilibrium, the equilibrium will shift in a direction to partially counteract the stress and once again reach equilibrium. Le Chatelier's principle is not an explanation of what happens on the molecular level to cause the equilibrium shift, it is simply a quick way to determine which way the reaction will run in response to a stress applied to the system at equilibrium.
Le Chatelier's Principle
If a system at equilibrium is subjected to an external stress, the equilibrium will shift to minimize the effects of that stress.
Equilibrium is all about rates—the rate of the forward reaction is equal to the rate of the reverse reaction. External stresses are factors that will cause the rate of either the forward or reverse reaction to change, throwing the system out of balance. Le Chatelier's Principle allows us to predict how this will affect our system.
In our unit on Kinetics we examined factors that influenced reaction rates. Recall these factors:
1. concentration
2. pressure and volume
3. temperature
4. catalysts
We will see how changing these factors affects a system at equilibrium.
Effect of Concentration Changes on a System at Equilibrium
For instance, if a stress is applied by increasing the concentration of a reactant, the reaction will adjust in such a way that the reactants and products can get back to equilibrium. In the case of too much reactant, the reaction will use up some of the reactant to make more product. It is said in this scenario that the reaction "shifts to the products" or "shifts to the right". If the concentration of a product is increased, there is an opposite effect. The reaction will use up some of the product to make more reactant. The reaction "shifts to the reactants" or "shifts to the left".
What if some reactant or product is removed? If a stress is applied by lowering a reactant concentration, the reaction will try to replace some of the missing reactant. It uses up some of the product to make more reactant, and the reaction "shifts to the reactants". If a stress is applied by reducing the concentration of a product, the equilibrium position will shift toward the products. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.07%3A_Disturbing_a_Reaction_at_Equilibrium-_Le_Chateliers_Principle.txt |
Consider the following system under equilibrium:
$\underbrace{\ce{Fe^{3+}(aq)}}_{\text{colorless}} + \underbrace{ \ce{SCN^{-}(aq)}}_{\text{colorless}} \rightleftharpoons \underbrace{\ce{FeSCN^{2+}(aq)}}_{\text{red}} \nonumber$
If more $Fe^{3+}$ is added to the reaction, what will happen?
According to Le Chatelier's Principle, the system will react to minimize the stress. Since Fe3+ is on the reactant side of this reaction, the rate of the forward reaction will increase in order to "use up" the additional reactant. This will cause the equilibrium to shift to the right, producing more FeSCN2+. For this particular reaction, we will be able to see that this has happened, as the solution will become a darker red color.
There are a few different ways to state what happens here when more Fe3+ is added, all of which have the same meaning:
• equilibrium shifts to the right
• equilibrium shifts to the product side
• the forward reaction is favored
What changes does this cause in the concentrations of the reaction participants?
Changes within reaction participants
$\ce{Fe^{3+}}$
Since this is what was added to cause the stress, the concentration of $\ce{Fe^{3+}}$ will increase. (A shorthand way to indicate this: $\ce{[Fe]^{3+}\: \uparrow}$ (Reminder: the square brackets represent "concentration")
$\ce{SCN^{-}(aq)}$ Equilibrium will shift to the right, which will use up the reactants. The concentration of $\ce{SCN^{-}(aq)}$ will decrease $\ce{[SCN]^{-}\: \downarrow}$ as the rate of the forward reaction increases.
$\ce{FeSCN^{2+}}$
When the forward reaction rate increases, more products are produced, and the concentration of $\ce{FeSCN^{2+}}$ will increase. $\ce{[FeSCN]^{2+}} \uparrow$
How about the value of Keq? Notice that the concentration of some reaction participants have increased, while others have decreased. Once equilibrium has re-established itself, the value of Keq will be unchanged.
The value of Keq does not change when changes in concentration cause a shift in equilibrium.
What if more FeSCN2+ is added?
Again, equilibrium will shift to use up the added substance. In this case, equilibrium will shift to favor the reverse reaction, since the reverse reaction will use up the additional FeSCN2+.
• equilibrium shifts to the left
• equilibrium shifts to the reactant side
• the reverse reaction is favored
How do the concentrations of reaction participants change?
Change of concentrations of reaction participants when adding substance
$\ce{Fe^{3+}}$ $\ce{[Fe]^{3+}\: \uparrow}$ as the reverse reaction is favored
$\ce{SCN^{-}(aq)}$ $\ce{[SCN]^{-}\: \uparrow}$ as the reverse reaction is favored
$\ce{FeSCN^{2+}}$ $\ce{[FeSCN]^{2+}} \uparrow$ because this is the substance that was added
Concentration can also be changed by removing a substance from the reaction. This is often accomplished by adding another substance that reacts (in a side reaction) with something already in the reaction.
Let's remove SCN- from the system (perhaps by adding some Pb2+ ions—the lead(II) ions will form a precipitate with SCN-, removing them from the solution). What will happen now? Equilibrium will shift to replace SCN-—the reverse reaction will be favored because that is the direction that produces more SCN-.
• equilibrium shifts to the left
• equilibrium shifts to the reactant side
• the reverse reaction is favored
How do the concentrations of reaction participants change?
Change of concentrations of reaction participants when removing a substance
$\ce{Fe^{3+}}$ $\ce{[Fe]^{3+}\: \uparrow}$ as the reverse reaction is favored
$\ce{SCN^{-}}$ $\ce{[SCN]^{-}\: \uparrow}$ as the reverse reaction is favored
(but also ↓ because it was removed)
$\ce{FeSCN^{2+}}$ $\ce{[FeSCN]^{2+}} \uparrow$ because this is the substance that was added | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.08%3A_The_Effect_of_a_Concentration_Change_on_Equilibrium.txt |
Changing the pressure or volume of a container enclosing an equilibrium system will only affect the reaction if gases are present.
You may remember from earlier chemistry classes that equal volumes of gases contain an equal number of particles and, under standard conditions of temperature and pressure (STP), one mole of gas occupies a volume of 22.4 L. This is known as the molar volume of gases. So, two moles of any gas will occupy a volume of 44.8 L and one-half mole would occupy 11.2 L.
How does changing pressure and volume affect equilibrium systems?
• If you increase the pressure of a system at equilibrium (typically by reducing the volume of the container), the stress will best be reduced by reaction that favors the side with the fewest moles of gas, since fewer moles will occupy the smallest volume.
• Conversely, if you decrease the pressure (by increasing the volume of the container), equilibrium will shift to favor the side with the most moles of gas, since more moles will occupy a greater volume.
• If both sides of the equation have the same number of moles of gas, then there will be no change in the position of equilibrium.
When considering the effect of changing volume or pressure on equilibrium systems, be sure to only count the number of moles of gases on each side of the equation. Solids, liquids, and aqueous solutions will not be affected by changing pressure and volume.
Example $1$
Predict the effect on equilibrium when the pressure is increased for the following reaction:
$\ce{N_2O_4(g) <=> 2NO_2(g)} \nonumber$
Solution
The reactant side of the equation has 1 mole of a gas; the product side has 2 moles. Increasing the pressure favors the side with the fewest moles of gas, so the equilibrium will shift to the left (the reverse reaction will be favored).
15.10: The Effect of Temperature Changes on Equilibrium
Learning Objectives
• Explain how temperature changes affect a system at equilibrium.
When temperature is the stress that affects a system at equilibrium, there are two important consequences:
• an increase in temperature will favor that reaction direction that absorbs heat (i.e. the endothermic reaction)
• the value of Keq will change
Consider the following equilibrium system:
$\ce{N_2O_4(g) \leftrightarrow 2NO_2(g)} \nonumber$
with $\ce{\Delta H^{\circ}={58.0}\:kJ}$
We see by the sign of ΔH° that the forward reaction is endothermic. Heat is absorbed (required as a reactant) when the reaction proceeds as
$\ce{N_2O_4(g) \rightarrow 2NO_2 (g)} \nonumber$
By adding more heat, equilibrium will shift to use up the additional heat, thus favoring this forward direction.
Why will Keq change, when it did not change when concentration, pressure, and volume were the applied stresses?
When temperature changes cause an equilibrium to shift, one entire side of the reaction equation is favored over the other side. Mathematically, this will alter the value of Keq as follows:
$\ce{K_{eq}=\dfrac{[products]}{[reactants]}} \nonumber$
$\ce{K_{eq}$ forward and reverse reactions
if the forward reaction is favored
more products are produced; fewer reactants
Keq will increase
if the reverse reaction is favored
fewer products; more reactants
Keq will decrease
So in our example given above, increasing the temperature will favor the forward direction. The value of Keq will increase. Removing heat (making the system colder) will favor the exothermic reaction—the exothermic reaction releases heat to the surroundings, thus "replacing" the heat that has been removed. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.09%3A_The_Effect_of_a_Volume_Change_on_Equilibrium.txt |
Learning Objectives
• Understand the basic concept of solution-solid equilibria.
We will now return to an important mathematical relationship that we first learned about in our unit on Equilibrium, the equilibrium constant expression. Recall that for any general reaction:
$aA + bB \rightleftharpoons cD + dD \nonumber$
an equilibrium constant expression can be defined as:
$K_{eq} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \nonumber$
Since saturated solutions are equilibrium systems, we can apply this mathematical relationship to solutions. We will refer to our equilibrium constant as $K_{sp}$, where "sp" stands for "solubility product"
For our silver sulfate saturated solution,
$Ag_2SO_{4 (s)} \rightleftharpoons 2Ag^+_{(aq)} + SO^{2-}_{4(aq)} \nonumber$
we can write a solubility product constant expression as
$K_{sp} = \dfrac{ [Ag^+_{(aq)}]^2[SO_{4(aq)}^{2-}] }{ [Ag_2SO_{4(s)}] } \nonumber$
However, remember from our earlier introduction to the equilibrium constant expression that the concentrations of solids and liquids are NOT included in the expression because while their amounts will change during a reaction, their concentrations will remain constant. Therefore, we will write our solubility product constant expression for our saturated silver sulfate solution as:
$K_{sp}= [Ag^+]^2[SO_4^{2-}] \nonumber$
Please note that it is VERY IMPORTANT to include the ion charges when writing this equation.
Example $1$
Write the expression for the solubility product constant, Ksp, for Ca3(PO4)2.
Solution
Step 1: Begin by writing the balanced equation for the reaction. Remember that polyatomic ions remain together as a unit and do not break apart into separate elements.
$Ca_3(PO_4)_{2 (s)} \rightleftharpoons 3 Ca^{2+}_{(aq)} + 2 PO^{3-}_{4(aq)} \nonumber$
Step 2: Write the expression for $K_{sp}$:
$K_{sp}= [Ca^{2+}]^3[PO_4^{3-}]^2 \nonumber$
Solubility Product Tables that give Ksp values for various ionic compounds are available. Because temperature affects solubility, values are given for specific temperatures (usually 25°C). Recall what we learned about Keq:
• If Keq is very large, the concentration of the products must be much greater than the concentration of the reactants. The reaction essentially "goes to completion", which means that all—or most of—the reactants are used up to form the products.
• If Keq is very small, the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent—most of the reactants remain unchanged, and there are few products produced.
• When Keq is not very large or very small, then roughly equal amounts of reactants and products are present at equilibrium.
$K_{sp}$, which again is just a special case of $K_{eq}$, provides us with the same useful information:
A low value of Ksp indicates a substance that has a low solubility (it is generally insoluble); for ionic compounds this means that there will be few ions in solution.
• Iron(II) sulfide, FeS, is an example of a low Ksp : Ksp = 4 ×10-19. In a saturated solution of FeS there would be few Fe2+ or S2- ions.
A large value of Ksp indicates a soluble substance; for ionic compounds it tells us that there will be many ions in solution.
• An example of a relatively large Ksp would be for lead(II) chloride, PbCl2 which has a Ksp of 1.8 ×10-4. A saturated solution of PbCl2 would have a relatively high concentration of Pb 2+ and Cl - ions.
There are several types of problems we can solve:
Example $1$: Calculating Ksp of a Saturated Solution
The concentration of a saturated solution of BaSO4 is 3.90 × 10-5M. Calculate Ksp for barium sulfate at 25°C.
Solution
Always begin problems involving Ksp by writing a balanced equation:
BaSO4 (s) Ba2+(aq) + SO42- (aq)
Next, write the Ksp expression:
Ksp= [Ba2+][SO42-]
The question provides us with the concentration of the solution, BaSO4 . We need to find the concentration of the individual ions for our equation.
Recall from Section 2.5: since 1 mole of BaSO4 produces 1 mole of Ba2+ and also 1 mole of SO42-, then...
• [BaSO4] = 3.90 × 10-5M
• [Ba2+] = 3.90 × 10-5M
• [SO42-] = 3.90 × 10-5M
Substitute values into the Ksp expression and solve for the unknown:
$K_{sp} = [Ba^{2+}][SO_4^{2-}] \nonumber$
$= (3.90 \times 10^{-5})(3.90 \times 10^{-5}) \nonumber$
$= 1.52 \times 10^{-9} \nonumber$
Example $3$: Calculating Ion Concentrations When Ksp is Known
The $K_{sp}$ for MgCO3 at 25°C is 2.0 × 10-8. What are the ion concentrations in a saturated solution at this temperature?
Solution
As always, begin with a balanced equation:
MgCO3(s) Mg2+(aq) + CO32-(aq)
Write the Ksp expression:
$K_{sp}= [Mg^{2+}][CO_3^{2-}] \nonumber$
For this example, we are given the value for Ksp and need to find the ion concentrations.
We will let our unknown ion concentrations equal x.
The balanced equation tells us that both Mg2+ and CO32- will have the same concentration.
Substitute values into the equation and solve for the unknown:
Ksp = [Mg2+][CO32-]
x2 = 2.0 × 10-8
x = √(2.0 × 10-8)
find the square root of 2.0 × 10-8:
= 1.4× 10-4M
x = [Mg2+]
= 1.4× 10-4M
AND x = [CO32-]
= 1.4× 10-4M
15.12: The Path of a Reaction and the Effect of a Catalyst
The addition of a catalyst to an equilibrium system is a final stress factor. We consider how adding a catalyst affects the following:
$\ce{N_2(g) + O_2(g) \leftrightharpoons 2NO(g)} \nonumber$
Adding a catalyst to this, or any other equilibrium system, will not affect the position of an equilibrium. A catalyst speeds up both the forward and the reverse reactions, so there is no uneven change in reaction rates. Generally, a catalyst will help a reaction to reach the point of equilibrium sooner, but it will not affect the equilibrium otherwise. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/15%3A_Chemical_Equilibrium/15.11%3A_The_Solubility-Product_Constant.txt |
The rusting of an old car. A burning campfire. A toy battery-operated car. The chemical processes in your body that break down carbohydrates to produce water, carbon dioxide and energy. The ripening of fruit. It's not easy to see what all of these types of reactions have in common, but they all belong to a very important category of chemical reactions known as oxidation-reduction (or redox) reactions.
• 16.2: Oxidation and Reduction- Some Definitions
"Redox" is short for "oxidation and reduction", two complimentary types of chemical reactions. The term oxidation originally referred to substances combining with oxygen, as happens when an iron bar rusts or a campfire log burns. We often refer to these two examples as corrosion and combustion. Reduction originally referred to the process of converting metal ores to pure metals, a process that is accompanied by a reduction in the mass of the ore.
• 16.3: Oxidation States- Electron Bookkeeping
Redox reactions are all about electrons being transferred from one substance to another, so it is useful to have a system for keeping track of what gains and what loses electrons, and how many electrons are involved. The record-keeping system is called Oxidation Numbers.
• 16.4: Balancing Redox Equations
Another way to balance redox reactions is by the half-reaction method. This technique involves breaking an equation into its two separate components—the oxidation reaction and the reduction reaction. Since neither oxidation nor reduction can actually occur without the other, we refer to the separate equations as half-reactions.
• 16.5: The Activity Series- Predicting Spontaneous Redox Reactions
Single-replacement reactions only occur when the element that is doing the replacing is more reactive than the element that is being replaced. Therefore, it is useful to have a list of elements in order of their relative reactivities. The activity series is a list of elements in decreasing order of their reactivity. Since metals replace other metals, while nonmetals replace other nonmetals, they each have a separate activity series.
• 16.6: Batteries- Using Chemistry to Generate Electricity
Electrochemical cells used for power generation are called batteries. Although batteries come in many different shapes and sizes, there are a few basic types. You won't be required to remember details of the batteries, but some general information and features of each type are presented here. Batteries are one way of producing electricity. Many important chemical reactions involve the exchange of one or more electrons, and we can use this movement of electrons as electricity.
• 16.7: Electrolysis- Using Electricity to Do Chemistry
Galvanic cells produce electricity from chemical reactions. Some reactions will, instead, use electricity to get a reaction to occur. In these reactions, electrical energy is given to the reactants, causing them to react to form the products. These reactions have many uses. For example, electrolysis is a process that involves forcing electricity through a liquid or solution to cause a reaction to occur. Electrolysis reactions will not run unless energy is added to the system.
• 16.8: Corrosion- Undesirable Redox Reactions
Corrosion of metals is a serious economic problem. Corrosion occurs as a result of spontaneous electrochemical reaction as metal undergoes oxidation.
16: Oxidation and Reduction
"Redox" is short for "oxidation and reduction", two complimentary types of chemical reactions. The term oxidation originally referred to substances combining with oxygen, as happens when an iron bar rusts or a campfire log burns. We often refer to these two examples as corrosion and combustion. Reduction originally referred to the process of converting metal ores to pure metals, a process that is accompanied by a reduction in the mass of the ore.
These two terms have broader meanings now. In all oxidation-reduction reactions, an exchange of electrons occurs—one substance loses electrons while another gains them. That is the key to understanding redox reactions. We'll define these terms below.
A simple demonstration of a redox reaction involves placing a solid piece of copper wire in a silver nitrate solution. Within minutes, the wire begins to look fuzzy or furry as small silver crystals begin to form on the wire. Meanwhile, the originally clear silver nitrate solution begins to take on a pale bluish tint. Furthermore, if the crystals are shaken off of the wire, we see that the wire partially disintegrated.
The overall equation for our demonstration describes the events:
$\ce{Cu(s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2 Ag(s)} \nonumber$
Remember that when we have aqueous solutions of ionic compounds, the ions are really present as separate ions, not as bonded particles. So we can write an expanded equation:
$\ce{Cu(s) + 2Ag^{+}(aq) + 2NO^{-}3(aq) → Cu^{2+}(aq) + 2NO^{-}3(aq) + 2Ag(s)} \nonumber$
Finally we remove spectator ions from the equation. Remember that spectator ions are ions that undergo no change during a reaction. Notice the nitrate ions, $NO_3^-$, they start as aqueous ions and end up exactly the same.
Removing the spectator ions gives us our net ionic equation:
$\ce{Cu(s) + 2Ag^{+}(aq) → Cu^{2+}(aq) + 2Ag(s)} \nonumber$
We can now see a bit more clearly what changes are occurring with this reaction.
Oxidation of Copper Metal to Make Copper Ions
Copper began as a neutral atom with no charge, but changed into an ion with a charge of +2. An atom becomes a positive ion by losing electrons:
$\ce{Cu(s) → Cu^{2+}(aq) + 2e^{-}} \label{oxeq}$
Notice that copper began as a solid, but is converted into aqueous ions—this is why the copper wire disintegrates. We say that copper was oxidized because it has lost electrons (i.e., electrons appear on the product side of the Equation $\ref{oxeq}$).
Reduction of Silver Ions to Make Silver Metal
Silver was converted from an ion with a charge of +1, Ag+, to a neutral atom, Ag. The only way an ion can undergo this change is to gain an electron:
$\ce{Ag^{+}(aq) + e^{-} → Ag(s)} \label{redeq}$
Notice that solid silver is formed—this is what causes the fuzzy appearance to begin appearing on the wire—solid silver crystals. Silver has gained electrons, it has been reduced (i.e., electrons appear on the reactant side of Equation $\ref{redeq}$).
The electrons that silver gained had to come from somewhere—they came from copper. Conversely, a substance such as copper can only lose electrons if there is something else that will take them up, the silver ions. One cannot occur without the other. This exchange of electrons is what defines an oxidation-reduction reaction.
Definition: Oxidation
Oxidation is the loss of electrons.
Definition: Reduction
Reduction is the gain of electrons.
You will be learning several new terms in this chapter and it is important that you learn them very quickly. You may find it useful to have some tricks to help you remember these terms, including the LEO/GER.
LEO the Lion Says GER
• LEO: Loss of Electrons is Oxidation.
• GER: Gain of Electrons is Reduction. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/16%3A_Oxidation_and_Reduction/16.02%3A_Oxidation_and_Reduction-_Some_Definitions.txt |
Redox reactions are all about electrons being transferred from one substance to another, so it is useful to have a system for keeping track of what gains and what loses electrons, and how many electrons are involved. The record-keeping system for redox reactions is called Oxidation Numbers. You may also remember something called electronegativity from earlier chemistry classes.
Electronegativity
The ability of a bonded atom to attract shared electrons toward itself.
When two atoms of different elements are bonded together by a covalent bond (sharing electrons), the electrons are generally not shared equally between the two atoms due to differences in their electronegativities. Think of this as a tug-of-war for electrons. Sometimes both atoms pull with equal strength on shared electrons; other times there is clearly a stronger player that will pull the electrons closer to itself.
Consider the bond between a hydrogen atom (with one valence electron) and an oxygen atom (with its six valence electrons):
+
=
Because oxygen has a higher electronegativity than hydrogen, the shared electrons are closer to the oxygen atom than to the hydrogen atom. This is not the total transfer of electrons that would create an ion, but partial charges do form. The hydrogen end of the bond is partially positive (+1) because it has partially lost one electron, and the oxygen end of the H–O is partially negative (-1) because it has partially gained the one electron from hydrogen:
Our molecule is incomplete, however, because there is a lone electron around oxygen. Let's add one more hydrogen to complete our water molecule:
We see that each of the two hydrogens has "lost" one electron to oxygen. Oxygen has "gained" two electrons—one from each hydrogen. Again, these are not true ions, but it is useful to think of them in the same way.
Charges given to atoms in a molecule in this way are called oxidation numbers. We can use oxidation numbers to keep track of where electrons are in a molecule, and how they move during a reaction. In our water example, hydrogen is assigned an oxidation number of +1 because each individual hydrogen has "lost" one electron. Oxygen has an oxidation number of +2 because the single oxygen atom has "gained" a total of two electrons, one from each hydrogen.
Here is another molecule involving hydrogen and oxygen—hydrogen peroxide, H2O2:
In hydrogen peroxide, each hydrogen still has an oxidation number of +1 because each hydrogen "gives up" a single electron to oxygen. Oxygen, however, now has an oxidation number of -1 because each oxygen gains just one electron from its neighboring hydrogen. The electrons between the two identical oxygen atoms are shared equally, so there is no partial charge resulting from that bond.
Oxidation Number
A positive or negative number assigned to an atom in a molecule or ion that reflects a partial gain or loss of electrons.
Knowing the oxidation number of each individual element in a molecule will be a key step in our understanding of redox reactions. Fortunately it will not usually involve drawing electron dot diagrams. Instead, there are a series of rules that we can use to determine oxidation numbers. Here are the main rules:
1. The oxidation number of a pure element (by itself, and not an ion) is zero.
Element
Oxidation
Number
Na
0
H2
0
O2
0
P4
0
2.
The oxidation number of a monatomic ion (by itself or as part of an ionic compound) is equal to its charge.
Alkali metals—elements in the first column of the periodic table—will always have an oxidation number of +1; Alkaline earth metals (column 2) are almost always +2.
Ionic
Compound
Ions
Charge
Oxidation
Number
NaCl
Na+
+1
+1
Cl-
-1
-1
Mg3N2
Mg+2
+2
+2
N-3
-3
-3
3. The oxidation number of hydrogen is almost always +1 when it is in a compound.
Compound
Element
Oxidation
Number
HCl
H
+1
Cl
-1
H2S
H
+1
S
-2
4.
The oxidation number of oxygen is almost always -2 when it is in a compound.
The exceptions:
• Peroxides, such as hydrogen peroxide. In peroxides, oxygen has an oxidation number of -1.
• When oxygen is combined with fluorine, its oxidation number is +2.
Compound
Element
Oxidation
Number
MgO
magnesium oxide
Mg
+2
O
-2
Na2O
sodium oxide
Na
+1
O
-2
Na2O2
sodium peroxide
Na
+1
O
-1
5.
The sum of the oxidation numbers in a compound is zero.
To determine the oxidation number of Mn in Mn2O7, we must work backwards:
• We know each oxygen is -2 (Rule 4).
• 7 oxygen gives a total of:
-2 × 7 atoms = -14 total
Since the sum of oxidation numbers must be zero, the total oxidation number of Mn must be +14 to cancel out oxygen's -14, but since there are 2 Mn atoms, each individual atom will have an oxidation number of +7:
+14 total
2 Mn atoms
= +7
Compound
Element
Oxidation
Number
Number of
Atoms
Total
Mg3N2
Mg
+2
3
+6
N
-3
2
-6
SUM
0
Mn2O7
Mn
+7
2
+14
O
-2
7
-14
SUM
0
Cl2O3
Cl
+3
2
+6
O
-2
3
-6
SUM
0
6.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge on that ion.
Again, work backwards to determine the oxidation number of any non-oxygen or non-hydrogen atom.
To determine the oxidation number of Cr in Cr2O72- :
• Oxygen will be -2 (Rule 4), for a total of:
-2 × 7 = -14
• Since the sum of the oxidation numbers will be -2 (the charge on the entire ion), the total for all Cr must be +12 because:
+12 + (-14) = -2
• Since there are two Cr, each Cr will have an oxidation number of +6.
= +7
+12
2
= +6
Compound
Element
Oxidation
Number
Number of
Atoms
Total
NO3-
N
+5
1
+5
O
-2
3
-6
SUM
-1
Cr2O72-
Cr
+6
2
+12
O
-2
7
-14
SUM
-2
SO42-
S
+6
1
+6
O
-2
4
-8
SUM
-2
It is important to note that oxidation number always refers to each individual atom in the compound, not to the total for that element.
For example, in H2O, the total positive "charge" for both hydrogen atoms will be +2 (which balances with the -2 from oxygen), but each hydrogen has an oxidation number of +1. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/16%3A_Oxidation_and_Reduction/16.03%3A_Oxidation_States-_Electron_Bookkeeping.txt |
Balancing Redox Equations Using Half-Reactions
Another way to balance redox reactions is by the half-reaction method. This technique involves breaking an equation into its two separate components—the oxidation reaction and the reduction reaction. Since neither oxidation nor reduction can actually occur without the other, we refer to the separate equations as half-reactions.
The general technique involves the following:
• The overall equation is broken down into two half-reactions. If there are any spectator ions, they are removed from the equations.
• Each half-reaction is balanced separately, first for atoms and then for charge. Electrons are added to one side of the equation or the other in order to balance charge. For example, if the reactant side of the equation has a total charge of +3, the product side must also equal +3.
• Next, the two equations are compared to make sure electrons lost equal electrons gained. One of the half reactions will be an oxidation reaction, the other will be a reduction reaction.
• Finally, the two half-reactions are added together, and any spectator ions that were removed are placed back into the equation.
Consider the following reaction:
$Mg_{(s)} + Cl_{2 (g)} → MgCl_{2 (s)} \nonumber$
In this reaction, Mg is oxidized and Cl is reduced. You may find it useful to use oxidation numbers to help you determine this. Mg changes from 0 to +2; Cl changes from 0 to -1.
When we write the half-reactions, we break apart compounds that contain either of the key elements (elements undergoing oxidation or reduction). Oxidation numbers are written as if they were ion charges. Notice that the chlorine from MgCl2 is written as two separate ions, not combined, as is Cl2. Balance the two reactions for atoms.
Mg → Mg+2 Cl2 → 2 Cl-
Next balance the equations for charge by adding electrons. Remember, one half-reaction will be an oxidation reaction (electrons on the product side) and the other will be reduction (electrons will be on the reactant side).
Mg → Mg+2 + 2 e- Cl2 + 2 e- → 2 Cl-
oxidation reduction
In this example, balancing for charge results in both sides, of both equations, having net charges of 0. That won't always be the case. Be sure you see in this example how charges are balanced.
We then compare the two equations for numbers of electrons. We see that both equations have 2 electrons so we do not need to make any adjustments for that. Finally, add the two equations together:
$Mg + Cl_2 → Mg^{+2} + 2 Cl^- \nonumber$
and reform any compounds broken apart in the earlier steps:
$Mg + Cl_2 → MgCl_2 \nonumber$
We see that the original equation was already balanced, not just for atoms, but for electrons as well.
Example 2
$Cu_{(s)} + AgNO_{3 (aq)} → Cu(NO_3){2 (aq)} + Ag_{(s)} \nonumber$
Solution
Identify the elements undergoing oxidation (Cu) and reduction (Ag). The nitrate group (NO3) is a spectator ion which we will not include in our half-reactions.
Cu → Cu+2 + 2 e- Ag+ + 1 e- → Ag
oxidation reduction
After balancing for atoms and for charge, we see that the two equations do not have the same number of electrons—there are 2 in the copper reaction, but only one in the silver reaction. Multiply everything in the silver reaction by 2, then we will add the equations together:
Step 1 Step 2 Step 3
Write the balanced
half-reactions
Balance electrons Add half-reactions
Cu → Cu+2 + 2 e- Cu → Cu+2 + 2 e-
Ag + 1 e- → Ag- × 2 2 Ag+ + 2e- → 2 Ag
Add equations together Cu + 2 Ag+ → Cu+2 + 2 Ag
Reform compound/return spectator ions Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
Here is a reaction occurring in an acid solution, which accounts for the presence of the H+ions. This example adds a little more complexity to our problem.
MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O
In this example, spectator ions have already been removed. Even though hydrogen and oxygen do not undergo changes in oxidation number, they are not spectators, and we need to work with them in our half-reactions.
We determine that Mn undergoes reduction (+7 to +2) while Fe undergoes oxidation (+2 to +3). The iron half-reaction is straight forward, but the manganese reaction is more complex—we must include hydrogen and oxygen in its half-reaction:
Fe2+→ Fe +3 + 1e- MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O
oxidation reduction
To balance the manganese half-reaction, first balance for Mn and O atoms. Next balance the H atoms, and finally add enough electrons to balance the charge on both sides of the equation. Be sure you see what has been done, so that you can do it on your own.
Step 1 Step 2 Step 3
Write the balanced
half-reactions
Add half-reactions
Fe2+→ Fe +3 + 1e- × 5 5Fe2+→ 5Fe +3 + 5e-
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O MnO4- + 8 H+ + 5e- → Mn2+ + 4 H2O
Add equations together MnO4- + 5 Fe2+ + 8 H+ → Mn2+ + 5 Fe3+ + 4H2O
Example 3
$HNO_3 + Cu + H^+ → NO_2 + Cu^{2+} + H_2O \nonumber$
Solution
1. Determine what is oxidized, what is reduced, and write the two balanced half-reactions (Step 1).
2. Balance for electrons lost = electrons gained (Step 2).
3. Add equations together.
Step 1 Step 2 Step 3
Write the balanced
half-reactions
Add half-reactions
Cu → Cu+2 + 2e- Cu → Cu+2 + 2e-
HNO3 + H+ + 1 e- → NO2 + H2O × 2 2HNO3 + 2H+ + 2e- → 2NO2 + 2H2O
Add equations together 2HNO3 + Cu + 2H+ → 2NO2 + Cu2+ + 2H2O
When balancing redox reactions, either the oxidation number method or the half-reaction method may be used. Often you'll find that one method works best for some equations, while the other method is more suited for other reactions. Or you may find one method just easier to use. The practice exercises and assignments tell you which method to use for a reaction, but as you get get more experience you'll be able to make your own decision as to which method to use.
Writing half-reactions, however, is a skill you will need for our final topic in this course—Electrochemistry—so be sure you can write balanced half-reactions.
Check your understanding with the practice questions, set 3, then complete Assignment 2 . | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/16%3A_Oxidation_and_Reduction/16.04%3A_Balancing_Redox_Equations.txt |
Learning Objectives
• Use the activity series to predict if a reaction will occur.
We see two metals below that can be exposed to water. The picture on the left is of sodium, which gives a violent reaction when it comes in contact with water. The picture on the right is of silver, a metal so unreactive with water that it can be made into drinking vessels.
The Activity Series
Single-replacement reactions only occur when the element that is doing the replacing is more reactive than the element that is being replaced. Therefore, it is useful to have a list of elements in order of their relative reactivities. The activity series is a list of elements in decreasing order of their reactivity. Since metals replace other metals, while nonmetals replace other nonmetals, they each have a separate activity series. The table below is an activity series of most common metals and of the halogens.
Table $1$: Activity Series of Metals in Aqueous Solutions
Most Active (Easily Oxidized—Readily Lose Electrons)
lithium Li These metals displace hydrogen from water Ca(s) + 2H2O(l)→ Ca(OH)2 + H2 (g). These elements are very reactive and react readily to form compounds.
potassium K
barium Ba
calcium Ca
sodium Na
These metals displace hydrogen from acids Zn(s) + HCl(aq)→ ZnCl2 + H2(g)
magnesium Mg
aluminum Al
zinc Zn
chromium Cr
iron Fe
cadmium Cd
nickel Ni
tin Sn
lead Pb
hydrogen H
copper Cu These metals do not displace hydrogen from acids or water. These elements are more stable, and form compounds less readily than do those higher in the table.
silver Ag
mercury Hg
platinum Pt
gold Au
Least Active
For a single-replacement reaction, a given element is capable of replacing an element that is below it in the activity series. This can be used to predict if a reaction will occur. Suppose that small pieces of the metal nickel were placed into two separate aqueous solutions: one of iron (III) nitrate and one of lead (II) nitrate. Looking at the activity series, we see that nickel is below iron, but above lead. Therefore, the nickel metal will be capable of replacing the lead in a reaction, but will not be capable of replacing iron.
$\ce{Ni} \left( s \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow \ce{Ni(NO_3)_2} \left( aq \right) + \ce{Pb} \left( s \right) \nonumber$
$\ce{Ni} \left( s \right) + \ce{Fe(NO_3)_3} \left( aq \right) \rightarrow \text{NR (no reaction)} \nonumber$
In the descriptions that accompany the activity series of metals, a given metal is also capable of undergoing the reactions described below that section. For example, lithium will react with cold water, replacing hydrogen. It will also react with steam and with acids, since that requires a lower degree of reactivity.
Steps for Problem Solving
Example $1$
Use the activity series to predict if the reaction below will occur. If not, write $\text{NR}$. If the reaction does occur, write the products of the reaction and balance the equation.
$\ce{Al} \left( s \right) + \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow$
Example $2$
Use the activity series to predict if the reaction below will occur. If not, write $\text{NR}$. If the reaction does occur, write the products of the reaction and balance the equation.
$\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow$
Plan the problem. Compare the placements of aluminum (the element doing the replacing) and zinc (the element being replaced) on the activity series. Compare the placements of silver (the element doing the replacing) and hydrogen (the element being replaced) on the activity series.
Solve.
Since aluminum is above zinc, it is capable of replacing it and a reaction will occur. The products of the reaction will be aqueous aluminum nitrate and solid zinc. Take care to write the correct formulas for the products before balancing the equation. Aluminum adopts a $+3$ charge in an ionic compound, so the formula for aluminum nitrate is $\ce{Al(NO_3)_3}$. The balanced equation is:
$2 \ce{Al} \left( s \right) + 3 \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{Al(NO_3)_3} \left( aq \right) + 3 \ce{Zn} \left( s \right) \nonumber$
Since silver is below hydrogen, it is not capable of replacing hydrogen in a reaction with an acid.
$\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow \text{NR} \nonumber$
Exercise $1$
Use the activity series to predict if the following reactions will occur. If not, write $\text{NR}$. If the reaction does occur, write the products of the reaction and balance the equation.
a. Would it be possible to store a silver spoon in a zinc nitrate solution? That is, will the following reaction occur?
\Ag_{(s)} + Zn(NO_3)_{2 (aq)} →
b. Would it be possible to store a silver nitrate solution in a copper container? That is, will the following reaction occur?
\[Cu_{(s)} + AgNO_{3 (aq)} →
Summary
• Metals and halogens are ranked according to their ability to displace other metals or halogens below them in the activity series. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/16%3A_Oxidation_and_Reduction/16.05%3A_The_Activity_Series-_Predicting_Spontaneous_Redox_Reactions.txt |
Learning Objectives
• Identify the substance being oxidized and the substance being reduced in an oxidation-reduction reaction.
• Identify the anode and the cathode given a diagram of an electrolysis apparatus that includes the compound being electrolyzed.
• Describe how batteries can produce electrical energy.
Electricity is an important form of energy that you use every day. It runs your calculators, cell phones, dishwashers, and watches. This form of energy involves moving electrons through a wire and using the energy of these electrons. Electrochemical cells used for power generation are called batteries. Although batteries come in many different shapes and sizes, there are a few basic types. You won't be required to remember details of the batteries, but some general information and features of each type are presented here. Many important chemical reactions involve the exchange of one or more electrons, and we can use this movement of electrons as electricity; batteries are one way of producing this type of energy. The reactions that drive electricity are called oxidation-reduction (or "redox") reactions.
Oxidation and Reduction
Reactions in which electrons are transferred are called oxidation-reduction (or "redox") reactions. There are two parts to these changes: one atom must lose electrons and another atom must gain them. These two parts are described by the terms "oxidation" and "reduction".
Originally, a substance was said to be oxidized when it reacted with oxygen. Today, the word "oxidized" is still used for those situations, but now we have a much broader second meaning for these words. Today, the broader sense of the word oxidation is defined as losing electrons. When a substance loses electrons, its charge will increase. This may feel a bit backwards, but remember that electrons are negative. If an atom loses electrons, it is losing negative particles—so its charge will increase.
The other half of this process, the gaining of electrons, also needs a name. When an atom or an ion gains electrons, the charge on the particle goes down. For example, if a sulfur atom whose charge is zero $\left( 0 \right)$ gains two electrons, its charge becomes $\left( -2 \right)$; if an $\ce{Fe^{3+}}$ ion gains an electron, its charge changes from $+3$ to $+2$. In both cases, the charge on the particle is reduced by the gain of electrons. Remember that electrons have a negative charge, so gaining electrons will result in the charge decreasing. The word reduction is defined to mean gaining electrons and the reduction of charge.
In chemical systems, these two processes (oxidation and reduction) must occur simultaneously and the number of electrons lost in the oxidation must be the same as the number of electrons gained in the reduction. In oxidation-reduction reactions, electrons are transferred from one substance to another. Here is an example of an oxidation-reduction reaction:
$2 \ce{Ag^+} \left( aq \right) + \ce{Cu} \left( s \right) \rightarrow 2 \ce{Ag} \left( s \right) + \ce{Cu^{2+}} \left( aq \right) \nonumber$
In this reaction, the silver ions are gaining electrons to become silver atoms. Therefore, the silver ions are being reduced and the charge of silver is decreasing. The copper atoms are losing electrons to become copper $+2$ ions, are therefore being oxidized, and the charge of copper is increasing. Whenever a chemical reaction involves electrons being transferred from one substance to another, the reaction is an oxidation-reduction reaction (or a redox reaction).
Half-equations are very helpful in discussing and analyzing processes, but half-reactions cannot occur as they appear. The half-reactions for the reaction above would be:
$2 \ce{Ag^+} \left( aq \right) + 2 \ce{e^-} \rightarrow 2 \ce{Ag} \left( s \right) \nonumber$
$\ce{Cu} \left( s \right) \rightarrow \ce{Cu^{2+}} \left( aq \right) + 2 \ce{e^-} \nonumber$
Both oxidation and reduction must occur at the same time, so the electrons are donated and absorbed nearly simultaneously. The two half-reactions may be added together to represent a complete reaction. In order to add the half-reactions, the number of electrons donated and the number of electrons accepted must be equal.
Example 9.3.1
For each reaction, identify what is oxidized and what is reduced.
a) $\ce{Zn} + \ce{HCl} \rightarrow \ce{H_2} + \ce{ZnCl_2}$
b) $\ce{Fe} + \ce{O_2} \rightarrow \ce{Fe_2O_3}$
c) $\ce{NaBr} + \ce{I_2} \rightarrow \ce{NaI} + \ce{Br_2}$
Solution
In order to determine what is being oxidized and reduced, we must look at charges of atoms and see if they increase or decrease. (Remember, elements have no charge. In a compound, we can use our periodic table and what we learned in Chapter 4 to assign charges.) If the charge increases, the atom was oxidized. If the charge decreases, the atom was reduced.
a) This reaction written with charges is:
$\ce{Zn^0} + \ce{H^+Cl^-} \rightarrow \ce{H_2^0} + \ce{Zn^{2+}Cl_2^-} \nonumber$
$\ce{Zn}$ is oxidized because it went from $0$ to $+2$. $\ce{H}$ is reduced because it went from $+1$ to $0$. $\ce{Cl}$ was neither oxidized nor reduced.
b) This reaction written with charges is:
$\ce{Fe^0} + \ce{O_2^0} \rightarrow \ce{Fe_2^{3+}O_3^{2-}} \nonumber$
$\ce{Fe}$ is oxidized because it went from $0$ to $+3$. $\ce{O}$ is reduced because it went from $0$ to $-2$.
c) This reaction written with charges is:
$\ce{Na^+Br^-} + \ce{I_2^0} \rightarrow \ce{Na^+I^-} + \ce{Br_2^0} \nonumber$
$\ce{Br}$ is oxidized because it went from $-2$ to $0$. $\ce{I}$ is reduced because it went from $0$ to $-1$. $\ce{Na}$ was neither oxidized nor reduced as it stayed $+1$ the whole time.
Batteries
Batteries are devices that use chemical reactions to produce electrical energy. These reactions occur because the products contain less potential energy in their bonds than the reactants. The energy produced from excess potential energy not only allows the reaction to occur, but also often gives off energy to the surroundings. Some of these reactions can be physically arranged so that the energy given off is in the form of an electric current. These are the type of reactions that occur inside batteries. When a reaction is arranged to produce an electric current as it runs, the arrangement is called an electrochemical cell or a Galvanic Cell.
If a strip of copper is placed in a solution of silver nitrate, the following reaction takes place:
$2 \ce{Ag^+} \left( aq \right) + \ce{Cu} \left( s \right) \rightarrow 2 \ce{Ag} \left( s\right) + \ce{Cu^{2+}} \left( aq \right) \nonumber$
In this reaction, copper atoms are donating electrons to silver ions, so the silver ions are reduced to silver atoms and copper atoms are oxidized to copper (II) ions.
As the reaction occurs, an observer would see the solution slowly turn blue ($\ce{Cu^{2+}}$ ions are blue in solution) and a mass of solid silver atoms would build up on the copper strip.
The reaction we just described is not set up in such a way as to produce electricity. It is true that electrons are being transferred, but to produce electricity, we need electrons flowing through a wire so that we can use the energy of these electrons. This reaction, $2 \ce{Ag^+} \left( aq \right) + \ce{Cu} \left( s \right) \rightarrow 2 \ce{Ag} \left( s \right) + \ce{Cu^{2+}} \left( aq \right)$, is one that could be arranged to produce electricity. To do this, the two half-reactions (oxidation and reduction) must occur in separate compartments, and the separate compartments must remain in contact through an ionic solution and an external wire.
In this electrochemical cell, the copper metal must be separated from the silver ions to avoid a direct reaction. Each electrode in its solution could be represented by a half-reaction.
$\ce{Cu} \rightarrow \ce{Cu^{2+}} + 2 \ce{e^-} \nonumber$
$2 \ce{Ag^+} + 2 \ce{e^-} \rightarrow 2 \ce{Ag} \nonumber$
The wire connects the two halves of the reaction, allowing electrons to flow from one metal strip to the other. In this particular example, electrons will flow from the copper electrode (which is losing electrons) into the silver electrode (which is where the silver ions gain the electrons). The cell produces electricity through the wire and will continue to do so as long as there are sufficient reactants ($\ce{Ag^+}$ and $\ce{Cu}$) to continue the reaction.
Electrochemical cells will always have two electrodes—the pieces of metal where electrons are gained or lost. (In this example, the strip of $\ce{Ag}$ metal and $\ce{Cu}$ metal are the electrodes.) The electrode where reduction occurs and electrons are gained is called the cathode. The electrode where oxidation occurs and electrons are lost is called the anode. Electrons will always move from the anode to the cathode. The electrons that pass through the external circuit can do useful work such as lighting lights, running cell phones, and so forth.
If the light bulb is removed from the circuit with the electrochemical cell and replaced with a voltmeter, the voltmeter will measure the voltage (electrical potential energy per unit charge) of the combination of half-cells. The size of the voltage produced by a cell depends on the temperature, the metals used for electrodes, and the concentrations of the ions in the solutions. If you increase the concentration of the reactant ion (not the product ion), the reaction rate will increase and so will the voltage.
It may seem complicated to construct an electrochemical cell due to all of their complexities. Electrochemical cells are actually easy to make and sometimes even occur accidentally. If you take two coins of different denomination, push them part way through the peel of a whole lemon, and then connect the two coins with a wire, a small electric current will flow.
1. Primary Batteries (Dry Cell Batteries)
• Non-rechargeable.
• Electrolytes are present as a paste rather than as a liquid.
• General purpose battery used for flashlights, transistor radios, toys, etc.
• The basic dry cell battery consists of: zinc case as the anode (oxidation); a graphite rod as the cathode (reduction) surrounded by a moist paste of either MnO2, NH4Cl, and ZnCl2 (or, in alkaline dry cells, a KOH electrolytic paste).
• General reactions for the battery: manganese (IV) oxide-zinc cell (different batteries have different reactions—you don't need to remember any of these reactions).
cathode
2MnO2(s) + 2 NH4+(aq)+ 2e- → Mn2O3 (s) + H2O(l) + 2NH3 (aq)
anode
Zn(s) → Zn2+(aq) + 2e-
• Maximum voltage of 1.5V. By connecting several cells in a series, 90V can be achieved.
• Advantages of alkaline batteries: consistent voltage, increased capacity, longer shelf life, and reliable operation at temperatures as low as -40°C.
• Disadvantage: higher cost.
2. Secondary Batteries (Storage Batteries)
• Rechargeable
• An example: the lead-acid battery used in cars. The anode is a grid of lead-antimony or lead-calcium alloy packed with spongy lead; the cathode is lead (IV) oxide. The electrolyte is aqueous sulfuric acid. This battery consists of numerous small cells connected in parallels (anode to anode; cathode to cathode).
• General reaction:
cathode
PbO2(s) + 4H+(aq)+ SO42-(aq) + 2e- → PbSO4 (s) + 2H2O(l) + 2NH3 (aq)
anode
Pb(s) + SO42-(aq) → PbSO4 (s) + 2e-
• Secondary batteries are recharged by passing a current through the battery in the opposite direction. In a car battery, this occurs when the engine is running.
• Other examples include the nickel-iron alkaline battery, nickel-zinc battery, nickel-cadmium alkaline battery, silver-zinc battery, and silver-cadmium battery.
3. Fuel Cells
• Fuel cells are electrochemical cells that convert the energy of a redox combustion reaction directly into electrical energy. Fuel cells require a continuous supply of reactants and a constant removal of products.
• The cathode reactant is usually air or pure oxygen; the anode fuel is a gas such as hydrogen, methane, or propane. Carbon electrodes typically contain a catalyst. The electrolyte is typically KOH.
• General reaction:
cathode
O2(g) + 2H2O(l) + 4e- → 4OH-(aq)
anode
2H2 (g) + 4OH-(aq) → 4H2O(l) + 4e-
net 2H2 (g) + O2(g) → 2H2O(l)
• Advantages: no toxic waste products (water is the only product); very efficient energy conversion (70-80% efficient).
• Disadvantage: too expensive for large-scale use.
Summary
• A reaction in which there is a transfer of electrons is said to be an oxidation-reduction reaction, or a redox reaction.
• A substance that loses electrons is said to be oxidized, and the substance that gains electrons is said to be reduced.
• Redox reactions can be used in electrochemical cells to produce electricity.
• Electrochemical cells are composed of an anode and cathode in two separate solutions. These solutions are connected by a salt bridge and a conductive wire.
• An electric current consists of a flow of charged particles.
• The electrode where oxidation occurs is called the anode, and the electrode where reduction occurs is called the cathode.
Vocabulary
• Electrochemical cell - An arrangement of electrodes and ionic solutions in which a redox reaction is used to make electricity (a battery).
• Electrolysis - A chemical reaction brought about by an electric current.
• Electroplating - A process in which electrolysis is used as a means of coating an object with a layer of metal. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/16%3A_Oxidation_and_Reduction/16.06%3A_Batteries-_Using_Chemistry_to_Generate_Electricity.txt |
So far, we have discussed how electricity can be produced from chemical reactions in batteries. Some reactions will, instead, use electricity to get a reaction to occur. In these reactions, electrical energy is given to the reactants, causing them to react to form the products. These reactions have many uses. For example, electrolysis is a process that involves forcing electricity through a liquid or solution to cause a reaction to occur. Electrolysis reactions will not run unless energy is put into the system from outside. In the case of electrolysis reactions, the energy is provided by the battery. Think of electrolysis and electrolytic cells as the opposite of electrochemical cells:
Electrochemical Cells
Electrolytic Cells
Energy conversion
Chemical → Electrical
Electrical → Chemical
Spontaneous chemical reaction?
Yes
No
Value of E°
Positive
Negative
In an electrochemical cell, a spontaneous redox reaction is used to create an electric current; in an electrolytic cell, the reverse will occur—an electric current will be required in order to cause a non-spontaneous chemical reaction to occur. We will look at three examples of the electrolytic process, keeping our discussion on a very basic level—the electrolysis of molten sodium chloride, the electrolysis of water, and electroplating.
If electrodes connected to battery terminals are placed in liquid sodium chloride, the sodium ions will migrate toward the negative electrode and be reduced while the chloride ions migrate toward the positive electrode and are oxidized. The processes that occur at the electrodes can be represented by what are called half-equations.
Reduction occurs at the positive electrode:
$\ce{Na^+} + \ce{e^-} \rightarrow \ce{Na} \nonumber$
Oxidation occurs at the negative electrode:
$2 \ce{Cl^-} \rightarrow \ce{Cl_2} + 2 \ce{e^-} \nonumber$
The overall reaction for this reaction is:
$2 \ce{Na^+} + 2 \ce{Cl^-} \rightarrow 2 \ce{Na} + \ce{Cl_2} \nonumber$
With appropriate treatment from the battery, it is possible to get the metal being reduced in an electrolysis process to adhere strongly to the electrode. The use of electrolysis to coat one material with a layer of metal is called electroplating. Usually, electroplating is used to cover a cheap metal with a layer of more expensive and more attractive metal. Many people buy jewelry that is plated in gold. Sometimes, electroplating is used to get a surface metal that is a better conductor of electricity. When you wish to have the surface properties of gold (attractive, corrosion resistant, or good conductor), but you don't want to have the great cost of making an entire object out of solid gold, the answer may be to use cheap metal to make the object and then electroplate a thin layer of gold on the surface.
To silver plate an object like a spoon (plated silverware is less expensive than pure silver), the spoon is placed in the position of the cathode in an electrolysis set up with a solution of silver nitrate. When the current is turned on, the silver ions will migrate through the solution, touch the cathode (spoon) and adhere to it. With enough time and care, a layer of silver can be plated over the entire spoon. The anode for this operation would often be a large piece of silver, from which silver ions would be oxidized and these ions would enter the solution. This is a way of ensuring a steady supply of silver ions for the plating process.
• Half-reaction at the cathode:
$\ce{Ag^+} + \ce{e^-} \rightarrow \ce{Ag} \nonumber$
• Half-reaction at the anode:
$\ce{Ag} \rightarrow \ce{Ag^+} + \ce{e^-} \nonumber$
Some percentage of the gold and silver jewelry sold is electroplated. The connection points in electric switches are often gold plated to improve electrical conductivity, and most of the chromium pieces on automobiles are chromium plated.
Electrolysis of Molten Sodium Chloride
If we look at the latin roots of the word "electrolysis" we learn that it means, essentially, to "break apart" (lysis) using electricity. Our first example of an electrolytic cell will examine how an electric current can be used to break apart an ionic compound into its elements. The following equation represents the breaking apart of NaCl(l):
2NaCl(l) → 2Na(l) + Cl2 (g)
The half-reactions involved in this process are:
reduction 2Na+(l) + 2e- → Na(s)
-2.71 V
oxidation Cl-(l) → Cl2 (g) + 2 e-
-1.36V
net voltage required
- 4.07V
Notice that a negative voltage (-4.07V) results when we add up the half-reactions. This tells us that the overall reaction will NOT be spontaneous, and a minimum of 4.07 volts will be required for this reaction to occur.
As we shall see, our set-up will have a number of similarities to our electrochemical cells. We will need electrodes and an electrolyte to carry the electric current.
In our NaCl example, the electrodes will simply carry the current, but otherwise will not be directly involved in the reaction. The electrolyte will be the actual molten (melted) NaCl. The electrodes and electrolyte are both required to carry the electric current. Molten NaCl must be used because solid ionic compounds do not carry an electric charge.
Some key differences with an electrochemical cell set-up:
• The two half-reactions are not separated by a salt bridge.
• An electrochemical cell (or other source of electric current) will be required.
Other important items to note:
• The anode of the electrolytic cell is the site of oxidation and the cathode is the site of reduction, just as in an electrochemical cell.
• In an electrochemical cell, the anode is negative and cathode positive, but this is reversed in the electrolytic cell—the anode is positive and the cathode is negative.
Carefully study the diagram of our set up, taking special care to trace the path of the electrons. Unless electrons make a complete circuit, a reaction will not occur.
1. Electrons are "produced" in the battery at the anode, the site of oxidation.
2. The electrons leave the electrochemical cell through the external circuit.
3. These negative electrons create a negative electrode in the electrolytic cell, which attracts the positive Na+ ions in the electrolyte. Na+ ions combine with the free electrons and become reduced (2Na+ + 2e- → Na).
4. Meanwhile, the negative Cl- become attracted to the positive electrode of the electrolytic cell. At this electrode, chlorine is oxidized, releasing electrons (Cl-→ Cl2 + 2 e-).
5. These electrons travel through the external circuit, returning to the electrochemical cell.
Electrolysis of Water
Our second example of electrolysis and electrolytic cells involves the breakdown of water. We will find a situation very similar to the electrolysis of molten NaCl. The following equation represents the breaking apart of H2O(l):
2H2O(l) → 2H2(g) + O2 (g)
It may be more difficult to predict the half-reactions involved, but they are:
reduction 2H2O(l) + 2e- → H2 + 2 OH-
-0.83 V
oxidation 2H2O(l) → O2 + 4H+ + 4e-
-1.23V
(see note below for net equation)
net voltage required
- 2.06V
The set up will be very similar to our last example with some minor differences. Water does not carry a charge well, so an electrolyte is added to the water. Vinegar, a weak acid (acetic acid) may be used. To collect the hydrogen and oxygen gases produced, inverted test tubes are often added, as shown in our diagram below.
Again, take special care to trace the path of the electrons. Unless electrons make a complete circuit, a reaction will not occur.
2H2O(l) + 2e- → H2(g) + 2 OH-(aq)
2H2O(l) → O2 (g) + 4H+ (aq) + 4e-
1. Electrons are "produced" in the battery at the anode, the site of oxidation.
2. The electrons leave the electrochemical cell through the external circuit.
3. These negative electrons create a negative electrode in the electrolytic cell which causes the reduction of water.
Note that the area around this electrode will become basic as OH- ions are produced.
1. Meanwhile, the positive electrode water will undergo oxidation.
1. Electrons produced during this oxidation process will return to the electrochemical cell.
A note about the balanced equation for the electrolysis of water:
You may notice from the half reactions that adding up the equations doesn't initially give us our net equation of:
2H2O(l) → 2H2(g) + O2 (g)
Once you balance for electrons (multiply the reduction equation by 2), you'll find that the equations actually add up to:
6H2O(l) → 2H2(g) + O2 (g) + 4H+ (aq) + 4 OH-(aq)
The hydrogen and hydroxide ions will combine to form 4 moles of H2O(l). Finding our net amount of H2O(l) involved gives us our final equation:
2H2O(l) → 2H2(g) + O2 (g)
Summary
• Electrochemical cells are composed of an anode and cathode in two separate solutions. These solutions are connected by a salt bridge and a conductive wire.
• An electric current consists of a flow of charged particles.
• The electrode where oxidation occurs is called the anode and the electrode where reduction occurs is called the cathode.
• In electroplating, the object to be plated is made the cathode.
Vocabulary
• Electrochemical cell - An arrangement of electrodes and ionic solutions in which a redox reaction is used to make electricity (also known as a battery).
• Electrolysis - A chemical reaction brought about by an electric current.
• Electroplating - A process in which electrolysis is used as a means of coating an object with a layer of metal.
16.08: Corrosion- Undesirable Redox Reactions
Corrosion of metals is a serious economic problem. Corrosion occurs as a result of spontaneous electrochemical reaction as metal undergoes oxidation. For example, the rusting of iron begins with the oxidation of solid iron:
$\ce{ Fe(s) → Fe^{2+}(aq) + 2e^{-}}\nonumber$
The corresponding reduction reaction involves water:
$\ce{H2O(l) + 1/2 O2 (g) + 2e^{-} → 2OH^{-}(aq)} \nonumber$
The flaky brown solid that we call rust forms when Fe2+ undergoes additional oxidation to form Fe3+, then reacts with hydroxide ions to form iron (III) oxide, Fe2O3, and iron (III) hydroxide, Fe(OH)3.
The rate of corrosion can be affected by several factors. Some examples:
• Metals corrode faster when in contact with another metal. The Statue of Liberty, for example, has a skin made of copper, but is supported by iron ribs. Since iron is oxidized more readily than copper, it acts as the anode. Earlier repairs to strengthen the statue used iron bolts, which exacerbated the problem. More recent repairs have replaced the iron ribs with stainless steel alloys. Stainless steel resists corrosion.
• Salt water speeds up the corrosion process, because the ions in salt water form a salt bridge between the anodic and cathodic sites. Salt may be great for icy roads, but it is tough on cars.
There are a number of ways to slow down corrosion, if not prevent it:
• Prevent oxygen and water from contacting the metal. This can be accomplished by paint, grease, plastic, or other methods of covering the metal.
• Cathodic protection: pieces of zinc or magnesium metal may be bolted to the surface of iron. Because they are more readily oxidized than Fe, Zn and Mg become oxidized over time, thus sparing and protecting the iron. Propeller shafts of speedboats are often protected this way. Anode rods in water heaters also work this way (they are often called sacrificial anodes). Galvanized nails—nails coated with the more reactive zinc—provide yet another example.
• Metal alloys: an alloy is a mixture of metals, or a mixture of a nonmetal with a metal. An alloy such as stainless steel (chromium is added to steel—a mixture of iron and other elements such as carbon—to make stainless steel) is highly resistant to corrosion, but can be prohibitively expensive. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/16%3A_Oxidation_and_Reduction/16.07%3A_Electrolysis-_Using_Electricity_to_Do_Chemistry.txt |
In today’s society, the term radioactivity conjures up a variety of images:
• Nuclear power plants producing hydrocarbon-free energy, but with potentially deadly by-products that are difficult to store safely.
• Bombs with the capacity to use nuclear reactions that produce devastating explosions with horrible side effects on the earth as we know it, and on the surviving populations that inhabit it.
• Medical technology that utilizes nuclear chemistry to peer inside living things to detect disease, and the power to irradiate tissues to potentially cure these diseases.
• Fusion reactors that hold the promise of limitless energy with few toxic side products.
Radioactivity has a colorful history and clearly presents a variety of social and scientific dilemmas. In this chapter we will introduce the basic concepts of radioactivity, nuclear equations, and the processes involved in nuclear fission and nuclear fusion.
• 17.2: The Discovery of Radioactivity
Henri Becquerel, Marie Curie, and Pierre Curie shared the discovery of radioactivity.
• 17.3: Types of Radioactivity- Alpha, Beta, and Gamma Decay
The major types of radioactivity include alpha particles, beta particles, and gamma rays. Fission is a type of radioactivity in which large nuclei spontaneously break apart into smaller nuclei.
• 17.4: Detecting Radioactivity
Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters. Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube.
• 17.5: Natural Radioactivity and Half-Life
During natural radioactive decay, not all atoms of an element are instantaneously changed to atoms of another element. The decay process takes time and there is value in being able to express the rate at which a process occurs. A useful concept is half-life, which is the time required for half of the starting material to change or decay. Half-lives can be calculated from measurements on the change in mass of a nuclide and the time it takes to occur.
• 17.6: Radiocarbon Dating- Using Radioactivity to Measure the Age of Fossils and Other Artifacts
Radiocarbon dating (usually referred to simply as carbon-14 dating) is a radiometric dating method. It uses the naturally occurring radioisotope carbon-14 to estimate the age of carbon-bearing materials up to about 58,000 to 62,000 years old. Carbon has two stable, nonradioactive isotopes: carbon-12 and carbon-13. There are also trace amounts of the unstable radioisotope carbon-14 on Earth.
• 17.7: The Discovery of Fission and the Atomic Bomb
Nuclei that are larger than iron-56 may undergo nuclear reactions in which they break up into two or more smaller nuclei. These reactions are called fission reactions. When a neutron strikes a UU -235 nucleus and the nucleus captures a neutron, it undergoes fission, producing two lighter nuclei and three free neutrons. The production of the free neutrons makes it possible to have a self-sustaining fission process—a nuclear chain reaction.
• 17.8: Nuclear Power- Using Fission to Generate Electricity
Fission reactions can be used in the production of electricity if we control the rate at which the fission occurs. The great majority of all electrical generating systems (whether coal burning power plants, hydroelectric plants, or nuclear power plants) all follow a reasonably simple design.
• 17.9: Nuclear Fusion- The Power of the Sun
Fusion is a method for obtaining energy from nuclear reactions that lies in the fusing together of two light nuclei to form a heavier nucleus.
• 17.10: The Effects of Radiation on Life
We are constantly exposed to radiation from naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful type of radiation because it can ionize molecules or break chemical bonds, which damages the molecules and causes malfunctions in cell processes. Types of radiation differ in their ability to penetrate material and damage tissue.
• 17.11: Radioactivity in Medicine
The field of nuclear medicine has expanded greatly in the last twenty years. A great deal of the expansion has come in the area of imaging. This section will focus on nuclear medicine involving the types of nuclear radiation introduced in this chapter.
• Wikibooks
17: 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/Introductory_Chemistry/Introductory_Chemistry/17%3A_Radioactivity_and_Nuclear_Chemistry/17.02%3A_The_Discovery_of_Radioactivity.txt |
Learning Objectives
• Compare qualitatively the ionizing and penetration power of alpha particles $\left( \alpha \right)$, beta particles $\left( \beta \right)$, and gamma rays $\left( \gamma \right)$.
• Express the changes in the atomic number and mass number of a radioactive nuclei when an alpha, beta, or gamma particle is emitted.
• Write nuclear equations for alpha and beta decay reactions.
Many nuclei are radioactive; that is, they decompose by emitting particles and in doing so, become a different nucleus. In our studies up to this point, atoms of one element were unable to change into different elements. That is because in all other types of changes discussed, only the electrons were changing. In these changes, the nucleus, which contains the protons that dictate which element an atom is, is changing. All nuclei with 84 or more protons are radioactive, and elements with less than 84 protons have both stable and unstable isotopes. All of these elements can go through nuclear changes and turn into different elements.
In natural radioactive decay, three common emissions occur. When these emissions were originally observed, scientists were unable to identify them as some already known particles and so named them:
• alpha particles ($\alpha$)
• beta particles $\left( \beta \right)$
• gamma rays $\left( \gamma \right)$
These particles were named using the first three letters of the Greek alphabet. Some later time, alpha particles were identified as helium-4 nuclei, beta particles were identified as electrons, and gamma rays as a form of electromagnetic radiation like x-rays, except much higher in energy and even more dangerous to living systems.
The Ionizing and Penetration Power of Radiation
With all the radiation from natural and man-made sources, we should quite reasonably be concerned about how all the radiation might affect our health. The damage to living systems is done by radioactive emissions when the particles or rays strike tissue, cells, or molecules and alter them. These interactions can alter molecular structure and function; cells no longer carry out their proper function and molecules, such as DNA, no longer carry the appropriate information. Large amounts of radiation are very dangerous, even deadly. In most cases, radiation will damage a single (or very small number) of cells by breaking the cell wall or otherwise preventing a cell from reproducing.
The ability of radiation to damage molecules is analyzed in terms of what is called ionizing power. When a radiation particle interacts with atoms, the interaction can cause the atom to lose electrons and thus become ionized. The greater the likelihood that damage will occur by an interaction is the ionizing power of the radiation.
Much of the threat from radiation is involved with the ease or difficulty of protecting oneself from the particles. How thick of a wall do you need to hide behind to be safe? The ability of each type of radiation to pass through matter is expressed in terms of penetration power. The more material the radiation can pass through, the greater the penetration power and the more dangerous it is. In general, the greater mass present, the greater the ionizing power, and the lower the penetration power.
Comparing only the three common types of ionizing radiation, alpha particles have the greatest mass. Alpha particles have approximately four times the mass of a proton or neutron and approximately 8,000 times the mass of a beta particle. Because of the large mass of the alpha particle, it has the highest ionizing power and the greatest ability to damage tissue. That same large size of alpha particles, however, makes them less able to penetrate matter. They collide with molecules very quickly when striking matter, add two electrons, and become a harmless helium atom. Alpha particles have the least penetration power and can be stopped by a thick sheet of paper or even a layer of clothes. They are also stopped by the outer layer of dead skin on people. This may seem to remove the threat from alpha particles, but it is only from external sources. In a nuclear explosion or some sort of nuclear accident, where radioactive emitters are spread around in the environment, the emitters can be inhaled or taken in with food or water and once the alpha emitter is inside you, you have no protection at all.
Beta particles are much smaller than alpha particles and therefore, have much less ionizing power (less ability to damage tissue), but their small size gives them much greater penetration power. Most resources say that beta particles can be stopped by a one-quarter inch thick sheet of aluminum. Once again, however, the greatest danger occurs when the beta emitting source gets inside of you.
Gamma rays are not particles, but a high energy form of electromagnetic radiation (like x-rays, except more powerful). Gamma rays are energy that has no mass or charge. Gamma rays have tremendous penetration power and require several inches of dense material (like lead) to shield them. Gamma rays may pass all the way through a human body without striking anything. They are considered to have the least ionizing power and the greatest penetration power.
Table $1$ Comparison of Penetrating Power,Ionizing Power and Shielding of Alpha and Beta Particles, and Gamma Rays.
Particle Symbol Mass Penetrating Power Ionizing Power Shielding
Alpha $\alpha$ $4 \mathrm{amu}$ Very Low Very High Paper Skin
Beta $\beta$ $1 / 2000 \mathrm{amu}$ Intermediate Intermediate Aluminum
Gamma $\gamma$ 0 (energy only) Very High Very Low 2 inches lead
The safest amount of radiation to the human body is zero. It is impossible to completely avoid ionizing radiation, so the next best goal is to be exposed to as little as possible. The two best ways to minimize exposure are to limit time of exposure, and to increase distance from the source.
Alpha Decay
The nuclear disintegration process that emits alpha particles is called alpha decay. An example of a nucleus that undergoes alpha decay is uranium-238. The alpha decay of $\ce{U}$-238 is
$\ce{_{92}^{238}U} \rightarrow \ce{_2^4He} + \ce{_{90}^{234}Th} \label{alpha1}$
In this nuclear change, the uranium atom $\left( \ce{_{92}^{238}U} \right)$ transmuted into an atom of thorium $\left( \ce{_{90}^{234}Th} \right)$ and, in the process, gave off an alpha particle. Look at the symbol for the alpha particle: $\ce{_2^4He}$. Where does an alpha particle get this symbol? The bottom number in a nuclear symbol is the number of protons. That means that the alpha particle has two protons in it that were lost by the uranium atom. The two protons also have a charge of $+2$. The top number, 4, is the mass number or the total of the protons and neutrons in the particle. Because it has two protons, and a total of four protons and neutrons, alpha particles must also have two neutrons. Alpha particles always have this same composition: two protons and two neutrons.
Another alpha particle producer is thorium-230.
$\ce{_{90}^{230}Th} \rightarrow \ce{_2^4He} + \ce{_{88}^{226}Ra} \label{alpha2}$
These types of equations are called nuclear equations and are similar to the chemical equivalent discussed through the previous chapters.
Beta Decay
Another common decay process is beta particle emission, or beta decay. A beta particle is simply a high energy electron that is emitted from the nucleus. It may occur to you that we have a logically difficult situation here. Nuclei do not contain electrons and yet during beta decay, an electron is emitted from a nucleus. At the same time that the electron is being ejected from the nucleus, a neutron is becoming a proton. It is tempting to picture this as a neutron breaking into two pieces with the pieces being a proton and an electron. That would be convenient for simplicity, but unfortunately that is not what happens (more on this subject will be explained at the end of this section). For convenience, we will treat beta decay as a neutron splitting into a proton and an electron. The proton stays in the nucleus, increasing the atomic number of the atom by one. The electron is ejected from the nucleus and is the particle of radiation called beta.
To insert an electron into a nuclear equation and have the numbers add up properly, an atomic number and a mass number had to be assigned to an electron. The mass number assigned to an electron is zero (0), which is reasonable since the mass number is the number of protons plus neutrons, and an electron contains no protons and no neutrons. The atomic number assigned to an electron is negative one (-1), because that allows a nuclear equation containing an electron to balance atomic numbers. Therefore, the nuclear symbol representing an electron (beta particle) is
$\ce{_{-1}^0e}$ or $\ce{_{-1}^0\beta} \label{beta1}$
Thorium-234 is a nucleus that undergoes beta decay. Here is the nuclear equation for this beta decay:
$\ce{_{90}^{234}Th} \rightarrow \ce{_{-1}^0e} + \ce{_{91}^{234}Pa} \label{beta2}$
Gamma Radiation
Frequently, gamma ray production accompanies nuclear reactions of all types. In the alpha decay of $\ce{U}$-238, two gamma rays of different energies are emitted in addition to the alpha particle.
$\ce{_{92}^{238}U} \rightarrow \ce{_2^4He} + \ce{_{90}^{234}Th} + 2 \ce{_0^0\gamma} \nonumber$
Virtually all of the nuclear reactions in this chapter also emit gamma rays, but for simplicity the gamma rays are generally not shown. Nuclear reactions produce a great deal more energy than chemical reactions. Chemical reactions release the difference between the chemical bond energy of the reactants and products, and the energies released have an order of magnitude of $1 \times 10^3 \: \text{kJ/mol}$. Nuclear reactions release some of the binding energy and may convert tiny amounts of matter into energy. The energy released in a nuclear reaction has an order of magnitude of $1 \times 10^{18} \: \text{kJ/mol}$. That means that nuclear changes involve almost one million times more energy per atom than chemical changes!
Note
Virtually all of the nuclear reactions in this chapter also emit gamma rays, but for simplicity the gamma rays are generally not shown.
The essential features of each reaction are shown in Figure 17.3.2
Figure 17.3.2: Three most common modes of nuclear decay.
"Nuclear Accounting"
When writing nuclear equations, there are some general rules that will help you:
• The sum of the mass numbers (top numbers) on the reactant side equal the sum of the mass numbers on the product side.
• The atomic numbers (bottom numbers) on the two sides of the reaction will also be equal.
In the alpha decay of $\ce{^{238}U}$ (Equation $\ref{alpha1}$), both atomic and mass numbers are conserved:
• mass number: $238 = 4 + 234$
• atomic number: $92 = 2 + 90$
Confirm that this equation is correctly balanced by adding up the reactants' and products' atomic and mass numbers. Also, note that because this was an alpha reaction, one of the products is the alpha particle, $\ce{_2^4He}$.
Note that both the mass numbers and the atomic numbers add up properly for the beta decay of thorium-234 (Equation $\ref{beta2}$):
• mass number: $234 = 0 + 234$
• atomic number: $90 = -1 + 91$
The mass numbers of the original nucleus and the new nucleus are the same because a neutron has been lost, but a proton has been gained, and so the sum of protons plus neutrons remains the same. The atomic number in the process has been increased by one since the new nucleus has one more proton than the original nucleus. In this beta decay, a thorium-234 nucleus has one more proton than the original nucleus. In this beta decay, a thorium-234 nucleus has become a protactinium-234 nucleus. Protactinium-234 is also a beta emitter and produces uranium-234.
$\ce{_{91}^{234}Pa} \rightarrow \ce{_{-1}^0e} + \ce{_{92}^{234}U} \label{nuke1}$
Once again, the atomic number increases by one and the mass number remains the same; this confirms that the equation is correctly balanced.
What About Balancing Charge?
Both alpha and beta particles are charged, but nuclear reactions in Equations $\ref{alpha1}$, $\ref{beta2}$, and most of the other nuclear reactions above, are not balanced with respect to charge, as discussed when balancing redox reactions. When studying nuclear reactions in general, there is typically little information or concern about the chemical state of the radioactive isotopes, because the electrons from the electron cloud are not directly involved in the nuclear reaction (in contrast to chemical reactions).
So it is acceptable to ignore charge in balancing nuclear reactions, and concentrate on balancing mass and atomic numbers only.
Example $1$
Complete the following nuclear reaction by filling in the missing particle.
$\ce{_{86}^{210}Rn} \rightarrow \ce{_2^4He} + ? \nonumber$
Solution
This reaction is an alpha decay. We can solve this problem one of two ways:
Solution 1: When an atom gives off an alpha particle, its atomic number drops by 2 and its mass number drops by 4, leaving: $\ce{_{84}^{206}Po}$. We know the symbol is $\ce{Po}$, for polonium, because this is the element with 84 protons on the periodic table.
Solution 2: Remember that the mass numbers on each side must total up to the same amount. The same is true of the atomic numbers.
• Mass numbers: $210 = 4 + ?$
• Atomic numbers: $86 = 2 + ?$
We are left with $\ce{_{84}^{206}Po}$.
Example $2$
Write each of the following nuclear reactions.
a) Carbon-14, used in carbon dating, decays by beta emission.
b) Uranium-238 decays by alpha emission.
Solution
a) Beta particles have the symbol $\ce{_{-1}^0e}$. Emitting a beta particle causes the atomic number to increase by 1 and the mass number to not change. We get atomic numbers and symbols for elements using our periodic table. We are left with the following reaction:
$\ce{_6^{14}C} \rightarrow \ce{_{-1}^0e} + \ce{_7^{14}N} \nonumber$
b) Alpha particles have the symbol $\ce{_2^4He}$. Emitting an alpha particle causes the atomic number to decrease by 2 and the mass number to decrease by 4. We are left with:
$\ce{_{92}^{238}U} \rightarrow \ce{_2^4He} + \ce{_{90}^{234}Th} \nonumber$
Decay Series
The decay of a radioactive nucleus is a move toward becoming stable. Often, a radioactive nucleus cannot reach a stable state through a single decay. In such cases, a series of decays will occur until a stable nucleus is formed. The decay of $\ce{U}$-238 is an example of this. The $\ce{U}$-238 decay series starts with $\ce{U}$-238 and goes through fourteen separate decays to finally reach a stable nucleus, $\ce{Pb}$-206 (Figure 17.3.3). There are similar decay series for $\ce{U}$-235 and $\ce{Th}$-232. The $\ce{U}$-235 series ends with $\ce{Pb}$-207 and the $\ce{Th}$-232 series ends with $\ce{Pb}$-208.
Several of the radioactive nuclei that are found in nature are present there because they are produced in one of the radioactive decay series. For example, there may have been radon on the earth at the time of its formation, but that original radon would have all decayed by this time. The radon that is present now is present because it was formed in a decay series (mostly by U-238).
Summary
A nuclear reaction is one that changes the structure of the nucleus of an atom. The atomic numbers and mass numbers in a nuclear equation must be balanced. Protons and neutrons are made up of quarks. The two most common modes of natural radioactivity are alpha decay and beta decay. Most nuclear reactions emit energy in the form of gamma rays.
Vocabulary
• Alpha decay - A common mode of radioactive decay in which a nucleus emits an alpha particle (a helium-4 nucleus).
• Beta decay - A common mode of radioactive decay in which a nucleus emits beta particles. The daughter nucleus will have a higher atomic number than the original nucleus.
• Quark - Particles that form one of the two basic constituents of matter. Various species of quarks combine in specific ways to form protons and neutrons, in each case taking exactly three quarks to make the composite particle. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/17%3A_Radioactivity_and_Nuclear_Chemistry/17.03%3A_Types_of_Radioactivity-_Alpha_Beta_and_Gamma_Decay.txt |
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.
17.05: Natural Radioactivity and Half-Life
Learning Objectives
• Describe what is meant by the term half-life and what factors affect half-life.
• Calculate the amount of radioactive material that will remain after an integral number of half-lives.
• Find the half-life of an isotope given graphical or other data.
• Describe how carbon-14 is used to determine the age of carbon-containing objects.
Rate of Radioactive Decay
During natural radioactive decay, not all atoms of an element are instantaneously changed to atoms of another element. The decay process takes time and there is value in being able to express the rate at which a process occurs. A useful concept is half-life, which is the time required for half of the starting material to change or decay. Half-lives can be calculated from measurements on the change in mass of a nuclide and the time it takes to occur. The only thing we know is that in the time of that substance's half-life, half of the original nuclei will disintegrate. Although chemical changes are sped up or slowed down by changing factors such as temperature and concentration, these factors have no effect on half-life. Each radioactive isotope will have its own unique half-life that is independent of any of these factors.
The half-lives of many radioactive isotopes have been determined; they have been found to range from extremely long half-lives of 10 billion years, to extremely short half-lives of fractions of a second.
Table 17.5.1: Table of Selected Half-lives
Element Mass Number (A) Half-life Element Mass Number (A) Half Life
Uranium 238 4.5 Billion years Californium 251 800 years
Neptunium 240 1 hour Nobelium 254 3 seconds
Plutonium 243 5 hours Carbon 14 5730 years
Americium 245 25 minutes Carbon 16 740 milliseconds
The quantity of radioactive nuclei at any given time will decrease to half as much in one half-life. For example, if there are $100 \: \text{g}$ of $\ce{Cf}$-251 in a sample at some time, after 800 years, there will be $50 \: \text{g}$ of $\ce{Cf}$-251 remaining; after another 800 years (1600 years total), there will only be $25 \: \text{g}$ remaining.
Remember, the half-life is the time it takes for half of your sample—no matter how much you have—to remain. Each half-life will follow the same general pattern as $\ce{Cf}$-251. The only difference is the length of time it takes for half of a sample to decay.
Example $3$
What is the half-life of an isotope that produces the following graph of decay over time?
Solution
We know that the half-life is the time it takes for half of a sample to change. How long did it take for half of our isotope to change? It took approximately 200 years for $100\%$ of our sample to leave only $50\%$ (half of the original amount) remaining. The half-life is 200 years.
*Notice that after another 200 years (400 years total), $25\%$ remains (half of $50\%$).
Look carefully at the graph in the previous example. All types of radioactive decay make a graph of the same general shape. The only difference is the scale and units of the $x$-axis, as the half-life time will be different.
Example $2$
If there are 60 grams of $\ce{Np}$-240 present, how much $\ce{Np}$-240 will remain after 4 hours? ($\ce{Np}$-240 has a half-life of 1 hour.)
Solution
We create a table based on $\ce{Np}$-240's half-life of 1 hour.
After 4 hours, only $3.75 \: \text{g}$ of our original $60 \: \text{g}$ sample would remain of the radioactive isotope $\ce{Np}$-240.
Example $3$
A sample of $\ce{Ac}$-225 originally contained 80 grams, and after 50 days only 2.55 grams of the original $\ce{Ac}$-225 remain. What is the half life of $\ce{Ac}$-225?
Solution
We will tackle this problem similarly to the last problem. The difference is that we are looking for the half-life time. Let's set up a similar table, though:
We know that 50 days is the same as 5 half-lives. Therefore, 1 half-life is 10 days. The half-life of $\ce{Ac}$-225 is 10 days.
Radioactive Dating
An ingenious application of half-life studies established a new science of determining ages of materials by half-life calculations. For geological dating, the decay of $\ce{U}$-238 can be used. The half-life of $\ce{U}$-238 is $4.5 \times 10^9$ years. The end product of the decay of $\ce{U}$-238 is $\ce{Pb}$-206. After one half-life, a 1.00 gram sample of uranium will have decayed to 0.50 grams of $\ce{U}$-238 and 0.43 grams of $\ce{Pb}$-206. By comparing the amount of $\ce{U}$-238 to the amount of $\ce{Pb}$-206 in a sample of uranium mineral, the age of the mineral can be estimated. Present day estimates for the age of the Earth's crust from this method is at 4 billion years.
Organic material (material made from things that were once living, such as paper and fabric) is radioactively dated using the long-lived nuclide of carbon, carbon-14. This method of determining the age of organic material (or once living materials) was given the name radiocarbon dating. The carbon dioxide consumed by living systems contains a certain concentration of $\ce{^{14}CO_2}$. When an organism dies, the acquisition of carbon-14 stops, but the decay of the $\ce{C}$-14 in the body continues. As time goes by, the ratio of $\ce{C}$-14 to $\ce{C}$-12 decreases at a rate determined by the half-life of $\ce{C}$-14. Using half-life equations, the time since the organism died can be calculated. These procedures have been used to determine the age of organic artifacts and determine, for instance, whether art works are real or fake.
Summary and Vocabulary
The half-life of an isotope is used to describe the rate at which the isotope will decay and give off radiation. Using the half-life, it is possible to predict the amount of radioactive material that will remain after a given amount of time. $\ce{C}$-14 dating procedures have been used to determine the age of organic artifacts. Its half-life is approximately 5700 years.
• Background radiation - Radiation that comes from environmental sources including the earth's crust, the atmosphere, cosmic rays, and radioisotopes. These natural sources of radiation account for the largest amount of radiation received by most people.
• Half-life - The time interval required for a quantity of (radioactive) material to decay to half of its original value. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/17%3A_Radioactivity_and_Nuclear_Chemistry/17.04%3A_Detecting_Radioactivity.txt |
Learning Objectives
• Identify the age of materials that can be approximately determined using Radiocarbon dating.
When we speak of the element Carbon, we most often refer to the most naturally abundant stable isotope 12C. Although 12C is definitely essential to life, its unstable sister isotope 14C has become of extreme importance to the science world. Radiocarbon dating is the process of determining the age of a sample by examining the amount of 14C remaining against its known half-life, 5,730 years. The reason this process works is because when organisms are alive, they are constantly replenishing their 14C supply through respiration, providing them with a constant amount of the isotope. However, when an organism ceases to exist, it no longer takes in carbon from its environment and the unstable 14C isotope begins to decay. From this science, we are able to approximate the date at which the organism lived on Earth. Radiocarbon dating is used in many fields to learn information about the past conditions of organisms and the environments present on Earth.
The Carbon-14 Cycle
Radiocarbon dating (usually referred to simply as carbon-14 dating) is a radiometric dating method. It uses the naturally occurring radioisotope carbon-14 (14C) to estimate the age of carbon-bearing materials up to about 58,000 to 62,000 years old. Carbon has two stable, nonradioactive isotopes: carbon-12 (12C) and carbon-13 (13C). There are also trace amounts of the unstable radioisotope carbon-14 (14C) on Earth. Carbon-14 has a relatively short half-life of 5,730 years, meaning that the fraction of carbon-14 in a sample is halved over the course of 5,730 years due to radioactive decay to nitrogen-14. The carbon-14 isotope would vanish from Earth's atmosphere in less than a million years were it not for the constant influx of cosmic rays interacting with molecules of nitrogen (N2) and single nitrogen atoms (N) in the stratosphere. Both processes of formation and decay of carbon-14 are shown in Figure 1.
When plants fix atmospheric carbon dioxide (CO2) into organic compounds during photosynthesis, the resulting fraction of the isotope 14C in the plant tissue will match the fraction of the isotope in the atmosphere (and biosphere since they are coupled). After a plant dies, the incorporation of all carbon isotopes, including 14C, stops and the concentration of 14C declines due to the radioactive decay of 14C following.
$\ce{ ^{14}C -> ^{14}N + e^-} + \mu_e \label{E2}$
This follows first-order kinetics:
$N_t= N_o e^{-kt} \label{E3}$
where
• $N_0$ is the number of atoms of the isotope in the original sample (at time t = 0, when the organism from which the sample is derived was de-coupled from the biosphere).
• $N_t$ is the number of atoms left after time $t$.
• $k$ is the rate constant for the radioactive decay.
The half-life of a radioactive isotope (usually denoted by $t_{1/2}$) is a more familiar concept than $k$ for radioactivity, so although Equation $\ref{E3}$ is expressed in terms of $k$, it is more usual to quote the value of $t_{1/2}$. The currently accepted value for the half-life of 14C is 5,730 years. This means that after 5,730 years, only half of the initial 14C will remain; a quarter will remain after 11,460 years; an eighth after 17,190 years; and so on.
The equation relating rate constant to half-life for first order kinetics is
$k = \dfrac{\ln 2}{ t_{1/2} } \label{E4}$
so the rate constant is then
$k = \dfrac{\ln 2}{5.73 \times 10^3} = 1.21 \times 10^{-4} \text{year}^{-1} \label{E5}$
and Equation $\ref{E2}$ can be rewritten as
$N_t= N_o e^{-\ln 2 \;t/t_{1/2}} \label{E6}$
or
$t = \left(\dfrac{\ln \dfrac{N_o}{N_t}}{\ln 2} \right) t_{1/2} = 8267 \ln \dfrac{N_o}{N_t} = 19035 \log_{10} \dfrac{N_o}{N_t} \;\;\; (\text{in years}) \label{E7}$
The sample is assumed to have originally had the same 14C/12C ratio as the ratio in the atmosphere, and since the size of the sample is known, the total number of atoms in the sample can be calculated, yielding $N_0$, the number of 14C atoms in the original sample. Measurement of N, the number of 14C atoms currently in the sample, allows the calculation of $t$, the age of the sample, using the Equation $\ref{E7}$.
Note
Deriving Equation $\ref{E7}$ assumes that the level of 14C in the atmosphere has remained constant over time. However, the level of 14C in the atmosphere has varied significantly, so time estimated by Equation $\ref{E7}$ must be corrected by using data from other sources.
Example 1: Dead Sea Scrolls
In 1947, samples of the Dead Sea Scrolls were analyzed by carbon dating. It was found that the carbon-14 present had an activity (rate of decay) of d/min.g (where d = disintegration). In contrast, living material exhibit an activity of 14 d/min.g. Thus, using Equation $\ref{E3}$,
$\ln \dfrac{14}{11} = (1.21 \times 10^{-4}) t \nonumber$
Thus,
$t= \dfrac{\ln 1.272}{1.21 \times 10^{-4}} = 2 \times 10^3 \text{years} \nonumber$
From the measurement performed in 1947, the Dead Sea Scrolls were determined to be 2000 years old, giving them a date of 53 BC, and confirming their authenticity. This discovery is in contrast to the carbon dating results for the Turin Shroud that was supposed to have wrapped Jesus’ body. Carbon dating has shown that the cloth was made between 1260 and 1390 AD. Thus, the Turin Shroud was made over a thousand years after the death of Jesus.
Describes radioactive half-life and how to do some simple calculations using half-life.
History
The technique of radiocarbon dating was developed by Willard Libby and his colleagues at the University of Chicago in 1949. Emilio Segrè asserted in his autobiography that Enrico Fermi suggested the concept to Libby at a seminar in Chicago that year. Libby estimated that the steady-state radioactivity concentration of exchangeable carbon-14 would be about 14 disintegrations per minute (dpm) per gram. In 1960, Libby was awarded the Nobel Prize in chemistry for this work. He demonstrated the accuracy of radiocarbon dating by accurately estimating the age of wood from a series of samples for which the age was known, including an ancient Egyptian royal barge dating from 1850 BCE. Before Radiocarbon dating was discovered, someone had to find the existence of the 14C isotope. In 1940, Martin Kamen and Sam Ruben at the University of California, Berkeley Radiation Laboratory did just that. They found a form, an isotope, of Carbon that contained 8 neutrons and 6 protons. Using this finding, Willard Libby and his team at the University of Chicago proposed that Carbon-14 was unstable and underwent a total of 14 disintegrations per minute per gram. Using this hypothesis, the initial half-life he determined was 5568, give or take 30 years. The accuracy of this proposal was proven by dating a piece of wood from an Ancient Egyptian barge, the age of which was already known. From that point on, scientists have used these techniques to examine fossils, rocks, and ocean currents; as well as to determine age and event timing. Throughout the years, measurement tools have become more technologically advanced, allowing researchers to be more precise. We now use what is known as the Cambridge half-life of 5730+/- 40 years for Carbon-14. Although it may be seen as outdated, many labs still use Libby's half-life in order to stay consistent in publications and calculations within the laboratory. From the discovery of Carbon-14 to radiocarbon dating of fossils, we can see what an essential role Carbon has played and continues to play in our lives today.
Summary
The entire process of Radiocarbon dating depends on the decay of carbon-14. This process begins when an organism is no longer able to exchange Carbon with its environment. Carbon-14 is first formed when cosmic rays in the atmosphere allow for excess neutrons to be produced, which then react with Nitrogen to produce a constantly replenishing supply of carbon-14 to exchange with organisms.
• Carbon-14 dating can be used to estimate the age of carbon-bearing materials up to about 58,000 to 62,000 years old.
• The carbon-14 isotope would vanish from Earth's atmosphere in less than a million years were it not for the constant influx of cosmic rays interacting with atmospheric nitrogen.
• One of the most frequent uses of radiocarbon dating is to estimate the age of organic remains from archeological sites.
Problems
1. If, when a hippopotamus lived, there was a total of 25 grams of Carbon-14, how many grams will remain 5730 years after he is laid to rest? 12.5 grams, because one half-life has occurred.
2. How many grams of Carbon-14 will be present in the hippopotamus' remains after three half-lives have passed? 3.125 grams of Carbon-14 will remain after three half-lives.
17.07: The Discovery of Fission and the Atomic Bomb
Learning Objectives
• Define and give examples of fission and fusion.
• Classify nuclear reactions as fission or fusion.
• List some medical uses of nuclear energy.
Nuclei that are larger than iron-56 may undergo nuclear reactions in which they break up into two or more smaller nuclei. These reactions are called fission reactions. Conversely, nuclei that are smaller than iron-56 become larger nuclei in order to be more stable. These nuclei undergo a nuclear reaction in which smaller nuclei join together to form a larger nucleus. Such nuclear reactions are called fusion reactions.
Fission and Chain Reactions
In both fission and fusion, large amounts of energy are given off in the form of heat, light, and gamma radiation. Nuclear fission was discovered in the late 1930s when $\ce{U}$-235 nuclides were bombarded with neutrons and were observed to split into two smaller-mass nuclei.
$\ce{_0^1n} + \ce{_{92}^{235}U} \rightarrow \ce{_{56}^{141}Ba} + \ce{_{36}^{92}Kr} + 3 \ce{_0^1n} \nonumber$
The products shown are only one of many sets of products from the disintegration of a $\ce{U}$-235 nucleus. Over 35 different elements have been observed in the fission products of $\ce{U}$-235.
When a neutron strikes a $\ce{U}$-235 nucleus and the nucleus captures a neutron, it undergoes fission, producing two lighter nuclei and three free neutrons. The production of the free neutrons makes it possible to have a self-sustaining fission process—a nuclear chain reaction. If at least one of the neutrons goes on to cause another $\ce{U}$-235 disintegration, the fission will be self-sustaining.
Nuclear Weapons
It is unfortunate that when the topics of radioactivity and nuclear energy come up, most thoughts probably go to weapons of war. The second thought might be about the possibility of nuclear energy contributing to the solution of the energy crisis. Nuclear energy, however, has many applications beyond bombs and the generation of electricity. Radioactivity has huge applications in scientific research, several fields of medicine both in terms of imaging and in terms of treatment, industrial processes, some very useful appliances, and even in agriculture.
Summary and Vocabulary
Naturally radioactive elements exist in the earth and are either alpha or beta emitters. Artificial transmutation of elements can be accomplished by bombarding the nuclei of some elements with alpha or subatomic particles. Nuclear radiation also has many medical uses.
• Chain reaction - A multi-stage nuclear reaction that sustains itself in a series of fissions, in which the release of neutrons from the splitting of one atom leads to the splitting of others.
• Critical mass - The smallest mass of a fissionable material that will sustain a nuclear chain reaction at a constant level.
• Fission - A nuclear reaction in which a heavy nucleus splits into two or more smaller fragments, releasing large amounts of energy.
• Fusion - A nuclear reaction in which nuclei combine to form more massive nuclei with the simultaneous release of energy.
• Control rods - Rods made of chemical elements capable of absorbing many neutrons, that are used to control the rate of a fission chain reaction in a nuclear reactor. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/17%3A_Radioactivity_and_Nuclear_Chemistry/17.06%3A_Radiocarbon_Dating-_Using_Radioactivity_to_Measure_the_Age_of_Fossils_and_Other_Artifacts.txt |
Fission Reactors
Fission reactions can be used in the production of electricity if we control the rate at which the fission occurs. The great majority of all electrical generating systems (whether coal burning power plants, hydroelectric plants, or nuclear power plants) is that they follow a reasonably simple design. The electricity is produced by spinning a coil of wire inside a magnetic field. When a fluid (air, steam, water) is forced through the pipe, it spins the fan blades, which in turn spin the axle. To generate electricity, the axle of a turbine is attached to the loop of wire in a generator. When a fluid is forced through the turbine, the fan blades turn, the turbine axle turns, and the loop of wire inside the generator turns—thus generating electricity.
The essential difference in various kinds of electrical generating systems is the method used to spin the turbine. For a wind generator, the turbine is a windmill. In a geothermal generator, steam from a geyser is forced through the turbine. In hydroelectric generating plants, water falling over a dam passes through the turbine and spins it. In fossil fuel (coal, oil, natural gas) generating plants, the fossil fuel is burned and the heat is used to boil water into steam, and then the steam passes through the turbine to make it spin. In a fission reactor generating plant, a fission reaction is used to boil the water into steam, and the steam passes through the turbine to make it spin. Once the steam is generated by the fission reaction, a nuclear power plant is essentially the same as a fossil fuel plant.
Naturally occurring uranium is composed almost totally of two uranium isotopes. It contains more than \(99\%\) uranium-238 and less than \(1\%\) uranium-235. It is the uranium-235, however, that is fissionable (will undergo fission). In order for uranium to be used as fuel in a fission reactor, the percentage of uranium-235 must be increased, usually to about \(3\%\). (Uranium in which the \(\ce{U}\)-235 content is more than \(1\%\) is called enriched uranium.)
Once the supply of \(\ce{U}\)-235 is acquired, it is placed in a series of long cylindrical tubes called fuel rods. These fuel cylinders are bundled together with control rods made of neutron-absorbing material. The amount of \(\ce{U}\)-235 in all the fuel rods taken together is adequate to carry on a chain reaction, but is less than the critical mass. (In the United States, all public nuclear power plants contain less than a critical mass of \(\ce{U}\)-235 and therefore, could never produce a nuclear explosion.) The amount of heat generated by the chain reaction is controlled by the rate at which the nuclear reaction occurs. The rate of the nuclear reaction is dependent on how many neutrons are emitted by one \(\ce{U}\)-235 nuclear disintegration and strike a new \(\ce{U}\)-235 nucleus to cause another disintegration. The purpose of the control rods is to absorb some of the neutrons and thus stop them from causing further disintegration. The control rods can be raised or lowered into the fuel rod bundle. When the control rods are lowered all the way into the fuel rod bundle, they absorb so many neutrons that the chain reaction essentially stops. When more heat is desired, the control rods are raised so that they catch fewer neutrons, the chain reaction speeds up, and more heat is generated. The control rods are operated in a fail-safe system, so that power is necessary to hold them up; during a power failure, gravity will pull the control rods down into the shut off position.
\(\ce{U}\)-235 nuclei can capture neutrons and disintegrate more efficiently if the neutrons are moving slower than the speed at which they are released. Fission reactors use a moderator surrounding the fuel rods to slow down the neutrons. Water is not only a good coolant, but also a good moderator. A common type of fission reactor has the fuel core submerged in a huge pool of water.
You can follow the operation of an electricity-generating fission reactor in the figure below. The reactor core is submerged in a pool of water. The heat from the fission reaction heats the water and the water is pumped into a heat exchanger container where the heated water boils the water in the heat exchanger. The steam from there is forced through a turbine which spins a generator and produces electricity. After the water passes through the turbine, it is condensed back to liquid water and pumped back to the heat exchanger.
In the United States, heavy opposition to the use of nuclear energy was mounted in the late 1960's and early 1970's. Every environmentalist organization in the US opposed the use of nuclear energy; the constant pressure from environmentalist groups caused an increase of public fear and, therefore, opposition to nuclear energy. This is not true today; at least one environmental leader has published a paper in favor of nuclear-powered electricity generation.
In 1979, a reactor core meltdown at Pennsylvania's Three Mile Island nuclear power plant reminded the entire country of the dangers of nuclear radiation. The concrete containment structure (six feet thick walls of reinforced concrete), however, did what it was designed to do—prevent radiation from escaping into the environment. Although the reactor was shut down for years, there were no injuries or deaths among nuclear workers or nearby residents. Three Mile Island was the only serious accident in the entire history of 103 civilian power plants operating for 40 years in the United States. There has never been a single injury or death due to radiation in any public nuclear power plant in the U.S. The accident at Three Mile Island did, however, frighten the public so that there has not been a nuclear power plant built in the U.S. since the accident.
The 103 nuclear power plants operating in the U.S. deliver approximately \(19.4\%\) of American electricity with zero greenhouse gas emission. There are 600 coal-burning electric plants in the US delivering \(48.5\%\) of American electricity and producing 2 billion tons of \(\ce{CO_2}\) annually, accounting for \(40\%\) of U.S. \(\ce{CO_2}\) emissions and \(10\%\) of global emissions. These coal burning plants also produce \(64\%\) of the sulfur dioxide emissions, \(26\%\) of the nitrous oxide emissions, and \(33\%\) of mercury emissions.
Fusion
Nuclear reactions, in which two or more lighter-mass nuclei join together to form a single nucleus, are called fusion reactions or nuclear fusions. Of particular interest are fusion reactions in which hydrogen nuclei combine to form helium. Hydrogen nuclei are positively charged and repel each other. The closer the particles come, the greater the force of repulsion. In order for fusion reactions to occur, the hydrogen nuclei must have extremely high kinetic energies, so that the velocities can overcome the forces of repulsion. These kinetic energies only occur at extreme temperatures such as those that occur in the cores of the sun and other stars. Nuclear fusion is the power source for the stars, where the necessary temperature to ignite the fusion reaction is provided by massive gravitational pressure. In stars more massive than our sun, fusion reactions involving carbon and nitrogen are possible. These reactions produce more energy than hydrogen fusion reactions.
Intensive research is now being conducted to develop fusion reactors for electricity generation. The two major problems slowing the development are: finding a practical means for generating the intense temperature needed, and developing a container that will not melt under the conditions of a fusion reaction. Electricity-producing fusion reactors are still a distant dream.
Summary
• Nuclear fission refers to the splitting of atomic nuclei.
• Nuclear fusion refers to the joining together of two or more smaller nuclei to form a single nucleus.
• The fission of \(\ce{U}\)-235 or \(\ce{Pu}\)-239 is used in nuclear reactors.
17.09: Nuclear Fusion- The Power of the Sun
In addition to fission, a second possible method for obtaining energy from nuclear reactions lies in the fusing together of two light nuclei to form a heavier nucleus. As we see when discussing Figure 1 from Mass-Energy Relationships, such a process results in nucleons which are more firmly bonded to each other, and hence lower in potential energy. This is particularly true if ${}_{\text{2}}^{\text{4}}\text{He}$ is formed, because this nucleus is very stable. Such a reaction occurs between the nuclei of the two heavy isotopes of hydrogen, deuterium and tritium:
${}_{\text{1}}^{\text{2}}\text{D + }{}_{\text{1}}^{\text{3}}\text{T }\to \text{ }{}_{\text{2}}^{\text{4}}\text{He + }{}_{\text{0}}^{\text{1}}n \label{1}$
For this reaction, Δm = – 0.018 88 g mol–1 so that ΔHm = – 1700 GJ mol–1. Although very large quantities of energy are released by a reaction like Equation $\ref{1}$, such a reaction is very difficult to achieve in practice. This is because of the very high activation energy, about 30 GJ mol–1, which must be overcome to bring the nuclei close enough to fuse together. This barrier is created by coulombic repulsion between the positively charged nuclei. The only place where scientists have succeeded in producing fusion reactions on a large scale is in a hydrogen bomb. Here, the necessary activation energy is achieved by exploding a fission bomb to heat the reactants to a temperature of about 108 K. Attempts to carry out fusion in a more controlled way have met only limited success. At the very high temperatures required, all molecules dissociate and most atoms ionize. A new state of matter called a plasma is formed. It is neither solid, liquid, nor gas. Plasma behaves much like the universal solvent of the alchemists by converting any solid material that it contacts into vapor.
Two techniques for producing a controlled fusion reaction are currently being explored. The first is to restrict the plasma by means of a strong magnetic field, rather than the walls of a container. This has met some success, but has not yet been able to contain a plasma long enough for usable energy to be obtained. The second technique involves the sudden compression and heating of pellets of deuterium and tritium by means of a sharply focused laser beam. Again, only limited success has been obtained.
Though these attempts at a controlled fusion reaction have so far been only partially successful, they are nevertheless worth pursuing. Because of the much readier availability of lighter isotopes necessary for fusion, as opposed to the much rarer heavier isotopes required for fission, controlled nuclear fusion would offer the human race an essentially limitless supply of energy. There would still be some environmental difficulties with the production of isotopes such as tritium, but these would be nowhere near the seriousness of the problem caused by the production of the witches brew of radioactive isotopes in a fission reactor. It must be confessed, though, that at the present rate of progress, the prospect of limitless clean energy from fusion seems unlikely in the next decade or two. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/17%3A_Radioactivity_and_Nuclear_Chemistry/17.08%3A_Nuclear_Power-_Using_Fission_to_Generate_Electricity.txt |
Learning Objectives
• Describe the biological impact of ionizing radiation.
• Define units for measuring radiation exposure.
• Explain the operation of common tools for detecting radioactivity.
• List common sources of radiation exposure in the US.
The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure $1$).
Ionizing vs. Nonionizing Radiation
There is a large difference in the magnitude of the biological effects of nonionizing radiation (for example, light and microwaves) and ionizing radiation, emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure $2$).
Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H2O (the most abundant molecule in living organisms), which forms a H2O+ ion that reacts with water, forming a hydronium ion and a hydroxyl radical:
Biological Effects of Exposure to Radiation
Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy.
Different types of radiation have differing abilities to pass through material (Figure $4$). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays.
For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238, which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure $5$).
Radon is found in buildings across the country, with amounts dependent on location. The average concentration of radon inside houses in the US (1.25 pCi/L) is about three times the level found in outside air, and about one in six houses have radon levels high enough that remediation efforts to reduce the radon concentration are recommended. Exposure to radon increases one’s risk of getting cancer (especially lung cancer), and high radon levels can be as bad for health as smoking a carton of cigarettes a day. Radon is the number one cause of lung cancer in nonsmokers and the second leading cause of lung cancer overall. Radon exposure is believed to cause over 20,000 deaths in the US per year.
Measuring Radiation Exposure
Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure $6$). Probably the best-known radiation instrument, the Geiger counter (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A scintillation counter contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. Radiation dosimeters also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters.
A variety of units are used to measure various aspects of radiation (Table $1$). The SI unit for rate of radioactive decay is the becquerel (Bq), with 1 Bq = 1 disintegration per second. The curie (Ci) and millicurie (mCi) are much larger units and are frequently used in medicine (1 curie = 1 Ci = $3.7 \times 10^{10}$ disintegrations per second). The SI unit for measuring radiation dose is the gray (Gy), with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the radiation absorbed dose (rad) is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the sievert (Sv). This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose.
Table $1$: Units Used for Measuring Radiation
Measurement Purpose Unit Quantity Measured Description
activity of source becquerel (Bq) radioactive decays or emissions amount of sample that undergoes 1 decay/second
curie (Ci) amount of sample that undergoes $\mathrm{3.7 \times 10^{10}\; decays/second}$
absorbed dose gray (Gy) energy absorbed per kg of tissue 1 Gy = 1 J/kg tissue
radiation absorbed dose (rad) 1 rad = 0.01 J/kg tissue
biologically effective dose sievert (Sv) tissue damage Sv = RBE × Gy
roentgen equivalent for man (rem) Rem = RBE × rad
The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine (1 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy), along with a biological factor referred to as the RBE (for relative biological effectiveness), that is an approximate measure of the relative damage done by the radiation. These are related by:
$\text{number of rems}=\text{RBE} \times \text{number of rads} \label{Eq2}$
with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation.
Example $1$: Amount of Radiation
Cobalt-60 (t1/2 = 5.26 y) is used in cancer therapy since the $\gamma$ rays it emits can be focused in small areas where the cancer is located. A 5.00-g sample of Co-60 is available for cancer treatment.
1. What is its activity in Bq?
2. What is its activity in Ci?
Solution
The activity is given by:
$\textrm{Activity}=λN=\left( \dfrac{\ln 2}{t_{1/2} } \right) N=\mathrm{\left( \dfrac{\ln 2}{5.26\ y} \right) \times 5.00 \ g=0.659\ \dfrac{g}{y} \ of\ \ce{^{60}Co} \text{ that decay}} \nonumber$
And to convert this to decays per second:
$\mathrm{0.659\; \frac{g}{y} \times \dfrac{y}{365 \;day} \times \dfrac{1\; day}{ 24\; hours} \times \dfrac{1\; h}{3,600 \;s} \times \dfrac{1\; mol}{59.9\; g} \times \dfrac{6.02 \times 10^{23} \;atoms}{1 \;mol} \times \dfrac{1\; decay}{1\; atom}} \nonumber$
$\mathrm{=2.10 \times 10^{14} \; \frac{decay}{s}} \nonumber$
(a) Since $\mathrm{1\; Bq = 1\; \frac{ decay}{s}}$, the activity in Becquerel (Bq) is:
$\mathrm{2.10 \times 10^{14} \dfrac{decay}{s} \times \left(\dfrac{1\ Bq}{1 \; \frac{decay}{s}} \right)=2.10 \times 10^{14} \; Bq} \nonumber$
(b) Since $\mathrm{1\ Ci = 3.7 \times 10^{11}\; \frac{decay}{s}}$, the activity in curie (Ci) is:
$\mathrm{2.10 \times 10^{14} \frac{decay}{s} \times \left( \dfrac{1\ Ci}{3.7 \times 10^{11} \frac{decay}{s}} \right) =5.7 \times 10^2\;Ci} \nonumber$
Exercise $1$
Tritium is a radioactive isotope of hydrogen ($t_{1/2} = \mathrm{12.32\; years}$) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu. What is the activity of a sample containing 1.00mg of tritium (a) in Bq and (b) in Ci?
Answer a
$\mathrm{3.56 \times 10^{11} Bq}$
Answer b
$\mathrm{0.962\; Ci}$
Effects of Long-term Radiation Exposure on the Human Body
The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure $8$, the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including CAT scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131).
A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table $2$.
Table $2$: Health Effects of Radiation
Exposure (rem) Health Effect Time to Onset (Without Treatment)
5–10 changes in blood chemistry
50 nausea hours
55 fatigue
70 vomiting
75 hair loss 2–3 weeks
90 diarrhea
100 hemorrhage
400 possible death within 2 months
1000 destruction of intestinal lining
internal bleeding
death 1–2 weeks
2000 damage to central nervous system
loss of consciousness minutes
death hours to days
It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure.
Summary
We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating, but potentially most damaging, and gamma rays the most penetrating.
Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source of radiation, and limiting time of exposure.
Footnotes
1. 1 Source: US Environmental Protection Agency
Glossary
becquerel (Bq)
SI unit for rate of radioactive decay; 1 Bq = 1 disintegration/s.
curie (Ci)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
Geiger counter
Instrument that detects and measures radiation via the ionization produced in a Geiger-Müller tube.
gray (Gy)
SI unit for measuring radiation dose; 1 Gy = 1 J absorbed/kg tissue.
ionizing radiation
Radiation that can cause a molecule to lose an electron and form an ion.
millicurie (mCi)
Larger unit for rate of radioactive decay frequently used in medicine; 1 Ci = 3.7 × 1010 disintegrations/s.
nonionizing radiation
Radiation that speeds up the movement of atoms and molecules; it is equivalent to heating a sample, but is not energetic enough to cause the ionization of molecules.
radiation absorbed dose (rad)
SI unit for measuring radiation dose, frequently used in medical applications; 1 rad = 0.01 Gy.
radiation dosimeter
Device that measures ionizing radiation and is used to determine personal radiation exposure.
relative biological effectiveness (RBE)
Measure of the relative damage done by radiation.
roentgen equivalent man (rem)
Unit for radiation damage, frequently used in medicine; 1 rem = 1 Sv.
scintillation counter
Instrument that uses a scintillator—a material that emits light when excited by ionizing radiation—to detect and measure radiation.
sievert (Sv)
SI unit measuring tissue damage caused by radiation; takes energy and biological effects of radiation into account. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/17%3A_Radioactivity_and_Nuclear_Chemistry/17.10%3A_The_Effects_of_Radiation_on_Life.txt |
Learning Objectives
• Outline manifestations of nuclear medicine.
The field of nuclear medicine has expanded greatly in the last twenty years, particularly in the area of imaging. This section will focus on nuclear medicine involving the types of nuclear radiation introduced in this chapter. The x-ray imaging systems will not be covered.
Radioiodine $\left( \ce{I}-131 \right)$ Therapy involves imaging and treatment of the thyroid gland. The thyroid gland is a gland in the neck that produces two hormones that regulate metabolism. In some individuals, this gland becomes overactive and produces too much of these hormones. The treatment for this problem uses radioactive iodine $\left( \ce{I}-131 \right)$, which is produced for this purpose in research fission reactors, or by neutron bombardment of other nuclei.
The thyroid gland uses iodine in the process of its normal function. Any iodine in food that enters the bloodstream is usually removed by, and concentrated in, the thyroid gland. When a patient suffering from an overactive thyroid swallows a small pill containing radioactive iodine, the $\ce{I}$-131 is absorbed into the bloodstream just like non-radioactive iodine, and follows the same process to be concentrated in the thyroid. The concentrated emissions of nuclear radiation in the thyroid destroy some of the gland's cells and control the problem of the overactive thyroid.
Smaller doses of $\ce{I}$-131 (too small to kill cells) are also used for purposes of imaging the thyroid. Once the iodine is concentrated in the thyroid, the patient lays down on a sheet of film and the radiation from the $\ce{I}$-131 makes a picture of the thyroid on the film. The half-life of iodine-131 is approximately 8 days so after a few weeks, virtually all of the radioactive iodine is out of the patient's system. During that time, the patient is advised that they will set off radiation detectors in airports and will need to get special permission to fly on commercial flights.
Positron Emission Tomography or PET scan is a type of nuclear medicine imaging. Depending on the area of the body being imaged, a radioactive isotope is either injected into a vein, swallowed by mouth, or inhaled as a gas. When the radioisotope is collected in the appropriate area of the body, the gamma ray emissions are detected by a PET scanner (often called a gamma camera) which works together with a computer to generate special pictures, providing details on both the structure and function of various organs. PET scans are used to:
• Detect cancer.
• Determine the amount of cancer spread.
• Assess the effectiveness of treatment plans.
• Determine blood flow to the heart muscle.
• Determine the effects of a heart attack.
• Evaluate brain abnormalities, such as tumors and memory disorders.
• Map brain and heart function.
External Beam Therapy (EBT) is a method of delivering a high energy beam of radiation to the precise location of a patient's tumor. These beams can destroy cancer cells and, with careful planning, will not kill surrounding cells. The concept is to have several beams of radiation, each of which is sub-lethal, enter the body from different directions. The only place in the body where the beam would be lethal is at the point where all the beams intersect. Before the EBT process, the patient is three-dimensionally mapped using CT scans and x-rays. The patient receives small tattoos to allow the therapist to line up the beams exactly. Alignment lasers are used to precisely locate the target. The radiation beam is usually generated with a linear accelerator. EBT is used to treat the following diseases, as well as others:
• Breast cancer
• Colorectal cancer
• Head and neck cancer
• Lung cancer
• Prostate cancer | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/17%3A_Radioactivity_and_Nuclear_Chemistry/17.11%3A_Radioactivity_in_Medicine.txt |
• 18.1: What Do I Smell
• 18.2: Vitalism- the Difference Between Organic and Inorganic
• 18.3: Carbon- A Versitile Atom
• 18.4: 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.
• 18.5: Alkanes- Saturated Hydrocarbons
Simple alkanes exist as a homologous series, in which adjacent members differ by a CH2 unit.
• 18.6: Isomers- Same Formula, Different Structure
• 18.7: Naming Alkanes
• 18.8: 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:
• 18.9: 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.
• 18.10: Aromatic Hydrocarbons
Aromatic hydrocarbons contain ring structures with delocalized π electron systems.
• 18.11: 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
• 18.12: 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
• 18.13: 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
• 18.14: 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.
• 18.15: 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.
• 18.16: 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.
• 18.17: Polymers
18: 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/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.04%3A_Hydrocarbons-_Compounds_Containing_Only_Carbon_and_Hydrocarbon.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/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.05%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/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.08%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).
18.10: 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/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.09%3A_Hydrocarbon_Reactions.txt |
Learning Objectives
• To know the major classes of organic compounds and identify important functional groups.
You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO2H). The major families of organic compounds are characterized by their functional groups. Figure \(1\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group.
The first family listed in Figure \(1\) is the hydrocarbons. These include alkanes, with the general molecular formula CnH2n+2 where n is an integer; alkenes, represented by CnH2n; alkynes, represented by CnH2n−2; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO2H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO2 group.
The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH3)2C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde.
Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH3 group in dimethyl benzene is indicated with a 1, but the second CH3 group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(2\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH3 groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene.
We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached.
Summary
Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom.
Conceptual Problems
1. Can two substances have the same systematic name and be different compounds?
2. Is a carbon–carbon multiple bond considered a functional group? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.11%3A_Functional_Groups.txt |
Learning Objectives
• Identify the general structure for an alcohol.
• Identify the structural feature that classifies alcohols as primary, secondary, or tertiary.
• Name alcohols with both common names and IUPAC names
An alcohol is an organic compound with a hydroxyl (OH) functional group on an aliphatic carbon atom. Because OH is the functional group of all alcohols, we often represent alcohols by the general formula ROH, where R is an alkyl group. Alcohols are common in nature. Most people are familiar with ethyl alcohol (ethanol), the active ingredient in alcoholic beverages, but this compound is only one of a family of organic compounds known as alcohols. The family also includes such familiar substances as cholesterol and the carbohydrates. Methanol (CH3OH) and ethanol (CH3CH2OH) are the first two members of the homologous series of alcohols.
Nomenclature of Alcohols
Alcohols with one to four carbon atoms are frequently called by common names, in which the name of the alkyl group is followed by the word alcohol:
According to the International Union of Pure and Applied Chemistry (IUPAC), alcohols are named by changing the ending of the parent alkane name to -ol. Here are some basic IUPAC rules for naming alcohols:
1. The longest continuous chain (LCC) of carbon atoms containing the OH group is taken as the parent compound—an alkane with the same number of carbon atoms. The chain is numbered from the end nearest the OH group.
2. The number that indicates the position of the OH group is prefixed to the name of the parent hydrocarbon, and the -e ending of the parent alkane is replaced by the suffix -ol. (In cyclic alcohols, the carbon atom bearing the OH group is designated C1, but the 1 is not used in the name.) Substituents are named and numbered as in alkanes.
3. If more than one OH group appears in the same molecule (polyhydroxy alcohols), suffixes such as -diol and -triol are used. In these cases, the -e ending of the parent alkane is retained.
Figure $1$ shows some examples of the application of these rules.
Example $1$
Give the IUPAC name for each compound.
• HOCH2CH2CH2CH2CH2OH
Solution
1. Ten carbon atoms in the LCC makes the compound a derivative of decane (rule 1), and the OH on the third carbon atom makes it a 3-decanol (rule 2).
The carbon atoms are numbered from the end closest to the OH group. That fixes the two methyl (CH3) groups at the sixth and eighth positions. The name is 6,8-dimethyl-3-decanol (not 3,5-dimethyl-8-decanol).
2. Five carbon atoms in the LCC make the compound a derivative of pentane. Two OH groups on the first and fifth carbon atoms make the compound a diol and give the name 1,5-pentanediol (rule 3).
Exercise $1$
Give the IUPAC name for each compound.
Example $2$
Draw the structure for each compound.
1. 2-hexanol
2. 3-methyl-2-pentanol
Solution
1. The ending -ol indicates an alcohol (the OH functional group), and the hex- stem tells us that there are six carbon atoms in the LCC. We start by drawing a chain of six carbon atoms: –C–C–C–C–C–C–.
The 2 indicates that the OH group is attached to the second carbon atom.
Finally, we add enough hydrogen atoms to give each carbon atom four bonds.
• The numbers indicate that there is a methyl (CH3) group on the third carbon atom and an OH group on the second carbon atom.
Exercise $2$
Draw the structure for each compound.
1. 3-heptanol
• 2-methyl-3-hexanol
Classification of Alcohols
Some of the properties of alcohols depend on the number of carbon atoms attached to the specific carbon atom that is attached to the OH group. Alcohols can be grouped into three classes on this basis.
• A primary (1°) alcohol is one in which the carbon atom (in red) with the OH group is attached to one other carbon atom (in blue). Its general formula is RCH2OH.
• A secondary (2°) alcohol is one in which the carbon atom (in red) with the OH group is attached to two other carbon atoms (in blue). Its general formula is R2CHOH.
• A tertiary (3°) alcohol is one in which the carbon atom (in red) with the OH group is attached to three other carbon atoms (in blue). Its general formula is R3COH.
Table $1$ names and classifies some of the simpler alcohols. Some of the common names reflect a compound’s classification as secondary (sec-) or tertiary (tert-). These designations are not used in the IUPAC nomenclature system for alcohols. Note that there are four butyl alcohols in the table, corresponding to the four butyl groups: the butyl group (CH3CH2CH2CH2) discussed before, and three others:
Table $1$: Classification and Nomenclature of Some Alcohols
Condensed Structural Formula Class of Alcohol Common Name IUPAC Name
CH3OH methyl alcohol methanol
CH3CH2OH primary ethyl alcohol ethanol
CH3CH2CH2OH primary propyl alcohol 1-propanol
(CH3)2CHOH secondary isopropyl alcohol 2-propanol
CH3CH2CH2CH2OH primary butyl alcohol 1-butanol
CH3CH2CHOHCH3 secondary sec-butyl alcohol 2-butanol
(CH3)2CHCH2OH primary isobutyl alcohol 2-methyl-1-propanol
(CH3)3COH tertiary tert-butyl alcohol 2-methyl-2-propanol
secondary cyclohexyl alcohol cyclohexanol
Summary
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.
Learning Objectives
• Explain why the boiling points of alcohols are higher than those of ethers and alkanes of similar molar masses.
• Explain why alcohols and ethers of four or fewer carbon atoms are soluble in water while comparable alkanes are not soluble.
Alcohols can be considered derivatives of water (H2O; also written as HOH).
Like the H–O–H bond in water, the R–O–H bond is bent, and alcohol molecules are polar. This relationship is particularly apparent in small molecules and reflected in the physical and chemical properties of alcohols with low molar mass. Replacing a hydrogen atom from an alkane with an OH group allows the molecules to associate through hydrogen bonding (Figure $1$).
Recall that physical properties are determined to a large extent by the type of intermolecular forces. Table $1$ lists the molar masses and the boiling points of some common compounds. The table shows that substances with similar molar masses can have quite different boiling points.
Table $1$: Comparison of Boiling Points and Molar Masses
Formula Name Molar Mass Boiling Point (°C)
CH4 methane 16 –164
HOH water 18 100
C2H6 ethane 30 –89
CH3OH methanol 32 65
C3H8 propane 44 –42
CH3CH2OH ethanol 46 78
C4H10 butane 58 –1
CH3CH2CH2OH 1-propanol 60 97
Alkanes are nonpolar and are thus associated only through relatively weak dispersion forces. Alkanes with one to four carbon atoms are gases at room temperature. In contrast, even methanol (with one carbon atom) is a liquid at room temperature. Hydrogen bonding greatly increases the boiling points of alcohols compared to hydrocarbons of comparable molar mass. The boiling point is a rough measure of the amount of energy necessary to separate a liquid molecule from its nearest neighbors. If the molecules interact through hydrogen bonding, a relatively large quantity of energy must be supplied to break those intermolecular attractions. Only then can the molecule escape from the liquid into the gaseous state.
Alcohols can also engage in hydrogen bonding with water molecules (Figure $2$). Thus, whereas the hydrocarbons are insoluble in water, alcohols with one to three carbon atoms are completely soluble. As the length of the chain increases, however, the solubility of alcohols in water decreases; the molecules become more like hydrocarbons and less like water. The alcohol 1-decanol (CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2OH) is essentially insoluble in water. We frequently find that the borderline of solubility in a family of organic compounds occurs at four or five carbon atoms.
Summary
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 bonding with water molecules; comparable alkane molecules cannot engage in hydrogen bonding.
Learning Objectives
1. Give two major types of reactions of alcohols.
2. Describe the result of the oxidation of a primary alcohol.
3. Describe the result of the oxidation of a secondary alcohol.
Chemical reactions in alcohols occur mainly at the functional group, but some involve hydrogen atoms attached to the OH-bearing carbon atom or to an adjacent carbon atom. Of the three major kinds of alcohol reactions, which are summarized in Figure $1$, two—dehydration and oxidation—are considered here. The third reaction type—esterification—is covered elsewhere.
Dehydration
As noted in Figure $1$, an alcohol undergoes dehydration in the presence of a catalyst to form an alkene and water. The reaction removes the OH group from the alcohol carbon atom and a hydrogen atom from an adjacent carbon atom in the same molecule:
Under the proper conditions, it is possible for the dehydration to occur between two alcohol molecules. The entire OH group of one molecule and only the hydrogen atom of the OH group of the second molecule are removed. The two ethyl groups attached to an oxygen atom form an ether molecule.
(Ethers are discussed in elsewhere) Thus, depending on conditions, one can prepare either alkenes or ethers by the dehydration of alcohols.
Both dehydration and hydration reactions occur continuously in cellular metabolism, with enzymes serving as catalysts and at a temperature of about 37°C. The following reaction occurs in the "Embden–Meyerhof" pathway
Although the participating compounds are complex, the reaction is the same: elimination of water from the starting material. The idea is that if you know the chemistry of a particular functional group, you know the chemistry of hundreds of different compounds.
Oxidation
Primary and secondary alcohols are readily oxidized. We saw earlier how methanol and ethanol are oxidized by liver enzymes to form aldehydes. Because a variety of oxidizing agents can bring about oxidation, we can indicate an oxidizing agent without specifying a particular one by writing an equation with the symbol [O] above the arrow. For example, we write the oxidation of ethanol—a primary alcohol—to form acetaldehyde—an aldehyde—as follows:
We shall see that aldehydes are even more easily oxidized than alcohols and yield carboxylic acids. Secondary alcohols are oxidized to ketones. The oxidation of isopropyl alcohol by potassium dichromate ($\ce{K2Cr2O7}$) gives acetone, the simplest ketone:
Unlike aldehydes, ketones are relatively resistant to further oxidation, so no special precautions are required to isolate them as they form. Note that in oxidation of both primary (RCH2OH) and secondary (R2CHOH) alcohols, two hydrogen atoms are removed from the alcohol molecule, one from the OH group and other from the carbon atom that bears the OH group.
These reactions can also be carried out in the laboratory with chemical oxidizing agents. One such oxidizing agent is potassium dichromate. The balanced equation (showing only the species involved in the reaction) in this case is as follows:
$\ce{8H^{=} + Cr2O7^{2-} + 3CH3CH2OH -> 3CH3CHO + 2Cr^{3+} + 7H2O} \nonumber$
Alcohol oxidation is important in living organisms. Enzyme-controlled oxidation reactions provide the energy cells need to do useful work. One step in the metabolism of carbohydrates involves the oxidation of the secondary alcohol group in isocitric acid to a ketone group:
The overall type of reaction is the same as that in the conversion of isopropyl alcohol to acetone.
Tertiary alcohols (R3COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The oxidation reactions we have described involve the formation of a carbon-to-oxygen double bond. Thus, the carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore tertiary alcohols are not easily oxidized.
Example $1$
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
1. CH3CH2CH2CH2CH2OH
Solution
The first step is to recognize the class of each alcohol as primary, secondary, or tertiary.
1. This alcohol has the OH group on a carbon atom that is attached to only one other carbon atom, so it is a primary alcohol. Oxidation forms first an aldehyde and further oxidation forms a carboxylic acid.
2. This alcohol has the OH group on a carbon atom that is attached to three other carbon atoms, so it is a tertiary alcohol. No reaction occurs.
3. This alcohol has the OH group on a carbon atom that is attached to two other carbon atoms, so it is a secondary alcohol; oxidation gives a ketone.
Exercise $1$
Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow.
Summary
Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.12%3A_Alcohols.txt |
Learning Objectives
• Describe the structural difference between an alcohol and an ether that affects physical characteristics and reactivity of each.
• Name simple ethers.
• Describe the structure and uses of some ethers.
With the general formula ROR′, an ether may be considered a derivative of water in which both hydrogen atoms are replaced by alkyl or aryl groups. It may also be considered a derivative of an alcohol (ROH) in which the hydrogen atom of the OH group is been replaced by a second alkyl or aryl group:
$\mathrm{HOH\underset{H\: atoms}{\xrightarrow{replace\: both}}ROR'\underset{of\: OH\: group}{\xleftarrow{replace\: H\: atom}}ROH} \nonumber$
Simple ethers have simple common names, formed from the names of the groups attached to oxygen atom, followed by the generic name ether. For example, CH3–O–CH2CH2CH3 is methyl propyl ether. If both groups are the same, the group name should be preceded by the prefix di-, as in dimethyl ether (CH3–O–CH3) and diethyl ether CH3CH2–O–CH2CH3.
Ether molecules have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore there is no intermolecular hydrogen bonding between ether molecules, and ethers therefore have quite low boiling points for a given molar mass. Indeed, ethers have boiling points about the same as those of alkanes of comparable molar mass and much lower than those of the corresponding alcohols (Table $1$).
Table $1$: Comparison of Boiling Points of Alkanes, Alcohols, and Ethers
Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid?
CH3CH2CH3 propane 44 –42 no
CH3OCH3 dimethyl ether 46 –25 no
CH3CH2OH ethyl alcohol 46 78 yes
CH3CH2CH2CH2CH3 pentane 72 36 no
CH3CH2OCH2CH3 diethyl ether 74 35 no
CH3CH2CH2CH2OH butyl alcohol 74 117 yes
Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water).
Example $1$
What is the common name for each ether?
1. CH3CH2CH2OCH2CH2CH3
Solution
1. The carbon groups on either side of the oxygen atom are propyl (CH3CH2CH2) groups, so the compound is dipropyl ether.
2. The three-carbon group is attached by the middle carbon atom, so it is an isopropyl group. The one-carbon group is a methyl group. The compound is isopropyl methyl ether.
Exercise $1$
What is the common name for each ether?
1. CH3CH2CH2CH2OCH2CH2CH2CH3
To Your Health: Ethers as General Anesthetics
A general anesthetic acts on the brain to produce unconsciousness and a general insensitivity to feeling or pain. Diethyl ether (CH3CH2OCH2CH3) was the first general anesthetic to be used.
Diethyl ether is relatively safe because there is a fairly wide gap between the dose that produces an effective level of anesthesia and the lethal dose. However, because it is highly flammable and has the added disadvantage of causing nausea, it has been replaced by newer inhalant anesthetics, including the fluorine-containing compounds halothane, enflurane, and isoflurane. Unfortunately, the safety of these compounds for operating room personnel has been questioned. For example, female operating room workers exposed to halothane suffer a higher rate of miscarriages than women in the general population.
These three modern, inhalant, halogen-containing, anesthetic compounds are less flammable than diethyl ether.
Summary
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 the alcohol that is isomeric with it. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.13%3A_Ethers.txt |
Learning Objectives
• Identify the general structure for an aldehyde and a ketone.
• Use common names to name aldehydes and ketones.
• Use the IUPAC system to name aldehydes and ketones.
The next functional group we consider, the carbonyl group, has a carbon-to-oxygen double bond.
Carbonyl groups define two related families of organic compounds: the aldehydes and the ketones.
The carbonyl group is ubiquitous in biological compounds. It is found in carbohydrates, fats, proteins, nucleic acids, hormones, and vitamins—organic compounds critical to living systems.
In a ketone, two carbon groups are attached to the carbonyl carbon atom. The following general formulas, in which R represents an alkyl group and Ar stands for an aryl group, represent ketones.
In an aldehyde, at least one of the attached groups must be a hydrogen atom. The following compounds are aldehydes:
In condensed formulas, we use CHO to identify an aldehyde rather than COH, which might be confused with an alcohol. This follows the general rule that in condensed structural formulas H comes after the atom it is attached to (usually C, N, or O).
The carbon-to-oxygen double bond is not shown but understood to be present. Because they contain the same functional group, aldehydes and ketones share many common properties, but they still differ enough to warrant their classification into two families.
Naming Aldehydes and Ketones
Both common and International Union of Pure and Applied Chemistry (IUPAC) names are frequently used for aldehydes and ketones, with common names predominating for the lower homologs. The common names of aldehydes are taken from the names of the acids into which the aldehydes can be converted by oxidation.
The stems for the common names of the first four aldehydes are as follows:
• 1 carbon atom: form-
• 2 carbon atoms: acet-
• 3 carbon atoms: propion-
• 4 carbon atoms: butyr-
Because the carbonyl group in a ketone must be attached to two carbon groups, the simplest ketone has three carbon atoms. It is widely known as acetone, a unique name unrelated to other common names for ketones.
Generally, the common names of ketones consist of the names of the groups attached to the carbonyl group, followed by the word ketone. (Note the similarity to the naming of ethers.) Another name for acetone, then, is dimethyl ketone. The ketone with four carbon atoms is ethyl methyl ketone.
Example $1$
Classify each compound as an aldehyde or a ketone. Give the common name for each ketone.
Solution
1. This compound has the carbonyl group on an end carbon atom, so it is an aldehyde.
2. This compound has the carbonyl group on an interior carbon atom, so it is a ketone. Both alkyl groups are propyl groups. The name is therefore dipropyl ketone.
3. This compound has the carbonyl group between two alkyl groups, so it is a ketone. One alkyl group has three carbon atoms and is attached by the middle carbon atom; it is an isopropyl group. A group with one carbon atom is a methyl group. The name is therefore isopropyl methyl ketone.
Exercise $1$
Classify each compound as an aldehyde or a ketone. Give the common name for each ketone.
Here are some simple IUPAC rules for naming aldehydes and ketones:
• The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group.
• For an aldehyde, drop the -e from the alkane name and add the ending -al. Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde.
• For a ketone, drop the -e from the alkane name and add the ending -one. Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone.
• To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number.
• To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1.
Example $2$
Give the IUPAC name for each compound.
Solution
1. There are five carbon atoms in the LCC. The methyl group (CH3) is a substituent on the second carbon atom of the chain; the aldehyde carbon atom is always C1. The name is derived from pentane. Dropping the -e and adding the ending -al gives pentanal. The methyl group on the second carbon atom makes the name 2-methylpentanal.
2. There are five carbon atoms in the LCC. The carbonyl carbon atom is C3, and there are methyl groups on C2 and C4. The IUPAC name is 2,4-dimethyl-3-pentanone.
3. There are six carbon atoms in the ring. The compound is cyclohexanone. No number is needed to indicate the position of the carbonyl group because all six carbon atoms are equivalent.
Exercise
Give the IUPAC name for each compound.
Example $3$
Draw the structure for each compound.
1. 7-chlorooctanal
2. 4-methyl–3-hexanone
Solution
1. The octan- part of the name tells us that the LCC has eight carbon atoms. There is a chlorine (Cl) atom on the seventh carbon atom; numbering from the carbonyl group and counting the carbonyl carbon atom as C1, we place the Cl atom on the seventh carbon atom.
2. The hexan- part of the name tells us that the LCC has six carbon atoms. The 3 means that the carbonyl carbon atom is C3 in this chain, and the 4 tells us that there is a methyl (CH3) group at C4:
Exercise $3$
Draw the structure for each compound.
1. 5-bromo-3-iodoheptanal
2. 5-bromo-4-ethyl-2-heptanone
Summary
The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone. Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an -al ending for an aldehydes and an -one ending for a ketone.
Learning Objectives
• Explain why the boiling points of aldehydes and ketones are higher than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols.
• Compare the solubilities in water of aldehydes and ketones of four or fewer carbon atoms with the solubilities of comparable alkanes and alcohols.
• Name the typical reactions take place with aldehydes and ketones.
• Describe some of the uses of common aldehydes and ketones.
The carbon-to-oxygen double bond is quite polar, more polar than a carbon-to-oxygen single bond. The electronegative oxygen atom has a much greater attraction for the bonding electron pairs than does the carbon atom. The carbon atom has a partial positive charge, and the oxygen atom has a partial negative charge:
In aldehydes and ketones, this charge separation leads to dipole-dipole interactions that are great enough to significantly affect the boiling points. Table $1$ shows that the polar single bonds in ethers have little such effect, whereas hydrogen bonding between alcohol molecules is even stronger.
Table $1$: Boiling Points of Compounds Having Similar Molar Masses but Different Types of Intermolecular Forces
Compound Family Molar Mass Type of Intermolecular Forces Boiling Point (°C)
CH3CH2CH2CH3 alkane 58 dispersion only –1
CH3OCH2CH3 ether 60 weak dipole 6
CH3CH2CHO aldehyde 58 strong dipole 49
CH3CH2CH2OH alcohol 60 hydrogen bonding 97
Formaldehyde is a gas at room temperature. Acetaldehyde boils at 20°C; in an open vessel, it boils away in a warm room. Most other common aldehydes are liquids at room temperature.
Although the lower members of the homologous series have pungent odors, many higher aldehydes have pleasant odors and are used in perfumes and artificial flavorings. As for the ketones, acetone has a pleasant odor, but most of the higher homologs have rather bland odors.
The oxygen atom of the carbonyl group engages in hydrogen bonding with a water molecule.
The solubility of aldehydes is therefore about the same as that of alcohols and ethers. Formaldehyde, acetaldehyde, and acetone are soluble in water. As the carbon chain increases in length, solubility in water decreases. The borderline of solubility occurs at about four carbon atoms per oxygen atom. All aldehydes and ketones are soluble in organic solvents and, in general, are less dense than water.
Oxidation of Aldehydes and Ketones
Aldehydes and ketones are much alike in many of their reactions, owing to the presence of the carbonyl functional group in both. They differ greatly, however, in one most important type of reaction: oxidation. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation.
The aldehydes are, in fact, among the most easily oxidized of organic compounds. They are oxidized by oxygen ($\ce{O2}$) in air to carboxylic acids.
$\ce{2RCHO + O_2 -> 2RCOOH} \nonumber$
The ease of oxidation helps chemists identify aldehydes. A sufficiently mild oxidizing agent can distinguish aldehydes not only from ketones but also from alcohols. Tollens’ reagent, for example, is an alkaline solution of silver ($\ce{Ag^{+}}$) ion complexed with ammonia (NH3), which keeps the $\ce{Ag^{+}}$ ion in solution.
$\ce{H_3N—Ag^{+}—NH_3} \nonumber$
When Tollens’ reagent oxidizes an aldehyde, the $\ce{Ag^{+}}$ ion is reduced to free silver ($\ce{Ag}$).
$\underbrace{\ce{RCHO(aq)}}_{\text{an aldehyde}} + \ce{2Ag(NH3)2^{+}(aq) -> RCOO^{-} +} \underbrace{\ce{2Ag(s)}}_{\text{free silver}} + \ce{4NH3(aq) + 2H2O} \nonumber$
Deposited on a clean glass surface, the silver produces a mirror (Figure $1$). Ordinary ketones do not react with Tollens’ reagent.
Although ketones resist oxidation by ordinary laboratory oxidizing agents, they undergo combustion, as do aldehydes.
Some Common Carbonyl Compounds
Formaldehyde has an irritating odor. Because of its reactivity, it is difficult to handle in the gaseous state. For many uses, it is therefore dissolved in water and sold as a 37% to 40% aqueous solution called formalin. Formaldehyde denatures proteins, rendering them insoluble in water and resistant to bacterial decay. For this reason, formalin is used in embalming solutions and in preserving biological specimens.
Aldehydes are the active components in many other familiar substances. Large quantities of formaldehyde are used to make phenol-formaldehyde resins for gluing the wood sheets in plywood and as adhesives in other building materials. Sometimes the formaldehyde escapes from the materials and causes health problems in some people. While some people seem unaffected, others experience coughing, wheezing, eye irritation, and other symptoms.
The odor of green leaves is due in part to a carbonyl compound, cis-3-hexenal, which with related compounds is used to impart a “green” herbal odor to shampoos and other products.
Acetaldehyde is an extremely volatile, colorless liquid. It is a starting material for the preparation of many other organic compounds. Acetaldehyde is formed as a metabolite in the fermentation of sugars and in the detoxification of alcohol in the liver. Aldehydes are the active components of many other familiar materials (Figure $2$).
Acetone is the simplest and most important ketone. Because it is miscible with water as well as with most organic solvents, its chief use is as an industrial solvent (for example, for paints and lacquers). It is also the chief ingredient in some brands of nail polish remover.
To Your Health: Acetone in Blood, Urine, and Breath
Acetone is formed in the human body as a by-product of lipid metabolism. Normally, acetone does not accumulate to an appreciable extent because it is oxidized to carbon dioxide and water. The normal concentration of acetone in the human body is less than 1 mg/100 mL of blood. In certain disease states, such as uncontrolled diabetes mellitus, the acetone concentration rises to higher levels. It is then excreted in the urine, where it is easily detected. In severe cases, its odor can be noted on the breath.
Ketones are also the active components of other familiar substances, some of which are noted in the accompanying figure.
Certain steroid hormones have the ketone functional group as a part of their structure. Two examples are progesterone, a hormone secreted by the ovaries that stimulates the growth of cells in the uterine wall and prepares it for attachment of a fertilized egg, and testosterone, the main male sex hormone. These and other sex hormones affect our development and our lives in fundamental ways.
Summary
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. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.14%3A_Aldehydes_and_Ketones.txt |
Aldehydes and Ketones
There are a number of functional groups that contain a carbon-oxygen double bond, which is commonly referred to as a carbonyl. Ketones and aldehydes are two closely related carbonyl-based functional groups that react in very similar ways. In a ketone, the carbon atom of a carbonyl is bonded to two other carbons. In an aldehyde, the carbonyl carbon is bonded on one side to a hydrogen, and on the other side to a carbon. The exception to this definition is formaldehyde, in which the carbonyl carbon has bonds to two hydrogens.
Molecules with carbon-nitrogen double bonds are called imines, or Schiff bases.
Carboxylic acids and acid derivatives
If a carbonyl carbon is bonded on one side to a carbon (or hydrogen) and on the other side to a heteroatom (in organic chemistry, this term generally refers to oxygen, nitrogen, sulfur, or one of the halogens), the functional group is considered to be one of the ‘carboxylic acid derivatives’, a designation that describes a grouping of several functional groups. The eponymous member of this grouping is the carboxylic acid functional group, in which the carbonyl is bonded to a hydroxyl (OH) group.
As the name implies, carboxylic acids are acidic, meaning that they are readily deprotonated to form the conjugate base form, called a carboxylate (much more about carboxylic acids in the acid-base chapter!).
In amides, the carbonyl carbon is bonded to a nitrogen. The nitrogen in an amide can be bonded either to hydrogens, to carbons, or to both. Another way of thinking of an amide is that it is a carbonyl bonded to an amine.
In esters, the carbonyl carbon is bonded to an oxygen which is itself bonded to another carbon. Another way of thinking of an ester is that it is a carbonyl bonded to an alcohol. Thioesters are similar to esters, except a sulfur is in place of the oxygen.
In an acyl phosphate, the carbonyl carbon is bonded to the oxygen of a phosphate, and in an acid chloride, the carbonyl carbon is bonded to a chlorine.
Finally, in a nitrile group, a carbon is triple-bonded to a nitrogen. Nitriles are also often referred to as cyano groups.
A single compound often contains several functional groups. The six-carbon sugar molecules glucose and fructose, for example, contain aldehyde and ketone groups, respectively, and both contain five alcohol groups (a compound with several alcohol groups is often referred to as a ‘polyol’).
Capsaicin, the compound responsible for the heat in hot peppers, contains phenol, ether, amide, and alkene functional groups.
The male sex hormone testosterone contains ketone, alkene, and secondary alcohol groups, while acetylsalicylic acid (aspirin) contains aromatic, carboxylic acid, and ester groups.
While not in any way a complete list, this section has covered most of the important functional groups that we will encounter in biological and laboratory organic chemistry. The table on the inside back cover provides a summary of all of the groups listed in this section, plus a few more that will be introduced later in the text.
18.16: Amines
Learning Objectives
• Name the typical reactions that take place with amines.
• Describe heterocyclic amines.
Recall that ammonia (NH3) acts as a base because the nitrogen atom has a lone pair of electrons that can accept a proton. Amines also have a lone electron pair on their nitrogen atoms and can accept a proton from water to form substituted ammonium (NH4+) ions and hydroxide (OH) ions:
As a specific example, methylamine reacts with water to form the methylammonium ion and the OH ion.
Nearly all amines, including those that are not very soluble in water, will react with strong acids to form salts soluble in water.
Amine salts are named like other salts: the name of the cation is followed by the name of the anion.
Example $1$
What are the formulas of the acid and base that react to form [CH3NH2CH2CH3]+CH3COO?
Solution
The cation has two groups—methyl and ethyl—attached to the nitrogen atom. It comes from ethylmethylamine (CH3NHCH2CH3). The anion is the acetate ion. It comes from acetic acid (CH3COOH).
Exercise $1$
What are the formulas of the acid and base that react to form $\ce{(CH3CH2CH2)3NH^{+}I^{−}}$?
To Your Health: Amine Salts as Drugs
Salts of aniline are properly named as anilinium compounds, but an older system, still in use for naming drugs, identifies the salt of aniline and hydrochloric acid as “aniline hydrochloride.” These compounds are ionic—they are salts—and the properties of the compounds (solubility, for example) are those characteristic of salts. Many drugs that are amines are converted to hydrochloride salts to increase their solubility in aqueous solution.
Heterocyclic Amines
Looking back at the various cyclic hydrocarbons discussed previously, we see that all the atoms in the rings of these compounds are carbon atoms. In other cyclic compounds, called heterocyclic compounds (Greek heteros, meaning “other”), nitrogen, oxygen, sulfur, or some other atom is incorporated in the ring. Many heterocyclic compounds are important in medicine and biochemistry. Some compose part of the structure of the nucleic acids, which in turn compose the genetic material of cells and direct protein synthesis.
Many heterocyclic amines occur naturally in plants. Like other amines, these compounds are basic. Such a compound is an alkaloid, a name that means “like alkalis.” Many alkaloids are physiologically active, including the familiar drugs caffeine, nicotine, and cocaine.
To Your Health: Three Well-Known Alkaloids
Caffeine is a stimulant found in coffee, tea, and some soft drinks. Its mechanism of action is not well understood, but it is thought to block the activity of adenosine, a heterocyclic base that acts as a neurotransmitter, a substance that carries messages across a tiny gap (synapse) from one nerve cell (neuron) to another cell. The effective dose of caffeine is about 200 mg, corresponding to about two cups of strong coffee or tea.
Nicotine acts as a stimulant by a different mechanism; it probably mimics the action of the neurotransmitter acetylcholine. People ingest this drug by smoking or chewing tobacco. Its stimulant effect seems transient, as this initial response is followed by depression. Nicotine is highly toxic to animals. It is especially deadly when injected; the lethal dose for a human is estimated to be about 50 mg. Nicotine has also been used in agriculture as a contact insecticide.
Cocaine acts as a stimulant by preventing nerve cells from taking up dopamine, another neurotransmitter, from the synapse. High levels of dopamine are therefore available to stimulate the pleasure centers of the brain. The enhancement of dopamine action is thought to be responsible for cocaine’s “high” and its addictive properties. After the binge, dopamine is depleted in less than an hour. This leaves the user in a pleasureless state and (often) craving more cocaine.
Cocaine is used as the salt cocaine hydrochloride and in the form of broken lumps of the free (unneutralized) base, which is called crack cocaine.
$\underbrace{\ce{C17H21O4N}}_{\text{cocaine (freebase)}} + \ce{HCl ->} \underbrace{\ce{C17H21O4NH^{+}Cl^{-}}}_{\text{cocaine hydrochloride}} \nonumber$
Because it is soluble in water, cocaine hydrochloride is readily absorbed through the watery mucous membranes of the nose when it is snorted. Crack cocaine is more volatile than cocaine hydrochloride. It vaporizes at the temperature of a burning cigarette. When smoked, cocaine reaches the brain in 15 s.
Summary
Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.15%3A_Carboxylic_Acids_and_Esters.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/Introductory_Chemistry/Introductory_Chemistry/18%3A_Organic_Chemistry/18.17%3A_Polymers.txt |
• 19.1: The Human Genome Project
• 19.2: The Cell and Its Main Chemical Components
• 19.3: Carbohydrates
All carbohydrates consist of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes or ketones or are compounds that can be broken down to form such compounds. Examples of carbohydrates include starch, fiber, the sweet-tasting compounds called sugars, and structural materials such as cellulose. The term carbohydrate had its origin in a misinterpretation of the molecular formulas of many of these substances.
• 19.4: Lipids
Compounds isolated from body tissues are classified as lipids if they are more soluble in organic solvents, such as dichloromethane, than in water. Fatty acids are carboxylic acids that are the structural components of many lipids. They may be saturated or unsaturated. Lipids are important components of biological membranes. These lipids have dual characteristics: part of the molecule is hydrophilic, and part of the molecule is hydrophobic.
• 19.5: Proteins
The proteins in all living species are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids.
• 19.6: Protein Structure
Proteins can be divided into two categories: fibrous, which tend to be insoluble in water, and globular, which are more soluble in water. A protein may have up to four levels of structure. The primary structure consists of the specific amino acid sequence. The peptide chain can form an α-helix or β-pleated sheet, which is known as secondary structure and are incorporated into the tertiary structure of the folded polypeptide. The quaternary structure describes the arrangements of subunits.
• 19.7: Nucleic Acids- Blueprints for Proteins
Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose.
19: Biochemistry
Learning Objectives
• To recognize carbohydrates and classify them as mono-, di-, or polysaccharides.
All carbohydrates consist of carbon, hydrogen, and oxygen atoms and are polyhydroxy aldehydes or ketones or are compounds that can be broken down to form such compounds. Examples of carbohydrates include starch, fiber, the sweet-tasting compounds called sugars, and structural materials such as cellulose. The term carbohydrate had its origin in a misinterpretation of the molecular formulas of many of these substances. For example, because its formula is C6H12O6, glucose was once thought to be a “carbon hydrate” with the structure C6·6H2O.
Example $1$
Which compounds would be classified as carbohydrates?
Solution
1. This is a carbohydrate because the molecule contains an aldehyde functional group with OH groups on the other two carbon atoms.
2. This is not a carbohydrate because the molecule does not contain an aldehyde or a ketone functional group.
3. This is a carbohydrate because the molecule contains a ketone functional group with OH groups on the other two carbon atoms.
4. This is not a carbohydrate; although it has a ketone functional group, one of the other carbons atoms does not have an OH group attached.
Exercise $1$
Which compounds would be classified as carbohydrates?
Green plants are capable of synthesizing glucose (C6H12O6) from carbon dioxide (CO2) and water (H2O) by using solar energy in the process known as photosynthesis:
$\ce{6CO_2 + 6H_2O} + \text{686 kcal} \rightarrow \ce{C_6H_{12}O_6 + 6O_2} \label{$1$}$
(The 686 kcal come from solar energy.) Plants can use the glucose for energy or convert it to larger carbohydrates, such as starch or cellulose. Starch provides energy for later use, perhaps as nourishment for a plant’s seeds, while cellulose is the structural material of plants. We can gather and eat the parts of a plant that store energy—seeds, roots, tubers, and fruits—and use some of that energy ourselves. Carbohydrates are also needed for the synthesis of nucleic acids and many proteins and lipids.
Animals, including humans, cannot synthesize carbohydrates from carbon dioxide and water and are therefore dependent on the plant kingdom to provide these vital compounds. We use carbohydrates not only for food (about 60%–65% by mass of the average diet) but also for clothing (cotton, linen, rayon), shelter (wood), fuel (wood), and paper (wood).
The simplest carbohydrates—those that cannot be hydrolyzed to produce even smaller carbohydrates—are called monosaccharides. Two or more monosaccharides can link together to form chains that contain from two to several hundred or thousand monosaccharide units. Prefixes are used to indicate the number of such units in the chains. Disaccharide molecules have two monosaccharide units, trisaccharide molecules have three units, and so on. Chains with many monosaccharide units joined together are called polysaccharides. All these so-called higher saccharides can be hydrolyzed back to their constituent monosaccharides.
Compounds that cannot be hydrolyzed will not react with water to form two or more smaller compounds.
Summary
Carbohydrates are an important group of biological molecules that includes sugars and starches. Photosynthesis is the process by which plants use energy from sunlight to synthesize carbohydrates. A monosaccharide is the simplest carbohydrate and cannot be hydrolyzed to produce a smaller carbohydrate molecule. Disaccharides contain two monosaccharide units, and polysaccharides contain many monosaccharide units. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/19%3A_Biochemistry/19.03%3A_Carbohydrates.txt |
On July 11, 2003, the Food and Drug Administration amended its food labeling regulations to require that manufacturers list the amount of trans fatty acids on Nutrition Facts labels of foods and dietary supplements, effective January 1, 2006. This amendment was a response to published studies demonstrating a link between the consumption of trans fatty acids and an increased risk of heart disease. Trans fatty acids are produced in the conversion of liquid oils to solid fats, as in the creation of many commercial margarines and shortenings. They have been shown to increase the levels of low-density lipoproteins (LDLs)—complexes that are often referred to as bad cholesterol—in the blood. In this chapter, you will learn about fatty acids and what is meant by a trans fatty acid, as well as the difference between fats and oils. You will also learn what cholesterol is and why it is an important molecule in the human body.
Fats and oils, found in many of the foods we eat, belong to a class of biomolecules known as lipids. Gram for gram, they pack more than twice the caloric content of carbohydrates: the oxidation of fats and oils supplies about 9 kcal of energy for every gram oxidized, whereas the oxidation of carbohydrates supplies only 4 kcal/g. Although the high caloric content of fats may be bad news for the dieter, it says something about the efficiency of nature’s designs. Our bodies use carbohydrates, primarily in the form of glucose, for our immediate energy needs. Our capacity for storing carbohydrates for later use is limited to tucking away a bit of glycogen in the liver or in muscle tissue. We store our reserve energy in lipid form, which requires far less space than the same amount of energy stored in carbohydrate form. Lipids have other biological functions besides energy storage. They are a major component of the membranes of the 10 trillion cells in our bodies. They serve as protective padding and insulation for vital organs. Furthermore, without lipids in our diets, we would be deficient in the fat-soluble vitamins A, D, E, and K.
Lipids are not defined by the presence of specific functional groups, as carbohydrates are, but by a physical property—solubility. Compounds isolated from body tissues are classified as lipids if they are more soluble in organic solvents, such as dichloromethane, than in water. By this criterion, the lipid category includes not only fats and oils, which are esters of the trihydroxy alcohol glycerol and fatty acids, but also compounds that incorporate functional groups derived from phosphoric acid, carbohydrates, or amino alcohols, as well as steroid compounds such as cholesterol (Figure \(1\) presents one scheme for classifying the various kinds of lipids). We will discuss the various kinds of lipids by considering one subclass at a time and pointing out structural similarities and differences as we go.
Learning Objectives
• To recognize the structures of common fatty acids and classify them as saturated, monounsaturated, or polyunsaturated.
Fatty acids are carboxylic acids that are structural components of fats, oils, and all other categories of lipids, except steroids. More than 70 have been identified in nature. They usually contain an even number of carbon atoms (typically 12–20), are generally unbranched, and can be classified by the presence and number of carbon-to-carbon double bonds. Thus, saturated fatty acids contain no carbon-to-carbon double bonds, monounsaturated fatty acids contain one carbon-to-carbon double bond, and polyunsaturated fatty acids contain two or more carbon-to-carbon double bonds.
Table \(1\) lists some common fatty acids and one important source for each. The atoms or groups around the double bonds in unsaturated fatty acids can be arranged in either the cis or trans isomeric form. Naturally occurring fatty acids are generally in the cis configuration.
Table \(1\): Some Common Fatty Acids Found in Natural Fats
Name Abbreviated Structural Formula Condensed Structural Formula Melting Point (°C) Source
lauric acid C11H23COOH CH3(CH2)10COOH 44 palm kernel oil
myristic acid C13H27COOH CH3(CH2)12COOH 58 oil of nutmeg
palmitic acid C15H31COOH CH3(CH2)14COOH 63 palm oil
palmitoleic acid C15H29COOH CH3(CH2)5CH=CH(CH2)7COOH 0.5 macadamia oil
stearic acid C17H35COOH CH3(CH2)16COOH 70 cocoa butter
oleic acid C17H33COOH CH3(CH2)7CH=CH(CH2)7COOH 16 olive oil
linoleic acid C17H31COOH CH3(CH2)3(CH2CH=CH)2(CH2)7COOH −5 canola oil
α-linolenic acid C17H29COOH CH3(CH2CH=CH)3(CH2)7COOH −11 flaxseed
arachidonic acid C19H31COOH CH3(CH2)4(CH2CH=CH)4(CH2)2COOH −50 liver
Two polyunsaturated fatty acids—linoleic and α-linolenic acids—are termed essential fatty acids because humans must obtain them from their diets. Both substances are required for normal growth and development, but the human body does not synthesize them. The body uses linoleic acid to synthesize many of the other unsaturated fatty acids, such as arachidonic acid, a precursor for the synthesis of prostaglandins. In addition, the essential fatty acids are necessary for the efficient transport and metabolism of cholesterol. The average daily diet should contain about 4–6 g of the essential fatty acids.
To Your Health: Prostaglandins
Prostaglandins are chemical messengers synthesized in the cells in which their physiological activity is expressed. They are unsaturated fatty acids containing 20 carbon atoms and are synthesized from arachidonic acid—a polyunsaturated fatty acid—when needed by a particular cell. They are called prostaglandins because they were originally isolated from semen found in the prostate gland. It is now known that they are synthesized in nearly all mammalian tissues and affect almost all organs in the body. The five major classes of prostaglandins are designated as PGA, PGB, PGE, PGF, and PGI. Subscripts are attached at the end of these abbreviations to denote the number of double bonds outside the five-carbon ring in a given prostaglandin.
The prostaglandins are among the most potent biological substances known. Slight structural differences give them highly distinct biological effects; however, all prostaglandins exhibit some ability to induce smooth muscle contraction, lower blood pressure, and contribute to the inflammatory response. Aspirin and other nonsteroidal anti-inflammatory agents, such as ibuprofen, obstruct the synthesis of prostaglandins by inhibiting cyclooxygenase, the enzyme needed for the initial step in the conversion of arachidonic acid to prostaglandins.
Their wide range of physiological activity has led to the synthesis of hundreds of prostaglandins and their analogs. Derivatives of PGE2 are now used in the United States to induce labor. Other prostaglandins have been employed clinically to lower or increase blood pressure, inhibit stomach secretions, relieve nasal congestion, relieve asthma, and prevent the formation of blood clots, which are associated with heart attacks and strokes.
Although we often draw the carbon atoms in a straight line, they actually have more of a zigzag configuration (Figure \(\PageIndex{2a}\)). Viewed as a whole, however, the saturated fatty acid molecule is relatively straight (Figure \(\PageIndex{2b}\)). Such molecules pack closely together into a crystal lattice, maximizing the strength of dispersion forces and causing fatty acids and the fats derived from them to have relatively high melting points. In contrast, each cis carbon-to-carbon double bond in an unsaturated fatty acid produces a pronounced bend in the molecule, so that these molecules do not stack neatly. As a result, the intermolecular attractions of unsaturated fatty acids (and unsaturated fats) are weaker, causing these substances to have lower melting points. Most are liquids at room temperature.
Waxes are esters formed from long-chain fatty acids and long-chain alcohols. Most natural waxes are mixtures of such esters. Plant waxes on the surfaces of leaves, stems, flowers, and fruits protect the plant from dehydration and invasion by harmful microorganisms. Carnauba wax, used extensively in floor waxes, automobile waxes, and furniture polish, is largely myricyl cerotate, obtained from the leaves of certain Brazilian palm trees. Animals also produce waxes that serve as protective coatings, keeping the surfaces of feathers, skin, and hair pliable and water repellent. In fact, if the waxy coating on the feathers of a water bird is dissolved as a result of the bird swimming in an oil slick, the feathers become wet and heavy, and the bird, unable to maintain its buoyancy, drowns.
Summary
Fatty acids are carboxylic acids that are the structural components of many lipids. They may be saturated or unsaturated. Most fatty acids are unbranched and contain an even number of carbon atoms. Unsaturated fatty acids have lower melting points than saturated fatty acids containing the same number of carbon atoms.
Learning Objectives
• Identify the distinguishing characteristics of membrane lipids.
• Describe membrane components and how they are arranged.
All living cells are surrounded by a cell membrane. Plant cells (Figure \(\PageIndex{1A}\)) and animal cells (Figure \(\PageIndex{1B}\)) contain a cell nucleus that is also surrounded by a membrane and holds the genetic information for the cell. Everything between the cell membrane and the nuclear membrane—including intracellular fluids and various subcellular components such as the mitochondria and ribosomes—is called the cytoplasm. The membranes of all cells have a fundamentally similar structure, but membrane function varies tremendously from one organism to another and even from one cell to another within a single organism. This diversity arises mainly from the presence of different proteins and lipids in the membrane.
The lipids in cell membranes are highly polar but have dual characteristics: part of the lipid is ionic and therefore dissolves in water, whereas the rest has a hydrocarbon structure and therefore dissolves in nonpolar substances. Often, the ionic part is referred to as hydrophilic, meaning “water loving,” and the nonpolar part as hydrophobic, meaning “water fearing” (repelled by water). When allowed to float freely in water, polar lipids spontaneously cluster together in any one of three arrangements: micelles, monolayers, and bilayers (Figure \(2\)).
Micelles are aggregations in which the lipids’ hydrocarbon tails—being hydrophobic—are directed toward the center of the assemblage and away from the surrounding water while the hydrophilic heads are directed outward, in contact with the water. Each micelle may contain thousands of lipid molecules. Polar lipids may also form a monolayer, a layer one molecule thick on the surface of the water. The polar heads face into water, and the nonpolar tails stick up into the air. Bilayers are double layers of lipids arranged so that the hydrophobic tails are sandwiched between an inner surface and an outer surface consisting of hydrophilic heads. The hydrophilic heads are in contact with water on either side of the bilayer, whereas the tails, sequestered inside the bilayer, are prevented from having contact with the water. Bilayers like this make up every cell membrane (Figure \(3\)).
In the bilayer interior, the hydrophobic tails (that is, the fatty acid portions of lipid molecules) interact by means of dispersion forces. The interactions are weakened by the presence of unsaturated fatty acids. As a result, the membrane components are free to mill about to some extent, and the membrane is described as fluid.
The lipids found in cell membranes can be categorized in various ways. Phospholipids are lipids containing phosphorus. Glycolipids are sugar-containing lipids. The latter are found exclusively on the outer surface of the cell membrane, acting as distinguishing surface markers for the cell and thus serving in cellular recognition and cell-to-cell communication. Sphingolipids are phospholipids or glycolipids that contain the unsaturated amino alcohol sphingosine rather than glycerol. Diagrammatic structures of representative membrane lipids are presented in Figure \(4\).
Phosphoglycerides (also known as glycerophospholipids) are the most abundant phospholipids in cell membranes. They consist of a glycerol unit with fatty acids attached to the first two carbon atoms, while a phosphoric acid unit, esterified with an alcohol molecule (usually an amino alcohol, as in part (a) of Figure \(5\)) is attached to the third carbon atom of glycerol (part (b) of Figure \(5\)). Notice that the phosphoglyceride molecule is identical to a triglyceride up to the phosphoric acid unit (part (b) of Figure \(5\)).
There are two common types of phosphoglycerides. Phosphoglycerides containing ethanolamine as the amino alcohol are called phosphatidylethanolamines or cephalins. Cephalins are found in brain tissue and nerves and also have a role in blood clotting. Phosphoglycerides containing choline as the amino alcohol unit are called phosphatidylcholines or lecithins. Lecithins occur in all living organisms. Like cephalins, they are important constituents of nerve and brain tissue. Egg yolks are especially rich in lecithins. Commercial-grade lecithins isolated from soybeans are widely used in foods as emulsifying agents. An emulsifying agent is used to stabilize an emulsion—a dispersion of two liquids that do not normally mix, such as oil and water. Many foods are emulsions. Milk is an emulsion of butterfat in water. The emulsifying agent in milk is a protein called casein. Mayonnaise is an emulsion of salad oil in water, stabilized by lecithins present in egg yolk.
Sphingomyelins, the simplest sphingolipids, each contain a fatty acid, a phosphoric acid, sphingosine, and choline (Figure \(6\)). Because they contain phosphoric acid, they are also classified as phospholipids. Sphingomyelins are important constituents of the myelin sheath surrounding the axon of a nerve cell. Multiple sclerosis is one of several diseases resulting from damage to the myelin sheath.
Most animal cells contain sphingolipids called cerebrosides (Figure \(7\)). Cerebrosides are composed of sphingosine, a fatty acid, and galactose or glucose. They therefore resemble sphingomyelins but have a sugar unit in place of the choline phosphate group. Cerebrosides are important constituents of the membranes of nerve and brain cells.
The sphingolipids called gangliosides are more complex, usually containing a branched chain of three to eight monosaccharides and/or substituted sugars. Because of considerable variation in their sugar components, about 130 varieties of gangliosides have been identified. Most cell-to-cell recognition and communication processes (e.g., blood group antigens) depend on differences in the sequences of sugars in these compounds. Gangliosides are most prevalent in the outer membranes of nerve cells, although they also occur in smaller quantities in the outer membranes of most other cells. Because cerebrosides and gangliosides contain sugar groups, they are also classified as glycolipids.
Membrane Proteins
If membranes were composed only of lipids, very few ions or polar molecules could pass through their hydrophobic “sandwich filling” to enter or leave any cell. However, certain charged and polar species do cross the membrane, aided by proteins that move about in the lipid bilayer. The two major classes of proteins in the cell membrane are integral proteins, which span the hydrophobic interior of the bilayer, and peripheral proteins, which are more loosely associated with the surface of the lipid bilayer (Figure \(3\)). Peripheral proteins may be attached to integral proteins, to the polar head groups of phospholipids, or to both by hydrogen bonding and electrostatic forces.
Small ions and molecules soluble in water enter and leave the cell by way of channels through the integral proteins. Some proteins, called carrier proteins, facilitate the passage of certain molecules, such as hormones and neurotransmitters, by specific interactions between the protein and the molecule being transported.
Summary
Lipids are important components of biological membranes. These lipids have dual characteristics: part of the molecule is hydrophilic, and part of the molecule is hydrophobic. Membrane lipids may be classified as phospholipids, glycolipids, and/or sphingolipids. Proteins are another important component of biological membranes. Integral proteins span the lipid bilayer, while peripheral proteins are more loosely associated with the surface of the membrane. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/19%3A_Biochemistry/19.04%3A_Lipids.txt |
Learning Objectives
• To recognize amino acids and classify them based on the characteristics of their side chains.
The proteins in all living species, from bacteria to humans, are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids. However, two additional amino acids have been found in limited quantities in proteins: Selenocysteine was discovered in 1986, while pyrrolysine was discovered in 2002.
The amino acids are colorless, nonvolatile, crystalline solids, melting and decomposing at temperatures above 200°C. These melting temperatures are more like those of inorganic salts than those of amines or organic acids and indicate that the structures of the amino acids in the solid state and in neutral solution are best represented as having both a negatively charged group and a positively charged group. Such a species is known as a zwitterion.
Classification
In addition to the amino and carboxyl groups, amino acids have a side chain or R group attached to the α-carbon. Each amino acid has unique characteristics arising from the size, shape, solubility, and ionization properties of its R group. As a result, the side chains of amino acids exert a profound effect on the structure and biological activity of proteins. Although amino acids can be classified in various ways, one common approach is to classify them according to whether the functional group on the side chain at neutral pH is nonpolar, polar but uncharged, negatively charged, or positively charged. The structures and names of the 20 amino acids, their one- and three-letter abbreviations, and some of their distinctive features are given in Table \(1\).
Table \(1\): Common Amino Acids Found in Proteins
Common Name Abbreviation Structural Formula (at pH 6) Molar Mass Distinctive Feature
Amino acids with a nonpolar R group
glycine gly (G) 75 the only amino acid lacking a chiral carbon
alanine ala (A) 89
valine val (V) 117 a branched-chain amino acid
leucine leu (L) 131 a branched-chain amino acid
isoleucine ile (I) 131 an essential amino acid because most animals cannot synthesize branched-chain amino acids
phenylalanine phe (F) 165 also classified as an aromatic amino acid
tryptophan trp (W) 204 also classified as an aromatic amino acid
methionine met (M) 149 side chain functions as a methyl group donor
proline pro (P) 115 contains a secondary amine group; referred to as an α-imino acid
Amino acids with a polar but neutral R group
serine ser (S) 105 found at the active site of many enzymes
threonine thr (T) 119 named for its similarity to the sugar threose
cysteine cys (C) 121 oxidation of two cysteine molecules yields cystine
tyrosine tyr (Y) 181 also classified as an aromatic amino acid
asparagine asn (N) 132 the amide of aspartic acid
glutamine gln (Q) 146 the amide of glutamic acid
Amino acids with a negatively charged R group
aspartic acid asp (D) 132 carboxyl groups are ionized at physiological pH; also known as aspartate
glutamic acid glu (E) 146 carboxyl groups are ionized at physiological pH; also known as glutamate
Amino acids with a positively charged R group
histidine his (H) 155 the only amino acid whose R group has a pKa (6.0) near physiological pH
lysine lys (K) 147
arginine arg (R) 175 almost as strong a base as sodium hydroxide
The first amino acid to be isolated was asparagine in 1806. It was obtained from protein found in asparagus juice (hence the name). Glycine, the major amino acid found in gelatin, was named for its sweet taste (Greek glykys, meaning “sweet”). In some cases an amino acid found in a protein is actually a derivative of one of the common 20 amino acids (one such derivative is hydroxyproline). The modification occurs after the amino acid has been assembled into a protein.
Configuration
Notice in Table \(1\) that glycine is the only amino acid whose α-carbon is not chiral. Therefore, with the exception of glycine, the amino acids could theoretically exist in either the D- or the L-enantiomeric form and rotate plane-polarized light. As with sugars, chemists used L-glyceraldehyde as the reference compound for the assignment of absolute configuration to amino acids. Its structure closely resembles an amino acid structure except that in the latter, an amino group takes the place of the OH group on the chiral carbon of the L-glyceraldehyde and a carboxylic acid replaces the aldehyde. Modern stereochemistry assignments using the Cahn-Ingold-Prelog priority rules used ubiquitously in chemistry show that all of the naturally occurring chiral amino acids are S except Cys which is R.
We learned that all naturally occurring sugars belong to the D series. It is interesting, therefore, that nearly all known plant and animal proteins are composed entirely of L-amino acids. However, certain bacteria contain D-amino acids in their cell walls, and several antibiotics (e.g., actinomycin D and the gramicidins) contain varying amounts of D-leucine, D-phenylalanine, and D-valine.
Summary
Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids.
Learning Objectives
• Describe the four levels of protein structure.
• Identify the types of attractive interactions that hold proteins in their most stable three-dimensional structure.
• Explain what happens when proteins are denatured.
• Identify how a protein can be denatured.
Each of the thousands of naturally occurring proteins has its own characteristic amino acid composition and sequence that result in a unique three-dimensional shape. Since the 1950s, scientists have determined the amino acid sequences and three-dimensional conformation of numerous proteins and thus obtained important clues on how each protein performs its specific function in the body.
Proteins are compounds of high molar mass consisting largely or entirely of chains of amino acids. Because of their great complexity, protein molecules cannot be classified on the basis of specific structural similarities, as carbohydrates and lipids are categorized. The two major structural classifications of proteins are based on far more general qualities: whether the protein is (1) fiberlike and insoluble or (2) globular and soluble. Some proteins, such as those that compose hair, skin, muscles, and connective tissue, are fiberlike. These fibrous proteins are insoluble in water and usually serve structural, connective, and protective functions. Examples of fibrous proteins are keratins, collagens, myosins, and elastins. Hair and the outer layer of skin are composed of keratin. Connective tissues contain collagen. Myosins are muscle proteins and are capable of contraction and extension. Elastins are found in ligaments and the elastic tissue of artery walls.
Globular proteins, the other major class, are soluble in aqueous media. In these proteins, the chains are folded so that the molecule as a whole is roughly spherical. Familiar examples include egg albumin from egg whites and serum albumin in blood. Serum albumin plays a major role in transporting fatty acids and maintaining a proper balance of osmotic pressures in the body. Hemoglobin and myoglobin, which are important for binding oxygen, are also globular proteins.
Levels of Protein Structure
The structure of proteins is generally described as having four organizational levels. The first of these is the primary structure, which is the number and sequence of amino acids in a protein’s polypeptide chain or chains, beginning with the free amino group and maintained by the peptide bonds connecting each amino acid to the next. The primary structure of insulin, composed of 51 amino acids, is shown in Figure \(1\).
A protein molecule is not a random tangle of polypeptide chains. Instead, the chains are arranged in unique but specific conformations. The term secondary structure refers to the fixed arrangement of the polypeptide backbone. On the basis of X ray studies, Linus Pauling and Robert Corey postulated that certain proteins or portions of proteins twist into a spiral or a helix. This helix is stabilized by intrachain hydrogen bonding between the carbonyl oxygen atom of one amino acid and the amide hydrogen atom four amino acids up the chain (located on the next turn of the helix) and is known as a right-handed α-helix. X ray data indicate that this helix makes one turn for every 3.6 amino acids, and the side chains of these amino acids project outward from the coiled backbone (Figure \(2\)). The α-keratins, found in hair and wool, are exclusively α-helical in conformation. Some proteins, such as gamma globulin, chymotrypsin, and cytochrome c, have little or no helical structure. Others, such as hemoglobin and myoglobin, are helical in certain regions but not in others.
Another common type of secondary structure, called the β-pleated sheet conformation, is a sheetlike arrangement in which two or more extended polypeptide chains (or separate regions on the same chain) are aligned side by side. The aligned segments can run either parallel or antiparallel—that is, the N-terminals can face in the same direction on adjacent chains or in different directions—and are connected by interchain hydrogen bonding (Figure \(3\)). The β-pleated sheet is particularly important in structural proteins, such as silk fibroin. It is also seen in portions of many enzymes, such as carboxypeptidase A and lysozyme.
Tertiary structure refers to the unique three-dimensional shape of the protein as a whole, which results from the folding and bending of the protein backbone. The tertiary structure is intimately tied to the proper biochemical functioning of the protein. Figure \(4\) shows a depiction of the three-dimensional structure of insulin.
Four major types of attractive interactions determine the shape and stability of the tertiary structure of proteins. You studied several of them previously.
1. Ionic bonding. Ionic bonds result from electrostatic attractions between positively and negatively charged side chains of amino acids. For example, the mutual attraction between an aspartic acid carboxylate ion and a lysine ammonium ion helps to maintain a particular folded area of a protein (part (a) of Figure \(5\)).
2. Hydrogen bonding. Hydrogen bonding forms between a highly electronegative oxygen atom or a nitrogen atom and a hydrogen atom attached to another oxygen atom or a nitrogen atom, such as those found in polar amino acid side chains. Hydrogen bonding (as well as ionic attractions) is extremely important in both the intra- and intermolecular interactions of proteins (part (b) of Figure \(5\)).
3. Disulfide linkages. Two cysteine amino acid units may be brought close together as the protein molecule folds. Subsequent oxidation and linkage of the sulfur atoms in the highly reactive sulfhydryl (SH) groups leads to the formation of cystine (part (c) of Figure \(5\)). Intrachain disulfide linkages are found in many proteins, including insulin (yellow bars in Figure \(1\)) and have a strong stabilizing effect on the tertiary structure.
1. Dispersion forces. Dispersion forces arise when a normally nonpolar atom becomes momentarily polar due to an uneven distribution of electrons, leading to an instantaneous dipole that induces a shift of electrons in a neighboring nonpolar atom. Dispersion forces are weak but can be important when other types of interactions are either missing or minimal (part (d) of Figure \(5\)). This is the case with fibroin, the major protein in silk, in which a high proportion of amino acids in the protein have nonpolar side chains. The term hydrophobic interaction is often misused as a synonym for dispersion forces. Hydrophobic interactions arise because water molecules engage in hydrogen bonding with other water molecules (or groups in proteins capable of hydrogen bonding). Because nonpolar groups cannot engage in hydrogen bonding, the protein folds in such a way that these groups are buried in the interior part of the protein structure, minimizing their contact with water.
When a protein contains more than one polypeptide chain, each chain is called a subunit. The arrangement of multiple subunits represents a fourth level of structure, the quaternary structure of a protein. Hemoglobin, with four polypeptide chains or subunits, is the most frequently cited example of a protein having quaternary structure (Figure \(6\)). The quaternary structure of a protein is produced and stabilized by the same kinds of interactions that produce and maintain the tertiary structure. A schematic representation of the four levels of protein structure is in Figure \(7\).
The primary structure consists of the specific amino acid sequence. The resulting peptide chain can twist into an α-helix, which is one type of secondary structure. This helical segment is incorporated into the tertiary structure of the folded polypeptide chain. The single polypeptide chain is a subunit that constitutes the quaternary structure of a protein, such as hemoglobin that has four polypeptide chains.
Denaturation of Proteins
The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job. (Sometimes denaturation is equated with the precipitation or coagulation of a protein; our definition is a bit broader.) A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation (Figure \(1\)).
Figure \(1\): Protein Denaturation Methods
Method Effect on Protein Structure
Heat above 50°C or ultraviolet (UV) radiation Heat or UV radiation supplies kinetic energy to protein molecules, causing their atoms to vibrate more rapidly and disrupting relatively weak hydrogen bonding and dispersion forces.
Use of organic compounds, such as ethyl alcohol These compounds are capable of engaging in intermolecular hydrogen bonding with protein molecules, disrupting intramolecular hydrogen bonding within the protein.
Salts of heavy metal ions, such as mercury, silver, and lead These ions form strong bonds with the carboxylate anions of the acidic amino acids or SH groups of cysteine, disrupting ionic bonds and disulfide linkages.
Alkaloid reagents, such as tannic acid (used in tanning leather) These reagents combine with positively charged amino groups in proteins to disrupt ionic bonds.
Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates. No one has yet reversed that process. However, given the proper circumstances and enough time, a protein that has unfolded under sufficiently gentle conditions can refold and may again exhibit biological activity (Figure \(8\)). Such evidence suggests that, at least for these proteins, the primary structure determines the secondary and tertiary structure. A given sequence of amino acids seems to adopt its particular three-dimensional arrangement naturally if conditions are right.
The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin.
Summary
Proteins can be divided into two categories: fibrous, which tend to be insoluble in water, and globular, which are more soluble in water. A protein may have up to four levels of structure. The primary structure consists of the specific amino acid sequence. The resulting peptide chain can form an α-helix or β-pleated sheet (or local structures not as easily categorized), which is known as secondary structure. These segments of secondary structure are incorporated into the tertiary structure of the folded polypeptide chain. The quaternary structure describes the arrangements of subunits in a protein that contains more than one subunit. Four major types of attractive interactions determine the shape and stability of the folded protein: ionic bonding, hydrogen bonding, disulfide linkages, and dispersion forces. A wide variety of reagents and conditions can cause a protein to unfold or denature. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/19%3A_Biochemistry/19.05%3A_Proteins.txt |
Learning Objectives
• Describe the four levels of protein structure.
• Identify the types of attractive interactions that hold proteins in their most stable three-dimensional structure.
• Explain what happens when proteins are denatured.
• Identify how a protein can be denatured.
Each of the thousands of naturally occurring proteins has its own characteristic amino acid composition and sequence that result in a unique three-dimensional shape. Since the 1950s, scientists have determined the amino acid sequences and three-dimensional conformation of numerous proteins and thus obtained important clues on how each protein performs its specific function in the body.
Proteins are compounds of high molar mass consisting largely or entirely of chains of amino acids. Because of their great complexity, protein molecules cannot be classified on the basis of specific structural similarities, as carbohydrates and lipids are categorized. The two major structural classifications of proteins are based on far more general qualities: whether the protein is (1) fiberlike and insoluble or (2) globular and soluble. Some proteins, such as those that compose hair, skin, muscles, and connective tissue, are fiberlike. These fibrous proteins are insoluble in water and usually serve structural, connective, and protective functions. Examples of fibrous proteins are keratins, collagens, myosins, and elastins. Hair and the outer layer of skin are composed of keratin. Connective tissues contain collagen. Myosins are muscle proteins and are capable of contraction and extension. Elastins are found in ligaments and the elastic tissue of artery walls.
Globular proteins, the other major class, are soluble in aqueous media. In these proteins, the chains are folded so that the molecule as a whole is roughly spherical. Familiar examples include egg albumin from egg whites and serum albumin in blood. Serum albumin plays a major role in transporting fatty acids and maintaining a proper balance of osmotic pressures in the body. Hemoglobin and myoglobin, which are important for binding oxygen, are also globular proteins.
Levels of Protein Structure
The structure of proteins is generally described as having four organizational levels. The first of these is the primary structure, which is the number and sequence of amino acids in a protein’s polypeptide chain or chains, beginning with the free amino group and maintained by the peptide bonds connecting each amino acid to the next. The primary structure of insulin, composed of 51 amino acids, is shown in Figure \(1\).
A protein molecule is not a random tangle of polypeptide chains. Instead, the chains are arranged in unique but specific conformations. The term secondary structure refers to the fixed arrangement of the polypeptide backbone. On the basis of X ray studies, Linus Pauling and Robert Corey postulated that certain proteins or portions of proteins twist into a spiral or a helix. This helix is stabilized by intrachain hydrogen bonding between the carbonyl oxygen atom of one amino acid and the amide hydrogen atom four amino acids up the chain (located on the next turn of the helix) and is known as a right-handed α-helix. X ray data indicate that this helix makes one turn for every 3.6 amino acids, and the side chains of these amino acids project outward from the coiled backbone (Figure \(2\)). The α-keratins, found in hair and wool, are exclusively α-helical in conformation. Some proteins, such as gamma globulin, chymotrypsin, and cytochrome c, have little or no helical structure. Others, such as hemoglobin and myoglobin, are helical in certain regions but not in others.
Another common type of secondary structure, called the β-pleated sheet conformation, is a sheetlike arrangement in which two or more extended polypeptide chains (or separate regions on the same chain) are aligned side by side. The aligned segments can run either parallel or antiparallel—that is, the N-terminals can face in the same direction on adjacent chains or in different directions—and are connected by interchain hydrogen bonding (Figure \(3\)). The β-pleated sheet is particularly important in structural proteins, such as silk fibroin. It is also seen in portions of many enzymes, such as carboxypeptidase A and lysozyme.
Tertiary structure refers to the unique three-dimensional shape of the protein as a whole, which results from the folding and bending of the protein backbone. The tertiary structure is intimately tied to the proper biochemical functioning of the protein. Figure \(4\) shows a depiction of the three-dimensional structure of insulin.
Four major types of attractive interactions determine the shape and stability of the tertiary structure of proteins. You studied several of them previously.
1. Ionic bonding. Ionic bonds result from electrostatic attractions between positively and negatively charged side chains of amino acids. For example, the mutual attraction between an aspartic acid carboxylate ion and a lysine ammonium ion helps to maintain a particular folded area of a protein (part (a) of Figure \(5\)).
2. Hydrogen bonding. Hydrogen bonding forms between a highly electronegative oxygen atom or a nitrogen atom and a hydrogen atom attached to another oxygen atom or a nitrogen atom, such as those found in polar amino acid side chains. Hydrogen bonding (as well as ionic attractions) is extremely important in both the intra- and intermolecular interactions of proteins (part (b) of Figure \(5\)).
3. Disulfide linkages. Two cysteine amino acid units may be brought close together as the protein molecule folds. Subsequent oxidation and linkage of the sulfur atoms in the highly reactive sulfhydryl (SH) groups leads to the formation of cystine (part (c) of Figure \(5\)). Intrachain disulfide linkages are found in many proteins, including insulin (yellow bars in Figure \(1\)) and have a strong stabilizing effect on the tertiary structure.
1. Dispersion forces. Dispersion forces arise when a normally nonpolar atom becomes momentarily polar due to an uneven distribution of electrons, leading to an instantaneous dipole that induces a shift of electrons in a neighboring nonpolar atom. Dispersion forces are weak but can be important when other types of interactions are either missing or minimal (part (d) of Figure \(5\)). This is the case with fibroin, the major protein in silk, in which a high proportion of amino acids in the protein have nonpolar side chains. The term hydrophobic interaction is often misused as a synonym for dispersion forces. Hydrophobic interactions arise because water molecules engage in hydrogen bonding with other water molecules (or groups in proteins capable of hydrogen bonding). Because nonpolar groups cannot engage in hydrogen bonding, the protein folds in such a way that these groups are buried in the interior part of the protein structure, minimizing their contact with water.
When a protein contains more than one polypeptide chain, each chain is called a subunit. The arrangement of multiple subunits represents a fourth level of structure, the quaternary structure of a protein. Hemoglobin, with four polypeptide chains or subunits, is the most frequently cited example of a protein having quaternary structure (Figure \(6\)). The quaternary structure of a protein is produced and stabilized by the same kinds of interactions that produce and maintain the tertiary structure. A schematic representation of the four levels of protein structure is in Figure \(7\).
The primary structure consists of the specific amino acid sequence. The resulting peptide chain can twist into an α-helix, which is one type of secondary structure. This helical segment is incorporated into the tertiary structure of the folded polypeptide chain. The single polypeptide chain is a subunit that constitutes the quaternary structure of a protein, such as hemoglobin that has four polypeptide chains.
Denaturation of Proteins
The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job. (Sometimes denaturation is equated with the precipitation or coagulation of a protein; our definition is a bit broader.) A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation (Figure \(1\)).
Figure \(1\): Protein Denaturation Methods
Method Effect on Protein Structure
Heat above 50°C or ultraviolet (UV) radiation Heat or UV radiation supplies kinetic energy to protein molecules, causing their atoms to vibrate more rapidly and disrupting relatively weak hydrogen bonding and dispersion forces.
Use of organic compounds, such as ethyl alcohol These compounds are capable of engaging in intermolecular hydrogen bonding with protein molecules, disrupting intramolecular hydrogen bonding within the protein.
Salts of heavy metal ions, such as mercury, silver, and lead These ions form strong bonds with the carboxylate anions of the acidic amino acids or SH groups of cysteine, disrupting ionic bonds and disulfide linkages.
Alkaloid reagents, such as tannic acid (used in tanning leather) These reagents combine with positively charged amino groups in proteins to disrupt ionic bonds.
Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates. No one has yet reversed that process. However, given the proper circumstances and enough time, a protein that has unfolded under sufficiently gentle conditions can refold and may again exhibit biological activity (Figure \(8\)). Such evidence suggests that, at least for these proteins, the primary structure determines the secondary and tertiary structure. A given sequence of amino acids seems to adopt its particular three-dimensional arrangement naturally if conditions are right.
The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin.
Summary
Proteins can be divided into two categories: fibrous, which tend to be insoluble in water, and globular, which are more soluble in water. A protein may have up to four levels of structure. The primary structure consists of the specific amino acid sequence. The resulting peptide chain can form an α-helix or β-pleated sheet (or local structures not as easily categorized), which is known as secondary structure. These segments of secondary structure are incorporated into the tertiary structure of the folded polypeptide chain. The quaternary structure describes the arrangements of subunits in a protein that contains more than one subunit. Four major types of attractive interactions determine the shape and stability of the folded protein: ionic bonding, hydrogen bonding, disulfide linkages, and dispersion forces. A wide variety of reagents and conditions can cause a protein to unfold or denature.
19.07: Nucleic Acids- Blueprints for Proteins
Learning Objectives
• To identify the different molecules that combine to form nucleotides.
The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides. The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 109 nucleotides. Nucleotides can be further broken down to phosphoric acid (H3PO4), a pentose sugar (a sugar with five carbon atoms), and a nitrogenous base (a base containing nitrogen atoms).
$\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar} \nonumber$
If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide, and the resulting nucleic acid is ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide, and the nucleic acid is DNA.
The nitrogenous bases found in nucleotides are classified as pyrimidines or purines. Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure $1$).
The formation of a bond between C1′ of the pentose sugar and N1 of the pyrimidine base or N9 of the purine base joins the pentose sugar to the nitrogenous base. In the formation of this bond, a molecule of water is removed. Table $1$ summarizes the similarities and differences in the composition of nucleotides in DNA and RNA.
The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring.
Table $1$: Composition of Nucleotides in DNA and RNA
Composition DNA RNA
purine bases adenine and guanine adenine and guanine
pyrimidine bases cytosine and thymine cytosine and uracil
pentose sugar 2-deoxyribose ribose
inorganic acid phosphoric acid (H3PO4) H3PO4
The names and structures of the major ribonucleotides and one of the deoxyribonucleotides are given in Figure $2$.
Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure $3$, have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD+), and coenzyme A, contain adenine nucleotides as structural components.
Summary
Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry/19%3A_Biochemistry/19.06%3A_Protein_Structure.txt |
How does soap relate to chemistry?
Chemistry affects every area of our lives. Here is just one example of chemistry in action – the making of soaps. Soap was once made by boiling animal fat in ashes – the product was hard on the skin and not very pleasant to use. Today, soap manufacture involves complicated chemical processes to provide a wide variety of soaps for different skin types. Colors and odors can be custom-made for that individual experience.
"What is this made of? How can we produce this material quickly and at a low cost? Will this product harm us or help us?"—these are all questions that can be answered using the science of chemistry.
What is Chemistry?
Science is a general term used to describe the principled, rigorous study of the natural world. Many interconnected disciplines fall under this broader concept. For example, physics is the study of motion and forces. Biology is the study of living things. Geology is the study of the Earth and the rocks and minerals of which it is comprised. Chemistry is the study of the composition of matter and the changes that matter undergoes. Matter is anything that has mass and takes up space. Virtually everything around us is matter, including both living and nonliving things. Chemistry affects nearly everything we see and every action we take. Chemistry explains why milk that is left in the refrigerator for too long turns sour. Chemistry explains why certain pollutants called chlorofluorocarbons have done lasting damage to the ozone layer of our planet. Chemistry explains why the leaves of deciduous trees turn from green in the summer to various shades of red and yellow in the autumn (Figure below).
Chemistry touches every area of our lives. The medicines we take, the food we eat, the clothes we wear—all of these materials and more are, in some way or another, products of chemistry.
Chemists look at the world in two ways, often simultaneously. The two worlds of the chemist are the macroscopic world and the microscopic world. Macroscopic refers to substances and objects that can be seen, touched, and measured directly. Microscopic refers to the small particles that make up all matter. Chemists must observe matter and do experiments macroscopically; then make generalizations and propose explanations that are microscopic in nature. For example, anyone can observe the physical change in appearance that occurs as an iron object, such as a tractor, is left out in the elements and gradually turns to rust. However, a chemist looking at the rusting tractor considers the individual atoms that make up the iron, and how they are changing as a result of exposure to oxygen in the air, and water from rain. Throughout the study of chemistry, there is often a switch back and forth between the macroscopic and microscopic worlds.
Summary
• Chemistry is the study of matter and the changes it undergoes.
• Chemistry considers both macroscopic and microscopic information.
Review
1. Give two examples of chemistry in your everyday life.
2. What is the macroscopic world?
3. What is the microscopic world?
Explore More
1. Read the label on a prepared food product (for example: bread, cereal, dessert). List all of the ingredients in the product. Look up each ingredient on the Internet and write down what that material is doing in the food product.
2. Select your favorite hobby or activity. List all of the items you use in that activity or hobby. For each item, find out how chemistry has contributed to the creation or better operation of that item. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.01%3A_Defining_Chemistry.txt |
Where did chemistry come from? Early "chemists" focused on practical problems—how to make dyes and perfumes, soap manufacture, uses of metals, and glass production, among others. The goal was not to understand the physical world—that came later. People just wanted to make things that would improve their lives in some way.
History of Chemistry
The history of chemistry is an interesting and challenging one. Very early chemists were often motivated mainly by the achievement of a specific goal or product. Making perfume and soaps did not require much theory, just a good recipe and careful attention to detail. There was no standard way of naming materials (and no periodic table that everyone could agree on). However, the science developed over the centuries.
Major progress was made in putting chemistry on a solid foundation when Robert Boyle (1637-1691) began his research in chemistry. He developed the basic ideas for the behavior of gases; gases could thereafter be described mathematically. Boyle also helped pioneer the idea that small particles could combine to form molecules. Many years later, John Dalton used these ideas to develop the atomic theory.
The field of chemistry began to develop rapidly in the 1700s. Joseph Priestley (1733-1804) isolated and characterized several gases: oxygen, carbon monoxide, and nitrous oxide. It was later discovered that nitrous oxide ("laughing gas") worked as an anesthetic. This gas was used for that purpose for the first time in 1844 during a tooth extraction. Other gases discovered during that time were chlorine, by C.W. Scheele (1742-1786) and nitrogen, by Antoine Lavoisier (1743-1794). Lavoisier has been considered by many scholars to be the "father of chemistry".
Chemists continued to discover new compounds in the 1800s. The science also began to develop a more theoretical foundation. John Dalton (1766-1844) put forth his atomic theory in 1807. This idea allowed scientists to think about chemistry in a much more systematic way. Amadeo Avogadro (1776-1856) laid the groundwork for a more quantitative approach to chemistry by calculating the number of particles in a given amount of a gas. A lot of effort was put forth in studying chemical reactions. These efforts led to new materials being produced. Following the invention of the battery by Alessandro Volta (1745-1827), the field of electrochemistry (both theory and application) developed through major contributions by Humphry Davy (1778-1829) and Michael Faraday (1791-1867). Other areas of the discipline also progressed rapidly.
It would take a large book to cover developments in chemistry during the twentieth century and up to today. One major area of expansion was in the area of the chemistry of living processes. Research in photosynthesis in plants, the discovery and characterization of enzymes as biochemical catalysts, elucidation of the structures of biomolecules such as insulin and DNA —these efforts gave rise to an explosion of information in the field of biochemistry.
The practical aspects of chemistry were not ignored. The work of Volta, Davy, and Faraday eventually led to the development of batteries that provided a source of electricity to power a number of devices.
Charles Goodyear (1800-1860) discovered the process of vulcanization, allowing a stable rubber product to be produced for the tires of all the vehicles that we have today. Louis Pasteur (1822-1895) pioneered the use of heat sterilization to eliminate unwanted microorganisms in wine and milk. Alfred Nobel (1833-1896) invented dynamite. After his death, the fortune he made from this product was used to fund the Nobel Prizes in science and the humanities. J.W. Hyatt (1837-1920) developed the first plastic. Leo Baekeland (1863-1944) developed the first synthetic resin, which is widely used for inexpensive and sturdy dinnerware.
Summary
• Many civilizations contributed to the growth of chemistry.
• A lot of early chemical research focused on practical uses.
• Basic chemistry theories were developed during the nineteenth century.
• New materials and batteries are products of modern chemistry.
Review
1. Who invented the first battery?
2. What contribution to chemistry did Robert Boyle make?
3. Who invented dynamite?
4. What was the first synthetic resin used for?
Explore More
Use this resource to answer the following questions: http://www.columbia.edu/itc/chemistr.../chemhist.html
1. Who published the atomic theory?
2. What elements are proposed by Aristotle?
3. Who discovered the proton?
4. What new elements did Marie Curie discover?
5. Who is called the "Father of Modern Chemistry"? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.02%3A_History_of_Chemistry.txt |
A common area of technology to societies of the past was metallurgy. Useful tools could be made with metallurgy that would last for a long time. Weapons could stay sharp longer with improved metals. Precious metals, such as gold and silver, could be refined and used for jewelry or for money. Because it was fairly rare, gold was considered to be very valuable, and became a common means of paying for goods and services.
However, mining for gold is a slow, dirty, and dangerous process. Not everyone owns a gold mine—in both the ancient Egyptian society and during the Roman Empire, the gold mines were the property of the state, not an individual or group. There were few ways for most people to legally get any gold for themselves.
Alchemy
Chemistry is a subject that has its roots in the ancient tradition known as alchemy, from which it derives its name. Alchemy was a combination of philosophy and science that had both practical and mystical aspects. The goals of alchemy were varied and difficult to summarize. In many ways, the alchemists sought to achieve perfection, through such actions as the pursuit of the philosopher’s stone and the elixir of life (Figure below). The philosopher’s stone, it was believed, was a substance that was capable of being used to turn base metals (such as lead) into gold. It was also believed that it could be used to achieve rejuvenation and perhaps immortality. While alchemists did not ultimately succeed in these quests, their work provided the foundation for the modern study of chemistry.
Many of the specific approaches that alchemists used when they tried changing lead into gold are vague and unclear. Each alchemist had their own code for recording data. The processes were kept secret so that others could not profit from them. Different scholars developed their own set of symbols as they recorded the information they came up with. Many alchemists were not very honest, and sometimes took advantage of noblemen by taking money and claiming to be able to make gold from lead, then left town in the middle of the night. On occasion, the nobleman would detect the fraud and have the alchemist hung. By the 1300s, several European rules had declared alchemy to be illegal, and set out strict punishments for those practicing the alchemical arts.
Alchemist Contributions to Chemistry
Alchemists laid the groundwork for many chemical processes, such as the refining of ores, the production of gunpowder, the manufacture of glass and ceramics, leather tanning, and the production of inks, dyes, and paints. Alchemists also made the first attempts at organizing and classifying substances so that they could better understand their reactions and be able to predict the products of their experiments. This eventually led to the modern periodic table, which you will learn about in a later chapter. Alchemy began to fully evolve into chemistry in the 17th century, with a greater emphasis on rational thought and experimentation and less emphasis on spirituality and mysticism.
The alchemists were never successful in changing lead into gold, but modern nuclear physics can accomplish this task. Lead is subjected to nuclear bombardment in a particle accelerator. A small amount of gold can be obtained by this process. However, the cost of the procedure is far more than that of the amount of gold obtained. So, the dream of the alchemists has never (and will never) come true.
Summary
• Gold has been considered valuable by all civilizations.
• The alchemists tried to find the philosopher's stone that would allow them to make gold from lead.
• The alchemists did not successfully transform lead into gold.
• The alchemists laid the ground work for many advances to the new science of chemistry.
Review
1. Why is gold considered to be valuable?
2. Who owned the gold mines during the ancient Egyptian and Roman civilizations?
3. What is the elixir of life?
4. What contributions to modern chemistry were made by the alchemists?
1.04: Areas of Chemistry
The study of modern chemistry has many branches, but can generally be broken down into five main disciplines, or areas of study:
• Physical chemistry
• Organic chemistry
• Inorganic chemistry
• Analytical chemistry
• Biochemistry
Physical Chemistry
Physical chemistry is the study of macroscopic properties, atomic properties, and phenomena in chemical systems. As examples, a physical chemist may study the rates of chemical reactions, the energy transfers that occur in reactions, or the physical structure of materials at the molecular level.
Organic Chemistry
Organic chemistry is the study of chemicals containing carbon. Carbon is one of the most abundant elements on Earth and is capable of forming a tremendously vast number of chemicals (over twenty million so far). Most chemicals found in every living organism are based on carbon.
Inorganic Chemistry
Inorganic chemistry is the study of chemicals that do not, in general, contain carbon. Inorganic chemicals are commonly found in rocks and minerals. One current important area of inorganic chemistry deals with the design and properties of materials involved in energy and information technology.
Analytical Chemistry
Analytical chemistry is the study of the composition of matter. It focuses on separating, identifying, and quantifying chemicals in samples of matter. An analytical chemist may use complex instruments to analyze an unknown material, in order to determine its various components.
Biochemistry
Biochemistry is the study of chemical processes that occur in living things. Research may cover basic cellular processes up to understanding disease states, so that better treatments can be developed.
In practice, chemical research is often not limited to just one of the five major disciplines. A chemist may use biochemistry to isolate a particular chemical found in the human body such as hemoglobin, the oxygen carrying component of red blood cells. He or she may then proceed to analyze the hemoglobin using methods that would pertain to the areas of physical or analytical chemistry. Many chemists specialize in areas that are combinations of the main disciplines, such as bioinorganic chemistry or physical organic chemistry.
Summary
• Five areas of chemistry are described:
• Physical chemistry
• Organic chemistry
• Inorganic chemistry
• Analytical chemistry
• Biochemistry
Review
Match the project with the proper chemistry discipline.
Match the project with the proper chemistry discipline.
a. measuring mercury in seawater 1. biochemistry
b. studying enzymes in cells 2. organic chemistry
c. measuring the electrical properties of solutions 3. inorganic chemistry
d. synthesizing new carbon compounds 4. physical chemistry
e. making new compounds for energy processes 5. analytical chemistry
Explore More
1. Do an internet search using one of the five chemistry areas as a search term. List two significant contributions made to chemistry by that area. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.03%3A_Alchemy.txt |
"How did chemistry develop? What is happening in the field of chemistry today? What can I do with a chemistry degree?"—these are all good questions, and should be asked by students interested in chemistry. Research in chemistry (or any other field, for that matter) is interesting and challenging. But there are different directions a student can take as they explore research opportunities.
Pure and Applied Research
The study of modern chemistry can be split into two types of research: pure and applied. Chemists who study pure chemistry do research primarily to advance mankind's understanding of chemistry. Pure chemistry is concerned with a greater understanding of the theories behind how matter is changing in chemical reactions. Pure chemists tend to be less concerned with direct applications of the research that they are doing. That is not to say that pure chemistry can never lead to a real-world application but, rather, that a potential application is not the primary motivation for doing the research in the first place. Applied chemistry is chemistry that is directed toward a specific practical goal or application. The video below further describes applied research.
Pure Research Examples
The early history of chemistry contains many examples of pure research. The ancient Greek philosophers debated the composition of matter (earth? air? fire? water? all of the above?). They weren’t going to do anything with their knowledge – they just wanted to know. Studies on the elements (especially after Mendeleev’s periodic table was published) were primarily “pure” research types of experiments. Does this element exist? What are its properties? The scientists did not have any practical application in mind, but were curious about the world around them. Below are some examples of questions that where pure research would be used:
• How was the universe formed?
• Is there life on Mars?
• What are protons, neutrons, and electrons composed of?
• What are the properties of boron?
Applied Research Examples
There is a great deal of "applied" research taking place today. In general, no new science principles are discovered, but existing knowledge is used to develop a new product. Research on laundry detergents will probably not give us any new concepts about soap, but will help us to develop materials that get our clothes cleaner, use less water, and create lower amounts of pollution. Petroleum companies use applied research to find better ways to power vehicles, better lubricants to cut down on engine wear, and better ways to lower air pollution. These companies will use information that is readily available to come up with new products.
"In-Between" Examples
The line between pure chemistry and applied chemistry is not always distinct. What may start out as simply asking a question may result is some very useful information. If scientists are studying the biochemistry of a microorganism that causes a disease, they may soon find information that would suggest a way to make a chemical that would inactivate the microorganism. The compound could be used to learn more about the biochemistry, but could also be used to cure the disease.
Hemoglobin is a protein in red blood cells that transports oxygen in the bloodstream. Scientists studied hemoglobin simply to learn how it worked. Out of this research came an understanding of how the protein changes shape when oxygen attaches to it. This information was then applied to help patients with sickle cell anemia, a disorder caused by an abnormal hemoglobin structure that makes hemoglobin molecules clump up when oxygen leaves the protein. Basic knowledge of protein structure led to an improved understanding of a wide-spread disease and opened the door for development of treatments.
Summary
• Pure research focuses on understanding basic properties and processes.
• Applied research focuses on the use of information to create useful materials.
• Sometimes there is no clear line between pure and applied research.
Review
1. What is pure research?
2. What is applied research?
3. Give one example of pure research.
4. Give on example of applied research.
5. Is it always easy to classify research as pure or applied? Explain your answer. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.05%3A_Pure_and_Applied_Chemistry.txt |
What are some types of chemical reactions?
Some chemical reactions can be very calm and boring, while other reactions release a great deal of energy. Dynamite is a chemical that can explode violently. Here we see dynamite being used to move boulders to clear a path for a road. The chemical reactions involved here release tremendous amounts of energy very quickly.
Dynamite was invented by Alfred Nobel in 1866. Nitroglycerin, a very unstable explosive, was already known. Nobel mixed the nitroglycerin with silica to stabilize it and form a solid material. He made a fortune with this discovery and established the Nobel Foundation, which funds the Nobel Prizes every year.
Energy in Chemical Bonds
Chemical reactions either require energy or release energy. The amount of energy needed or released depends upon the structure of the molecules that are involved in the reaction. Some reactions need to be heated for long periods of time in order for change to take place. Other reactions release energy, allowing heat to be given off to the surroundings. This energy can be used in a variety of ways.
Heating
Coal, natural gas, oil—these materials can be burned to produce heat. The reaction with oxygen releases a great deal of energy that can warm homes and offices. Wood is another example of a chemical (yes, a very complicated one) that will release energy when burned.
Transportation
A major use for petroleum products is fuel for cars, trucks, airplanes, trains, and other vehicles. The chemicals used are usually a mixture of compounds containing several carbon atoms in a chain. When the material is ignited, a massive amount of gas is created almost instantaneously. This increase in volume will move the pistons in an internal combustion engine to provide power. A jet plane works on a similar principle—air comes into the front of the engine and mixes with the jet fuel. This mixture is ignited and the gases formed create a lot of pressure to push the plane forward. The idea can be seen even more clearly in the case of a rocket launch. The ignition of the fuel (either solid-state or liquid) creates gases produced under great pressure that push the rocket up.
Batteries
A major source of energy produced by chemical reactions involves batteries. There are many types of batteries that operate using a variety of chemical reactions. The general principle behind these reactions is the release of electrons that can then flow through a circuit, producing an electrical current.
Batteries are used in a wide variety of applications, among them:
1. flashlights
2. watches
3. computers
4. cars
5. hybrid vehicles (provide some power to wheels)
6. cell phones
Batteries in cars, computers, cell phones, and other devices are usually rechargeable. An electric current is passed through the battery to provide electrons that reverse (at least partially) the chemical reactions originally used to create the electric current. Eventually, however, the system can no longer be recharged and the battery has to be discarded.
Hand-Warmers
Hikers, campers, and other outdoor recreationists take advantage of chemical reactions to keep their hands warm. Small containers of chemicals can undergo reactions to generate heat that can be used to avoid frostbite. Some products contain iron filings that will react with air to release thermal energy. These types of warmers cannot be reused. Other systems rely on heat being released when certain chemicals crystallize. If the warmer is placed in very hot water after use, the system can be regenerated.
Summary
• Some chemical reactions release energy, some require energy.
• Energy released by chemical reactions can be used in a variety of ways.
Review
1. Who invented dynamite?
2. How was the nitroglycerin made more stable?
3. What kind of energy is released when we burn natural gas?
4. How does burning gasoline power a car?
5. How do batteries create energy?
6. Explain how a hand-warmer works. | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.06%3A_Energy_in_Chemistry.txt |
How has chemistry made a contributions to medicine?
Diabetes mellitus is a disease characterized by the body's inability to use glucose (a component of table sugar). Glucose is needed to provide biochemical energy for all the cells of the body. When the body cannot make energy using glucose, it begins to break down fat and protein to provide the needed energy, eventually leading to death. Diabetes is the result of the pancreas losing the ability to make insulin, a protein that helps glucose to enter the body's cells and be used for biochemical energy. A key piece of the puzzle surrounding our understanding of diabetes came when Frederick Sanger, a British biochemist, carried out experiments that gave him the structure of the insulin molecule. Sanger used basic chemistry techniques and reactions, and took twelve years to complete his research. Today, automated instruments based on his approach can perform the same analysis in a matter of days. Sanger was awarded the Nobel Prize in Chemistry in 1958 for his insulin research.
Chemistry in Medicine
Major contributions to health care have been made by chemistry. The development of new drugs involves chemical analysis and synthesis of new compounds. Many recent television programs advertise the large number of new drugs produced by chemists.
The development of a new drug for any disease is long and complicated. The chemistry of the disease must be studied, as well as how the drug affects the human body. A drug may work well in animals, but not in humans. Out of one hundred drugs that offer the possibility of treating disease, only a small handful actually turn out to be both safe and effective.
Chemistry contributes to the preparation and use of materials for surgery (sutures, artificial skin, and sterile materials). The sutures used in many surgeries today do not have to be removed, because they simply dissolve in the body after a period of time. Replacement blood vessels for heart and other types of surgery are often made of chemicals that do not react with the tissues, so they will not be rejected by the body. Artificial skin can be used to replace human skin for burn patients.
Clinical laboratory testing uses a wide variety of chemical techniques and instrumentation for analysis. Clinical laboratory testing allows us to answer commonly asked questions such as "is your cholesterol high?" and "do you have diabetes?" Some laboratory tests use simple techniques. Other processes involve complex equipment and computer analysis data, in order to perform measurements on large numbers of patient samples.
Laboratory testing has come to the local drug store or grocery store because of developments in chemistry. You can test your blood glucose using a simple portable device that runs a chemical test on the blood sample and tells you how much glucose is present, allowing a diabetic patient to regulate how much insulin to administer (chemistry is also used to produce the insulin and the disposable syringe that administers the drug).
Science Friday: The Medical Wonders of Worm Spit
How useful is worm spit? It turns out that worm spit, also known as silk, is a very useful material in medicine. In this video by Science Friday, Dr. David Kaplan describes how silk is used in a variety of medical applications.
Summary
• Chemistry finds many applications in the healthcare field.
• Development of medicines involves many complicated chemistry processes.
• Chemistry is used to create materials used in surgery.
• Much of laboratory testing is based on chemistry techniques.
Review
1. What chemical is missing in the diabetic patient?
2. Who discovered the structure of insulin?
3. What two things need to be studied to develop a new drug?
4. List two areas where chemistry has helped surgical patients
5. What blood test can be performed using material purchased from your local drugstore? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.07%3A_Medicine.txt |
How does chemistry aid the success of crops?
In the spring, many people begin to plant their gardens. They see ads in catalogs or shop the gardening section of a local store to get ideas. The right place in the garden is selected, seeds or plants are put in the soil, and then they wait. Whether it is a small home garden or a large thousand acre farm, chemistry contributes greatly to the success of the crop.
Agriculture
Crops need four things for good growth: sunlight, water, nutrients from the soil, and protection from predators such as insects. Chemistry has made major contributions within water usage, nutrient availability, and plant protection. Water purification uses a number of chemical and physical techniques to remove salts and contaminants that would pollute the soil. Chemical analysis of soil allows the grower to see what nutrients are lacking, so that they can be added. In the spring, grocery stores, hardware stores, and gardening centers have high stacks of bags containing fertilizers and weed killers that enrich the soil and keep down unwanted plants. These same stores also provide a number of sprays or solid treatments for insects that might otherwise eat the plants.
Water Purification
Fresh water is essential for good crops. In some areas of the world, there is enough rain to accomplish this task. In other locales, water must be provided so that the crops will grow. Worldwide, irrigation covers about \(18\%\) of farm land and produces some \(40\%\) of crops. Obtaining clean water in many parts of the world is made possible by the process of desalinization.
Desalinization involves to the treatment of sea water to remove salts; the resulting water can then be used for irrigation without contaminating the soil with materials that harm the growing plants.
Soil Nutrients
In many areas of the world, the soil is deficient in essential nutrients. A number of minerals such as phosphorus, potassium, calcium, and magnesium may not be present in large enough amounts for plants to grow well. Nitrogen is also extremely important for good crops.
Soil analysis is available from a variety of labs. Local university extension services can provide valuable information as to the composition of a soil, and will also make suggestions as to the types and amounts of needed nutrients. Fertilizers can be purchased and added to the soil to enrich it and ensure better yield of crops.
Insect Control
Even if the crop grows well, there is still the possibility of insect or pest damage. The insect or pest can consume the crop or can damage it to the point where it will not grow well. Infestations of army worms can do major damage to corn and grain crops. Aphids and boll weevils are major predators of cotton crops. Failure to control these pests results in widespread crop damage and financial loss for the farmer.
A wide variety of pesticides have been developed by chemists and other scientists to deal with pests. The basic approach is to develop pesticides that interfere with some biochemical process in the pest. Ideally, the pesticide will not affect other living organisms, but this is not always the case. It is very important to read the labels and observe all precautions when using pesticides.
Summary
• Obtaining clean water in many parts of the world is made possible by the process of desalinization.
• Plant nutrients are very important for good plant growth.
• Chemical analysis of soil can tell he farmer or gardener what nutrients are needed.
• Chemists have developed many pesticides that will kill plant predators such as the army worm and the boll weevil.
Review
1. List three things crops need for good growth.
2. How much of the water used in farming is provided by irrigation?
3. What fraction of crops are grown using irrigation?
4. Why do nutrients need to be added to the soil?
5. How do pesticides work? | textbooks/chem/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/01%3A_Introduction_to_Chemistry/1.08%3A_Agriculture.txt |
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