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Chemical Equilibrium: Previous page: Mass Action Law
Learning Objectives
• Write the equilibrium constant expression when the solvent is one of the products or reactants.
• Write the equilibrium constant expression for heterogeneous equilibria.
We treat chemical equilibria in which the solvent is a reactant or product and heterogeneous equilibria as special chemical equilibria. The expressions for equilibrium constants of these systems are discussed. The mass action law is valid for the case when the solvent is a reactant or product. However, due to the large amount of solvent present, the equilibrium constant expression can be simplified. Special consideration also applies to heterogeneous equilibria, in which solids or liquids are involved. Phases such as liquid and solid are not sensitive to pressure, and their "concentrations" are constant as long as these phases are present. Their "concentrations" are not defined, and that is why we use a " " to mark the special meaning of these phrases. Perhaps "activity" or "property" is a better term to use than "concentration" in these cases.
Equilibrium Constant for Reactions Involving Solvents
As a solvent in the system, the concentration of the solvent is so large that its concentration rarely changes during the course of the reaction. For example, when water is used as a solvent, the concentration of water is almost always a constant. Its concentration can be calculated to be 55.6 M. (1 L of water has 1000 g, and molecular weight 18 g/mol; and thus $\mathrm{[H_2O] = \dfrac{1000}{18} = 55.6\: mol/L}$)
For example, consider the following reaction,
$\ce{CH3COOH + C2H5OH \rightleftharpoons CH3COOC2H5 + H2O}$
According to the Mass Action Law, we can use either
$K' = \mathrm{\dfrac{[CH_3COOC_2H_5] {\color{Red} [H_2O]}}{[CH_3COOH] [C_2H_5OH]}}$
or
$K = \ce{\dfrac{[CH3COOC2H5]}{[CH3COOH] [C2H5OH]}}$
Of course,
$K = \mathrm{\dfrac{\mathit K\,' }{ {\color{Red} [H_2O]}}} = \dfrac{K' }{ {\color{Red} 55.6}}$
Heterogeneous Equilibria
Heterogeneous reactions involve at least two phases. That is two of gas, liquid, and solid are present as reactants or products. In heterogeneous equilibria, the activities (or concentrations) of solid and liquid but not gas are always a constant. In these cases, their activities or concentrations are omitted in the expression of Q or K. Let us look at these expressions for the equilibria:
\begin{align} \ce{CaCO_{3\large{(s)}} &\rightleftharpoons CaO_{\large{(s)}} + CO_{2\large{(g)}}} &&K_{\ce p} = \ce{P(CO2)}\ \ce{2 H2O_{\large{(l)}} &\rightleftharpoons 2 H2O_{\large{(g)}}} &&K_{\ce p} = \ce{P^2(H2O)}\ \ce{2 H2O_{\large{(l)}} &\rightleftharpoons 2 H_{2\large{(g)}} + O_{2\large{(g)}}} &&K_{\ce p} = \ce{P^2(H2) P(O2)}\ \mathrm{AgCl_{\large{(s)}}} &\mathrm{\rightleftharpoons \sideset{ }{_{\large{(aq)}}^{+}}{Ag} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl}} &&K = \ce{[Ag^+] [Cl^- ]} \end{align}
Heterogeneous equilibria discusses this type of system in more detail.
Questions
1. What is the concentration of water in pure water?
2. Give the Kp expression for the reaction
$\ce{HgO_{\large{(s)}} \rightleftharpoons Hg_{\large{(s)}} + O_{2\large{(g)}}}$
3. For the phase transition,
$\ce{H2O_{\large{(l)}} \rightleftharpoons H2O_{\large{(g)}}}$,
the vapor pressure of water depends on temperature. At 298 K, the saturated vapor pressure is 23.8 torr. What is the equilibrium constant based on pressure Kp in atm? (1 atm = 760 torr)
4. At some low temperature, the reaction of ethyl acetate with water,
$\ce{CH3COOC2H5_{\large{(aq)}} + H2O_{\large{(l)}} \rightleftharpoons CH3COOH_{\large{(aq)}} + C2H5OH_{\large{(aq)}}}$
has an equilibrium constant of 0.10. If $\mathrm{[CH_3COOC_2H_5] = 0.90\: M}$ in a system that is at equilibrium, what is $\ce{[C2H5OH]}$?
5. The value of Kp is 81 Pa at some temperature for the reaction,
$\ce{NH4Cl_{\large{(s)}} \rightleftharpoons NH_{3\large{(g)}} + HCl_{\large{(g)}}}$.
What is the partial pressure (in Pa) of $\ce{HCl}$ in an enclosed system containing solid $\ce{NH4Cl}$ and its vapor?
Solutions
1. 55.6
Consider...
For most solutions, $\mathrm{[H_2O] = \dfrac{1000}{18} =\: ?}$
2. $\ce{P(O2)}$
Consider...
This reaction is used to produce $\ce{O2}$ from $\ce{HgO}$.
3. 0.031 atm
Consider...
$K_{\ce p} = \ce{P(H2O)}$ in this case. Just want you to consider one of the common equilibrium states. The amount of water present does not affect the vapor pressure.
4. 0.3
Consider...
\begin{alignat}{2} \ce{&CH3COOC2H5_{\large{(aq)}} + H2O_{\large{(l)}} \rightleftharpoons \; &&CH3COOH_{\large{(aq)}} +\; &&C2H5OH_{\large{(aq)}}}\ &\hspace{55px}0.9 &&\hspace{40px}x &&\hspace{30px}x \end{alignat}
$\dfrac{x^2}{0.90} = K = 0.1$; $x =\: ?$
What was the original $\ce{[CH3COOC2H5]}$ before any hydration reaction takes place? (Answer, 1.2)
5. 9
Consider...
$\ce{P(HCl)} = \sqrt{K_{\ce p}}$
\begin{alignat}{2} \ce{NH4Cl_{\large{(s)}} \rightleftharpoons \; &NH_{3\large{(g)}} + \; &&HCl_{\large{(g)}}}\ &\:\:\:x &&\:\:\:x \end{alignat}
$K_{\ce p} = x^2$
The partial pressure of $\ce{NH3}$ is the same as that of $\ce{HCl}$.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Special_Equilibria.txt
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The Bends is an illness that arises from the rapid release of nitrogen gas from the bloodstream and is caused by bubbles forming in the blood and other tissues when a diver ascends to the surface of the ocean too rapidly. It is also referred to as Caisson sickness, decompression sickness (DCS), and Divers' Disease.
Introduction
As divers descend into the ocean, the external pressure on their bodies increases by about 1 atm every 10.06 m. To balance this it is necessary to increase the pressure of the air they breathe from tanks or pumped to them from the surface so that their chests and lungs do not collapse. Unfortunately, our bodies aren't used to the pressurized air (because we normally breathe air under normal atmospheric conditions). With higher air pressure in the lungs Henry's Law tells us that gases such as nitrogen, helium (when used in diving gas mixtures) and oxygen become increasingly soluble in the blood. Unlike oxygen which is metabolized, nitrogen and helium build up throughout the body When divers want to emerge from the water, they have to make sure they don't ascend to the surface level too quickly because they risk numerous bubbles forming as the nitrogen/helium re-equilibrates, much as when a pressurized bottle of soda is suddenly opened. When nitrogen (N2) gas forms bubbles, it accumulates and saturates the muscles and blood, causing pain. Called the Bends, this condition can also cause injuries involving the nervous system.
The solubility of a gas is the ability for the gas to dissolve in a solvent (in our case, blood, which although it contains organic components is essentially an aqueous solution). Both temperature and pressure affect the solubility of a gas.
The Solubility as a Function of Temperature
• In water solvents, the higher the temperature, the less soluble the gas is.
• In organic solvents, the higher the temperature, the more soluble the gas is.
The Solubility as a Function of Pressure
English chemist William Henry discovered that as the pressure increases, the solubility of a gas increases. Henry's Law is then:
\[ C =k P_{gas} \]
where
• C=solubility of a gas in a solvent at a specific temperature,
• Pgas is the partial pressure of the gas, and
• k is Henry's Law Constant
In the case of The Bends:
• If a diver goes deeper and deeper into the water, more nitrogen builds up in the bloodstream and other tissues.
• Following Henry's Law; as the pressure increases, the solubility of nitrogen in the diver's bloodstream increases.
• As a result, nitrogen from the compressed air stays in the bloodstream and other tissues
• However, since the diver is in a highy-pressurized environment, the excess N2 can only be relieved when the diver ascends to levels with lower external pressure
• Ideally, this should happen during the diver's gradual rise to the surface
• Unfortunately, sometimes, the diver ascends too quickly, resulting in the rapid formation of bubbles, which interfers with nerves, blood and lymphatic vessels and leads to excruciating joint pain and clotting.
Example \(1\):
Determine Henry's Law Constant, k, with the information that the aqueous solubility of N2 at 10 degrees Celsius is 11.5 mL N2 / L and 1 atm.
\( k= \dfrac {11.5 mL N_{2}/ L}{\ 1 atm} \)
Now if the Pgas of N2 increases to 5 atm:
\[ P_{N2}=\dfrac {C}{\dfrac {11.5 mL N_{2}/ L}{\ 1 atm}} \]
\[ 5 atm=\dfrac {C}{\dfrac {11.5 mL N_{2}/ L}{\ 1 atm}} \]
Solve for C: C= 57.5 mL N2 /L
Therefore, both examples show that as the the pressure increases from 1 atm to 5 atm, the solubility of the N2 gas increases from 11.5 to 57.5 mL N2 / L. This supports Henry's Law.
Symptoms of the Bends
• Joint pain
• Fatigue
• Itching and rashes
• Coughing and chest pain
• Dizziness and paralysis
• Unconsciousness
• Death
Most symptoms occur 24 hours after decompression, but can occur up to 3 days after.
Prevention
• Ascending to the surface slowly (rate of 60 ft/min.)
• The slower the diver surfaces, the more slowly the excess nitrogen is equilibrated and the lower the impact on the diver
• Spending time in a decompression chamber
• Chambers that high-pressured divers are placed in.
• Once in the chamber, the diver is immersed in a high pressure environment which is slowly reduced, minimizing any effect.
• Breathing a compressed air mixture of helium and oxygen with no nitrogen.
• For deep dives, in addition to the Bends, excess nitrogen can lead to decreased mental function. This is called nitrogen narcosis.
• Helium is less soluble in the blood stream and thus does not build up as much, providing a smaller threat to divers and is used for deep dives.
References
1. Petrucci, et al. General Chemistry: Principles & Modern Applications: Custom Edition for CHEM 2 (Hardcover). Upper Saddle River: Pearson Education, Inc., 2011.
2. Phatak, Uday. "Decompression Syndrome (Caisson Disease) in an Indian Diver." Annals of Indian Academy of Neurology. Online-Only Journal. 13.3. 2010 n.pag. web. 30 May 2011.
Problems
1. What is Henry's law?
2. How does Henry's law relate to the Bends illness?
3. How does temperature affect solubility?
4. How does ascending to the surface make the bends less prevalent?
5. Why is breathing the compressed helium/oxygen mixture better than air with N2?
Answers
1. \(C =k *P_{gas}\) (where C=solubility of a gas in a solvent at a specific temperature, Pgas is the partial pressure of the gas, and k is Henry's Law Constant)
2. As the pressure increases, the solubility of gases in the diver's bloodstream increases. Henry's law states that the solubility of a gas increases when the pressure increases.
3. As temperature increases, the solubility of gases decrease in aqueous solutions. In organic solutions, the solubility of gases increases at higher temperatures.
4. When diver surfaces slowly, he/she will have a reduced impact of pain from the bubbles that form. Instead of rapidly forming and causing joint pain, the slow rise to the surface creates a steady loss of pressure, resulting in pain that is not as severe.
5. Helium is less soluble in the blood stream, providing a smaller threat to divers when they come up to the surface of the ocean. Fewer bubbles are formed, meaning that the divers encounter less pain as they ascend.
Contributors and Attributions
• Dhara Shah (UCD)
• Josh Halpern (LibreTexts)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/The_Bends.txt
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Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position.
• Case Study: The Manufacture of Ethanol from Ethene
This page describes the manufacture of ethanol by the direct hydration of ethene, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture and the rate of the reaction.
• Effect of Temperature on Equilibrium
A temperature change occurs when temperature is increased or decreased by the flow of heat. This shifts chemical equilibria toward the products or reactants, which can be determined by studying the reaction and deciding whether it is endothermic or exothermic.
• ICE Tables
An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation).
• Le Chatelier's Principle and Dynamic Equilbria
This page looks at Le Châtelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. It covers changes to the position of equilibrium if you change concentration, pressure or temperature. It also explains very briefly why catalysts have no effect on the position of equilibrium.
• Le Chatelier's Principle Fundamentals
Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change.
• The Contact Process
The Contact Process is used in the manufacture of sulfuric acid. This Modules explain the reasons for the conditions used in the process by considering the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process.
• The Effect of Changing Conditions
This page looks at the relationship between equilibrium constants and Le Châtelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same.
• The Haber Process
This page describes the Haber Process for the manufacture of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process.
Le Chateliers Principle
This page describes the manufacture of ethanol by the direct hydration of ethene, and then goes on to explain the reasons for the conditions used in the process. It looks at the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture and the rate of the reaction.
A brief summary of the manufacture of ethanol
Ethanol is manufactured by reacting ethene with steam. The reaction is reversible, and the formation of the ethanol is exothermic.
Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion. A flow scheme for the reaction looks like this:
Explaining the conditions
The proportions of ethene and steam
The equation shows that the ethene and steam react 1 : 1. In order to get this ratio, you would have to use equal volumes of the two gases.
Because water is cheap, it would seem sensible to use an excess of steam in order to move the position of equilibrium to the right according to Le Chatelier's Principle. In practice, an excess of ethene is used.
This is very surprising at first sight. Even if the reaction was one-way, you couldn't possibly convert all the ethene into ethanol. There isn't enough steam to react with it.
The reason for this oddity lies with the nature of the catalyst. The catalyst is phosphoric(V) acid coated onto a solid silicon dioxide support. If you use too much steam, it dilutes the catalyst and can even wash it off the support, making it useless.
The temperature
You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible amount of ethanol in the equilibrium mixture. The forward reaction (the production of ethanol) is exothermic.
According to Le Chatelier's Principle, this will be favoured if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. In order to get as much ethanol as possible in the equilibrium mixture, you need as low a temperature as possible. However, 300°C isn't particularly low.
Rate considerations
The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much ethanol as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of ethanol if it takes several years for the reaction to reach that equilibrium. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor.
The compromise
300°C is a compromise temperature producing an acceptable proportion of ethanol in the equilibrium mixture, but in a very short time. Under these conditions, about 5% of the ethene reacts to give ethanol at each pass over the catalyst.
The pressure
Equilibrium considerations
Notice that there are 2 molecules on the left-hand side of the equation, but only 1 on the right. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favoring the reaction which produces fewer molecules. That will cause the pressure to fall again. In order to get as much ethanol as possible in the equilibrium mixture, you need as high a pressure as possible. High pressures also increase the rate of the reaction. However, the pressure used isn't all that high.
Problems with high pressures
There are two quite separate problems in this case:
• High pressures are expensive. It costs more to build the original plant because you need extremely strong pipes and containment vessels. It also needs a lot of energy to produce the high pressures. That can make the ethanol uneconomic to produce.
• At high pressures, the ethene polyerizes to make poly(ethene). Apart from wasting ethene, this could also clog up the plant.
The catalyst
The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of ethanol in the equilibrium mixture. Its only function is to speed up the reaction.
Rate considerations
In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/Case_Study%3A_The_Manufacture_of_Ethanol_from_Ethene.txt
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A temperature change occurs when temperature is increased or decreased by the flow of heat. This shifts chemical equilibria toward the products or reactants, which can be determined by studying the reaction and deciding whether it is endothermic or exothermic.
Introduction
Le Chatelier's principle states that a change in temperature, pressure, or concentration of reactants in an equilibrated system will stimulate a response that partially off-sets the change to establish a new equilibrium. In the case of changing temperature, adding or removing of heat shifts the equilibrium. Typically chemical reactions are written to not explicitly address the flow of heat in the reaction. For example, the below chemical equation describing the oxidation of carbon to make carbon monoxide contains all the information regarding matter and bonding:
$\ce{2C (s) + O_2 (g) -> 2CO (g)} \nonumber$
However, reactions invariably involve changes in enthalpy, with energy (typically in the form of thermal energy via heat) either being absorbed or released during the reaction. The more complete reaction would be written as
$\ce{2C (s) + O2 (g) -> 2CO(g) + heat } \nonumber$
Heat of Reaction
The Heat of Reaction is the change in the enthalpy of a chemical reaction. $ΔH>0$) thermal energy is absorbed via the reaction. Anther way to view endothermic reactions is that more (thermal) energy is needed to overcome the forces of attraction between molecules and to separate them from one another (the activation energy) than (thermal) released when new bonds are formed.
$\ce{ heat + 6CO2(g) + 6H2O(l) <=> C6H12O6(aq) + 6O2(g)} \nonumber$
In exothermic reactions, ($ΔH<0$) thermal energy is general with reaction. When new bonds are generated, more thermal energy is released that needed to break bonds in the reactants. In this chemical reaction
$\ce{CaO (s) + H2O(l) <=> Ca(OH)2(s) + heat}$
the forward reaction is exothermic because energy is released when $\ce{CaO(s)}$ and $\ce{H2O(l)}$ combine to form $\ce{Ca(OH)2(s)}$. The energy to break the bonds in $\ce{CaO(s)}$ and $\ce{H2O(l)}$ on the left side of the equation is less than the energy released from forming the $\ce{Ca(OH)2\;(s)}$ on the right side of the equation; the net difference is observed as heat on the right side of the equation.
Example $1$
In the oxidation reaction
$\ce{CaO(s) + H2O(l) <=> Ca(OH)2(s) + heat} \nonumber$
Raising the temperature favors the reverse reaction (endothermic) and similarly Lowering the temperature favors the forward reaction (exothermic)
Example $2$
In the reaction
$\ce{2C(s) + O2 (g) <=> 2CO(g) + heat} \nonumber$
Le Chatelier's principle explains that the reaction will proceed in such a way as to counteract the temperature change. The exothermic reaction will favor the reverse reaction, opposite the side heat is (the opposite is true in endothermic reactions; the reaction will proceed in the forward reaction)
Although it is not technically correct to do so, if heat is treated as product in the above reaction, then it becomes clear that if the temperature is increased the equilibrium will shift to the left (using Le Chatelier's principle). If temperature is decreased, the reaction will proceed forward to produce more heat (which is lacking). The effect of temperature on equilibrium will also change the value of the equilibrium constant.
Problems
1. If heat is added to a phase change equation at equilibrium from solid to liquid, which way will the reaction proceed?
2. Which side is heat on in this reaction (photosynthesis): $6CO_{2(g)} + 6H_2O_{(l)} \rightleftharpoons C_{6}H_{12}O_{6(aq)} + 6O_{2(g)}$
3. In a combustion reaction is heat absorbed or released?
4. In this reaction: $H_{2}O_{(l)} \rightleftharpoons H_{2}O_{(g)}$, how could conditions be manipulated to create more $H_2O_{(l)}$?
5. Explain how to determine if a reaction is exothermic or endothermic.
Solutions
1. The reaction will proceed towards the liquid phase.
2. Heat is on the reactant side of the equation.
3. Heat is released in a combustion reaction.
4. Lowering temperature will shift equilibrium left, creating more liquid water.
5. A reaction that is exothermic releases heat, while an endothermic reaction absorbs heat.
Contributors and Attributions
• Karissa Pulido (UCD), Carlynn Chappell (UCD), Aileen McDuff (UCD)
Effect Of Temperature On Equilibrium Composition
An exothermic reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released into the surroundings, resulting in an overall negative quantity for the heat of reaction ($q_{rxn} < 0$). An endothermic reaction occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat. Endothermic reactions result in an overall positive heat of reaction ($q_{rxn} > 0$).
Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy ($ΔH$), the sum of all potential and kinetic energies. ΔH is determined by the system, not the surrounding environment in a reaction. A system that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system.
$\ce{C(s) + O2(g) -> CO2 (g)} \tag{ΔH = –393.5 kJ}$
$\ce{ H2 (g) + 1/2 O2 (g) -> H2O(l)} \tag{ΔH = –285.8 kJ}$
The enthalpies of these reactions are less than zero, and are therefore exothermic reactions. A system of reactants that absorbs heat from the surroundings in an endothermic reaction has a positive $ΔH$, because the enthalpy of the products is higher than the enthalpy of the reactants of the system.
$\ce{N2(g) + O2(g) -> 2NO(g)} \tag{ΔH = +180.5 kJ > 0}$
$\ce{ C(s) + 2S(s) -> CS2(l)} \tag{ΔH = +92.0 kJ > 0}$
Because the enthalpies of these reactions are greater than zero, they are endothermic reactions.
The Equilibrium Constant
The equilibrium constant ($K_c$) defines the relationship among the concentrations of chemical substances involved in a reaction at equilibrium. The Le Chatelier's principle states that if a stress, such as changing temperature, pressure, or concentration, is inflicted on an equilibrium reaction, the reaction will shift to restore the equilibrium. For exothermic and endothermic reactions, this added stress is a change in temperature. The equilibrium constant shows how far the reaction will progress at a specific temperature by determining the ratio of products to reactions using equilibrium concentrations.
The equilibrium expression for the following equation
$aA + bB \rightleftharpoons cC + dD$
is given below:
$K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Equation:Kc}$
where
• Kc is the equilibrium constant (for concentrations)
• [A], [B], [C], [D] are concentrations
• a, b, c, and d are the stoichiometric coefficients of the balanced equation
Exothermic Reactions Endothermic Reactions
If Kc decreases with an increase in temperature, the reaction shifts to the left. If Kc increases with an increase in temperature, the reaction to shifts to the right.
If Kc increases with a decreases in temperature, the reaction to shifts to the right. If Kc decreases with a decrease in temperature, the reaction to shifts to the left.
If the products dominate in a reaction, the value for K is greater than 1. The larger the K value, the more the reaction will tend toward the right and thus to completion.
• If K=1, neither the reactants nor the products are favored. Note that this is not the same as both being favored.
• If the reactants dominate in a reaction, then K< 1. The smaller the K value, the more the reaction will tend toward the left.
Example $1$ : The Haber Process
Suppose that the following reaction is at equilibrium and that the concentration of N2 is 2 M, the concentration of H2 is 4 M, and the concentration of NH3 is 3 M. What is the value of Kc?
$\ce{ N2 + 3H2 <=> 2NH3} \nonumber$
The coefficients and the concentrations are plugged into the $K_c$ expression (Equation \ref{Equation:Kc}) to calculate its value.
\begin{align*} K_c &= \dfrac{[NH_3]^2}{[N_2]^1[H_2]^3} \[4pt] &= \dfrac{[3]^2}{[2]^1[4]^3} \[4pt] &= \dfrac{9}{128} \[4pt] &= 0.07 \end{align*}
Exercise $1$
Determine $K_c$ for the following chemical reaction at equilibrium if the molar concentrations of the molecules are:
• 0.20 M $\ce{H2}$,
• 0.10 M $\ce{NO}$,
• 0.20 M $\ce{H2O}$, and
• 0.10M $\ce{N2}$.
$\ce{2H2 (g) + 2NO (g) <=> 2H2O (g) + N2 (g)} \nonumber$
Answer
Using the $K_c$ expression (Equation \ref{Equation:Kc}) and plugging in the concentration values of each molecule:
\begin{align*} K_c &= \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \[4pt] &= \dfrac{[\ce{H2O}]^2[\ce{N2}]^1}{[\ce{H2}]^2[\ce{NO}]^2} \[4pt] &= \dfrac{0.20^2\, 0.1}{0.20^2 \, 0.10 ^2 }\[4pt] &= 10 \end{align*}
Exercise $2$
For the previous equation, does the equilibrium favor the products or the reactants?
Answer
Because $K_c = 10 > 1$, the reaction favors the products.
Exercise $3$
In the following reaction, the temperature is increased and the $K_ c$ value decreases from 0.75 to 0.55. Is this an exothermic or endothermic reaction?
$\ce{N_2 (g) + 3H_2 <=>2NH_3 (g) } \nonumber$
Answer
Because the K value decreases with an increase in temperature, the reaction is an exothermic reaction.
Exercise $4$
In the following reaction, in which direction will the equilibrium shift if there is an increase in temperature and the enthalpy of reaction is given such that $ΔH$ is -92.5 kJ?
$\ce{PCl3(g) + Cl2(g) <=> PCl_5(g)} \nonumber$
Answer
In the initial reaction, the energy given off is negative and thus the reaction is exothermic. However, an increase in temperature allows the system to absorb energy and thus favor an endothermic reaction; the equilibrium will shift to the left.
Contributors and Attributions
• Alyson Salmon, Nikita Patel (UCD), Deepak Nallur (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/Effect_Of_Temperature_On_Equilibrium_Composition/Exothermic_Versus_Endothermic_A.txt
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An ICE (Initial, Change, Equilibrium) table is simple matrix formalism that used to simplify the calculations in reversible equilibrium reactions (e.g., weak acids and weak bases or complex ion formation).
Introduction
ICE tables are composed of the concentrations of molecules in solution in different stages of a reaction, and are usually used to calculate the K, or equilibrium constant expression, of a reaction (in some instances, K may be given, and one or more of the concentrations in the table will be the unknown to be solved for). ICE tables automatically set up and organize the variables and constants needed when calculating the unknown.
ICE is a simple acronym for the titles of the first column of the table.
• I stands for initial concentration. This row contains the initial concentrations of products and reactants.
• C stands for the change in concentration. This is the concentration change required for the reaction to reach equilibrium. It is the difference between the equilibrium and initial rows. The concentrations in this row are, unlike the other rows, expressed with either an appropriate positive (+) or negative (-) sign and a variable; this is because this row represents an increase or decrease (or no change) in concentration.
• E is for the concentration when the reaction is at equilibrium. This is the summation of the initial and change rows. Once this row is completed, its contents can be plugged into the equilibrium constant equation to solve for $K_c$.
The procedure for filling out an ICE table is best illustrated through example.
Example 1
Use an ICE table to determine $K_c$ for the following balanced general reaction:
$\ce{ 2X(g) <=> 3Y(g) + 4Z(g)} \nonumber$
where the capital letters represent the products and reactants.
• This equation will be placed horizontally above the table, with each product and reactant having a separate column.
A sample consisting of 0.500 mol of x is placed into a system with a volume of 0.750 liters.
• This statement implies that there are no initial amounts of Y and Z. For the I row of the Y and Z columns, 0.000 mol will be entered.
• Notice that the initial composition is given in moles. The amounts can either be converted to concentrations before putting them into the ICE table or after the equilibrium amounts have been calculated. This example uses moles for the ICE table, and calculates concentrations later.
At equilibrium, the amount of sample x is known to be 0.350 mol.
• For the equilibrium row of X, 0.350 mol will be entered.
Desired Unknown
$K_c = ? \nonumber$
Solution
The equilibrium constant expression is expressed as products over reactants, each raised to the power of their respective stoichiometric coefficients:
$K_c = \dfrac{[Y]^3[Z]^4}{[X]^2} \nonumber$
The equilibrium concentrations of Y and Z are unknown, but they can be calculated using the ICE table.
STEP 1: Fill in the given amounts
Reaction: 2X 3Y 4Z
Initial amounts 0.500 mol 0.000 mol 0.000 mol
Change in amount ? ? ?
Equilibrium amount 0.350 mol ? ?
This is the first step in setting up the ICE table. As mentioned above, the ICE mnemonic is vertical and the equation heads the table horizontally, giving the rows and columns of the table, respectively. The numerical amounts were given. Any amount not directly given is unknown.
STEP 2: Fill in the amount of change for each compound
Reaction 2X 3Y 4Z
Initial amounts 0.500 mol 0.000 mol 0.000 mol
Change in amount -0.150 mol +0.225 mol +0.300 mol
Equilibrium amounts 0.350 mol ? ?
Notice that the equilibrium in this equation is shifted to the right, meaning that some amount of reactant will be taken away and some amount of product will be added (for the Change row).
The change in amount ($x$) can be calculated using algebra:
$Equilibrium \; Amount = Initial \; Amount + Change \; in \; Amount \nonumber$
Solving for the Change in the amount of $2x$ gives:
$0.350 \; mol - 0.500 \; mol = -0.150 \; mol \nonumber$
The change in reactants and the balanced equation of the reaction is known, so the change in products can be calculated. The stoichiometric coefficients indicate that for every 2 mol of x reacted, 3 mol of Y and 4 mol of Z are produced. The relationship is as follows:
$\begin{eqnarray} Change \; in \; Product &=& -\left(\dfrac{\text{Stoichiometric Coefficient of Product}}{\text{Stoichiometric Coefficient of Reactant}}\right)(\text{Change in Reactant}) \ Change \; in \; Y &=& -\left(\dfrac{3}{2}\right)(-0.150 \; mol) \ &= +0.225 \; mol \end{eqnarray} \nonumber$
Try obtaining the change in Z with this method (the answer is already in the ICE table).
STEP 3: Solve for the equilibrium amounts
Reaction 2X 3Y 4Z
Initial amounts 0.500 mol 0.000 mol 0.000 mol
Change in amounts -0.150 mol +0.225 mol +0.300 mol
Equilibrium amounts 0.350 mol 0.225 mol 0.300 mol
If the initial amounts of Y and/or Z were nonzero, then they would be added together with the change in amounts to determine equilibrium amounts. However, because there was no initial amount for the two products, the equilibrium amount is simply equal to the change:
$\begin{eqnarray} Equilibrium \; Amount &=& Initial \; Amount + Change \; in \; Amount \ Equilibrium \; Amount \; of \; Y &=& 0.000 \; mol\; + 0.225 \; mol \ &=& +0.225 \; mol \end{eqnarray} \nonumber$
Use the same method to find the equilibrium amount of Z. Convert the equilibrium amounts to concentrations. Recall that the volume of the system is 0.750 liters.
$[Equilibrium \; Concentration \; of \; Substance] = \dfrac{Amount \; of \; Substance}{Volume \; of \; System}\nonumber$
$[X] = \dfrac{0.350 \; mol}{0.750 \; L} = 0.467 \; M \nonumber$
$[Y] = \dfrac{0.225 \; mol}{0.750 \; L} = 0.300 \; M \nonumber$
$[Z] = \dfrac{0.300 \; mol}{0.750 \; L} = 0.400 \; M \nonumber$
Use the concentration values to solve the $K_c$ equation:
$\begin{eqnarray} K_c &=& \dfrac{[Y]^3[Z]^4}{[X]^2} \ &=& \dfrac{[0.300]^3[0.400]^4}{[0.467]^2} \ K_c &=& 3.17 \times 10^{-3} \end{eqnarray}\nonumber$
Example 2: Using an ICE Table with Concentrations
n this example an ICE table is used to find the equilibrium concentration of the reactants and products. (This example will be less in depth than the previous example, but the same concepts are applied.) These calculations are often carried out for weak acid titrations.
Find the concentration of A- for the generic acid dissociation reaction:
$\ce{HA(aq) + H_2O(l) <=> A^{-}(aq) + H_3O^{+}(aq)} \nonumber$
with $[HA (aq)]_{initial} = 0.150 M$ and $K_a = 1.6 \times 10^{-2}$
Solution
This equation describes a weak acid reaction in solution with water. The acid (HA) dissociates into its conjugate base ($A^-$) and protons (H3O+). Notice that water is a liquid, so its concentration is not relevant to these calculations.
STEP 1: Fill in the given concentrations
Reaction: HA A- H3O+
I 0.150 M 0.000 M 0.000 M
C ? ? ?
E ? ? ?
• The contents of the leftmost column column are shortened for convenience.
STEP 2: Calculate the change concentrations by using a variable 'x'
Reaction: HA A- H3O+
I 0.150 M 0.000 M 0.000 M
C -x M +x M +x M
E ? ? ?
• The change in concentration is unknown, so the variable x is used to denote the change. x is the same for both products and reactants because equal stoichiometric amounts of A- and H3O+ are generated when HA dissociates in water.
STEP 3: Calculate the concentrations at equilibrium
Reaction: HA A- H3O+
I 0.150 M 0.000 M 0.000 M
C -x M +x M +x M
E 0.150 - x M x M x M
• To find the equilibrium amounts the I row and the C row are added. Use these values and Ka (the equilibrium constant for acids) to find the concentration x.
STEP 4: Use the ICE table to calculate concentrations with $K_a$
The expression for Ka is written by dividing the concentrations of the products by the concentrations of the reactants. Plugging in the values at equilibrium into the equation for Ka gives the following:
$K_a = \dfrac{x^2}{0.150-x} = 1.6 \times 10^{-2} \nonumber$
To find the concentration x, rearrange this equation to its quadratic form, and then use the quadratic formula to find x:
\begin{align*} (1.6 \times 10^{-2})({0.150-x}) &= {x^2} \[4pt] x^2+(1.6 \times 10^{-2})x-(0.150)(1.6 \times 10^{-2}) &= 0 \end{align*}
This is the typical form for a quadratic equation:
$Ax^{2}+Bx+C=0\nonumber$
where, in this case:
• $A = 1$
• $B = 1.6 \times 10^{-2}$
• $C =( -0.150)( 1.6 \times 10^{-2}) = -2.4 \times 10^{-3}$
The quadratic formula gives two solutions (but only one physical solution) for x:
$x = \dfrac{-B+\sqrt{B^2-4AC}}{2A}\nonumber$
and
$x = \dfrac{-B-\sqrt{B^2-4AC}}{2A}\nonumber$
Intuition must be used in determining which solution is correct. If one gives a negative concentration, it can be eliminated, because negative concentrations are unphysical.
The x value can be used to calculate the equilibrium concentrations of each product and reactant by plugging it into the elements in the E row of the ice table.
[Solution: x = 0.0416, -0.0576. x = 0.0416 makes chemical sense and is therefore the correct answer.]
For some problems like example 2, if x is significantly less than the value for Ka, then the x of the reactants (in the denominator) can be omitted and the concentration for x should not be greatly affected. This will make calculations faster by eliminating the necessity of the quadratic formula.
Checklist for ICE tables
• Make sure the reversible equation is balanced at the start of the problem; otherwise, the wrong amounts will be used in the table.
• The given data should be in amounts, concentrations, partial pressures, or somehow able to be converted to such. If it is not, then an ICE table will not help solve the problem.
• If the ICE table has the equilibrium in amounts, make sure to convert equilibrium values to concentrations before plugging in to solve for $K_c$.
• If the given data is in amounts or concentrations, use the ICE table to find $K_c$. If the given data is in partial pressures, use the ICE table to find $K_p$. If you desire to convert from one to the other, remember that $K_p = K_c(RT)^{\Delta n_{gas}}\nonumber$ It is simpler to use the ICE table with the appropriate givens and convert at the end of the problem.
• Enter in known data first, and then calculate the unknown data.
• If there is a negative value in the "initial" or "equilibrium" rows, reexamine the calculation. A negative concentration, amount, or partial pressure is physically impossible. Obviously, the "change" row can contain a negative value.
• Pay attention to the state of each reactant and product. If a compound is a solid or a liquid, its concentrations are not relevant to the calculations. Only concentrations of gaseous and aqueous compounds are used.
• In the "change" row the values will usually be a variable, denoted by x. It must first be understood which direction the equation is going to reach equilibrium (from left to right or from right to left). The value for "change" in the "from" direction of the reaction will be the opposite of x and the "to" direction will be the positive of x (adding concentration to one side and take away an equal amount from the other side).
• Know the direction of the reaction. This knowledge will affect the "change" row of the ICE table (for our example, we knew the reaction would proceed forward, as there was no initial products). Direction of reaction can be calculated using Q, the reaction quotient, which is then compared to a known K value.
• It is easiest to use the same units every time an ICE table is used (molarity is usually preferred). This will minimize confusion when calculating the equilibrium constants. ICE tables are usually used for weak acid or weak base reactions because all of the nature of these solutions. The amount of acid or base that will dissociate is unknown (for strong acids and strong bases it can be assumed that all of the acid or base will dissociate, meaning that the concentration of the strong acid or base is the same as its dissociated particles).
Partial pressures may also be substituted for concentrations in the ICE table, if desired (i.e., if the concentrations are not known, $K_p$ instead of $K_c$ is desired, etc.). "Amount" is also acceptable (the ICE table may be done in amounts until the equilibrium amounts are found, after which they will be converted to concentrations). For simplicity, assume that the word "concentration" can be replaced with "partial pressure" or "amounts" when formulating ICE tables.
Exercise $1$
0.200 M acetic acid is added to water. What is the concentration of H3O+ in solution if $K_c = 1.8 \times 10^{-6}$?
Answer
5.99×10-4
Exercise $2$
If the initial concentration of NH3 is 0.350 M and the concentration at equilibrium is 0.325 M, what is $K_c$ for this reaction?
Answer
1.92×10-3
Exercise $3$
How is $K_c$ derived from $K_p$?
Answer
$K_p = K_c(RT)^{\Delta n}$ then solve for $K_c$
Exercise $4$
Complete this ICE table:
Reaction: [HA] [A-] [H3O+]
I 0.650 mol ? ?
C ? ? ?
E 0.250 mol ? ?
Answer
Reaction: HA A- H3O+
I 0.650 mol 0.000 mol 0.000 mol
C -0.400 mol +0.400 mol +0.400 mol
E 0.250 mol 0.400 mol 0.400 mol
Contributors and Attributions
• Alexander Shei (UCD), Aileen McDuff (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/Ice_Tables.txt
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Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position.
Introduction
An action that changes the temperature, pressure, or concentrations of reactants in a system at equilibrium stimulates a response that partially offsets the change while a new equilibrium condition is established (2). Hence, Le Châtelier's principle states that any change to a system at equilibrium will adjust to compensate for that change. In 1884 the French chemist and engineer Henry-Louis Le Châtelier proposed one of the central concepts of chemical equilibria, which describes what happens to a system when something briefly removes it from a state of equilibrium.
It is important to understand that Le Châtelier's principle is only a useful guide to identify what happens when the conditions are changed in a reaction in dynamic equilibrium; it does not give reasons for the changes at the molecular level (e.g., timescale of change and underlying reaction mechanism).
Concentration Changes
Le Châtelier's principle states that if the system is changed in a way that increases the concentration of one of the reacting species, it must favor the reaction in which that species is consumed. In other words, if there is an increase in products, the reaction quotient, \(Q_c\), is increased, making it greater than the equilibrium constant, \(K_c\). Consider an equilibrium established between four substances, \(A\), \(B\), \(C\), and \(D\):
\[ A + 2B \rightleftharpoons C + D\]
Increasing a concentration
What happens if conditions are altered by increasing the concentration of A?
According to Le Châtelier, the position of equilibrium will move in such a way as to counteract the change. In this case, the equilibrium position will move so that the concentration of A decreases again by reacting it with B to form more C and D. The equilibrium moves to the right (indicated by the green arrow below).
In a practical sense, this is a useful way of converting the maximum possible amount of B into C and D; this is advantageous if, for example, B is a relatively expensive material whereas A is cheap and plentiful.
Decreasing a concentration
In the opposite case in which the concentration of A is decreased, according to Le Châtelier, the position of equilibrium will move so that the concentration of A increases again. More C and D will react to replace the A that has been removed. The position of equilibrium moves to the left.
This is essentially what happens if one of the products is removed as soon as it is formed. If, for example, C is removed in this way, the position of equilibrium would move to the right to replace it. If it is continually removed, the equilibrium position shifts further and further to the right, effectively creating a one-way, irreversible reaction.
Pressure Changes
This only applies to reactions involving gases, although not necessarily all species in the reaction need to be in the gas phase. A general homogeneous gaseous reaction is given below:
\[ A(g) + 2B(g) \rightleftharpoons C(g) + D(g)\]
Increasing the pressure
According to Le Châtelier, if the pressure is increased, the position of equilibrium will move so that the pressure is reduced again. Pressure is caused by gas molecules hitting the sides of their container. The more molecules in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way as to produce fewer molecules.
In this case, there are three moles on the left-hand side of the equation, but only two on the right. By forming more C and D, the system causes the pressure to reduce. Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer moles of gas molecules.
Example 1: Haber Process
\[ N_2 + 3H_2 \rightleftharpoons 2NH_3 \]
If this mixture is transferred from a 1.5 L flask to a 5 L flask, in which direction does a net change occur to return to equilibrium?
Solution
Because the volume is increased (and therefore the pressure reduced), the shift occurs in the direction that produces more moles of gas. To restore equilibrium the shift needs to occur to the left, in the direction of the reverse reaction.
Decreasing the pressure
The equilibrium will move in such a way that the pressure increases again. It can do that by producing more gaseous molecules. In this case, the position of equilibrium will move towards the left-hand side of the reaction.
What happens if there are the same number of molecules on both sides of the equilibrium reaction?
In this case, increasing the pressure has no effect on the position of the equilibrium. Because there are equal numbers of molecules on both sides, the equilibrium cannot move in any way that will reduce the pressure again. Again, this is not a rigorous explanation of why the position of equilibrium moves in the ways described. A mathematical treatment of the explanation can be found on this page.
Summary of Pressure Effects
Three ways to change the pressure of an equilibrium mixture are: 1. Add or remove a gaseous reactant or product, 2. Add an inert gas to the constant-volume reaction mixture, or 3. Change the volume of the system.
1. Adding products makes \(Q_c\) greater than \(K_c\). This creates a net change in the reverse direction, toward reactants. The opposite occurs when adding more reactants.
2. Adding an inert gas into a gas-phase equilibrium at constant volume does not result in a shift. This is because the addition of a non-reactive gas does not change the partial pressures of the other gases in the container. While the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant.
3. When the volume of a mixture is reduced, a net change occurs in the direction that produces fewer moles of gas. When volume is increased the change occurs in the direction that produces more moles of gas.
Temperature Changes
To understand how temperature changes affect equilibrium conditions, the sign of the reaction enthalpy must be known. Assume that the forward reaction is exothermic (heat is evolved):
In this reaction, 250 kJ is evolved (indicated by the negative sign) when 1 mole of A reacts completely with 2 moles of B. For reversible reactions, the enthalpy value is always given as if the reaction was one-way in the forward direction. The back reaction (the conversion of C and D into A and B) would be endothermic, absorbing the same amount of heat.
The main effect of temperature on equilibrium is in changing the value of the equilibrium constant.
Temperature is Neither a Reactant nor Product
It is not uncommon that textbooks and instructors to consider heat as a independent "species" in a reaction. While this is rigorously incorrect because one cannot "add or remove heat" to a reaction as with species, it serves as a convenient mechanism to predict the shift of reactions with changing temperature. For example, if heat is a "reactant" (\(\Delta{H} > 0 \)), then the reaction favors the formation of products at elevated temperature. Similarly, if heat is a "product" (\(\Delta{H} < 0 \)), then the reaction favors the formation of reactants. A more accurate, and hence preferred, description is discussed below.
Increasing the temperature
If the temperature is increased, then the position of equilibrium will move so that the temperature is reduced again. Suppose the system is in equilibrium at 300°C, and the temperature is increased 500°C. To cool down, it needs to absorb the extra heat added. In the case, the back reaction is that in which heat is absorbed. The position of equilibrium therefore moves to the left. The new equilibrium mixture contains more A and B, and less C and D.
If the goal is to maximize the amounts of C and D formed, increasing the temperature on a reversible reaction in which the forward reaction is exothermic is a poor approach.
Decreasing the temperature?
The equilibrium will move in such a way that the temperature increases again. Suppose the system is in equilibrium at 500°C and the temperature is reduced to 400°C. The reaction will tend to heat itself up again to return to the original temperature by favoring the exothermic reaction. The position of equilibrium will move to the right with more \(A\) and \(B\) converted into \(C\) and \(D\) at the lower temperature:
Example 2
Consider the formation of water
\[O_2 + 2H_2 \rightleftharpoons 2H_2O\;\;\; \Delta{H}= -125.7\, kJ\]
1. What side of the reaction is favored? Because the heat is a product of the reaction, the reactants are favored.
2. Would the conversion of \(O_2\) and \(H_2\) to \(H_2O\) be favored with heat as a product or as a reactant? Heat as a product would shift the reaction forward, creating \(H_2O\). The more heat added to the reaction, the more \(H_2O\) created
Summary of Temperature Effects
• Increasing the temperature of a system in dynamic equilibrium favors the endothermic reaction. The system counteracts the change by absorbing the extra heat.
• Decreasing the temperature of a system in dynamic equilibrium favors the exothermic reaction. The system counteracts the change by producing more heat.
Catalysts
Adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Châtelier's principle does not apply. This is because a catalyst speeds up the forward and back reaction to the same extent and adding a catalyst does not affect the relative rates of the two reactions, it cannot affect the position of equilibrium.
However, catalysts have some application to equilibrium systems. For a dynamic equilibrium to be set up, the rates of the forward reaction and the back reaction must be equal. This does not happen instantly and for very slow reactions, it may take years! A catalyst speeds up the rate at which a reaction reaches dynamic equilibrium.
Example 3
You might try imagining how long it would take to establish a dynamic equilibrium if you took the visual model on the introductory page and reduced the chances of the colors changing by a factor of 1000 - from 3 in 6 to 3 in 6000 and from 1 in 6 to 1 in 6000. Starting with blue squares, by the end of the time taken for the examples on that page, you would most probably still have entirely blue squares. Eventually, though, you would end up with the same sort of patterns as before - containing 25% blue and 75% orange squares.
Problems
1. Varying Concentration
What will happen to the equilibrium when more 2SO2 (g) is added to the following system?
\[2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3 (g) \]
Solution:
Adding more reactants shifts the equilibrium in the direction of the products; therefore, the equilibrium shifts to the right.
Overall, the concentration of \(2SO_2\) from initial equilibrium to final equilibrium will increase because only a portion of the added amount of \(2SO_2\) will be consumed.
The concentration of \(O_2\) will decrease because as the equilibrium is reestablished, \(O_2\) is consumed with the \(2SO_2\) to create more \(2SO_3\). The concentration of \(2SO_3\) will be greater because none of it is lost and more is being generated.
2. Varying Pressure
What will happen to the equilibrium when the volume of the system is decreased?
\[2SO_{2(g)} + O_{2 (g)} \rightleftharpoons 2SO_{3 (g)}\]
Solution:
Decreasing the volume leads to an increase in pressure which will cause the equilibrium to shift towards the side with fewer moles. In this example there are 3 moles on the reactant side and 2 moles on the product side, so the new equilibrium will shift towards the products (to the right).
3. Varying Temperature
What will happen to the equilibrium when the temperature of the system is decreased?
\[N_{2(g)} + O_{2 (g)} \rightleftharpoons 2NO_{(g)} \;\;\;\; \Delta{H} = 180.5\; kJ\]
Solution
Because \(\Delta{H}\) is positive, the reaction is endothermic in the forward direction. Removing heat from the system forces the equilibrium to shift towards the exothermic reaction, so the reverse reaction will occur and more reactants will be produced.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/Le_Chatelier%27s_Principle_Fundamentals.txt
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This page looks at Le Chatelier's Principle and explains how to apply it to reactions in a state of dynamic equilibrium. It covers changes to the position of equilibrium if you change concentration, pressure or temperature. It also explains very briefly why catalysts have no effect on the position of equilibrium. It is important in understanding everything on this page to realize that Le Chatelier's Principle is no more than a useful guide to help you work out what happens when you change the conditions in a reaction in dynamic equilibrium. It does not explain why the system moved to equilibrium.
Definition: Le Chatelier's Principle
If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
Using Le Chatelier's Principle with a Change of Concentration
Suppose you have an equilibrium established between four substances A, B, C and D.
$\ce{ A + 2B <=> C +D} \nonumber$
What would happen if you changed the conditions by increasing the concentration of A?
According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of A decreases again - by reacting it with B and turning it into C + D. The position of equilibrium moves to the right.
This is a useful way of converting the maximum possible amount of B into C and D. You might use it if, for example, B was a relatively expensive material whereas A was cheap and plentiful.
What would happen if you changed the conditions by decreasing the concentration of A?
According to Le Chatelier, the position of equilibrium will move so that the concentration of A increases again. That means that more C and D will react to replace the A that has been removed. The position of equilibrium moves to the left.
This is essentially what happens if you remove one of the products of the reaction as soon as it is formed. If, for example, you removed C as soon as it was formed, the position of equilibrium would move to the right to replace it. If you kept on removing it, the equilibrium position would keep on moving rightwards - turning this into a one-way reaction.
This is not in any way an explanation of why the position of equilibrium moves in the ways described. All Le Chatelier's Principle gives you is a quick way of working out what happens.
Using Le Chatelier's Principle with a Change of Pressure
This only applies to reactions involving gases:
$\ce{A(g) + 2B(g) <=> C(g) + D(g)} \nonumber$
What would happen if you changed the conditions by increasing the pressure?
According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the pressure is reduced again. Pressure is caused by gas molecules hitting the sides of their container. The more molecules you have in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way as to produce fewer molecules.
In this case, there are 3 molecules on the left-hand side of the equation, but only 2 on the right. By forming more C and D, the system causes the pressure to reduce. Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules.
What would happen if you changed the conditions by decreasing the pressure?
The equilibrium will move in such a way that the pressure increases again. It can do that by producing more molecules. In this case, the position of equilibrium will move towards the left-hand side of the reaction.
What happens if there are the same number of molecules on both sides of the equilibrium reaction?
In this case, increasing the pressure has no effect whatsoever on the position of the equilibrium. Because you have the same numbers of molecules on both sides, the equilibrium can't move in any way that will reduce the pressure again.
Again, this is not an explanation of why the position of equilibrium moves in the ways described.
Using Le Chatelier's Principle with a Change of Temperature
For this, you need to know whether heat is given out or absorbed during the reaction. Assume that our forward reaction is exothermic (heat is evolved):
This shows that 250 kJ is evolved (hence the negative sign) when 1 mole of A reacts completely with 2 moles of B. For reversible reactions, the value is always given as if the reaction was one-way in the forward direction. The back reaction (the conversion of C and D into A and B) would be endothermic by exactly the same amount.
What would happen if you changed the conditions by increasing the temperature?
According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the temperature is reduced again. Suppose the system is in equilibrium at 300°C, and you increase the temperature to 500°C. How can the reaction counteract the change you have made? How can it cool itself down again? To cool down, it needs to absorb the extra heat that you have just put in. In the case we are looking at, the back reaction absorbs heat. The position of equilibrium therefore moves to the left. The new equilibrium mixture contains more A and B, and less C and D.
If you were aiming to make as much C and D as possible, increasing the temperature on a reversible reaction where the forward reaction is exothermic is not a good idea!
What would happen if you changed the conditions by decreasing the temperature?
The equilibrium will move in such a way that the temperature increases again. Suppose the system is in equilibrium at 500°C and you reduce the temperature to 400°C. The reaction will tend to heat itself up again to return to the original temperature. It can do that by favoring the exothermic reaction. The position of equilibrium will move to the right. More A and B are converted into C and D at the lower temperature.
Le Chatelier's Principle and catalysts
Catalysts have sneaked onto this page under false pretenses, because adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Chatelier's Principle does not apply to them. This is because a catalyst speeds up the forward and back reaction to the same extent. Because adding a catalyst does not affect the relative rates of the two reactions, it can't affect the position of equilibrium. So why use a catalyst? For a dynamic equilibrium to be set up, the rates of the forward reaction and the back reaction have to become equal. This does not happen instantly. For a very slow reaction, it could take years! A catalyst speeds up the rate at which a reaction reaches dynamic equilibrium.
Summary
Increasing the temperature of a system in dynamic equilibrium favors the endothermic reaction. The system counteracts the change you have made by absorbing the extra heat. Decreasing the temperature of a system in dynamic equilibrium favors the exothermic reaction. The system counteracts the change you have made by producing more heat. Again, this is not in any way an explanation of why the position of equilibrium moves in the ways described. It is only a way of helping you to work out what happens.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/Le_Chatelier%27s_Principle_and_Dynamic_Equilbria.txt
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The Contact Process is used in the manufacture of sulfuric acid. This Modules explain the reasons for the conditions used in the process by considering the effect of proportions, temperature, pressure and catalyst on the composition of the equilibrium mixture, the rate of the reaction and the economics of the process. The Contact Process:
• Step 1: Make sulfur dioxide
• Step 2: Convert sulfur dioxide into sulfur trioxide (the reversible reaction at the heart of the process)
• Step 3: Convert sulfur trioxide into concentrated sulfuric acid
Step 1: Making sulfur dioxide
This can either be made by burning sulfur in an excess of air:
$S_{(s)} + O_2 (g) \rightarrow SO_{2\; (g)} \label{1}$
or by heating sulfide ores like pyrite in an excess of air:
$4FeS_2 (s) + 11O_2 (g) \rightarrow 2Fe_2O_3 (s) + 8SO_2 (g) \label{2}$
In either case, an excess of air is used so that the sulfur dioxide produced is already mixed with oxygen for the next stage.
Step 2: Converting sulfur dioxide into sulfur trioxide
This is a reversible reaction and exothermic.
$2SO_2 (g) + O_2(g) \rightleftharpoons 2SO_3 (g) \;\;\; \Delta{H}=-196\; kJ/mol \label{3}$
A flow scheme for this part of the process looks like this:
The reasons for all these conditions will be explored in detail further down the page.
Step 3: Converting sulfur trioxide into sulfuric acid
This cannot be done by simply adding water to the sulfur trioxide; the reaction is so uncontrollable that it creates a fog of sulfuric acid. Instead, the sulfur trioxide is first dissolved in concentrated sulfuric acid:
$H_2SO_{4 (l)} + SO_{3(g)} \rightarrow H_2S_2O_{7 (l)} \label{4}$
The product is known as fuming sulfuric acid or oleum, which can then be reacted safely with water to produce concentrated sulfuric acid - twice as much originally used to make the fuming sulfuric acid.
$H_2S_2O_{7 (l)} + H_2O_{(l)} \rightarrow 2H_2SO_{4 (l)} \label{5}$
Explaining the conditions
The mixture of sulfur dioxide and oxygen going into the reactor is in equal proportions by volume. Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. That means that the gases are going into the reactor in the ratio of 1 molecule of sulfur dioxide to 1 of oxygen.
That is an excess of oxygen relative to the proportions demanded by the equation.
$2SO_{2 (g)} + O_{2(g)} \rightleftharpoons 2SO_{3 (g)} \;\;\; \Delta{H}=-196\;kJ/mol$
According to Le Chatelier's Principle, increasing the concentration of oxygen in the mixture causes the position of equilibrium to shift towards the right. Since the oxygen comes from the air, this is a very cheap way of increasing the conversion of sulfur dioxide into sulfur trioxide.
Why not use an even higher proportion of oxygen? This is easy to see if you take an extreme case. Suppose you have a million molecules of oxygen to every molecule of sulfur dioxide. The equilibrium is going to be tipped very strongly towards sulfur trioxide - virtually every molecule of sulfur dioxide will be converted into sulfur trioxide. However, you aren't going to produce much sulfur trioxide every day. The vast majority of what you are passing over the catalyst is oxygen which has nothing to react with.
By increasing the proportion of oxygen you can increase the percentage of the sulfur dioxide converted, but at the same time decrease the total amount of sulfur trioxide made each day. The 1:1 mixture results in the best possible overall yield of sulfur trioxide.
Temperature
Equilibrium considerations: You need to shift the position of the equilibrium as far as possible to the right in order to produce the maximum possible amount of sulfur trioxide in the equilibrium mixture. The forward reaction (the production of sulfur trioxide) is exothermic.
$2SO_{3 (g)} + O_{2 (g)} \rightleftharpoons 2SO_{3(g)} \;\;\; \Delta{H} = -196\, kJ/mol$
According to Le Chatelier's Principle, this will be favored if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. To get as much sulfur trioxide as possible in the equilibrium mixture, you need as low a temperature as possible. However, 400 - 450°C is not a low temperature!
Rate considerations: The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much sulfur trioxide as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of sulfur trioxide if it takes several years for the reaction to reach that equilibrium. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor.
The compromise: 400 - 450°C is a compromise temperature producing a fairly high proportion of sulfur trioxide in the equilibrium mixture, but in a very short time.
Pressure
Equilibrium considerations:
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \;\;\; \Delta H = -196\;kJ/mol$
Notice that there are three molecules on the left-hand side of the equation, but only two on the right. According to Le Châtelier's Principle, if you increase the pressure the system will respond by favoring the reaction which produces fewer molecules. That will cause the pressure to fall again. To get as much sulfur trioxide as possible in the equilibrium mixture, you need as high a pressure as possible. High pressures also increase the rate of the reaction. However, the reaction is done at pressures close to atmospheric pressure!
Economic considerations: Even at these relatively low pressures, there is a 99.5% conversion of sulfur dioxide into sulfur trioxide. The very small improvement that you could achieve by increasing the pressure isn't worth the expense of producing those high pressures.
Catalyst
Equilibrium considerations: The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of sulfur trioxide in the equilibrium mixture. Its only function is to speed up the reaction.
Rate considerations: In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor.
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This page looks at the relationship between equilibrium constants and Le Chatelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same.
Changing concentrations
Equilibrium constants are not changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.
Suppose you have an equilibrium established between four substances A, B, C and D.
$A + 2B \rightleftharpoons C + D$
According to Le Chatelier's Principle, if you decrease the concentration of C, for example, the position of equilibrium will move to the right to increase the concentration again.
Explanation in terms of the constancy of the equilibrium constant
The equilibrium constant, $K_c$ for this reaction looks like this:
$K_c = \dfrac{[C][D]}{[A][B]^2}$
If you have moved the position of the equilibrium to the right (and so increased the amount of $C$ and $D$), why hasn't the equilibrium constant increased? This is actually the wrong question to ask! We need to look at it the other way round.
Let's assume that the equilibrium constant must not change if you decrease the concentration of $C$ - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?
If you decrease the concentration of $C$, the top of the $K_c$ expression gets smaller. That would change the value of $K_c$. In order for that not to happen, the concentrations of $C$ and $D$ will have to increase again, and those of $A$ and $B$ must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before. The position of equilibrium moves - not because Le Chatelier says it must - but because of the need to keep a constant value for the equilibrium constant.
If you decrease the concentration of $C$:
Changing pressure
This only applies to systems involving at least one gas. Equilibrium constants are not changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium may be changed if you change the pressure. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.
That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favoring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium.
Case 1: Differing Numbers of Gaseous Species on each side of the Equation
Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are three molecules on the left, but only two on the right. An increase in pressure would move the position of equilibrium to the right.
$A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
Because this is an all-gas equilibrium, it is much easier to use $K_p$:
$K_p = \dfrac{P_c\;P_D}{P_A\;P_B^2} \label{EqC1}$
Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure, $K_p$ will increase as well. Not so! To understand why, you need to modify the $K_p$ expr ession. Remember the relationship between partial pressure, mole fraction and total pressure?
$P_A = (\text{mole fraction of A} )( \text{total pressure})$
$P_A = \chi_A P_{tot}$
Replacing all the partial pressure terms in $\ref{EqC1}$ by mole fractions$\chi_A$ and total pressure ($P_{tot}$) g ives you this:
$K_p \dfrac{(\chi_C P_{tot} )( \chi_D P_{tot} )}{(\chi_A P_{tot} )(\chi_B P_{tot})^2}$
Most of the "P"s cancel out, with one left at the bottom of the expression.
$K_p \dfrac{\chi_C \chi_D}{\chi_A\chi_B^2 P_{tot}}$
Now, remember that $K_p$ has got to stay constant because the temperature is unchanged. How can that happen if you increase P? To compensate, you would have to increase the terms on the top, $\chi_C$ and $\chi_D$, and decrease the terms on the bottom, $\chi_A$ and $\chi_B$.
Increasing the terms on the top means that you have increased the mole fractions of the molecules on the right-hand side. Decreasing the terms on the bottom means that you have decreased the mole fractions of the molecules on the left. That is another way of saying that the position of equilibrium has moved to the right - exactly what Le Chatelier's Principle predicts. The position of equilibrium moves so that the value of $K_p$ is kept constant.
Case 2: Same Numbers of Gaseous Species on each side of the Equation
There are the same numbers of molecules on each side of the equation. In this case, the position of equilibrium is not affected by a change of pressure. Why not?
$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
Let's go through the same process as in Case 1:
$K_p = \dfrac{ P_C P_D}{P_A P_B}$
Substituting mole fractions and total pressure:
$K_p = \dfrac{ (\chi_C P_{tot}) (\chi_D P_{tot}) }{ (\chi_A P_{tot}) (\chi_B P_{tot})}$
Cancelling out as far as possible:
$K_p = \dfrac{ \chi_C \chi_D }{ \chi_A \chi_B }$
There is not a single $P_{tot}$ left in the expression so changing the pressure makes no difference to the $K_p$ expression. The position of equilibrium doesn't need to move to keep $K_p$ constant.
Changing temperature
Equilibrium constants are changed if you change the temperature of the system. $K_c$ or $K_p$ are constant at constant temperature, but they vary as the temperature changes. Look at the equilibrium involving hydrogen, iodine and hydrogen iodide:
$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)} \label{EqHI}$
with $\Delta H = -10.4\; kJ/mol$. The $K_p$ expression is:
$P_p =\dfrac{P_{HI}^2}{P_{H_2}P_{(I_2)}}$
Two values for $K_p$ a re:
temperature Kp
500 K 160
700 K 54
You can see that as the temperature increases, the value of $K_p$ falls. This is typical of what happens with any equilibrium where the forward reaction is exothermic. Increasing the temperature decreases the value of the equilibrium constant. Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant.
The position of equilibrium also changes if you change the temperature. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favoring the reaction which absorbs heat. In the equilibrium we've just looked at ($\ref{EqHI}$, that will be the back reaction because the forward reaction is exothermic.
So, according to Le Chatelier's Principle the position of equilibrium will move to the left with increasing temperature. Less hydrogen iodide will be formed, and the equilibrium mixture will contain more unreacted hydrogen and iodine. That is entirely consistent with a fall in the value of the equilibrium constant.
Adding a catalyst
Equilibrium constants are not changed if you add (or change) a catalyst. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium is not changed if you add (or change) a catalyst. A catalyst speeds up both the forward and back reactions by exactly the same amount. Dynamic equilibrium is established when the rates of the forward and back reactions become equal. If a catalyst speeds up both reactions to the same extent, then they will remain equal without any need for a shift in position of equilibrium.
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The Haber Process is used in the manufacturing of ammonia from nitrogen and hydrogen, and then goes on to explain the reasons for the conditions used in the process. The process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.
$\ce{ N2(g) + 3H2(g) <=> 2NH3 (g)} \label{eq1}$
with $ΔH=-92.4 kJ/mol$.
A flow scheme for the Haber Process looks like this:
General Conditions of the Process
• The catalyst: The catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added to it as a promoter - a substance that increases its efficiency.
• The pressure: The pressure varies from one manufacturing plant to another, but is always high. You cannot go far wrong in an exam quoting 200 atmospheres.
• Recycling: At each pass of the gases through the reactor, only about 15% of the nitrogen and hydrogen converts to ammonia. (This figure also varies from plant to plant.) By continual recycling of the unreacted nitrogen and hydrogen, the overall conversion is about 98%.
Composition
The proportions of nitrogen and hydrogen: The mixture of nitrogen and hydrogen going into the reactor is in the ratio of 1 volume of nitrogen to 3 volumes of hydrogen. Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. That means that the gases are going into the reactor in the ratio of 1 molecule of nitrogen to 3 of hydrogen. That is the proportion demanded by the equation.
In some reactions you might choose to use an excess of one of the reactants. You would do this if it is particularly important to use up as much as possible of the other reactant - if, for example, it was much more expensive. That does not apply in this case. There is always a down-side to using anything other than the equation proportions. If you have an excess of one reactant there will be molecules passing through the reactor which cannot possibly react because there is not anything for them to react with. This wastes reactor space - particularly space on the surface of the catalyst.
Temperature
• Equilibrium considerations: You need to shift the position of the equilibrium (Equation $\ref{eq1}$) as far as possible to the right in order to produce the maximum possible amount of ammonia in the equilibrium mixture. The forward reaction is exothermic with $ΔH=-92.4 kJ/mol$. According to Le Chatelier's Principle, this will be favored if you lower the temperature. The system will respond by moving the position of equilibrium to counteract this - in other words by producing more heat. To get as much ammonia as possible in the equilibrium mixture, you need as low a temperature as possible. However, 400 - 450° C is not a low temperature!
• Rate considerations: The lower the temperature you use, the slower the reaction becomes. A manufacturer is trying to produce as much ammonia as possible per day. It makes no sense to try to achieve an equilibrium mixture which contains a very high proportion of ammonia if it takes several years for the reaction to reach that equilibrium. You need the gases to reach equilibrium within the very short time that they will be in contact with the catalyst in the reactor.
• The compromise: 400 - 450°C is a compromise temperature producing a reasonably high proportion of ammonia in the equilibrium mixture (even if it is only 15%), but in a very short time.
Pressure
Notice that there are 4 molecules on the left-hand side of Equation $\ref{eq1}$, but only 2 on the right. According to Le Chatelier's Principle, if you increase the pressure the system will respond by favoring the reaction which produces fewer molecules. That will cause the pressure to fall again. In order to get as much ammonia as possible in the equilibrium mixture, you need as high a pressure as possible. 200 atmospheres is a high pressure, but not amazingly high.
• Rate considerations: Increasing the pressure brings the molecules closer together. In this particular instance, it will increase their chances of hitting and sticking to the surface of the catalyst where they can react. The higher the pressure the better in terms of the rate of a gas reaction.
• Economic considerations: Very high pressures are very expensive to produce on two counts. You have to build extremely strong pipes and containment vessels to withstand the very high pressure. That increases your capital costs when the plant is built. High pressures cost a lot to produce and maintain. That means that the running costs of your plant are very high.
• The compromise: 200 atmospheres is a compromise pressure chosen on economic grounds. If the pressure used is too high, the cost of generating it exceeds the price you can get for the extra ammonia produced.
Catalyst
• Equilibrium considerations: The catalyst has no effect whatsoever on the position of the equilibrium. Adding a catalyst doesn't produce any greater percentage of ammonia in the equilibrium mixture. Its only function is to speed up the reaction.
• Rate considerations: In the absence of a catalyst the reaction is so slow that virtually no reaction happens in any sensible time. The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor.
• Separating the ammonia: When the gases leave the reactor they are hot and at a very high pressure. Ammonia is easily liquefied under pressure as long as it is not too hot, and so the temperature of the mixture is lowered enough for the ammonia to turn to a liquid. The nitrogen and hydrogen remain as gases even under these high pressures, and can be recycled.
By mixing one part ammonia to nine parts air with the use of a catalyst, the ammonia will get oxidized to nitric acid.
\begin{align*} \ce{4 NH_3} + \ce{5 O_2} &\rightarrow \ce{4 NO} + \ce{6 H_2O} \[4pt] \ce{2 NO} + \ce{O_2} &\rightarrow \ce{2 NO_2} \[4pt] \ce{2 NO_2} + \ce{2 H_2O} &\rightarrow \ce{2 HNO_3} + \ce{H_2} \end{align*}
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An introduction to saturated vapor pressure, Raoult's Law, and to several types of phase diagrams - including phase diagrams for pure substances, solutions of non-volatile solutes, eutectic mixtures, systems of two miscible liquids (including fractional distillation), and systems of two immiscible liquids (including steam distillation).
• Fractional Distillation of Ideal Mixtures
This page explains how the fractional distillation (both in the lab and industrially) of an ideal mixture of liquids relates to their phase diagram.
• Fractional Distillation of Non-ideal Mixtures (Azeotropes)
Remember that a large positive deviation from Raoult's Law produces a vapor pressure curve with a maximum value at some composition other than pure A or B. If a mixture has a high vapor pressure it means that it will have a low boiling point. The molecules are escaping easily and you won't have to heat the mixture much to overcome the intermolecular attractions completely.
• Immiscible Liquids and Steam Distillation
This page looks at systems containing two immiscible liquids. Immiscible liquids are those which won't mix to give a single phase. Oil and water are examples of immiscible liquids - one floats on top of the other. It explains the background to steam distillation and looks at a simple way of carrying this out.
• Liquid-Solid Phase Diagrams: Salt Solutions
This page looks at the phase diagram for mixtures of salt and water - how the diagram is built up, and how to interpret it. It includes a brief discussion of solubility curves.
• Liquid-Solid Phase Diagrams: Tin and Lead
This page explains the relationship between the cooling curves for liquid mixtures of tin and lead, and the resulting phase diagram. It also offers a simple introduction to the idea of a eutectic mixture.
• Non-Ideal Mixtures of Liquids
This page looks at the phase diagrams for non-ideal mixtures of liquids, and introduces the idea of an azeotropic mixture (also known as an azeotrope or constant boiling mixture). It goes on to explain how this complicates the process of fractionally distilling such a mixture.
• Phase Diagrams for Pure Substances
This page explains how to interpret the phase diagrams for simple pure substances - including a look at the special cases of the phase diagrams of water and carbon dioxide.
• Raoult's Law and Ideal Mixtures of Liquids
This page deals with Raoult's Law and how it applies to mixtures of two volatile liquids. It covers cases where the two liquids are entirely miscible in all proportions to give a single liquid - NOT those where one liquid floats on top of the other (immiscible liquids). The page explains what is meant by an ideal mixture and looks at how the phase diagram for such a mixture is built up and used.
Physical Equilibria
This page explains how the fractional distillation (both in the lab and industrially) of an ideal mixture of liquids relates to their phase diagram.
Using the phase diagram
On the last page, we looked at how the phase diagram for an ideal mixture of two liquids was built up. I want to start by looking again at material from the last part of that page. The next diagram is new - a modified version of diagrams from the previous page.
• If you boil a liquid mixture C1, you will get a vapor with composition C2, which you can condense to give a liquid of that same composition (the pale blue lines).
• If you reboil that liquid C2, it will give a vapor with composition C3. Again you can condense that to give a liquid of the same new composition (the red lines).
• Reboiling the liquid C3 will give a vapor still richer in the more volatile component B (the green lines). You can see that if you were to do this once or twice more, you would be able to collect a liquid which was virtually pure B.
The secret of getting the more volatile component from a mixture of liquids is obviously to do a succession of boiling-condensing-reboiling operations. It is not quite so obvious how you get a sample of pure A out of this.
Fractional Distillation in the lab
A typical lab fractional distillation would look like this:
The fractionating column is packed with glass beads (or something similar) to give the maximum possible surface area for vapor to condense on. Some fractionating columns have spikes of glass sticking out from the sides which serve the same purpose. If you sketch this, make sure that you do not completely seal the apparatus. There has to be a vent in the system otherwise the pressure build-up when you heat it will blow the apparatus apart. In some cases, where you are collecting a liquid with a very low boiling point, you may need to surround the collecting flask with a beaker of cold water or ice. The mixture is heated at such a rate that the thermometer is at the temperature of the boiling point of the more volatile component. Notice that the thermometer bulb is placed exactly at the outlet from the fractionating column.
Relating what happens in the fractionating column to the phase diagram
Suppose you boil a mixture with composition C1. The vapor over the top of the boiling liquid will be richer in the more volatile component, and will have the composition C2.
That vapor now starts to travel up the fractionating column. Eventually it will reach a height in the column where the temperature is low enough that it will condense to give a liquid. The composition of that liquid will, of course, still be C2. So what happens to that liquid now? It will start to trickle down the column where it will meet new hot vapor rising. That will cause the already condensed vapor to reboil.
Some of the liquid of composition C2 will boil to give a vapor of composition C3. Let's concentrate first on that new vapor and think about the unvaporized part of the liquid afterwards.
The vapor
This new vapor will again move further up the fractionating column until it gets to a temperature where it can condense. Then the whole process repeats itself. Each time the vapor condenses to a liquid, this liquid will start to trickle back down the column where it will be reboiled by up-coming hot vapor. Each time this happens the new vapor will be richer in the more volatile component. The aim is to balance the temperature of the column so that by the time vapor reaches the top after huge numbers of condensing and reboiling operations, it consists only of the more volatile component - in this case, B.
Whether or not this is possible depends on the difference between the boiling points of the two liquids. The closer they are together, the longer the column has to be.
The liquid
So what about the liquid left behind at each reboiling? Obviously, if the vapor is richer in the more volatile component, the liquid left behind must be getting richer in the other one. As the condensed liquid trickles down the column constantly being reboiled by up-coming vapor, each reboiling makes it richer and richer in the less volatile component - in this case, A. By the time the liquid drips back into the flask, it will be very rich in A indeed. So, over time, as B passes out of the top of the column into the condenser, the liquid in the flask will become richer in A. If you are very, very careful over temperature control, eventually you will have separated the mixture into B in the collecting flask and A in the original flask. Finally, what is the point of the packing in the column?
To make the boiling-condensing-reboiling process as effective as possible, it has to happen over and over again. By having a lot of surface area inside the column, you aim to have the maximum possible contact between the liquid trickling down and the hot vapor rising. If you didn't have the packing, the liquid would all be on the sides of the condenser, while most of the vapor would be going up the middle and never come into contact with it.
Fractional distillation industrially
There is no difference whatsoever in the theory involved. All that is different is what the fractionating column looks like. The diagram shows a simplified cross-section through a small part of a typical column.
The column contains a number of trays that the liquid collects on as the vapor condenses. The up-coming hot vapor is forced through the liquid on the trays by passing through a number of bubble caps. This produces the maximum possible contact between the vapor and liquid. This all makes the boiling-condensing-reboiling process as efficient as possible. The overflow pipes are simply a controlled way of letting liquid trickle down the column.
If you have a mixture of lots of liquids to separate (such as in petroleum fractionation), it is possible to tap off the liquids from some of the trays rather than just collecting what comes out of the top of the column. That leads to simpler mixtures such as gasoline, kerosene and so on.
Example 1
1. Use the phase diagram below to explain how you can obtain a pure sample of B from a mixture M by successively boiling and condensing the liquid mixture.
1. Why is it important to carefully control how strongly the original mixture is heated during the separation?
2. Explain briefly how the separation occurs, making use of the phase diagram above if you think it helps.
Solution
1. If you boil the mixture M, it will boil at a temperature T1. The vapor above the liquid at this temperature will be richer in the more volatile substance B. If you condense that vapor, it will give a liquid of the composition M1. If you reboil that, it will boil at a temperature T2. The vapor over that liquid will have a composition M2, still richer in B. If you go on doing that, reboiling and recondensing, then the vapor becomes richer and richer in B until it eventually becomes pure B. When you finally get to that point and condense the vapor, then you will have pure B liquid.
2. You have to be sure that only the vapor of the more volatile of the two liquids passes into the condenser. That means that the thermometer has to read exactly the boiling point of the more volatile liquid. If it is below that, then nothing is going to pass out into the condenser. If it is above that, then your distillate will still contain some of the less volatile component.
3. B is the more volatile liquid; A is the less volatile one. The vapor over the boiling liquid in the flask will be richer in B than the original liquid is. That vapor will pass up the column until the temperature falls enough for it to condense to give a liquid richer in B than the one in the flask (equivalent to M1 in the diagram). This will start to trickle down the column. Hot vapor coming up from the flask will reboil the condensed liquid, giving a vapor which will be even richer in B (M2 on the diagram). This will condense to a liquid, trickle down the column and then be reboiled. This continuous process will go on until the vapor is entirely B. The column is heated so that this is finally complete right at the top of the column. Meanwhile, the liquids trickling down the column get richer and richer in A as the B is removed and carried up the column. Eventually, the liquid in the flask will end up as pure A.
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Positive Deviation from Raoult's Law
Remember that a large positive deviation from Raoult's Law produces a vapor pressure curve with a maximum value at some composition other than pure A or B. If a mixture has a high vapor pressure it means that it will have a low boiling point. The molecules are escaping easily and you won't have to heat the mixture much to overcome the intermolecular attractions completely. The implication of this is that the boiling point / composition curve will have a minimum value lower than the boiling points of either A or B.
In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures.
Suppose you are going to distil a mixture of ethanol and water with composition C1 as shown on the next diagram. It will boil at a temperature given by the liquid curve and produce a vapor with composition C2.
When that vapor condenses it will, of course, still have the composition C2. If you reboil that, it will produce a new vapor with composition C3.
You can see that if you carried on with this boiling-condensing-reboiling sequence, you would eventually end up with a vapor with a composition of 95.6% ethanol. If you condense that you obviously get a liquid with 95.6% ethanol.
What happens if you reboil that liquid? The liquid curve and the vapor curve meet at that point. The vapor produced will have that same composition of 95.6% ethanol. If you condense it again, it will still have that same composition. You have hit a barrier. It is impossible to get pure ethanol by distilling any mixture of ethanol and water containing less than 95.6% of ethanol. This particular mixture of ethanol and water boils as if it were a pure liquid. It has a constant boiling point, and the vapor composition is exactly the same as the liquid. It is known as a constant boiling mixture or an azeotropic mixture or an azeotrope.
The implications of this for fractional distillation of dilute solutions of ethanol are obvious. The liquid collected by condensing the vapor from the top of the fractionating column cannot be pure ethanol. The best you can produce by simple fractional distillation is 95.6% ethanol. What you can get (although it isn't very useful!) from the mixture is pure water. As ethanol rich vapor is given off from the liquid boiling in the distillation flask, it will eventually lose all the ethanol to leave just water.
To Summarize
Distilling a mixture of ethanol containing less than 95.6% of ethanol by mass lets you collect:
• a distillate containing 95.6% of ethanol in the collecting flask (provided you are careful with the temperature control, and the fractionating column is long enough);
• pure water in the boiling flask.
What if you distil a mixture containing more than 95.6% ethanol? Work it out for yourself using the phase diagram, and starting with a composition to the right of the azeotropic mixture. You should find that you get:
• a distillate containing 95.6% of ethanol in the collecting flask (provided you are careful with the temperature control and the fractionating column is long enough);
• pure ethanol in the boiling flask.
A negative deviation from Raoult's Law
Nitric acid and water form mixtures in which particles break away to form the vapor with much more difficulty than in either of the pure liquids. You can see this from the vapor pressure / composition curve discussed further up the page. That means that mixtures of nitric acid and water can have boiling points higher than either of the pure liquids because it needs extra heat to break the stronger attractions in the mixture.
In the case of mixtures of nitric acid and water, there is a maximum boiling point of 120.5°C when the mixture contains 68% by mass of nitric acid. That compares with the boiling point of pure nitric acid at 86°C, and water at 100°C. Notice the much bigger difference this time due to the presence of the new ionic interactions (see above). The phase diagram looks like this:
Distilling dilute nitric acid. Start with a dilute solution of nitric acid with a composition of C1 and trace through what happens.
The vapor produced is richer in water than the original acid. If you condense the vapor and reboil it, the new vapor is even richer in water. Fractional distillation of dilute nitric acid will enable you to collect pure water from the top of the fractionating column. As the acid loses water, it becomes more concentrated. Its concentration gradually increases until it gets to 68% by mass of nitric acid. At that point, the vapor produced has exactly the same concentration as the liquid, because the two curves meet. You produce a constant boiling mixture (or azeotropic mixture or azeotrope) and if you distil dilute nitric acid, that's what you will eventually be left with in the distillation flask. You cannot produce pure nitric acid from the dilute acid by distilling it.
You cannot produce pure nitric acid from the dilute acid (<68%) by distilling it.
Distilling nitric acid more concentrated than 68% by mass
This time you are starting with a concentration C2 to the right of the azeotropic mixture.
The vapor formed is richer in nitric acid. If you condense and reboil this, you will get a still richer vapor. If you continue to do this all the way up the fractionating column, you can get pure nitric acid out of the top. As far as the liquid in the distillation flask is concerned, it is gradually losing nitric acid. Its concentration drifts down towards the azeotropic composition. Once it reaches that, there cannot be any further change, because it then boils to give a vapor with the same composition as the liquid. Distilling a nitric acid / water mixture containing more than 68% by mass of nitric acid gives you pure nitric acid from the top of the fractionating column and the azeotropic mixture left in the distillation flask.
You can produce pure nitric acid from the concetrated acid (>68%) by distilling it.
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This page looks at systems containing two immiscible liquids. Immiscible liquids are those which won't mix to give a single phase. Oil and water are examples of immiscible liquids - one floats on top of the other. It explains the background to steam distillation and looks at a simple way of carrying this out.
The vapor pressure of mixtures of immiscible liquids
Obviously if you have two immiscible liquids in a closed flask and keep everything still, the vapor pressure you measure will simply be the vapor pressure of the one which is floating on top. There is no way that the bottom liquid can turn to vapor. The top one is sealing it in. For the purposes of the rest of this topic, we always assume that the mixture is being stirred or agitated in some way so that the two liquids are broken up into drops. At any one time there will be drops of both liquids on the surface. That means that both of them contribute to the overall vapor pressure of the mixture.
Total vapor pressure of the mixture
Assuming that the mixture is being agitated, then both of the liquids will be in equilibrium with their vapors. The total vapor pressure is then simply the sum of the individual vapor pressures:
$\text{Total vapor pressure} = p_A^o + p_B^o$
where $p^o$ refers to the saturated vapor pressure of the pure liquid. Notice that this is independent of the amount of each sort of liquid present. All you need is enough of each so that both can exist in equilibrium with their vapor.
For example, phenylamine and water can be treated as if they were completely immiscible. (That isn't actually true, but they are near enough immiscible to be usable as an example.)
At 98°C, the saturated vapor pressures of the two pure liquids are:
phenylamine 7.07 kPa
water 94.30 kPa
The total vapor pressure of an agitated mixture would just be the sum of these - in other words, 101.37 kPa
Boiling point of the mixture
Liquids boil when their vapor pressure becomes equal to the external pressure. Normal atmospheric pressure is 101.325 kPa. Compare that with the figure we have just got for the total vapor pressure of a mixture of water and phenylamine at 98°C. Its total vapor pressure is fractionally higher than the normal external pressure. This means that such a mixture would boil at a temperature just a shade less than 98°C - in other words lower than the boiling point of pure water (100°C) and much lower than the phenylamine (184°C).
Exactly the same sort of argument could be applied to any other mixture of immiscible liquids. I've chosen the phenylamine-water mixture just because I happen to have some figures for it!
Important conclusion
Agitated mixtures of immiscible liquids will boil at a temperature lower than the boiling point of either of the pure liquids. Their combined vapor pressures are bound to reach the external pressure before the vapor pressure of either of the individual components get there.
Steam distillation
Notice that in the presence of water, phenylamine (or any other liquid which is immiscible with water) boils well below its normal boiling point. This has an important advantage in separating molecules like this from mixtures. Normal distillation of these liquids would need quite high temperatures. On the whole these tend to be big molecules we are talking about. Quite a lot of molecules of this sort will be broken up by heating at high temperatures. Distilling them in the presence of water avoids this by keeping the temperature low. That's what steam distillation achieves.
We will carry on with the phenylamine example for now. During the preparation of phenylamine it is produced as a part of a mixture containing a solution of all sorts of inorganic compounds. It is removed from this by steam distillation.
Steam is blown through the mixture and the water and phenylamine turn to vapor. This vapor can be condensed and collected.
The steam can be generated by heating water in another flask (or something similar). As the hot steam passes through the mixture it condenses, releasing heat. This will be enough to boil the mixture of water and phenylamine at 98°C provided the volume of the mixture isn't too great. For large volumes, it is better to heat the flask as well to avoid having to condense too much steam and increase the volume of liquid in the flask too much.
The condensed vapor will consist of both water and phenylamine. If these were truly immiscible, they would form two layers which could be separated using a separating funnel. In fact, the phenylamine has a slight solubility in water and various other techniques have to be used in this particular case to get the maximum yield of phenylamine. These aren't relevant to this topic.
Some other applications of steam distillation
Steam distillation can be used to extract some natural products - for example, to extract eucalyptus oil from eucalyptus, citrus oils from lemon or orange peel, and to extract oils used in perfumes from various plant materials.
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This page looks at the phase diagram for mixtures of salt and water - how the diagram is built up, and how to interpret it. It includes a brief discussion of solubility curves.
Two Typical Solubility Curves
A solubility curve shows how the solubility of a salt like sodium chloride or potassium nitrate varies with temperature. The solubility is often (although not always) measured as the mass of salt which would saturate 100 grams of water at a particular temperature. A solution is saturated if it will not dissolve any more of the salt at that particular temperature - in the presence of crystals of the salt.
For most, but not all, substances, solubility increases with temperature. For some (like potassium nitrate), the increase is quite fast. For others (like sodium chloride), there is only a small change in solubility with temperature.
Obviously, a solubility curve shows you the solubility of a substance at a particular temperature. For phase diagram purposes, one important way of looking at this is to examine what happens if you decrease the temperature of a solution with some given concentration. For example, suppose you have a near-boiling solution of potassium nitrate in water. We'll take a solution containing 100 g of potassium nitrate and 100 g of water. Now let the solution cool.
At all temperatures above that marked on the graph (about 57°C), 100 g of water will dissolve more than 100 g of potassium nitrate. All the potassium nitrate will stay in solution.
At 57°C, you hit the solubility curve. This is the temperature at which 100 g of water will dissolve 100 g of potassium nitrate to give a saturated solution. If the temperature falls even the tiniest bit below 57°C, the water will no longer dissolve as much potassium nitrate - and so some crystallises out. The lower the temperature falls, the more potassium nitrate crystallizes, because the water will dissolve less and less of it.
You can think of this as a simple phase diagram. If you have a mixture of 100 g of potassium nitrate and 100 g of water and the temperature is above 57°C, you have a single phase - a solution of potassium nitrate. If the temperature is below 57°C for this mixture, you will have a mixture of two phases - the solution and some solid potassium nitrate. The solubility curve represents the boundary between these two different situations. We'll look now at the phase diagram for sodium chloride solution in some detail.
The phase diagram for sodium chloride solution
We'll talk through what everything means in detail, but first of all notice two things compared with the similar tin-lead phase diagram I hope you have already read about.
First we are only looking at a very restricted temperature range. The top of this particular diagram is only about 50°C - although it could go higher than that. Secondly, all the action takes place in the left-hand side of the diagram. That's why a lot of the horizontal scale is missing (represented by the broken zigzag in the scale).
The labeled areas in the phase diagram
These areas all show what you would see if you had a particular mixture of salt and water at a given temperature. For example, if the temperature was below -21.1°C, you would always see a mixture of solid salt and ice. There would never be any liquid whatever proportions of salt and water you had. At temperatures above this, what you would see would depend on where your particular set of conditions fell in the diagram. You just need to notice what area the conditions fall into.
What the lines mean
Let's take the lines one at a time. The first one to look at is the one in bold green in the next diagram.
This line represents the effect of increasing amounts of salt on the freezing point of water. Up to the point where there is 23.3% of salt in the mixture, the more salt the lower the freezing point of the water. The second line is actually a solubility curve for salt in water - although it doesn't look quite like the one we described earlier on this page.
The reason that it looks different is because the axes have been reversed. On a normal solubility curve, temperature is the horizontal axis and the vertical axis shows the solubility in grams per 100 g of water - a measure of concentration.
This time the measure of concentration is the horizontal axis and temperature the vertical one.
If you have already read about the tin-lead system, you might think it strange to include a solubility curve in the phase diagram. In the tin-lead system, both of the sloping lines showed the effect of one of the components on the freezing point of the other. But if you think what this means, it isn't so odd.
The freezing point is the temperature at which crystals start to appear when you cool a liquid mixture. In the case of a salt solution with concentrations of salt greater than 23.3%, the solubility curve shows the temperature at which crystals of salt will appear when you cool a solution of a given concentration. If you aren't clear about that, go back and re-read the beginning of this page.
There is also a hidden difference between this line and the corresponding line on the tin-lead diagram. In that case, both of the sloping lines eventually reached the the left-hand or the right-hand axis. It was possible to talk about the case where you had 100% lead or 100% tin. In this case it is impossible to get to 100% salt. This line eventually comes to an end. At 1 atmosphere pressure it won't get very far above 100°C because the water will boil and you won't have a solution any more.
Even if you raise the pressure, the maximum temperature you could achieve would be 374°C - the critical temperature of the water. Water doesn't exist as a liquid above this temperature. To get to the right-hand side of the graph where you have 100% salt, you would have to get the temperature up to 801°C - the melting point of the salt.
After all this hassle, the final line is easy!
This line is simply drawn across at the lowest temperature at which a mixture of salt and water can contain any liquid. This is known as the eutectic temperature. The particular mixture of salt and water (containing 23.3% salt) which freezes at this temperature is known as a eutectic mixture.
Using Phase Diagrams
The phase diagram is used to find out what happens if you cool salt solution of a particular concentration. We need to look at three separate cases.
Cooling salt solution with the eutectic composition
This is the easy one!
Nothing happens until you get down to the eutectic temperature. At that point, you will start to get both ice crystals and salt crystals forming. If you keep cooling it, you will obviously end up with a solid mixture of ice and salt. You are moving straight from the "salt solution" area of the phase diagram into the "solid salt + ice" area.
Cooling salt solution more dilute than the eutectic composition
When the temperature drops enough so that it reaches the boundary between the two areas of the phase diagram, ice crystals start to form - in other words, the solution starts to freeze. As the solution continues to cool, it moves down into the "ice + salt solution" region. Obviously, the composition of the solution has changed because it contains less water - some of it has frozen to give ice. But the composition of the system as a whole is still the same, and so we continue down the same line.
To find out what is actually in the mixture, you draw a horizontal tie line through the point showing the temperature you have got to, and look at what it hits at either end.
• On the left it hits the vertical axis showing 100% water - in this case, that's the pure ice crystals.
• On the other end, it hits the sloping line - this tells you the composition of the remaining salt solution.
As the mixture continues to cool, it will eventually reach the eutectic temperature, and at this point all of the rest of the solution will turn to solid - a mixture of ice and salt crystals.
Cooling salt solution more concentrated than the eutectic composition
The phase diagram is interpreted in just the same way - except that this time, salt crystals are formed at first rather than ice crystals. In the next diagram, the first salt crystals will form when the temperature hits the boundary line. As the temperature continues to drop into the "solid salt + salt solution" area, more and more salt will crystallize out. To find out exactly what is present at any temperature, you can again draw a tie line and look at what is at either end.
The diagram shows the tie line when the temperature has dropped to some random place in the "solid salt + salt solution" area. Again, by looking at the ends of this tie line, you can see that the mixture contains solid salt (the 100% salt at the right-hand end of the line) and a solution whose concentration can be found by looking at the left-hand end of the line. As the temperature continues to fall, you will eventually reach the eutectic temperature, when everything left - salt and water - will turn to solid giving you a mixture of solid salt and ice.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
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This page explains the relationship between the cooling curves for liquid mixtures of tin and lead, and the resulting phase diagram. It also offers a simple introduction to the idea of a eutectic mixture.
Cooling Curves for Pure Substances
Suppose you have some pure molten lead and allow it to cool down until it has all solidified, plotting the temperature of the lead against time as you go. You would end up with a typical cooling curve for a pure substance.
Throughout the whole experiment, heat is being lost to the surroundings - and yet the temperature doesn't fall at all while the lead is freezing. This is because the freezing process liberates heat at exactly the same rate that it is being lost to the surroundings. Energy is released when new bonds form - in this case, the strong metallic bonds in the solid lead. If you repeated this process for pure liquid tin, the shape of the graph would be exactly the same, except that the freezing point would now be at 232°C (the graph for this is further down the page.)
Cooling Curves for Tin-Lead Mixtures
If you add some tin to the lead, the shape of the cooling curve changes. The next graph shows what happens if you cool a liquid mixture containing about 67% lead and 33% tin by mass.
There are lots of things to look at:
• Notice that nothing happens at all at the normal freezing point of the lead. Adding the tin to it lowers its freezing point.
• Freezing starts for this mixture at about 250°C. You would start to get some solid lead formed - but no tin. At that point the rate of cooling slows down - the curve gets less steep.
• However, the graph doesn't go horizontal yet. Although energy is being given off as the lead turns to a solid, there isn't anything similar happening to the tin. That means that there isn't enough energy released to keep the temperature constant.
• The temperature does stop falling at 183°C. Now both tin and lead are freezing. Once everything has solidified, the temperature continues to fall.
Changing the Proportions of Tin and Lead
If you had less tin in the mixture, the overall shape of the curve stays much the same, but the point at which the lead first starts to freeze changes. The less tin there is, the smaller the drop in the freezing point of the lead. For a mixture containing only 20% of tin, the freezing point of the lead is about 275°C. That's where the graph would suddenly become less steep. BUT . . . you will still get the graph going horizontal (showing the freezing of both the tin and lead) at exactly the same temperature: 183°C.
As you increase the proportion of tin, the first signs of solid lead appear at lower and lower temperatures, but the final freezing of the whole mixture still happens at 183°C. That continues until you have added enough tin that the mixture contains 62% tin and 38% lead. At that point, the graph changes.
This particular mixture of lead and tin has a cooling curve which looks exactly like that of a pure substance rather than a mixture. There is just the single horizontal part of the graph where everything is freezing. However, it is still a mixture (not a solution). If you use a microscope to look at the solid formed after freezing, you can see the individual crystals of tin and lead.
This particular mixture is known as a eutectic mixture. The word "eutectic" comes from Greek and means "easily melted". The eutectic mixture has the lowest melting point (which is, of course, the same as the freezing point) of any mixture of lead and tin. The temperature at which the eutectic mixture freezes or melts is known as the eutectic temperature.
What happens if there is more than 62% of tin in the mixture?
You can trace it through in exactly the same way, by imagining starting with pure tin and then adding lead to it. The cooling curve for pure liquid tin looks like this:
It's just like the pure lead cooling curve except that tin's freezing point is lower. If you add small amounts of lead to the tin, so that you have perhaps 80% tin and 20% lead, you will get a curve like this:
Notice the lowered freezing point of the tin. Notice also the final freezing of the whole mixture again takes place at 183°C. As you increase the amount of lead (or decrease the amount of tin - same thing!) until there is 62% of tin and 38% of lead, you will again get the eutectic mixture with the curve we've already looked at.
The phase diagram
Constructing the phase diagram
You start from data obtained from the cooling curves. You draw a graph of the temperature at which freezing first starts against the proportion of tin and lead in the mixture. The only unusual thing is that you draw the temperature scale at each end of the diagram instead of only at the left-hand side.
Notice that at the left-hand side and right-hand sides of the curves you have the freezing points (melting points) of the pure lead and tin.
To finish off the phase diagram, all you have to do is draw a single horizontal line across at the eutectic temperature. Then you label each area of the diagram with what you would find under the various different conditions.
Using the Phase Diagram
Suppose you have a mixture of 67% lead and 33% tin. That's the mixture from the first cooling curve plotted above. Suppose it is at a temperature of 300°C. That corresponds to a set of conditions in the area of the phase diagram labeled as molten tin and lead.
Now consider what happens if you cool that mixture. Eventually the temperature will drop to a point where it crosses the line into the next region of the diagram. At that point, the mixture will start to produce some solid lead - in other words, the lead (but not the tin) starts to freeze. That happens at a temperature of about 250°C.
Now, it is the next bit that needs careful thinking about, because there are two different ways you can look at it. If you have been taught to do it one way, then stick with that - otherwise you risk getting very confused!
Thinking about changes in the composition of the liquid
When the first of the lead freezes, the composition of the remaining liquid changes. It obviously becomes proportionally richer in tin. That lowers the freezing point of the lead a bit more, and so the next bit of lead freezes at a slightly lower temperature - leaving a liquid still richer in tin.
This process goes on. The liquid gets richer and richer in tin, and the temperature needed to freeze the next lot of lead continues to fall. The set of conditions of temperature and composition of the liquid essentially moves down the curve - until it reaches the eutectic point.
Once it has reached the eutectic point, if the temperature continues to fall, you obviously just move into the region of a mixture of solid lead and solid tin - in other words, all the remaining liquid freezes.
Thinking about the composition of the system as a whole
We've seen that as the liquid gradually freezes, its composition changes. But if you look at the system as a whole, obviously the proportions of lead and tin remain constant - you aren't taking anything away or adding anything. All that is happening is that things are changing from liquids to solids. So suppose we continue the cooling beyond the temperature that the first solid lead appears and the temperature drops to the point shown in the next diagram - a point clearly in the "solid lead and molten mixture" area.
What would you see in the mixture? To find out, you draw a horizontal tie line through that point, and then look at the ends of it.
At the left-hand end, you have 100% lead. That represents the solid lead that has frozen from the mixture. At the right-hand end, you have the composition of the liquid mixture. This is now much richer in tin than the whole system is - because obviously a fair bit of solid lead has separated out. As the temperature continues to fall, the composition of the liquid mixture (as shown by the right-hand end of the tie line) will get closer and closer to the eutectic mixture.
It will finally reach the eutectic composition when the temperature drops to the eutectic temperature - and the whole lot then freezes. At a temperature lower than the eutectic temperature, you are obviously in the solid lead plus solid tin region. If you cooled a liquid mixture on the right-hand side of the phase diagram (to the right of the eutectic mixture), everything would work exactly the same except that solid tin would be formed instead of solid lead. If you have understood what has gone before, it isn't at all difficult to work out what happens.
Finally . . . what happens if you cool a liquid mixture which has exactly the eutectic composition? It simply stays as a liquid mixture until the temperature falls enough that it all solidifies. You never get into the awkward areas of the phase diagram.
Tin-lead mixtures as solder
Traditionally, tin-lead mixtures have been used as solder, but these are being phased out because of health concerns over the lead. This is especially the case where the solder is used to join water pipes where the water is used for drinking. New non-lead solders have been developed as safer replacements.
Typical old-fashioned solders include:
• 60% tin and 40% lead. This is close to the eutectic composition (62% tin and 38% lead), giving a low melting point. It will also melt and freeze cleanly over a very limited temperature range. This is useful for electrical work.
• 50% tin and 50% lead. This will melt and freeze over a wider range of temperatures. When it is molten it will start to freeze at about 220°C and finally solidify at the eutectic temperature of 183°C. That means that it stays workable for a useful amount of time. That's helpful if it is being used for plumbing joints.
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This page looks at the phase diagrams for non-ideal mixtures of liquids, and introduces the idea of an azeotropic mixture (also known as an azeotrope or constant boiling mixture). It goes on to explain how this complicates the process of fractionally distilling such a mixture.
Ideal Solutions follow Raoult's Law and Real Solutions Do Not
Due to Raoult's Law, a plot the vapor pressure of an ideal mixture of two liquids against their composition will result in a straight line:
In this case, pure A has the higher vapor pressure and so is the more volatile component. Raoult's Law only works for ideal mixtures. In these, the forces between the particles in the mixture are exactly the same as those in the pure liquids. The tendency for the particles to escape is the same in the mixture and in the pure liquids. That's not true in non-ideal mixtures.
In mixtures showing a positive deviation from Raoult's Law, the vapor pressure of the mixture is always higher than you would expect from an ideal mixture. The deviation can be small - in which case, the straight line in the last graph turns into a slight curve.
Notice that the highest vapor pressure anywhere is still the vapor pressure of pure A. Cases like this, where the deviation is small, behave just like ideal mixtures as far as distillation is concerned, and we do not need to say anything more about them. But some liquid mixtures have very large positive deviations from Raoult's Law, and in these cases, the curve becomes very distorted.
Notice that mixtures over a range of compositions have higher vapor pressures than either pure liquid. The maximum vapor pressure is no longer that of one of the pure liquids. This has important consequences when we look at boiling points and distillation further down the page. The fact that the vapor pressure is higher than ideal in these mixtures means that molecules are breaking away more easily than they do in the pure liquids. That is because the intermolecular forces between molecules of A and B are less than they are in the pure liquids.
You can see this when you mix the liquids. Less heat is evolved when the new attractions are set up than was absorbed to break the original ones. Heat will therefore be absorbed when the liquids mix. The enthalpy change of mixing is endothermic. The classic example of a mixture of this kind is ethanol and water. This produces a highly distorted curve with a maximum vapor pressure for a mixture containing 95.6% of ethanol by mass.
In exactly the same way, you can have mixtures with vapor pressures which are less than would be expected by Raoult's Law. In some cases, the deviations are small, but in others they are much greater giving a minimum value for vapor pressure lower than that of either pure component.
These are cases where the molecules break away from the mixture less easily than they do from the pure liquids. New stronger forces must exist in the mixture than in the original liquids. You can recognize this happening because heat is evolved when you mix the liquids - more heat is given out when the new stronger bonds are made than was used in breaking the original weaker ones. Many (although not all) examples of this involve actual reaction between the two liquids. The example of a major negative deviation is a mixture of nitric acid and water. These two covalent molecules react to give hydroxonium ions and nitrate ions.
$\ce{H2O(l) + HNO3(l) <=> H3O^{+} (aq) + NO3(aq)^{-}} \label{1}$
You now have strong ionic attractions involved which increases the intermolecular interactions.
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This page explains how to interpret the phase diagrams for simple pure substances - including a look at the special cases of the phase diagrams of water and carbon dioxide.
The Basic Phase Diagram
At its simplest, a phase can be just another term for solid, liquid or gas. If you have some ice floating in water, you have a solid phase present and a liquid phase. If there is air above the mixture, then that is another phase. But the term can be used more generally than this. For example, oil floating on water also consists of two phases - in this case, two liquid phases. If the oil and water are contained in a bucket, then the solid bucket is yet another phase. In fact, there might be more than one solid phase if the handle is attached separately to the bucket rather than molded as a part of the bucket.
You can recognize the presence of the different phases because there is an obvious boundary between them - a boundary between the solid ice and the liquid water, for example, or the boundary between the two liquids. A phase diagram lets you work out exactly what phases are present at any given temperature and pressure. In the cases we'll be looking at on this page, the phases will simply be the solid, liquid or vapor (gas) states of a pure substance. This is the phase diagram for a typical pure substance.
These diagrams (including this one) are nearly always drawn highly distorted in order to see what is going on more easily. There are usually two major distortions. We'll discuss these when they become relevant. If you look at the diagram, you will see that there are three lines, three areas marked "solid", "liquid" and "vapor", and two special points marked "C" and "T".
The Three Areas
These are easy! Suppose you have a pure substance at three different sets of conditions of temperature and pressure corresponding to 1, 2 and 3 in the next diagram.
Under the set of conditions at 1 in the diagram, the substance would be a solid because it falls into that area of the phase diagram. At 2, it would be a liquid; and at 3, it would be a vapor (a gas).
Phase Transitions
Moving from solid to liquid by changing the temperature
Suppose you had a solid and increased the temperature while keeping the pressure constant - as shown in the next diagram. As the temperature increases to the point where it crosses the line, the solid will turn to liquid. In other words, it melts.
If you repeated this at a higher fixed pressure, the melting temperature would be higher because the line between the solid and liquid areas slopes slightly forward.
So what actually is this line separating the solid and liquid areas of the diagram?
It simply shows the effect of pressure on melting point. Anywhere on this line, there is an equilibrium between solid and liquid. You can apply Le Chatelier's Principle to this equilibrium just as if it was a chemical equilibrium. If you increase the pressure, the equilibrium will move in such a way as to counter the change you have just made.
If it converted from liquid to solid, the pressure would tend to decrease again because the solid takes up slightly less space for most substances. That means that increasing the pressure on the equilibrium mixture of solid and liquid at its original melting point will convert the mixture back into the solid again. In other words, it will no longer melt at this temperature.
To make it melt at this higher pressure, you will have to increase the temperature a bit. Raising the pressure raises the melting point of most solids. That's why the melting point line slopes forward for most substances.
Moving from solid to liquid by changing the pressure
You can also play around with this by looking at what happens if you decrease the pressure on a solid at constant temperature.
Moving from liquid to vapor
In the same sort of way, you can do this either by changing the temperature or the pressure.
The liquid will change to a vapor - it boils - when it crosses the boundary line between the two areas. If it is temperature that you are varying, you can easily read off the boiling temperature from the phase diagram. In the diagram above, it is the temperature where the red arrow crosses the boundary line.
So, again, what is the significance of this line separating the two areas? Anywhere along this line, there will be an equilibrium between the liquid and the vapor. The line is most easily seen as the effect of pressure on the boiling point of the liquid. As the pressure increases, so the boiling point increases.
The critical point
You will have noticed that this liquid-vapor equilibrium curve has a top limit (labeled as C in the phase diagram), which is known as the critical point. The temperature and pressure corresponding to this are known as the critical temperature and critical pressure. If you increase the pressure on a gas (vapor) at a temperature lower than the critical temperature, you will eventually cross the liquid-vapor equilibrium line and the vapor will condense to give a liquid.
This works fine as long as the gas is below the critical temperature. What, though, if your temperature was above the critical temperature? There wouldn't be any line to cross! That is because, above the critical temperature, it is impossible to condense a gas into a liquid just by increasing the pressure. All you get is a highly compressed gas. The particles have too much energy for the intermolecular attractions to hold them together as a liquid. The critical temperature obviously varies from substance to substance and depends on the strength of the attractions between the particles. The stronger the intermolecular attractions, the higher the critical temperature.
Moving from solid to vapor
There's just one more line to look at on the phase diagram. This is the line in the bottom left-hand corner between the solid and vapor areas. That line represents solid-vapor equilibrium. If the conditions of temperature and pressure fell exactly on that line, there would be solid and vapor in equilibrium with each other - the solid would be subliming. (Sublimation is the change directly from solid to vapor or vice versa without going through the liquid phase.)
Once again, you can cross that line by either increasing the temperature of the solid, or decreasing the pressure. The diagram shows the effect of increasing the temperature of a solid at a (probably very low) constant pressure. The pressure obviously has to be low enough that a liquid can't form - in other words, it has to happen below the point labelled as T.
You could read the sublimation temperature off the diagram. It will be the temperature at which the line is crossed.
The Triple Point
Point T on the diagram is called the triple point. If you think about the three lines which meet at that point, they represent conditions of:
• solid-liquid equilibrium
• liquid-vapor equilibrium
• solid-vapor equilibrium
Where all three lines meet, you must have a unique combination of temperature and pressure where all three phases are in equilibrium together. That's why it is called a triple point.
If you controlled the conditions of temperature and pressure in order to land on this point, you would see an equilibrium which involved the solid melting and subliming, and the liquid in contact with it boiling to produce a vapor - and all the reverse changes happening as well. If you held the temperature and pressure at those values, and kept the system closed so that nothing escaped, that's how it would stay.
Normal melting and boiling points
The normal melting and boiling points are those when the pressure is 1 atmosphere. These can be found from the phase diagram by drawing a line across at 1 atmosphere pressure.
Example \(1\): Phase Diagram for Water
There is only one difference between this and the phase diagram that we've looked at up to now. The solid-liquid equilibrium line (the melting point line) slopes backwards rather than forwards.
In the case of water, the melting point gets lower at higher pressures. Why?
If you have this equilibrium and increase the pressure on it, according to Le Chatelier's Principle the equilibrium will move to reduce the pressure again. That means that it will move to the side with the smaller volume. Liquid water is produced. To make the liquid water freeze again at this higher pressure, you will have to reduce the temperature. Higher pressures mean lower melting (freezing) points.
Now lets put some numbers on the diagram to show the exact positions of the critical point and triple point for water.
Notice that the triple point for water occurs at a very low pressure. Notice also that the critical temperature is 374°C. It would be impossible to convert water from a gas to a liquid by compressing it above this temperature. The normal melting and boiling points of water are found in exactly the same way as we have already discussed - by seeing where the 1 atmosphere pressure line crosses the solid-liquid and then the liquid-vapor equilibrium lines.
Just one final example of using this diagram. Imagine lowering the pressure on liquid water along the line in the diagram below.
The phase diagram shows that the water would first freeze to form ice as it crossed into the solid area. When the pressure fell low enough, the ice would then sublime to give water vapor. In other words, the change is from liquid to solid to vapor.
Example \(2\): Phase Diagram for Carbon Dioxide
The only thing special about this phase diagram is the position of the triple point which is well above atmospheric pressure. It is impossible to get any liquid carbon dioxide at pressures less than 5.11 atmospheres.
That means that at 1 atmosphere pressure, carbon dioxide will sublime at a temperature of -78°C. This is the reason that solid carbon dioxide is often known as "dry ice". You can't get liquid carbon dioxide under normal conditions - only the solid or the vapor.
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The interactions between atoms and molecules arre divided into covalent bonding, ionic attraction, dipole-dipole interaction, London dispersion forces or van der Waal's force, hydrogen bond, and metallic bonding. These forces affect properties of solids, liquids, and phase transitions.
Phases and Their Transitions
Learning Objectives
• Define liquid.
• Explain viscosity, surface tension, and capillary action.
Phase Diagrams
Learning Objectives
• Explain the phase diagrams of water and carbon dioxide.
• Define and explain triple point, sublimation curve, vaporization curve, melting curve, and critical temperature and pressure.
Phase Transition - A Review
This quiz covers aspects related to phase transitions or phase changes. You will encounter calculations and short-answer questions involving the following terms and concepts:
• freezing point, melting point
• simulation curve, vaporization curve
• simulation, deposition
• boiling point, condensation point
• cooling curve, heating curve
• heat of vaporization, molar heat of vaporization
• heat of sublimation, molar heat of sublimation
• heat capacity
• phase transition
• packing of spheres
• calculations of density, unit cell edges, number of atoms,
• distances between atoms for simple cubic packing,
• f.c.c packing, and body-centered cubic structures
• hexagonal closest packing
Phases
Learning Objectives
• Explain states of matter.
• Define and explain what a phase is.
• Interpret these phase transitions:
sublimation, deposition,
vaporization, condensation, and
melting and freezing.
States: Gas, Liquid and Solid
Gas, liquid, and solid are known as the three states of matter or material, but each of solid and liquid states may exist in one or more forms. Thus, another term is required to describe the various forms, and the term phase is used. Each distinct form is called a phase; however, the concept of phase defined as a homogeneous portion of a system extends beyond a single material, because a phase may also involve several materials. For example, a homogeneous solution of any number of substances is a one-phase system. Phase is a concept used to explain many physical and chemical changes (reactions).
• A solid has a definite shape and volume. A liquid has a definite volume but it takes the shape of a container whereas a gas fills the entire volume of a container. You already know that diamond and graphite are solids made up of the element carbon; they are two phases of carbon, but both are solids. Solids are divided into subclasses of amorphous (or glassy) solids and crystalline solids. Arrangements of atoms or molecules in crystalline solids are repeated regularly over a very long range of millions of atoms, but their arrangements in amorphous solids are somewhat random or short range of say some tens or hundreds of atoms.
• In general, there is only one liquid phase of a material. However, there are two forms of liquid helium; each have some unique properties. Thus, the two forms are different (liquid) phases of helium. At a definite temperature and pressure, the two phases co-exist.
• So far, all gases behave alike as do mixtures of gases. Thus, a gas is usually considered as a phase.
The Concept of Phase
A phase is a distinct and homogeneous state of a system with no visible boundary separating it into parts.
Water, $\ce{H2O}$, is such a common substance that its gas (steam), liquid (water), and solid (ice) phases are widely known. An ice water mixture has two phases, as do systems containing ice-and-vapor, and water-and-vapor. To recognize the vapor system in these systems may require a keen observation, because the vapor usually blends with air, and is not detected directly.
You probably also know that several solids may exist for a substance, and each of the solid forms is also called a phase. Diamond and graphite are the most quoted examples; both are solid carbon, but they have different crystal shapes, colors, and structures. They represent two different phases of carbon. Ice is another example, under 1 atm, ice has hexagonal symmetry, while cubic ice is formed under high pressure. In fact, there are at least eight different types of ice, each being a solid phase.
When you mix water and alcohol, regardless of the relative amounts that you use, they are fully miscible. The resulting mixture has only one phase (a solution). However, water and oil are normally immiscible, and their boundary of separation is visible; they form a two-phase system. Sometimes you cannot "see" the boundary, and you will need scientific reasoning to realize the number of phases present in system.
Well, there is so much concept packed into one term that we can not make the definition any simpler for you. However, the term is useful because it can be used to explain many phenomena. There is no substitute for it. Learn it and use it to explain physical changes.
Phase Transitions
The conversion between these phases is called a phase transition. A state change of any material due to temperature or pressure change is a phase transition. A phase transition is a physical change (or reaction). The following diagram illustrates the key phase transitions: You should know the names of the process for these phase transitions.
$\mathrm{SOLID \xrightarrow{\large{sublimation\:}} GAS \xrightarrow{\large{deposition\:}} SOLID}$
$\mathrm{SOLID \xrightarrow{\large{melting\:}} LIQUID \underset{\large{(solidfication)}}{\xrightarrow{\large{\:\:\:freezing\:\:\:\:}}} SOLID}$
$\mathrm{GAS \xrightarrow{\large{condensation\:}} LIQUID \xrightarrow{\large{vaporization\:}} GAS}$
The concepts of phase and phase transition introduce you into fields of materials studies. For example, if you search the internet with the phrase "phase transition", you get thousands of websites; some are related to the concept we have discussed here, but some may be using "phase transition" as a catchy phrase. The concept of phase transition is also applied to the study of nuclear matter such as protons and neutrons.
Confidence Building Questions
1. How many phases are there in milk?
Hint: more than one
Skill -
Define phase and enumerate the number of phases in a system.
Milk is a colloid.
2. Which one of the following systems has three phases that are at equilibrium with one another?
1. a sealed container full of unsaturated $\ce{NaCl}$ solution.
2. a sealed container full of oversaturated sugar solution.
3. a sealed container full of colloid made by suspending solid in liquid.
4. a sealed container half full of water and oil.
5. an open container half full of oil and water and ice.
Hint: d
Skill -
Identify the number of phases in a system.
3. How many phases are present in a closed flask if it contains ice, water and air?
Hint: Three phases: solid, liquid, and vapor (gas solution).
Discussion -
What will be the temperature for a system describe here?
4. How many phases are present in a closed flask if it contains dry ice and air?
Hint: Two phases: solid and vapor (gas solution).
Skill -
Define phase and enumerate the number of phases in a system.
5. A sealed 5-L flask contains 1.0 L of solution, some $\ce{NaCl}$ crystals, and some ice crystals. The rest of the space is filled with air. How many phases are present in such a system?
1. 1
2. 2
3. 3
4. 4
5. 5
Hint: d. Four phases: liquid, $\ce{NaCl}$, ice, gas
Skill -
Enumerate the number of phases in a system.
Do you know what the temperature is for such a system to be at equilibrium? It's 0 degree F (the temperature of equal weight mixture of snow and salt).
6. A 750-mL bottle contains 300 mL clear colorless alcoholic beverage. What are the phases present in the bottle?
Hint: A liquid and a gas solution, 2 phases.
Skill -
Identify the phases in a system.
7. What is the name for the process of phase transition by which a gas is converted to a solid?
Hint: Deposition
Skill -
Know the names of all phase transition processes.
8. The vapor pressure of a solid
1. increases as the volume of the solid increases.
2. decreases as the temperature of the solid increases.
3. increases as the temperature of the solid increases.
4. decreases as the volume of the solid increases.
Hint: c. increases as the temperature of the solid increases.
Skill -
This is a test of common sense.
A scientific way of looking at the vapor pressure of a solid as a function of temperature is given by the Clausius-Clapeyron Equation:
$\mathrm{P_{vap} = A e^{-\mathit H_{\large{vap}}/\mathit{RT}}}$
where Hvap is the heat of vaporization, and R T are the gas constant and temperature respectively.
9. Water ALWAYS boils at 100 degree C: true or false?
Hint: false
Skill -
This is a test of common sense.
Only when the atmospheric pressure is 1.00 atm will the boiling temperature be 100 deg. C, which is the normal boiling point of water.
10. The volume of 18 grams of water increases as its temperature increases from 0 to 4 deg. C: true or false?
Hint: false!
Skill -
This is a test of common sense. You know that the density of water is the highest at 4 deg. C, and its volume for a mole is the lowest at 4 deg C.
One mole of water is 18 g, and its volume is called the molar volume. Water is a strange substance!
11. The volume of 18 grams of water increases as its temperature increases from 10 to 84 deg. C: true or false?
Hint: true!
Skill -
Analyze the problem carefully. Most substances expand when heated.
12. One mole water of which phase occupies the largest volume: solid, liquid or gas?
Hint: gas
Discussion -
Temperature is not specified here, but you may assume a temperature at which all three phases are stable. Under the circumstance, the pressure of vapor phase is much lower than 1 atm. For the same quantity of water, ice has a larger volume than water.
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This page deals with Raoult's Law and how it applies to mixtures of two volatile liquids. It covers cases where the two liquids are entirely miscible in all proportions to give a single liquid - NOT those where one liquid floats on top of the other (immiscible liquids). The page explains what is meant by an ideal mixture and looks at how the phase diagram for such a mixture is built up and used.
Ideal Mixtures
An ideal mixture is one which obeys Raoult's Law, but I want to look at the characteristics of an ideal mixture before actually stating Raoult's Law. The page will flow better if I do it this way around. There is actually no such thing as an ideal mixture! However, some liquid mixtures get fairly close to being ideal. These are mixtures of two very closely similar substances. Commonly quoted examples include:
• hexane and heptane
• benzene and methylbenzene
• propan-1-ol and propan-2-ol
In a pure liquid, some of the more energetic molecules have enough energy to overcome the intermolecular attractions and escape from the surface to form a vapor. The smaller the intermolecular forces, the more molecules will be able to escape at any particular temperature.
If you have a second liquid, the same thing is true. At any particular temperature a certain proportion of the molecules will have enough energy to leave the surface.
In an ideal mixture of these two liquids, the tendency of the two different sorts of molecules to escape is unchanged.
You might think that the diagram shows only half as many of each molecule escaping - but the proportion of each escaping is still the same. The diagram is for a 50/50 mixture of the two liquids. That means that there are only half as many of each sort of molecule on the surface as in the pure liquids. If the proportion of each escaping stays the same, obviously only half as many will escape in any given time. If the red molecules still have the same tendency to escape as before, that must mean that the intermolecular forces between two red molecules must be exactly the same as the intermolecular forces between a red and a blue molecule.
If the forces were any different, the tendency to escape would change. Exactly the same thing is true of the forces between two blue molecules and the forces between a blue and a red. They must also be the same otherwise the blue ones would have a different tendency to escape than before. If you follow the logic of this through, the intermolecular attractions between two red molecules, two blue molecules or a red and a blue molecule must all be exactly the same if the mixture is to be ideal.
This is why mixtures like hexane and heptane get close to ideal behavior. They are similarly sized molecules and so have similarly sized van der Waals attractions between them. However, they obviously are not identical - and so although they get close to being ideal, they are not actually ideal. For the purposes of this topic, getting close to ideal is good enough!
Ideal Mixtures and the Enthalpy of Mixing
When you make any mixture of liquids, you have to break the existing intermolecular attractions (which needs energy), and then remake new ones (which releases energy). If all these attractions are the same, there won't be any heat either evolved or absorbed. That means that an ideal mixture of two liquids will have zero enthalpy change of mixing. If the temperature rises or falls when you mix the two liquids, then the mixture is not ideal. You may have come cross a slightly simplified version of Raoult's Law if you have studied the effect of a non-volatile solute like salt on the vapor pressure of solvents like water. The definition below is the one to use if you are talking about mixtures of two volatile liquids.
Definition: Raoult's Law
The partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.
Raoult's Law only works for ideal mixtures. In equation form, for a mixture of liquids A and B, this reads:
$P_A = \chi_A P^o_A \label{1}$
$P_B = \chi_B P^o_B \label{2}$
In this equation, PA and PB are the partial vapor pressures of the components A and B. In any mixture of gases, each gas exerts its own pressure. This is called its partial pressure and is independent of the other gases present. Even if you took all the other gases away, the remaining gas would still be exerting its own partial pressure. The total vapor pressure of the mixture is equal to the sum of the individual partial pressures.
$\underset{\text{total vapor pressure}}{P_{total} } = P_A + P_B \label{3}$
The Po values are the vapor pressures of A and B if they were on their own as pure liquids. xA and xB are the mole fractions of A and B. That is exactly what it says it is - the fraction of the total number of moles present which is A or B. You calculate mole fraction using, for example:
$\chi_A = \dfrac{\text{moles of A}}{\text{total number of moles}} \label{4}$
Example $1$
Suppose you had a mixture of 2 moles of methanol and 1 mole of ethanol at a particular temperature. The vapor pressure of pure methanol at this temperature is 81 kPa, and the vapor pressure of pure ethanol is 45 kPa. What is total vapor pressure of this solution?
Solution
There are 3 moles in the mixture in total.
• 2 of these are methanol. The mole fraction of methanol is 2/3.
• Similarly, the mole fraction of ethanol is 1/3.
You can easily find the partial vapor pressures using Raoult's Law - assuming that a mixture of methanol and ethanol is ideal.
First for methanol:
$P_{methanol} = \dfrac{2}{3} \times 81\; kPa$
$= 54\; kPa$
Then for ethanol:
$P_{ethanol} = \dfrac{1}{3} \times 45\; kPa$
$= 15\; kPa$
You get the total vapor pressure of the liquid mixture by adding these together.
$P_{total} = 54\; kPa + 15 \; kPa = 69 kPa$
In practice, this is all a lot easier than it looks when you first meet the definition of Raoult's Law and the equations!
Vapor Pressure and Composition Diagrams
Suppose you have an ideal mixture of two liquids A and B. Each of A and B is making its own contribution to the overall vapor pressure of the mixture - as we've seen above. Let's focus on one of these liquids - A, for example. Suppose you double the mole fraction of A in the mixture (keeping the temperature constant). According to Raoult's Law, you will double its partial vapor pressure. If you triple the mole fraction, its partial vapor pressure will triple - and so on. In other words, the partial vapor pressure of A at a particular temperature is proportional to its mole fraction. If you plot a graph of the partial vapor pressure of A against its mole fraction, you will get a straight line.
Now we'll do the same thing for B - except that we will plot it on the same set of axes. The mole fraction of B falls as A increases so the line will slope down rather than up. As the mole fraction of B falls, its vapor pressure will fall at the same rate.
Notice that the vapor pressure of pure B is higher than that of pure A. That means that molecules must break away more easily from the surface of B than of A. B is the more volatile liquid. To get the total vapor pressure of the mixture, you need to add the values for A and B together at each composition. The net effect of that is to give you a straight line as shown in the next diagram.
Boiling point and Composition Diagrams
The relationship between boiling point and vapor pressure
• If a liquid has a high vapor pressure at a particular temperature, it means that its molecules are escaping easily from the surface.
• If, at the same temperature, a second liquid has a low vapor pressure, it means that its molecules are not escaping so easily.
What do these two aspects imply about the boiling points of the two liquids? There are two ways of looking at the above question:
Either:
• If the molecules are escaping easily from the surface, it must mean that the intermolecular forces are relatively weak. That means that you won't have to supply so much heat to break them completely and boil the liquid. Therefore, the liquid with the higher vapor pressure at a particular temperature is the one with the lower boiling point.
Or:
• Liquids boil when their vapor pressure becomes equal to the external pressure. If a liquid has a high vapor pressure at some temperature, you won't have to increase the temperature very much until the vapor pressure reaches the external pressure. On the other hand if the vapor pressure is low, you will have to heat it up a lot more to reach the external pressure. Therefore, the liquid with the higher vapor pressure at a particular temperature is the one with the lower boiling point.
For two liquids at the same temperature, the liquid with the higher vapor pressure is the one with the lower boiling point.
Constructing a boiling point / composition diagram
To remind you - we've just ended up with this vapor pressure / composition diagram:
We're going to convert this into a boiling point / composition diagram. We'll start with the boiling points of pure A and B. Since B has the higher vapor pressure, it will have the lower boiling point. If that is not obvious to you, go back and read the last section again!
For mixtures of A and B, you might perhaps have expected that their boiling points would form a straight line joining the two points we've already got. Not so! In fact, it turns out to be a curve.
To make this diagram really useful (and finally get to the phase diagram we've been heading towards), we are going to add another line. This second line will show the composition of the vapor over the top of any particular boiling liquid.
If you boil a liquid mixture, you would expect to find that the more volatile substance escapes to form a vapor more easily than the less volatile one. That means that in the case we've been talking about, you would expect to find a higher proportion of B (the more volatile component) in the vapor than in the liquid. You can discover this composition by condensing the vapor and analyzing it. That would give you a point on the diagram.
The diagram just shows what happens if you boil a particular mixture of A and B. Notice that the vapor over the top of the boiling liquid has a composition which is much richer in B - the more volatile component. If you repeat this exercise with liquid mixtures of lots of different compositions, you can plot a second curve - a vapor composition line.
This is now our final phase diagram.
Using the phase diagram
The diagram is used in exactly the same way as it was built up. If you boil a liquid mixture, you can find out the temperature it boils at, and the composition of the vapor over the boiling liquid. For example, in the next diagram, if you boil a liquid mixture C1, it will boil at a temperature T1 and the vapor over the top of the boiling liquid will have the composition C2.
All you have to do is to use the liquid composition curve to find the boiling point of the liquid, and then look at what the vapor composition would be at that temperature. Notice again that the vapor is much richer in the more volatile component B than the original liquid mixture was.
The beginnings of fractional distillation
Suppose that you collected and condensed the vapor over the top of the boiling liquid and reboiled it. You would now be boiling a new liquid which had a composition C2. That would boil at a new temperature T2, and the vapor over the top of it would have a composition C3.
You can see that we now have a vapor which is getting quite close to being pure B. If you keep on doing this (condensing the vapor, and then reboiling the liquid produced) you will eventually get pure B. This is obvious the basis for fractional distillation. However, doing it like this would be incredibly tedious, and unless you could arrange to produce and condense huge amounts of vapor over the top of the boiling liquid, the amount of B which you would get at the end would be very small. Real fractionating columns (whether in the lab or in industry) automate this condensing and reboiling process. How these work will be explored on another page.
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Solubility refers to the concentration of a saturated solution; it can be affected by the following factors.
• An Introduction to Solubility Products
This page discusses how solubility products are defined, including their units units. It also explores the relationship between the solubility product of an ionic compound and its solubility.
• Calculations Involving Solubility Products
This page is a brief introduction to solubility product calculations.
• Common Ion Effect
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.
• Pressure Effects On the Solubility of Gases
The solubility of gases depends on the pressure: an increase in pressure increases solubility, whereas a decrease in pressure decreases solubility. This statement is formalized in Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.
• Relating Solubility to Solubility Product
this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants ( Ksq ) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the Ksq of a slightly soluble solute from its solubility.
• Solubility
One of the general properties of ionic compounds is water solubility. The oceans are solutions of salt in water. In a mixture, two or more materials are mixed together but they remain essentially separate, like sand and water. The sand can be easily distinguished from the water, because even if a mixture of the two is shaken it will spontaneously separate over time.
• Solubility and Factors Affecting Solubility
• Solubility Product Constant, Ksp
• Solubility Rules
In order to predict whether a precipitate will form in a reaction, the solubility of the substances involved must be known. There are rules or guidelines determining solubility of substances. If a substance involved is not soluble, the reaction forms a precipitate.
• Temperature Effects on Solubility
The solubility of solutes is dependent on temperature. When a solid dissolves in a liquid, a change in the physical state of the solid analogous to melting takes place. Heat is required to break the bonds holding the molecules in the solid together. At the same time, heat is given off during the formation of new solute -- solvent bonds.
• Temperature Effects on the Solubility of Gases
The solubility of gases is not constant in all conditions. If temperatures differ, the solubility of gases differ. Additionally, the solvent (the substance that is mixed with a gas to form a solution) can affect the solubility of a gas (its ability to become dissolved and in turn contribute to a formed amount of concentration).
Solubilty
This page discusses how solubility products are defined, including their units units. It also explores the relationship between the solubility product of an ionic compound and its solubility.
Solubility products are equilibrium constants
Barium sulfate is almost insoluble in water. It is not totally insoluble—very small amounts do dissolve. This is true of any "insoluble" ionic compound. If solid barium sulfate is shaken with water, a small amount of barium ions and sulfate ions break away from the surface of the solid and go into solution. Over time, some of these return from solution to stick onto the solid again.
An equilibrium is established when the rate at which some i ons are breaking away from the solid lattice is exactly matched by the rate at which others are returning. Consider the balanced equation for the barium sulfate reaction:
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)\nonumber$
The position of this equilibrium lies very far to the left. The great majority of the barium sulfate is present as solid—there are no visible changes to the solid. However, the equilibrium does exist, and an equilibrium constant can be written. The equilibrium constant is called the solubility product, and is given the symbol Ksp:
$K_{sp} = [Ba^{2+} (aq)] [SO_4^{2-}(aq)]\nonumber$
For simplicity, solubility product expressions are often written without the state symbols:
$K_{sp} = [Ba^{2+}] [SO_4^{2-}]\nonumber$
Notice that there is no solid barium sulfate term. For many simple equilibria, the equilibrium constant expression has terms for the right side of the equation divided by terms for the left side. But in this case, there is no term for the concentration of the solid barium sulfate. This is a heterogeneous equilibrium, one which contains substances in more than one state. In a heterogeneous equilibrium, concentration terms for solids are left out of the expression.
Solubility products for more complicated solids
Here is the corresponding equilibrium for calcium phosphate, Ca3(PO4)2:
$Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)\nonumber$
Below is the solubility product expression:
$K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2\nonumber$
As with any other equilibrium constant, the concentrations are raised to the power of their respective stoichiometric coefficients in the equilibrium equation. Solubility products only apply to sparingly soluble ionic compounds, not to normally soluble compounds such as sodium chloride. Interactions between the ions in the solution interfere with the simple equilibrium.
The units for solubility products
The units for solubility products differ depending on the solubility product expression, and you need to be able to work them out each time.
Example $1$: Barium Sulfate
Below is the solubility product expression for barium sulfate (\ce{BaSO4}\)):
$K_{sp} = [Ba^{2+}] [SO_4^{2-}]\nonumber$
Each concentration has the unit mol dm-3. In this case, the units for the solubility product in this case are the following:
(mol dm-3) x (mol dm-3)
= mol2 dm-6
Example $2$: calcium phosphate
Recall the solubility product expression for calcium phosphate ($\ce{CaPO4}$)
$K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2\nonumber$
The units this time will be:
(mol dm-3)3 x (mol dm-3)2
= (mol dm-3)5
= mol5 dm-15
Solubility products apply only to saturated solutions
Recall the barium sulfate equation:
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)\nonumber$
The corresponding solubility product expression is the following:
$K_{sp}= [Ba^{2+}][SO_4^{2-}]\nonumber$
Ksp for barium sulfate at 298 K is 1.1 x 10-10 mol2 dm-6.
In order for this equilibrium constant (the solubility product) to apply, solid barium sulfate must be present in a saturated solution of barium sulfate. This is indicated by the equilibrium equation. If barium ions and sulfate ions exist in solution in the presence of some solid barium sulfate at 298 K, and multiply the concentrations of the ions together, the answer will be 1.1 x 10-10 mol2 dm-6. It is possible to multiply ionic concentrations and obtain a value less than this solubility product; in such cases the solution is too dilute, and no equilibrium exists. If the concentrations are lowered enough, no precipitate can form.
However, it is impossible to calculate a product greater than the solubility product. If solutions containing barium ions and sulfate ions are mixed such that the product of the concentrations would exceed Ksp, a precipitate forms. Enough solid is produced to reduce the concentrations of the barium and sulfate ions down to the value of the solubility product.
Summary
• The value of a solubility product relates only to a saturated solution.
• If the ionic concentrations give a value less than the solubility product, the solution is not saturated. No precipitate would be formed in such a case.
• If the ionic concentrations give a value more than the solubility product, enough precipitate would be formed to reduce the concentrations to give an answer equal to the solubility product.
Calculations Involving Solubility Products
This page is a brief introduction to solubility product calculations. The solubilities of the ionic compounds in the examples below are given in mol dm-3. If it is given in g dm-3, or another concentration unit, it must first be converted into mol dm-3.
Example $1$
The solubility of barium sulfate at 298 K is 1.05 x 10-5 mol dm-3. Calculate the solubility product. The equilibrium is given below:
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq) \nonumber$
Notice that each mole of barium sulfate dissolves to give 1 mole of barium ions and 1 mole of sulfate ions in solution. That means that:
$[Ba^{2+}] = 1.05 \times 10^{-5}\ mol\ dm^{-3} \nonumber$
$[SO_4^{2-}] = 1.05 \times 10^{-5}\ mol\ dm^{-3} \nonumber$
Substitute these values into the solubility product expression, and simplify:
$\begin{eqnarray} K_{sp} &=& [Ba^{2+}][SO_4^{2-}] \ &=& (1.05 \times 10^{-5}) \times (1.05 \times 10^{-5}) \ &=& 1.10 \times 10^{-10}\ mol^2\ dm^{-6} \end{eqnarray} \nonumber$
Verify that the units are correct.
These calculations are very simple if for compounds in which ratio of the numbers of positive and negative ions is 1:1. This next example illustrates the procedure for a different ratio.
Example $2$
The solubility of magnesium hydroxide at 298 K is 1.71 x 10-4 mol dm-3. Calculate the solubility product. The equilibrium is:
$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq) \nonumber$
For every mole of magnesium hydroxide that dissolves, one mole of magnesium ions is generated, but twice that number of hydroxide ions form. So the concentration of the dissolved magnesium ions is the same as the dissolved magnesium hydroxide:
$[Mg^{2+}] = 1.71 \times 10^{-4}\ mol\ dm^{-3} \nonumber$
The concentration of dissolved hydroxide ions is twice that:
$\begin{eqnarray} [OH^-] &=& 2 \times 1.71 \times 10^{-4} \ &=& 3.42 \times 10^{-4}\ mol\ dm^{-3} \end{eqnarray} \nonumber$
Substitute these values into the solubility product expression as before:
$\begin{eqnarray} K_{sp} &=& [Mg^{2+}][OH^-]^2 \ &=& (1.71 \times 10^{-4}) \times (3.42 \times 10^{-4})^2 \ &=& 2.00 \times 10^{-11}\ mol^3\ dm^{-9} \end{eqnarray} \nonumber$
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Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
The common-ion effect is used to describe the effect on an equilibrium when one or more species in the reaction is shared with another reaction. This results in a shifitng of the equilibrium properties.
Introduction
The solubility products Ksp's are equilibrium constants in hetergeneous equilibria (i.e., between two different phases). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}\nonumber$
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
\begin{align*} \ce{NaCl &\rightleftharpoons Na^{+}} + \color{Green} \ce{Cl^{-}}\[4pt] \ce{KCl &\rightleftharpoons K^{+}} + \color{Green} \ce{Cl^{-}} \[4pt] \ce{CaCl_2 &\rightleftharpoons Ca^{2+}} + \color{Green} \ce{2 Cl^{-}}\[4pt] \ce{AlCl_3 &\rightleftharpoons Al^{3+}} + \color{Green} \ce{3 Cl^{-}}\[4pt] \ce{AgCl & \rightleftharpoons Ag^{+}} + \color{Green} \ce{Cl^{-}} \end{align*}
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\ce{[Na^{+}] = [Ca^{2+}] = [H^{+}] = 0.10\, \ce M}. \nonumber$
but
\begin{align*} \ce{[Cl^{-}]} &= 0.10 \, \ce{(due\: to\: NaCl)}\[4pt] &+ 0.20\, \ce{(due\: to\: CaCl_2)} \[4pt] &+ 0.10\, \ce{(due\: to\: HCl)} \[4pt] &= 0.40\, \ce{M} \end{align*}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}$
Le Chatelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.
Example $2$
Consider the lead(II) ion concentration in this saturated solution of $\ce{PbCl2}$. The balanced reaction is
$\ce{ PbCl2 (s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \label{Ex1.1}$
Defining $s$ as the concentration of dissolved lead(II) chloride, then:
$[Pb^{2+}] = s \nonumber$
$[Cl^- ] = 2s\nonumber$
These values can be substituted into the solubility product expression, which can be solved for $s$:
\begin{align*} K_{sp} &= [Pb^{2+}] [Cl^{-}]^2 \[4pt] &= s \times (2s)^2 \[4pt] 1.7 \times 10^{-5} &= 4s^3 \[4pt] s^3 &= \dfrac{1.7 \times 10^{-5}}{4} \[4pt] &= 4.25 \times 10^{-6} \[4pt] s &= \sqrt[3]{4.25 \times 10^{-6}} \[4pt] &= 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{align*}
The concentration of lead(II) ions in the solution is 1.62 x 10-2 M. Consider what happens if sodium chloride is added to this saturated solution. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".
Look at the original equilibrium expression in Equation \ref{Ex1.1}. What happens to that equilibrium if extra chloride ions are added? According to Le Chatelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.
Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.
A Simple Example
If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.
$\ce{[Pb^{2+}]} = s \label{2}\nonumber$
The calculations are different from before. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.
In calculations like this, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.
So we assume:
$\ce{[Cl^{-} ]} = 0.100\; M \label{3}\nonumber$
The rest of the mathematics looks like this:
\begin{align*} K_{sp}& = [Pb^{2+}][Cl^-]^2 \[4pt] & = s \times (0.100)^2 \[4pt] 1.7 \times 10^{-5} & = s \times 0.00100 \end{align*}
therefore:
\begin{align*} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \[4pt] & = 1.7 \times 10^{-3} \, \text{M} \end{align*}
Finally, compare that value with the simple saturated solution:
Original solution:
$\ce{[Pb^{2+}]} = 0.0162 \, M \label{5}\nonumber$
Solution in 0.100 M $\ce{NaCl}$ solution:
$\ce{[Pb^{2+}]} = 0.0017 \, M \label{6}\nonumber$
The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.
Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant.
Common Ion Effect with Weak Acids and Bases
Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
Example $\PageIndex{3A}$
The common ion effect of $\ce{H3O^{+}}$ on the ionization of acetic acid
The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium.
Example $\PageIndex{3B}$
Consider the common ion effect of $\ce{OH^{-}}$ on the ionization of ammonia
Adding the common ion of hydroxide shifts the reaction towards the left to decrease the stress (in accordance with Le Chatelier's Principle), forming more reactants. This decreases the reaction quotient, because the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, $K_b=1.8 \times 10^{-5}$, does not change. The reaction is put out of balance, or equilibrium.
$Q_a = \dfrac{[\ce{NH_4^{+}}][\ce{OH^{-}}]}{[\ce{NH_3}]} \nonumber$
At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing $Q$ to decrease towards $K$.
Common Ion Effect on Solubility
Adding a common ion decreases solubility, as the reaction shifts toward the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, toward the reactants, causing precipitation.
Example $4$
Consider the reaction:
$\ce{ PbCl_2(s) <=> Pb^{2+}(aq) + 2Cl^{-}(aq)} \nonumber$
What happens to the solubility of $\ce{PbCl2(s)}$ when 0.1 M $\ce{NaCl}$ is added?
Solution
$K_{sp}=1.7 \times 10^{-5} \nonumber$
$Q_{sp}= 1.8 \times 10^{-5} \nonumber$
Identify the common ion: $\ce{Cl^{-}}$
Notice: $Q_{sp} > K_{sp}$ The addition of $\ce{NaCl}$ has caused the reaction to shift out of equilibrium because there are more dissociated ions. Typically, solving for the molarities requires the assumption that the solubility of $\ce{PbCl2(s)}$ is equivalent to the concentration of $\ce{Pb^{2+}}$ produced because they are in a 1:1 ratio.
Because $K_{sp}$ for the reaction is $1.7 \times 10^{-5}$, the overall reaction would be
$(s)(2s)^2= 1.7 \times 10^{-5}. \nonumber$
Solving the equation for $s$ gives $s= 1.62 \times 10^{-2}\, \text{M}$. The coefficient on $\ce{Cl^{-}}$ is 2, so it is assumed that twice as much $\ce{Cl^{-}}$ is produced as $\ce{Pb^{2+}}$, hence the '2s.' The solubility equilibrium constant can be used to solve for the molarities of the ions at equilibrium.
The molarity of Cl- added would be 0.1 M because $\ce{Na^{+}}$ and $\ce{Cl^{-}}$ are in a 1:1 ratio in the ionic salt, $\ce{NaCl}$. Therefore, the overall molarity of $\ce{Cl^{-}}$ would be $2s + 0.1$, with $2s$ referring to the contribution of the chloride ion from the dissociation of lead chloride.
\begin{align*} Q_{sp} &= [\ce{Pb^{2+}}][\ce{Cl^{-}}]^2 \[4pt] &= 1.8 \times 10^{-5} \[4pt] &= (s)(2s + 0.1)^2 \[4pt] s &= [Pb^{2+}] \[4pt] &= 1.8 \times 10^{-3} M \[4pt] 2s &= [\ce{Cl^{-}}] \[4pt] &\approx 0.1 M \end{align*}
Notice that the molarity of $\ce{Pb^{2+}}$ is lower when $\ce{NaCl}$ is added. The equilibrium constant remains the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that $[\ce{Cl^{-}}]$ is approximately 0.1 M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for $\ce{PbCl2(s)}$ is greater than the equilibrium constant because of the added $\ce{Cl^{-}}$. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.
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The solubility of gases depends on the pressure: an increase in pressure increases solubility, whereas a decrease in pressure decreases solubility. This statement is formalized in Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution. This can be expressed in the equation:
$C = k \times P_{gas}$
where $C$ is the the solubility of a gas in solvent, $k$is the proportionality constant, and $P_{gas}$ is the partial pressure of the gas above the solution.
Example $1$: Henry's Law
The aqueous solubility at 20 degrees Celsius of Ar at 1.00 atm is equivalent to 33.7 mL Ar(g), measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with the air at 1.00 atm and 20 degrees Celsius? Air contains 0.934% Ar by volume. Assume that the volume if the water does not change when it becomes saturated with air.
Solution
STP molar volume: (22.414 L = 22,414 mL)
First Determine Molarity
\begin{align*} k_{Ar} &= \dfrac{C}{P_{Ar}} \[4pt] &= \dfrac{\left(\dfrac{33.7 mL\ Ar}{1 L}\right)\left(\dfrac{1 mol\ Ar}{22,414 mL}\right)}{1 atm} \[4pt] &= 0.00150 M\ atm^{-1} \end{align*}
Solve for Concentration
\begin{align*} C &= k_{Ar}P_{Ar} \[4pt] &= 0.0015 M\ atm^{-1} \times 0.00934 atm \[4pt] &= 1.40 \times 10^{-5} M\ Ar \end{align*}
The example above illustrates the following:
1. At low pressure, a gas has a low solubility. Decreased pressure allows more gas molecules to be present in the air, with very little being dissolved in solution.
2. At high pressure, a gas has a high solubility. Increased pressure forces the gas molecules into the solution, relieving the pressure that is applied, causing there to be fewer gas molecules in the air and more of it in solution.
Common examples of pressure effects on gas solubility can be demonstrated with carbonated beverages, such as a bottle of soda (above). Once the pressure within the unopened bottle is released, $\ce{CO2(g)}$ is released from the solution as bubbles or fizzing.
Deep Sea Divers and "The Bends"
In order for deep sea divers to breathe underwater, they must inhale highly compressed air in deep water, resulting in more nitrogen dissolving in their blood, tissues, and other joints. If the diver returns to the surface too rapidly, the nitrogen gas diffuses out of the blood too quickly and causes pain and possibly death. This condition is known as "the bends."
To prevent the bends, one can return to the surface slowly so that the gases will diffuse slowly and adjust to the partial decrease in pressure or breathe a mixture of compressed helium and oxygen gas, because helium is only one-fifth as soluble in blood than nitrogen. Think of a human body under water as a soda bottle under pressure. Imagine dropping the bottle and trying to open it. In order to prevent the soda from fizzing out, you open the cap slowly to let the pressure decrease. The atmosphere is approximately 78% nitrogen and 21% oxygen, but the body primarily uses the oxygen. Under water, however, the high pressure of water surrounding our bodies causes nitrogen to form in our blood and tissue. And like the bottle of soda, if the body moves or ascends to the water surface the water too quickly, the nitrogen is released too quickly and creates bubbles in the blood.
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Considering the relation between solubility and $K_{sq}$ is important when describing the solubility of slightly ionic compounds. However, this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants ($K_{sq}$) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the $K_{sq}$ of a slightly soluble solute from its solubility.
Introduction
Solubility is the ability of a substance to dissolve. The two participants in the dissolution process are the solute and the solvent. The solute is the substance that is being dissolved, and the solvent is the substance that is doing the dissolving. For example, sugar is a solute and water is a solvent. Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place.
The solubility product constant ($K_{sq}$) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. For example, the higher the $K_{sq}$, the more soluble the compound is. $K_{sq}$ is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. $K_{sq}$ is used to describe the saturated solution of ionic compounds. (A saturated solution is in a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound.)
Example 1: Barium Carbonate
Consider the compound barium carbonate BaCO3 (an ionic compound that is not very soluble):
$\ce{BaCO_{3(s)} \rightleftharpoons Ba^{2+} (aq)+CO_3^{2-} (aq) } \nonumber$
Solution
First, write down the equilibrium constant expression:
$K_c = \dfrac{[\ce{Ba^{2+}}][\ce{CO_3^{2-}}]}{[\ce{BaCO_3}]} \nonumber$
The activity of solid $\ce{BaCO3}$ is 1, and considering that the concentrations of these ions are small, the activities of the ions are approximated to their molar concentrations. $K_{sq}$ is therefore equal to the product of the ion concentrations:
\begin{align*} K_{sp} &= [\ce{Ba^{2+}}][\ce{CO_3^{2-}}] \[4pt] &= 5.1 \times 10^{-9} \end{align*}
How are $K_{sp}$ and Solubility Related?
The relation between solubility and the solubility product constants is that one can be used to derive the other. In other words, there is a relationship between the solute's molarity and the solubility of the ions because $K_{sq}$ is the product of the solubility of each ion in moles per liter.
For example, to find the $K_{sq}$ of a slightly soluble compound from its solubility, the solubility of each ion must be converted from mass per volume to moles per liter to find the molarity of each ion. These numbers can then be substituted into the $K_{sq}$ formula, which is the product of the solubility of each ion. An example of this process is given below:
Example 2: Lead Iodide
Suppose the aqueous solubility for compound PbI2 is 0.54 grams/100 ml at 25 °C and calculate the $K_{sq}$ of PbI2 at 25°C.
$PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^-_{(aq)} \nonumber$
Solution
a. Convert 0.54 grams of $PbI_2$ to moles:
$0.54\ grams \times \dfrac{1\ mol\ PbI_2}{461.0\ grams} = 0.001171\ mol\ PbI_2 \nonumber$
b. Convert ml to L:
$\dfrac{100\ mL}{1000\ L} = 0.100\ L \nonumber$
c. Find the molarity:
$\dfrac{0.001171\ mol}{0.100\ L} = 0.01171\ M\ PbI_2 \nonumber$
d. Now find the molarity of each ion by using the stoichiometric ratio (remember there are two I- ions for each Pb2+ ion):
$\begin{eqnarray} [Pb^{2+}] &=& \dfrac{0.01171\ M}{1\ L} \times \dfrac{1\ mol\ Pb}{1\ mol\ PbI_2} \ &=& 0.011714\ M\ Pb^{2+} \ [I^-] &=& \dfrac{0.01171\ M}{1\ L} \times \dfrac{2\ mol\ I^-}{1\ mol\ PbI_2} \ &=& 0.23427\ M\ I^- \end{eqnarray}$
e. Finally, plug in the molarity to find $K_{sq}$:
$\begin{eqnarray} K_{sp} &=& [Pb^{2+}][I^-]^2 \ &=& (0.011714\ M)(0.023427\ M)^2 \ &=& 6.4 \times 10^{-6} \end{eqnarray}$
*This relation facilitates solving for the molar solubility of the ionic compounds when the $K_{sq}$ is given to us. The process involves working backwards from $K_{sq}$ to the molarity of the ionic compound.*
Example 3
Suppose the $K_{sq}$ at 25 0C is 8.5 x 10-17 for the compound AgI. What is the molar solubility?
$AgI_{(s)} \rightleftharpoons Ag^+_{(aq)} + I^-_{(aq)}\nonumber$
Solution
a. Let "g" represent the number of moles:
$\begin{eqnarray} K_{sp} &=& [Ag^{2+}][I^-] \ &=& g^2 \ &=& 8.5 \times 10^{-17} \end{eqnarray}$
b. Solve for "g":
$\begin{eqnarray} g^2 &=& 8.5 \times 10^{-17} \ g &=& (8.5 \times 10^{-17})^{\dfrac{1}{2}} \ &=& 9.0 \times 10^{-9} \end{eqnarray}$
The molar solubility of AgI is 9.0 x 10-9
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One of the general properties of ionic compounds is water solubility. The oceans are solutions of salt in water. In a mixture, two or more materials are mixed together but they remain essentially separate, like sand and water. The sand can be easily distinguished from the water, because even if a mixture of the two is shaken it will spontaneously separate over time.
In a suspension, one or more materials is mixed into a liquid, and the mixture becomes somewhat homogeneous. Instead of having easily identifiable layers, the liquid has a uniform appearance throughout. However, suspensions are generally cloudy liquids. Milk is a suspension containing water, fats and proteins. They may settle out into separate layers eventually, but it takes time.
In a solution, one or more materials is mixed into a liquid, and the mixture becomes a completely homogeneous liquid. Solutions are transparent, not cloudy, and can be colored or colorless. Saltwater, an example of a solution, is diagrammed in the figure below.
Pieces of salt are not visible in the solution; the salt particles are too small. The salt is separated into individual ions, surrounded by water molecules. This change from Figure IC4.3 to Figure IC4.2 is not instantaneous upon adding salt to water; stirring is required to produce the solution.
Eventually more of the salt dissolves in the water, as shown:
If enough salt is added, the system might come to "equilibrium": the water has dissolved all of the salt that it can, so the rest of the salt remains solid. This equilibrium is "dynamic": ions are dissolved in the water at the same time that ions are deposited from solution into the solid state. However, the overall ratio of dissolved ions to water stays the same.
Problem IC4.1.
Consider further the idea that a given amount of water is only able to dissolve a specific amount of salt.
1. In the diagram above, how many water molecules are there?
2. How many units of salt (an anion and a cation) are dissolved?
3. If there were only a dozen water molecules present, how many units of salt would dissolve?
4. If a hundred water molecules were present, how many units of salt would dissolve?
Why do salts dissolve in water? Water is a molecular compound; the atoms are directly attached to each other, rather than being ions that are attracted to each other. Because of electronegativity differences, the oxygen atom in water has a partial negative charge and the hydrogen atoms have partial positive charges. Ionic compounds can dissolve in polar liquids like water because the ions are attracted to either the positive or negative part of the molecule.
There is a sort of tug-of-war involved with species dissolved in water. The water pulls individual ions away from the solid. The solid is pulling individual ions back out of the water. There exists an equilibrium based on how strongly the water attracts the ions, versus how strong the ionic solid attracts the ions.
It is possible to predict varying degrees of solubility in water for different ionic compounds using the principles of Coulomb's law. The smaller the ions, the closer together they are, and the harder it is for the water molecules to pull the ions away from each other.
Problem IC4.2.
Predict which of the following pairs should be more soluble in water, based on the Coulombic attraction between ions.
1. LiF or NaF
2. NaK or KF
3. BeO or LiF
Problem IC4.3.
Although lithium fluoride and magnesium oxide contain cations and anions of roughly the same size, lithium fluoride is much more soluble in water (2.7 g/L) than magnesium oxide (0.087 g/L) at room temperature. Propose a reason why.
The trends in melting points in ionic compounds are more complicated with regard to solubility. The water solubility of alkali chlorides does not follow a simple trend (as shown in Table IC4.1).
Table IC4.1 Water solubility among alkali chlorides.
Compound Water Solubility in g/L at 20oC
LiCl 83
NaCl 359
KCl 344
Lithium chloride is the least water-soluble of the three compounds. This is feasible, as the lithium ions are small and the attraction for the chloride would be stronger over that shorter distance. However, potassium chloride would be expected to be the most soluble of the three compounds, and it is slightly less soluble than sodium chloride.
Problem IC4.4.
Propose an explanation for why the water solubility of the alkali chlorides does not simply increase as the cation gets larger.
If the halide is varied is used, similar trends are observed:.
Table IC4.2 Water solubility among lithium halides.
Compound Water Solubility in g/L at 20oC
LiCl 83
LiBr 166
LiI 150
Again, it is unsurprising that the lithium chloride is the least soluble, but the most soluble is the lithium bromide, not the lithium iodide.
This type of behavior indicates that there is more than one factor influencing the phenomenon of interest. In the case of solubility, there are several other factors, some of which are more complicated. One simply involves the fact that there are two interactions to consider. The dissolution of these compounds requires more than simply overcoming the attraction of the ionic solid for individual ions, as does melting: the attraction of the water to the ion must also be considered. That attraction is also governed by Coulomb's Law. At some stage, there is a tipping point, when the factors that increase attraction between the ions also increase the attraction between the ion and the water. One or the other of these factors becomes the dominant influence under different circumstances.
• Several interactions are involved in dissolution.
• Cation-anion attraction is just one of these interactions.
• Cation-water and anion-water interactions are important, too.
• Water-water interactions also play a role.
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The solubility product constant, $K_{sp}$, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the $K_{sp}$ value it has.
Consider the general dissolution reaction below (in aqueous solutions):
$\ce{aA(s) <=> cC(aq) + dD(aq)} \nonumber$
To solve for the $K_{sp}$ it is necessary to take the molarities or concentrations of the products ($\ce{cC}$ and $\ce{dD}$) and multiply them. If there are coefficients in front of any of the products, it is necessary to raise the product to that coefficient power(and also multiply the concentration by that coefficient). This is shown below:
$K_{sp} = [C]^c [D]^d \nonumber$
Note that the reactant, aA, is not included in the $K_{sp}$ equation. Solids are not included when calculating equilibrium constant expressions, because their concentrations do not change the expression; any change in their concentrations are insignificant, and therefore omitted. Hence, $K_{sp}$ represents the maximum extent that a solid that can dissolved in solution.
Exercise 1: Magnesium Floride
What is the solubility product constant expression for $MgF_2$?
Solution
The relavant equilibrium is
$MgF_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2F^-_{(aq)} \nonumber$
so the associated equilibrium constant is
$K_{sp} = [Mg^{2+}][F^-]^2\nonumber$
Exercise 2: Silver Chromate
What is the solubility product constant expression for $Ag_2CrO_4$?
Solution
The relavant equilibrium is
$Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{2-}_{4(aq)}\nonumber$
so the associated equilibrium constant is
$K_{sp} = [Ag^{+}]^2[CrO_4^{2-}]\nonumber$
Important effects
• For highly soluble ionic compounds the ionic activities must be found instead of the concentrations that are found in slightly soluble solutions.
• Common Ion Effect: The solubility of the reaction is reduced by the common ion. For a given equilibrium, a reaction with a common ion present has a lower $K_{sp}$, and the reaction without the ion has a greater $K_{sp}$.
• Salt Effect (diverse ion effect): Having an opposing effect on the $K_{sp}$ value compared to the common ion effect, uncommon ions increase the $K_{sp}$ value. Uncommon ions are ions other than those involved in equilibrium.
• Ion Pairs: With an ionic pair (a cation and an anion), the $K_{sp}$ value calculated is less than the experimental value due to ions involved in pairing. To reach the calculated $K_{sp}$ value, more solute must be added.
Solubility Rules
When a substance is mixed with a solvent, there are several possible results. The determining factor for the result is the solubility of the substance, which is defined as the maximum possible concentration of the solute. The solubility rules help determine which substances are soluble, and to what extent.
Solubility Effects on Reactions
Depending on the solubility of a solute, there are three possible results: 1) if the solution has less solute than the maximum amount that it is able to dissolve (its solubility), it is a dilute solution; 2) if the amount of solute is exactly the same amount as its solubility, it is saturated; 3) if there is more solute than is able to be dissolved, the excess solute separates from the solution. If this separation process includes crystallization, it forms a precipitate. Precipitation lowers the concentration of the solute to the saturation in order to increase the stability of the solution.
Solubility Rules
The following are the solubility rules for common ionic solids. If there two rules appear to contradict each other, the preceding rule takes precedence.
1. Salts containing Group I elements (Li+, Na+, K+, Cs+, Rb+) are soluble . There are few exceptions to this rule. Salts containing the ammonium ion (NH4+) are also soluble.
2. Salts containing nitrate ion (NO3-) are generally soluble.
3. Salts containing Cl -, Br -, or I - are generally soluble. Important exceptions to this rule are halide salts of Ag+, Pb2+, and (Hg2)2+. Thus, AgCl, PbBr2, and Hg2Cl2 are insoluble.
4. Most silver salts are insoluble. AgNO3 and Ag(C2H3O2) are common soluble salts of silver; virtually all others are insoluble.
5. Most sulfate salts are soluble. Important exceptions to this rule include CaSO4, BaSO4, PbSO4, Ag2SO4 and SrSO4 .
6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al3+ are insoluble. Thus, Fe(OH)3, Al(OH)3, Co(OH)2 are not soluble.
7. Most sulfides of transition metals are highly insoluble, including CdS, FeS, ZnS, and Ag2S. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.
8. Carbonates are frequently insoluble. Group II carbonates (CaCO3, SrCO3, and BaCO3) are insoluble, as are FeCO3 and PbCO3.
9. Chromates are frequently insoluble. Examples include PbCrO4 and BaCrO4.
10. Phosphates such as Ca3(PO4)2 and Ag3PO4 are frequently insoluble.
11. Fluorides such as BaF2, MgF2, and PbF2 are frequently insoluble.
{{media("www.youtube.com/watch?v=VJoKQ3ULCVs&NR=1")}}
References
1. Petrucci, Ralph H., F. Geoffrey Herring, Jeffrey D. Madura and Carey Bissonnette. General Chemistry: Principles and Modern Applications. 10th ed. Upper Saddle River, New Jersey: Pearson Education, 2011. Print.
2. Nathan, Harold D., and Charles Henrickson. Chemistry. New York: Wiley, 2001. Print.
Problems
1. Is FeCO3 soluble?
According to Rule #5, carbonates tend to be insoluble. Therefore, FeCO3 is likely to form a precipitate.
2. Does ClO4- tend to form a precipitate?
This is perchlorate, which according to Rule #2 is likely to be soluble. Therefore, it will not form a precipitate.
3. Which of these substances is likely to form a precipitate?
a) CaSO4 b) table salt c) AgBr
Letters a and c are both likely to form precipitates.
Concerning a) CaSO4, although sulfates tend to be soluble, Rule #5 indicates that calcium sulfate is an important exception to this rule.
For b), Rule #1 indicates that table salt (NaCl) is soluble because it is a salt of an alkali metal.
c) is an example of two rules contradicting each other. Rule #4 states that bromides are usually soluble, but Rule #3 states that salts of silver are insoluble. Because Rule #3 precedes Rule #4, the compound is insoluble and will form a precipitate.
4. Predict whether a precipitate will form as a result of this reaction:
$2AgNO_3 + Na_2S \rightarrow Ag_2S + 2NaNO_3$
The products of the reaction must be examined; if either of the substances formed in the reaction is insoluble, a precipitate will form.
Considering NaNO3, Rule #3 states that nitrates tend to be soluble. A precipitate of this compound will not form.
Next, consider Ag2S. According to Rule #5, that sulfides tend to be insoluble. Therefore, because of this compound, a precipitate will form in the course of this reaction.
5. Predict if a precipitate will form as a result of this reaction:
$2NaOH + K_2CrO_4 \rightarrow KOH + Na_2CrO_4$
Consider again the products of the reaction: if either is insoluble, a precipitate will form.
The first product, KOH, is an example of two rules contradicting each other. Although Rule #5 says that hydroxides tend to be insoluble, Rule #1 states that salts of alkali metal cations tend to be soluble, and Rule #1 precedes Rule #5. Therefore, this compound will not contribute to any precipitation being formed.
The second product, Na2CrO4, also adheres to Rule #1, which states that salts of alkali metals tend to be soluble.
Because both products are soluble, no precipitate form as a result of this reaction.
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Learning Objectives
• To understand how Temperature, Pressure, and the presence of other solutes affect the solubility of solutes in solvents.
Solubility is defined as the upper limit of solute that can be dissolved in a given amount of solvent at equilibrium. In such an equilibrium, Le Chatelier's principle can be used to explain most of the main factors that affect solubility. Le Châtelier's principle dictates that the effect of a stress upon a system in chemical equilibrium can be predicted in that the system tends to shift in such a way as to alleviate that stress.
Solute-Solvent Interactions Affect Solubility
The relation between the solute and solvent is very important in determining solubility. Strong solute-solvent attractions equate to greater solubility while weak solute-solvent attractions equate to lesser solubility. In turn, polar solutes tend to dissolve best in polar solvents while non-polar solutes tend to dissolve best in non-polar solvents. In the case of a polar solute and non-polar solvent (or vice versa), it tends to be insoluble or only soluble to a miniscule degree. A general rule to remember is, "Like dissolves like."
Common-Ion Effect
The common-ion effect is a term that describes the decrease in solubility of an ionic compound when a salt that contains an ion that already exists in the chemical equilibrium is added to the mixture. This effect best be explained by Le Chatelier's principle. Imagine if the slightly soluble ionic compound calcium sulfate, CaSO4, is added to water. The net ionic equation for the resulting chemical equilibrium is the following:
\[ CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
Calcium sulfate is slightly soluble; at equilibrium, most of the calcium and sulfate exists in the solid form of calcium sulfate.
Suppose the soluble ionic compound copper sulfate (CuSO4) were added to the solution. Copper sulfate is soluble; therefore, its only important effect on the net ionic equation is the addition of more sulfate (SO42-) ions.
\[ CuSO_{4(s)} \rightleftharpoons Cu^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \]
The sulfate ions dissociated from copper sulfate are already present (common to) in the mixture from the slight dissociation of calcium sulfate. Thus, this addition of sulfate ions places stress on the previously established equilibrium. Le Chatelier's principle dictates that the additional stress on this product side of the equilibrium results in the shift of equilibrium towards the reactants side in order to alleviate this new stress. Because of the shift toward the reactant side, the solubility of the slightly soluble calcium sulfate is reduced even further.
Temperature Affects Solubility
Temperature changes affect the solubility of solids, liquids and gases differently. However, those effects are finitely determined only for solids and gases.
Solids
The effects of temperature on the solubility of solids differ depending on whether the reaction is endothermic or exothermic. Using Le Chatelier's principle, the effects of temperature in both scenarios can be determined.
1. First, consider an endothermic reaction (\(\Delta{H_{solvation}}>0\)): Increasing the temperature results in a stress on the reactants side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the product side in order to alleviate this stress. By shifting towards the product side, more of the solid is dissociated when equilibrium is again established, resulting in increased solubility.
2. Second, consider an exothermic reaction ((\(\Delta{H_{solvation}}<0\)): Increasing the temperature results in a stress on the products side from the additional heat. Le Chatelier's principle predicts that the system shifts toward the reactant side in order to alleviate this stress. By shifting towards the reactant's side, less of the solid is dissociated when equilibrium is again established, resulting in decreased solubility.
Liquids
In the case of liquids, there is no defined trends for the effects of temperature on the solubility of liquids.
Gases
In understanding the effects of temperature on the solubility of gases, it is first important to remember that temperature is a measure of the average kinetic energy. As temperature increases, kinetic energy increases. The greater kinetic energy results in greater molecular motion of the gas particles. As a result, the gas particles dissolved in the liquid are more likely to escape to the gas phase and the existing gas particles are less likely to be dissolved. The converse is true as well. The trend is thus as follows: increased temperatures mean lesser solubility and decreased temperatures mean higher solubility.
Le Chatelier's principle allows better conceptualization of these trends. First, note that the process of dissolving gas in liquid is usually exothermic. As such, increasing temperatures result in stress on the product side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts towards the reactant side in order to alleviate this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase increases, resulting in lowered solubility.
Conversely, decreasing temperatures result in stress on the reactant side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts toward the product side in order to compensate for this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase would decrease, resulting in greater solubility.
Pressure Affects Solubility of Gases
The effects of pressure are only significant in affecting the solubility of gases in liquids.
• Solids & Liquids: The effects of pressure changes on the solubility of solids and liquids are negligible.
• Gases: The effects of pressure on the solubility of gases in liquids can best be described through a combination of Henry's law and Le Chatelier principle. Henry's law dictates that when temperature is constant, the solubility of the gas corresponds to it's partial pressure. Consider the following formula of Henry's law:
\[ p = k_h \; c \]
where:
• \(p\) is the partial pressure of the gas above the liquid,
• \(k_h\) is Henry's law constant, and
• \(c\) is the concentrate of the gas in the liquid.
This formula indicates that (at a constant temperature) when the partial pressure decreases, the concentration of gas in the liquid decreases as well, and consequently the solubility also decreases. Conversely, when the partial pressure increases in such a situation, the concentration of gas in the liquid will increase as well; the solubility also increases. Extending the implications from Henry's law, the usefulness of Le Chatelier's principle is enhanced in predicting the effects of pressure on the solubility of gases.
Consider a system consisting of a gas that is partially dissolved in liquid. An increase in pressure would result in greater partial pressure (because the gas is being further compressed). This increased partial pressure means that more gas particles will enter the liquid (there is therefore less gas above the liquid, so the partial pressure decreases) in order to alleviate the stress created by the increase in pressure, resulting in greater solubility.
The converse case in such a system is also true, as a decrease in pressure equates to more gas particles escaping the liquid to compensate.
Example 1
Consider the following exothermic reaction that is in equilibrium
\[ CO_2 (g) + H_2O (l) \rightleftharpoons H_2CO_3 (aq) \]
What will happen to the solubility of the carbon dioxide if:
1. Temperature is increased?
2. Pressure and temperature are increased?
3. Pressure is increased but temperature is decreased?
4. Pressure is increased?
Solution
1. The reaction is exothermic, so an increase in temperature means that solubility would decrease.
2. The change in solubility cannot be determined from the given information. Increasing pressure increased solubility, but increasing temperature decreases solubility
3. An increase in pressure and an increase in temperature in this reaction results in greater solubility.
4. An increase in pressure results in more gas particles entering the liquid in order to decrease the partial pressure. Therefore, the solubility would increase.
Example 2: The Common Ion Effect
Bob is in the business of purifying silver compounds to extract the actual silver. He is extremely frugal. One day, he finds a barrel containing a saturated solution of silver chloride. Bob has a bottle of water, a jar of table salt (NaCl(s)), and a bottle of vinegar (CH3COOH). Which of the three should Bob add to the solution to maximize the amount of solid silver chloride (minimizing the solubility of the silver chloride)?
Solution
Bob should add table salt to the solution. According to the common-ion effect, the additional Cl- ions would reduce the solubility of the silver chloride, which maximizes the amount of solid silver chloride.
Example 3:
Allison has always wanted to start her own carbonated drink company. Recently, she opened a factory to produce her drinks. She wants her drink to "out-fizz" all the competitors. That is, she wants to maximize the solubility of the gas in her drink. What conditions (high/low temperature, high/low pressure) would best allow her to achieve this goal?
Solution
She would be able to maximize the solubility of the gas, (\(CO_2\) in this case, in her drink (maximize fizz) when the pressure is high and temperature is low.
Example 4
Butters is trying to increase the solubility of a solid in some water. He begins to frantically stir the mixture. Should he continue stirring? Why or why not?
Solution
He stop stop stirring. Stirring only affects how fast the system will reach equilibrium and does not affect the solubility of the solid at all.
Example 5: Outgassing Soda
With respect to Henry's law, why is it a poor ideal to open a can of soda in a low pressure environment?
Solution
The fizziness of soda originates from dissolved \(CO_2\), partially in the form of carbonic acid. The concentration of \(CO_2\) dissolved in the soda depends on the amount of ambient pressure pressing down on the liquid. Hence, the soda can will be under pressure to maintain the desired \(CO_2\) concentration. When the can is opened to a lower pressure environment (e.g., the ambient atmosphere), the soda will quickly "outgas" (\(CO_2\) will come out of solution) at a rate depending on the surrounding atmospheric pressure. If a can of soda were opened under a lower pressure environment, this outgassing will be faster and hence more explosive (and dangerous) than under a high pressure environment.
Terms
• The solubility of a solute is the concentration of the saturated solution.
• A saturated solution a solution in which the maximum amount of solute has dissolved in the solvent at a given temperature.
• An unsaturated solution a solution in which the solute has completely dissolved in the solvent.
• A supersaturated solution is a solution in which the amount of solute dissolved under given conditions exceeds it's supposed upper limit.
• Le Châtelier's principle states that when a system in chemical equilibrium is stressed, the system will shift in a way that alleviates the stress.
• Endothermic reaction: a reaction in which heat is absorbed (ΔH>0)
• Exothermic reaction: a reaction in which heat is released (ΔH < 0)
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The solubility of solutes is dependent on temperature. When a solid dissolves in a liquid, a change in the physical state of the solid analogous to melting takes place. Heat is required to break the bonds holding the molecules in the solid together. At the same time, heat is given off during the formation of new solute -- solvent bonds.
• CASE I: Decrease in solubility with temperature: If the heat given off in the dissolving process is greater than the heat required to break apart the solid, the net dissolving reaction is exothermic (energy given off). The addition of more heat (increases temperature) inhibits the dissolving reaction since excess heat is already being produced by the reaction. This situation is not very common where an increase in temperature produces a decrease in solubility.
• CASE II: Increase in solubility with temperature: If the heat given off in the dissolving reaction is less than the heat required to break apart the solid, the net dissolving reaction is endothermic (energy required). The addition of more heat facilitates the dissolving reaction by providing energy to break bonds in the solid. This is the most common situation where an increase in temperature produces an increase in solubility for solids.
Figure: Temperature dependent solubilities of three salts in water.
The use of first-aid instant cold packs is an application of this solubility principle. A salt such as ammonium nitrate is dissolved in water after a sharp blow breaks the containers for each.
$NH_4NO_{3(s)} \rightarrow NH_{4(aq)}^+ + NO^-_{3(aq)}$
The dissolving reaction is endothermic and requires heat. Therefore the heat is drawn from the surroundings and the pack feels cold.
Contributors and Attributions
Charles Ophardt (Professor Emeritus, Elmhurst College); Virtual Chembook
Temperature Effects on the Solubility of Gases
As with all processes under constant pressure and constant temperature, dissolving a solution into solution will occur only if $\Delta G_{total} < 0$.
$\Delta G_{sol} = \Delta H_{sol} - T\Delta S_{sol} < 0\label{eq1}$
For dissolving solids in liquids, $\Delta S_{sol} > 0$, but for dissolving gases solutes, the entropy of solution is negative ($\Delta S _{sol} < 0$) since the entropy of the gas phase solute is appreciably greater than the entropy of that solute in solution. Consequently, the only way that $\Delta G_{sol}<0$ for a dissolving a gas in solution is if the solution process is exothermic (i.e., $\Delta H_{sol}<0$). This occurs due to the enthalpy differences from making and breaking intermolecular interactions in the solvent and solution. There are three basic steps involved in dissolving a solute from a condensed state (or a non-ideal gas) into a solution each with a corresponding enthalpy change.
Steps involved in Dissolving of a solute in a solvent
Dissolution can be viewed as occurring in three steps:
1. Breaking solute-solute attractions (endothermic), i.e., lattice energy in salts ($\Delta H_{\text{solute-solute}}>0$).
2. Breaking solvent-solvent attractions (endothermic), i.e., hydrogen bonding and dipole-dipole interactions in water ($\Delta H_{\text{solvent-solvent}}>0$.
3. Forming solvent-solute attractions (exothermic), i.e., solvation energy ($\Delta H_{\text{solute-solvent}}<0$).
The enthalpy of solution $\Delta H_{sol}$ is the sum of these three individual steps.
$\Delta H_{sol} = \Delta H_{\text{solute-solute}} + \Delta H_{\text{solvent-solvent}} + \Delta H_{\text{solute-solvent}} \label{eq2}$
For solids solutes, $\Delta H_{\text{solute-solute}}$ is just the lattice energy of the solute, but for gases that follow the ideal gas equation of state, the enthalpy change associated with Step 1 is zero.
$\Delta H_{\text{solute-solute}} (gas) =0$
This is because are no intermolecular interactions exist in ideal gases (van der Waals gases will differ as expected).
Solubility of Gases in Polar Solvents
In polar solvents like water, $\Delta H_{\text{solute-solvent}} > \Delta H_{\text{solvent-solvent}}$, so the dissolution of most gases is exothermic (i.e., $\Delta H_{sol} <0$). Hence, when a gas dissolves in a liquid solvent, thermal energy is released which warms both the system (the solution) and the surroundings.
$\ce{ solute (gas) + water (l) \rightleftharpoons solute (aq) + water (aq)} + \Delta \label{eq4}$
where $Delta$ is thermal enegy. Consequently, the solubility of a gas is dependent on temperature (Figure $1$). The solubility of gases in liquids decreases with increasing temperature. Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas; pushes the reaction in Equation \ref{eq4} to the left. The thermodynamic perspective is that at elevated temperatures, the negative entropy term in Equation \ref{eq1} will dominate the enthalpic term that is driving the dissolution process and make $ΔG_{soln}$ less negative and hence less spontaneous.
Example $1$
Determine the solubility of $\ce{N2(g)}$ when combined with $\ce{H2O}$ at 0.0345 °C the pressure of $\ce{N2}$ is 1.00 atm, and its solubility is 21.0 ml at STP.
Solution
Begin by determining the molarity (solubility) of $\ce{N2(g)}$ at 0 °C and STP.
At STP 1 mol=22L
\begin{align*} \text{Molarity of } \ce{N2} &= 21\,ml \left( \dfrac{1L}{1000\,ml}\right) \[4pt] &=0.021\,L \,\ce{N2} \[4pt] &= 0.021\,L \,\ce{N2} \left(\dfrac{1\,mol}{22\,L}\right)\[4pt] &=\dfrac{0 .000954\,mol}{1\, L} \[4pt] &= 9.5 \times 10^{-4}\, M\, \ce{N2} \end{align*}
Now that the molarity of $\ce{N2}$. $C$ has been attained, the Henry's law constant, $k$, can be evaluated.
$C = k_H P_{gas} \nonumber$
where $C$ is solubility, $k_H$ is Henry's constant, and $P_{gas}$ is the partial pressure of the gas being considered.
Rearranging the formula to solve for $k_H$
\begin{align} k_H&= \dfrac{C}{P_{gas}} \ &= 9.5 \times 10^{-4}\, M \,N_2/ 1\,atm \nonumber \end{align} \nonumber
Now substitute k and the partial pressure of $\ce{N2}$ into Henry's law:
\begin{align} C&= (9.5 \times 10^{-4}\,M\, N_2)(0.0345) \nonumber \[4pt] &= 3.29 \times 10^{-5}\, M\, N_2 \nonumber \end{align} \nonumber
The Survival of Fish
A fish kill can occur with rapid fluctuations in temperature or sustained high temperatures. Generally, cooler water has the potential to hold more oxygen, so a period of sustained high temperatures can lead to decreased dissolved oxygen in a body of water. A short period of hot weather can increase temperatures in the surface layer of water, as the warmer water tends to stay near the surface and be further heated by the air. In this case, the top warmer layer may have more oxygen than the lower, cooler layers because it has constant access to atmospheric oxygen.
Solubility of Gases in Organic Solvents
Dissolving gases in non-polar organic solvents is a different situation than in polar solvents like water discussed above. For non-polar solvents, both the solvent-solvent interactions ($\Delta H_{\text{solvent-solvent}}$) and the solvation enthalpies ($\Delta H_{\text{solute-solvent}}$) are considerably weaker than in polar liquids like water due to the absence of strong dipole-dipole intermolecular interactions (or hydrogen bonding). In many cases, the enthalpy needed to break solvent-solvent interactions is comparable to the enthalpy released in making solvent-gas interactions.
$\Delta H_{\text{solvent-solvent}} \approx \Delta H_{\text{solute-solvent}}$
which means the $\Delta H_{sol} \approx 0$ via Equation \ref{eq2}. In some solvent-solute combination
$\Delta H_{\text{solvent-solvent}} > \Delta H_{\text{solute-solvent}}$
so that $\Delta H_{sol} > 0$. In this case, we can write the solution reactions thusly
$\Delta + \ce{ solute (gas) + solvent (solvent) \rightleftharpoons solute (sol) + solvent (sol)} \label{eq5}$
where $Delta$ is thermal energy. In these cases, gases dissolved in organic solvents can actually be more soluble at higher temperatures. A Le Chatelier perspective, like that used above for water, can help understand why. Increasing the temperature will shift in the equilibrium to favor dissolution (i.e, shift Equation \ref{eq5} to the right). As a result, the solubilities of gases in organic solvents often increase with increasing temperature (Table $1$) in contrast to the trend observed in water (Figure $1$).
Table $1$: Mole Fraction Solubility of Hydrogen gas at a Hydrogen Partial Pressure of 101.325 kPa in selects solvents at select temperatures.
Temperature (ºC) $\ce{H2}$ in n-hexane $\ce{H2}$ in Toluene $\ce{H2}$ in Acetonitrile $\ce{O_2}$ in Water $\ce{O_2}$ in Ethanol $\ce{O_2}$ in Decane
25 $7.13 \times 10^4$ $3.15 \times 10^4$ $1.78 \times 10^4$ $0.249 \times 10^4$ $5.59 \times 10^4$ $21.78\times 10^4$
50 $8.20 \times 10^4$ $3.75 \times 10^4$ $2.16 \times 10^4$ $0.196 \times 10^4$ $5.598 \times 10^4$ $21.76\times 10^4$
100 $10.78 \times 10^4$ $5.05 \times 10^4$ $3.02 \times 10^4$ $0.086 \times 10^4$ $5.695 \times 10^4$ -
Exercise $1$
Based off of the three steps outlined above for dissolving a solute a liquid, explain why the solubility of hydrogen is smaller in acetonitrile than in toluene and that is less than in n-hexane.
Answer
Acetinitrile is a polar organic solvent and has stronger intermolecular bonds than toluene, which is a weakly polar solvent and has stronger intermolecular interactions than n-hexane.
Exercise $2$
A system of cyclopentane and oxygen gas are at equilibrium with an enthalpy of -1234 kJ; predict whether the solubility of oxygen gas will be greater when heat is added to the system or when a temperature decrease occurs.
Answer
Cyclopentane is an organic solvent. Oxygen gas and cyclopentane in a system at equilibrium, where the entropy is negative, will be be displaced from equilibrium when any type of temperature change is inflicted on the system. Because the dissolution of the gas is endothermic, more heat increases the solubility of the gas.
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• Chemical Change vs. Physical Change
In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring.
• Dexter Energy Transfer
Dexter energy transfer is sometimes called short-range, collisional or exchange energy transfer which is a non-radiative process with electron exchange. Dexter Energy transfer although similar to Förster energy transfer, differs greatly in length scale and underlying mechanism.
• De Broglie thermal wavelength
The thermal de Broglie wavelength is roughly the average de Broglie wavelength of the gas particles in an ideal gas at the specified temperature.
• Fluorescence Resonance Energy Transfer
Fluorescence Resonance Energy Transfer (FRET) is a special technique to gauge the distance between two chromophores, called a donor-acceptor pair. The limitation of FRET is that this transfer process is effective only when the separating distance of donor-acceptor pair is smaller than 10 nanometers. However, FRET is a highly distance-dependent phenomenon and thus has become a popular tool to measure the dynamic activities of biological molecules within nanoscale.
• Ladder Operators (Creation/Annihilation Operators)
Ladder Operators are operators that increase or decrease eigenvalue of another operator. There are two types; raising operators and lowering operators. In quantum mechanics the raising operator is called the creation operator because it adds a quantum in the eigenvalue and the annihilation operators removes a quantum from the eigenvalue.
• Pressure
Pressure ( pp ) is the force per unit area applied on a surface, in a direction perpendicular to that surface, i.e. the scalar part of the stress tensor under equilibrium/hydrosatic conditions.
• Quantum Tunneling
Quantum tunneling is a phenomenon where particles may "tunnel through" a barrier which they have insufficient kinetic energy to overcome according to classical mechanics. Tunneling is a result of the wavelike nature of quantum particles, and cannot be predicted by any classical system.
• Stirling’s Approximation
Stirling's approximation (or Stirling's formula) is an approximation for factorials. It is a good quality approximation, leading to accurate results even for small values of n.
• WKB Approximation
The WKB Approximation, named after scientists Wentzel–Kramers–Brillouin, is a method to approximate solutions to a time-independent linear differential equation or in this case, the Schrodinger Equation. Its principal applications are for calculating bound-state energies and tunneling rates through potential barriers. The WKB Approximation is most often applied to 1D problems, but also works for 3D spherically symmetric problems.
Fundamentals
The difference between a physical reaction and a chemical reaction is composition. In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition. Although we call them physical "reactions," no reaction is actually occurring. In order for a reaction to take place, there must be a change in the elemental composition of the substance in question. Thus, we shall simply refer to physical "reactions" as physical changes from now on.
Introduction
Physical changes are limited to changes that result in a difference in display without changing the composition. Some common changes (but not limited to) are:
• Texture
• Color
• Temperature
• Shape
• Change of State (Boiling Point and Melting Point are significant factors in determining this change.)
Physical properties include many other aspects of a substance. The following are (but not limited to) physical properties.
• Luster
• Malleability
• Ability to be drawn into a thin wire
• Density
• Viscosity
• Solubility
• Mass
• Volume
Any change in these physical properties is referred to as a physical change. For further information, please refer to Properties of Matter.
Chemical changes, on the other hand, are quite different. A chemical change occurs when the substance's composition is changed. When bonds are broken and new ones are formed a chemical change occurs. The following are indicators of chemical changes:
• Change in Temperature
• Change in Color
• Noticeable Odor (after reaction has begun)
• Formation of a Precipitate
• Formation of Bubbles
Note: When two or more reactants are mixed and a change in temperature, color, etc. is noticed, a chemical reaction is probably occurring. These are not definite indicators; a chemical reaction may not be occurring. A change in color is not always a chemical change. If one were to change the color of a substance in a non-chemical reaction scenario, such as painting a car, the change is physical and not chemical. This is because the composition of the car has not changed. Proceed with caution.
Common Physical Changes
Texture
The texture of a substance can differ with a physical change. For example, if a piece of wood was sanded, waxed, and polished, it would have a very different texture than it initially had as a rough piece of wood.
(left) Rough plank boardwalk, Quebec City, Canada (right) Finished mountain ash floor. (CC BY-SA 4.0; WikiPedant and CC BY-SA 2.5; MarkAnthonyBoyle, respectively).
As you can see, the texture of the finished wood is much smoother than the initial grainy wood.
Color
The changing of color of a substance is not necessarily an indicator of a chemical change. For example, changing the color of a metal does not change its physical properties. However, in a chemical reaction, a color change is usually an indicator that a reaction is occurring. Painting the metal car does not changing the composition of the metallic substance.
Robotic arm applying paint on car parts. Image use with permission (CC BY-SA 4.0l RoboGuru).
Temperature
Although we cannot see temperature change, unless if a change of state is occurring, it is a physical change.
Hot metalwork. (CC BY-SA-NC 2.0; flagstaffotos.com.au)
One cannot see the pan physically changing shape, color, texture, or any of the other physical properties. However, if one were to touch the pan, it would be incredibly hot and could cause a burn. Sitting idle in a cupboard, this pan would be cold. One cannot assess this change only through visual exposure; the use of a thermometer or other instrument is necessary.
Shape
The shape of an object can be changed and the object will still remain true to its chemical composition. For example, if one were to fold money, as shown by the figure below, the money is still chemically the same.
Origami Money
Change of State
The change of state is likewise a physical change. In this scenario, one can observe a number of physical properties changing, such as viscosity and shape. As ice turns into water, it does not retain a solid shape and now becomes a viscous fluid. The physical "reaction" for the change of ice into liquid water is:
$\ce{H_2O_{(s)} \rightarrow H_2O_{(l)}}\nonumber$
The following are the changes of state:
Solid → Liquid Melting
Liquid → Gas Vaporization
Liquid → Solid Freezing
Gas → Liquid Condensation
Solid → Gas Sublimation
• If heat is added to a substance, such as in melting, vaporization, and sublimation, the process is endothermic. In this instance, heat is increasing the speed of the molecules causing them move faster.
• If heat is removed from a substance, such as in freezing and condensation, then process is exothermic. In this instance, heat is decreasing the speed of the molecules causing them move slower.
Physical Properties
Luster
The luster of an element is defined as the way it reacts to light. Luster is a quality of a metal. Almost all of the metals, transition metals, and metalloids are lustrous. The non-metals and gases are not lustrous. For example, oxygen and bromine are not lustrous. Shown below is are lustrous paper clips.
Lustrous Paperclips
Malleability
Malleability is also a quality of metals. Metals are said to be malleable. This means that the metals can deform under an amount of stress. For example, if you can hit a metal with a mallet and it deforms, it is malleable. Also, a paperclip can be shaped with bare hands.
Bent Paperclip
The image shows the malleability of a certain metal as stress is applied to it.
Ability to be drawn into a thin wire
In materials science, this property is called ductility. For example, raw copper can be obtained and it can be purified and wrapped into a cord. Once again, this property is characteristic of mainly metals, nonmetals do not possess this quality.
Copper Wire
Density
The density of an object is its mass divided by its volume ($d=m/v$). A substance will have a higher density if it has more mass in a fixed amount of volume. For example, take a ball of metal, roughly the size of a baseball, compressed from raw metal. Compare this to a baseball made of paper. The baseball made of metal has a much greater weight to it in the same amount of volume. Therefore the baseball made out of metal has a much higher density. The density of an object will also determine whether it will sink or float in a particular chemical. Water for example has a density of 1 g/cm3. Any substance with a density lower than that will float, while any substance with a density above that will sink.
Viscosity
Viscosity is defined to be the resistance to deformation of a particular chemical substance when a force is applied to it. In the example below, one can see two cubes falling into two different test tubes. The upper substance shows a violent reaction to the dropping of the cube. The lower substance simply engulfs it slowly without much reaction. The upper substance has a lower viscosity relative to the lower substance, which has very high viscosity. One may even think of viscosity in terms of thickness. The substance with more thickness has higher viscosity than a substance that is deemed "thin." Water has a lower viscosity than honey or magma, which have relatively high viscosities.
Viscosity of Fluids
Common Chemical Changes
The follow are all indicators of chemical reactions. For further information on chemical reactions, please refer to Chemical Reactions.
Change in Temperature
A change in temperature is characteristic of a chemical change. During an experiment, one could dip a thermometer into a beaker or Erlenmeyer Flask to verify a temperature change. If temperature increases, as it does in most reactions, a chemical change is likely to be occurring. This is different from the physical temperature change. During a physical temperature change, one substance, such as water is being heated. However, in this case, one compound is mixed in with another, and these reactants produce a product. When the reactants are mixed, the temperature change caused by the reaction is an indicator of a chemical change.
As an example of a exothermic reaction, if $Fe_2O_3$ is mixed with Al and ignighted (often with burning Mg), then the thermite reaciton is initiated
$Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 + \text{Heat}\nonumber$
This reaction generates heat as a product and is (very) exothermic.
However, physical changes can be exothermic or endothermic. The melting of an ice cube, which is endothermic, is a change in a physical property and not composition. Thus, it is a physical change.
Change in Color
A change in color is also another characteristic of a chemical reaction taking place. For example, if one were to observe the rusting of metal over time, one would realized that the metal has changed color and turned orange. This change in color is evidence of a chemical reaction. However, one must be careful; sometimes a change in color is simply the mixing of two colors, but no real change in the composition of the substances in question.
The reaction above is that of the rusting of iron.
$\ce{4Fe + 3O_2 + 6H_2O \rightarrow 4Fe(OH)_3}\nonumber$
Noticeable Odor
When two or more compounds or elements are mixed and a scent or odor is present, a chemical reaction has taken place. For example, when an egg begins to smell, (a rotten egg) a chemical reaction has taken place. This is the result of a chemical decomposition.
Spoiled Egg
Formation of a Precipitate
The formation of a precipitate may be one of the most common signs of a chemical reaction taking place. A precipitate is defined to be a solid that forms inside of a solution or another solid. Precipitates should not be confused with suspensions, which are solutions that are homogeneous fluids with particles floating about in them. For instance, when a soluble carbonate reacts with Barium, a Barium Carbonate precipitate can be observed.
Test Tube
Reaction:
$\ce{Ba^{2+}(aq) + CO^{2-}3(aq) \rightarrow BaCO3(s)}\nonumber$
For further information, please refer to Classification of Matter.
Formation of Bubbles
The formation of bubbles, or rather a gas, is another indicator of a chemical reaction taking place. When bubbles form, a temperature change could also be taking place. Temperature change and formation of bubbles often occur together. For example, in the following image, one can see a gas spewing. This is the formation of a gas.
Gas Formation
However, most reactions are much more subtle. For instance, if the following reaction occurs, one may notice Carbon Dioxide bubbles forming. If there is enough Hydrochloric Acid, bubbles are visible. If there is not, one can't readily notice the change:
$\ce{Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2}\nonumber$
Exercise $1$
Which of the following is a chemical reaction?
1. Freezing liquid Mercury
2. Adding yellow to blue to make green
3. Cutting a piece of paper into two pieces
4. Dropping a sliced orange into a vat of Sodium Hydroxide
5. Filling a balloon with natural air
Answer
D
Exercise $2$
Which of the following is a physical reaction?
1. Shattering Glass with a baseball
2. Corroding Metal
3. Fireworks Exploding
4. Lighting a match
5. Baking a cake
Answer
A
Exercise $3$
Which of the following is a chemical reaction?
1. Painting a wall blue
2. A bicycle rusting
3. Ice cream melting
4. Scratching a key across a desk
5. Making a sand castle
Answer
B
Exercise $4$
Which of the following is a physical reaction?
1. Frying an egg
2. Digesting carrots
3. A Macbook falling out of a window
4. Creating ATP in the human body
5. Dropping a fizzy tablet into a glass of water
Answer
C
Exercise $5$
Write C for Chemical Reaction or P for Physical Reaction.
1. Burning Leaves
2. Cutting Diamonds
3. Crushing a pencil
4. The salivary amylase enzyme that breaks down food in the mouth
5. Salt mixing in with water
Answer
a) C
b) P
c) P
d) C
e) Neither. This is one of the gray areas of chemical change and physical change. Although the salt has dissociated into Sodium and Chloride ions, it is still salt in water. Salt, initially is actually just a conglomerate of sodium and chloride ions and by dissociating them, just the arrangement of the ions has changed. Please click here for more information
Outside Links
All images are courtesy of http://www.sxc.hu, which provides royalty free images that are free to be copied without restrictions. The viscosity image is also free to be duplicated as per permission of author on Wikipedia.com.
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The thermal de Broglie wavelength is roughly the average de Broglie wavelength of the gas particles in an ideal gas at the specified temperature. It is defined as
$\Lambda= \sqrt{\frac{h^2}{2\pi mk_BT}} \nonumber$
where
• $h$ is the Planck constant
• $m$ is the mass
• $k_B$ is the Boltzmann constant
• $T$ is the temperature.
Related reading
• Zijun Yan, "General thermal wavelength and its applications", Eur. J. Phys. 21 pp. 625-631 (2000)
Dexter Energy Transfer
Dexter energy transfer is sometimes called short-range, collisional or exchange energy transfer which is a non-radiative process with electron exchange. Dexter Energy transfer although similar to Förster energy transfer, differs greatly in length scale and underlying mechanism.
Introduction
The energy transfer could take place via the interaction between an excited chemical group, say D*, and a ground-state chemical group, say A, without emitting a photon when transferring energy. Förster provided a model saying that the energy released from an excited donor could simultaneously excite the ground-state acceptor based on the Coulombic interaction between these two chemical groups. Independently, David L. Dexter provided another mechanism that an excited donor group and an acceptor group might indeed exchange electrons to accomplish the non-radiative process. The Dexter exchange energy transfer is generally associated with quenching. The term “quenching” means any physical process or molecular state that decreases the molecular fluorescence. Previously, the Dexter energy transfer was treated a fundamental phenomena in photochemistry field. Recently, it is greatly applied to novel emitting materials, such as white organic light-emitting diodes and energy up-conversion systems. (Blue light emission and white light emission.
The big picture of Dexter Energy Transfer
Dexter energy transfer is a process that two molecules (intermolecular) or two parts of a molecule (intramolecular) bilaterally exchange their electrons. Unlike the sixth-power dependence of Förster energy transfer, the reaction rate constant of Dexter energy transfer exponentially decays as the distance between these two parties increases. On account of the exponential relationship to the distance, the exchange mechanism typically occurs within 10 Angstroms. Hence, the exchange mechanism is also called the short-range energy transfer.
The exchange mechanism is based on the Wigner spin conservation rule; thus, the spin-allow process could be:
Singlet-singlet energy transfer: $\ce{^{1}D^{*} + ^1A -> ^{1}D + ^{1}A^{*}}$
It could be understood by: a singlet group produces another singlet group.
Triplet-triplet energy transfer: $\ce{^{3}D^{*} + ^{1}A → ^{1}D + ^{3}A^{*}}$
It could be understood by: a triplet group produces another triplet group.
The singlet-singlet energy transfer can happen when undergoing the Coulombic interaction. However, the Coulombic interaction will not involve the triplet-triplet energy transfer because that violates the Wigner spin conservation law.
Important factors about Dexter Energy transfer
Wavefunction Overlap
As the figure 1 implies, the Dexter energy transfer is a process that the donor and the acceptor exchange their electron. In other words, the exchanged electrons should occupy the orbital of the other party. Hence, besides the overlap of emission spectra of $D$ and absorption spectra of $A$, the exchange energy transfer needs the overlap of wavefunctions. In the popular words, it needs the overlap of the electron cloud. The overlap of wavefunctions also implies that the excited donor and ground-state acceptor should be close enough so the exchange could happen. If $D$ and $A$ are different molecules, collisions can greatly make them bump into each other. The short distance that makes energy transfer happen is almost comparable to the collisional diameter. This is the reason why the exchange energy transfer is always referred to the term “collision.”
Rate constant
The rate constant of exchange energy transfer is given by
$k_{dexter}= KJ \; \text{exp} \left (\dfrac {-2R_{DA}}{L} \right )$
while the rate constant of exchange energy transfer is given by
$k_{dexter} \propto \dfrac {D_{D}^{2} D_{A}^{2}}{R_{DA}^{6}}$
$J$ is the normalized spectral overlap integral. The term “normalized” here means making the absorption spectra and the emission spectra on the same scale and have the same highest level (For real examples, please refer to this paper) $K$ is an experimental factor, $R_{DA}$ is the distance between $D$ and $A$, and $L$ is the sum of van der Waals radius; $D_D$ is the transition dipole of donor, $D_A$ is the transition dipole of acceptor. Comparing rate constants of different energy transfer models, we can find that the rate constant of exchange energy transfer decays steeply because of its intrinsic exponential relationship, which is the reason that the exchange energy transfer is also called the short-range energy transfer and the Förster mechanism is called the long-range energy transfer.
Preparing the excited triplet state of interest
According to the Figure $1$, by exchanging electrons, a triplet excited donor could also excite the acceptor to its triplet state. This process is called “sensitization” process; hence, the triplet excited-state donor is the sensitizer. Generally speaking, the time scale for fluorescing a photon is about several nanosecond. The intersystem crossing happens in this time scale, too. However, roughly speaking, fluorescence generally happens faster that the intersystem crossing, which means an electron would fluoresce before it proceeds to occupy a triplet state through thermal relaxation.
A special case: triplet-triplet annihilation
The triplet-triplet annihilation (TTA) is an important case of exchange energy transfer. Two triplet chemical groups, D* and A*, react to produce two singlet states. Generally, the energy gap between S0 and T1 is bigger than the energy gap between T1 and S1. Hence, if two triplet excited-state molecules encounter, the process might have enough energy to excite one of them to the higher singlet states, which can make the fluorescence happen in this system. After the annihilation process, the energy level that the electron occupies will be twice of the lowest triplet energy gap.
Conclusion
The Dexter energy transfer is a fundamental phenomenon in photochemistry. The difference between Förster and Dexter mechanism include:
1. Dexter mechanism involve the overlap of wavefunctions so that electrons can occupy the other’s molecular orbitals.
2. The reaction rate constant of Dexter energy transfer sharply decreases while the distance between $D$ and $A$ increase and the distance is generally smaller than 10 angstroms.
3. The Dexter mechanism can be applied to produce the triplet state of some molecules of interest.
4. The special case of exchange-triplet-triplet annihilation-can “push” the electron to upper singlet states by exchanging the electrons of two triplet molecules.
Problems
The figure above is the energy levels that stands for an upconversion fluorescence system. Surprisingly, the Dexter energy transfer and the triplet-triplet annihilation mechanisms provide a fantastic way to produce high-energy light simply by using medium-energy light. If choosing the donor and the acceptor properly, one can produce three kinds of wavelengths of fluorescence, say red, green, and blue. After mixing these light, this system might produce white light if the ingredient light is well mixed. Today, one can even simply predict th photophysical properties by using commercial quantum chemistry software, before starting to do the experiment. Below are some questions behind this figure.
Problems
(These question are adapted from Chemical Communication, 2009, 27, 4064 and Physical Review B, 2008, 78, 195112)
Question 1: What does the worm-like arrow "A" stand for?
Question 2: What does the worm-like arrow "B" stand for?
Question 3: What does the process "C" stand for?
Question 4: What does the process "D" stand for?
Question 5: The original compound absorbs the light with frequency v1 and ultimately emits the light
with frequency v2. Which frequency is bigger? So this research group used low-energy
light to produce high-energy light or use inverse idea? Does it make sense?
Answer 1: Internal conversion or vibrational relaxation.
Answer 2: Intersystem crossing.
Answer 3: Dexter exchange energy transfer.
Answer 4: Triplet-triplet annihilation.
Answer 5: v1 < v2.
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Fluorescence Resonance Energy Transfer (FRET) is a special technique to gauge the distance between two chromophores, called a donor-acceptor pair. The limitation of FRET is that this transfer process is effective only when the separating distance of donor-acceptor pair is smaller than 10 nanometers. However, FRET is a highly distance-dependent phenomenon and thus has become a popular tool to measure the dynamic activities of biological molecules within nanoscale.
Introduction
FRET is the acronym of the Förster (Flourescence) Resonance Energy Transfer. The Förster energy transfer is the phenomenon that an excited donor transfers energy (not an electron) to an acceptor group through a non-radiative process. This process is highly distance-dependent, thus allowing one to probe biological structures. One common application is simply to measure the distance between two positions of interest on a big molecule, generally a biological macromolecule, by attaching appropriate donor-acceptor groups to the big one. If the big molecule only involves one donor and one acceptor group, the distance between the donor and the acceptor can be easily measured if there is no conformational change within this process. Besides, if the molecule has a huge conformational change, one may also measure the dynamical activities between two sites on this macromolecule such as protein interactions. Today, this technique is widely applied in many fields such as single-molecule experiments, molecular motors, biosensors and DNA mechanical movements. FRET is also called the "Spectroscopic Ruler" because of its intrinsic convenience.
The theoretical analysis was well developed by Theodor Förster. This non-radiative transfer mechanism is schematically capsuled in Figure $1$. A donor group (D) is excited by a photon and then relaxes to the lowest excited singlet state, S1 (by Kasha’s rule). If the acceptor group is not too far, the energy released when the electron returns to the ground state (S0) may simultaneously excite the donor group. This non-radiative process is referred to as “resonance”. After excitation, the excited acceptor emits a photon and returns to the ground state, if the other quenching states do not exist.
The resonance mechanism is associated with the Coulombic interaction between electrons. Thus, the relative distance of Coulombic interaction between the donor-acceptor pair could be longer than the electron exchange energy transfer which needs the overlap of wavefunctions, namely the Dexter Energy Transfer. The Coulombic interaction only needs the overlap of the spectrum which means that the identity of resonance energy. Figure $2$ here shall give a sense about what the resonance mechanism is. (Note that the HOMO-LUMO gap does not equal the energy difference between the ground state and the lowest S1 excited state of the molecule.)
Factors that affect FRET
The FRET efficiency ($E$) is the quantum yield of the energy transfer transition; i.e., the fraction of energy transfer event occurring per donor excitation event. The FRET efficiency is given by
$\ E = \dfrac{k_{ET}}{k_f + k_{ET}+\sum k_i} \label{1}$
where $k_{ET}$ is the rate of FRET, $k_f$ is the rate of radiative relaxation (i.e., fluorescence), and $k_i$ are the non-radiative relaxation rates (e.g., internal conversion, intersystem crossing, external conversion etc). Within a point dipole-dipole approximation, the FRET efficiency can be related to the donar-acceptor distance via
$\ E = \dfrac{1}{1+\left( \dfrac{r}{R_0} \right)^6} \label{2}$
where $r$ is the distance between donor and acceptor chromophores and $R_o$ is the characteristic distance (the Förster distance or Förster radius) with a 50% transfer efficiency.
Overlap of Spectrum
To enhance the FRET efficiency, the donor group should have good abilities to absorb photons and emit photons. That means the donor group should have a high extinction coefficient and a high quantum yield. The overlap of emission spectrum of the donor and absorption spectrum of the acceptor means that the energy lost from excited donor to ground state could excite the acceptor group. The energy matching is called the resonance phenomenon. Thus, the more overlap of spectra, the better a donor can transfer energy to the acceptor. The overlap integral, $J(λ)$, between the donor and the acceptor stands for the overlap of spectra, as shown in Figure $3$.
The overlap integral is given by
$J = \int_o^{\infty} F_D(\lambda)\epsilon_A(\lambda) \lambda^4 \,d\lambda \label{3}$
where $F_D(λ)$ is the normalized emission spectrum of the donor, $\epsilon_{A}$ standards for the molar absorption coefficient of the acceptor, and $λ$ is the wavelength.
Orientation of Transition Dipoles
The resonance energy transfer mechanism is also affected by the orientations of the emission transition dipole of the donor and the absorption dipole of the acceptor. The orientation parameter $κ^2$ gives the quantitative value of interaction between two dipole moments. $κ^2$ can theoretically be values from 0 (when dipoles are perpendicular to each other) to 4 (when dipoles are collinear).
\begin{aligned} \kappa^{2} &=\left(\cos \theta_{\tau} \cdot 3 \cos \theta_{0} \cos \theta_{A}\right)^{2} \[4pt] &=\left(\sin \theta_{0} \sin \theta_{A} \cos \phi \cdot 2 \cos \theta_{0} \sin \theta_{A}\right)^{2} \end{aligned}
$κ^2$ is equal to 1 when these two transition dipoles are parallel. The orientation of transition dipoles is shown in Figure $4$. For a freely rotational donor and acceptor group, the average $κ^2$ is treated 2/3.
Förster Distance (R0)
A long $R_0$ can cause a high FRET efficiency. Based on Förster's analysis, $R_0$ is a function of quantum yield of the donor chromophore $\Phi_{D}$, spectral overlap of donor and acceptor $J(λ)$, directional relationship of transition dipoles $κ^2$ and the refractive index of the medium $n$.
$(R_0)^6 \propto \kappa^2 \Phi_{D} J(\lambda) n^{-4} \label{4}$
Table $1$: Förster distances of different donor-acceptor pairs. Abbreviations: BPE, B-phycoerythrin; CY5, carboxymethylindocyanine; Dansyl, just dansyl group; EM, eosin maleimide; FITC, fluorscein-5-isothiocyanate; LY, Lucifer yellow;ODR, octadecylrhodamine; TNP-ATP, trinitrophenyl-ATP.
Donor Acceptor
Förster distance ($R_0$, nm)
Naphthalene Dansyl 2.2
LY TNP-ATP 3.5
Dansyl ODR 4.3
LY EM 5.3
FITC EM 6.0
BPE CY5 7.2
Conclusion and the limitations of FRET
FRET provides an efficient way to measure the distance between a donor and an acceptor chromophore. The energy transfer efficiency is highly influenced by the ratio of $R$ and $R_0$ because of the exponent 6. Thus, by measuring the FRET efficiency, one can easily get the precise distance between the donor and the acceptor. If choosing the donor and acceptor properly, this experiment can also be carried out in vivo. However, the FRET only gives the information about distances. If a dramatic conformational change happens, such as lengthening or kink, it is unable to know the exact movement of donor and the acceptor. Besides, attaching the chromophores to precise sites of a macromolecule is also important, both in quantity of chromophores and in position of a macromolecule, or the FRET might produce noise signals. (Please refer to question 5)
Exercise $1$
The F-actin filament is composed by G-actin monomers. By attaching either a donor (D) or an acceptor (A) choromophore to the G-actin monomer and measuring the energy transfer efficiency to gauge the average distance between G-actin monomers in a F-actin filment (assuming that the monomers are well arranged in DADADADA....sequence), and one finds that the average energy transfer efficiencies is 23%. If the $R_0$ is 4.5 nm, what is the average distance between monomers in a filament?
Answer
5.5 nm.
Exercise $2$
Based on the question 1, if the filament sequence is composed by 8 monomers in the order of D-A-D-A-D-A-D-A, how many kinds of efficiencies might be detected if the filament does not bend and the $R_0$ is big enough to see all of them?
Answer
4 kinds of efficiencies will be detected.
Exercise $3$
The cy3-dornor and cy5-acceptor pair is attached onto the theminals of a DNA sequence. If ignoring the orientations of transition dipoles of donor and acceptor, please plot the relationship between the ratio ($R/R_0$) and the energy transfer efficiency.
Answer
Please refer to Professor Taekjip Ha's website. netfiles.uiuc.edu/tjha/www/newTechnique.html
Exercise $4$
Use the example in question 3, and now consider the orientations of transition dipoles. Please plot the relationship between the separating distance of donor-acceptor pair and the energy transfer efficiency. (Remember that when elongating the DNA by adding base pairs, the orientations of donor and acceptor chromophores will change) (Question 3 and 4 are revised from a paper which measured the orientation dependence in FRET process by using the DNA helix.
Answer
Please refer to PNAS, 2008 Aug, 105(32), 11176-11181, doi:10.1073/pnas.0801707105
Exercise $5$
One of the most challenging problems within ion channels field is to observe how the channel works in vivo and the conformational change of the channel which is embedded on the cell membrane . If a scientist wants to investigate the movement of the ion channel gate by using FRET, what kinds of factors should be considered? For example, how to attached a proper donor and acceptor to precise positions of the channel gate.
Answer
An open question. Please refer to this introductory paper about using FRET to investigate the ion channel movements. Biophysical Journal, 2003 Jan., 84(1), 1-2, doi:10.1016/S0006-3495(03)74827-9
Footnotes
1. PNAS, 2006 Dec., 103(49), 18458-18463, doi: 10.1073/pnas.0605422103
2. PNAS, 2008 Nov., 105(47), 18337-18342, doi: 10.1073/pnas.0800977105
3. Nature Biotechnology 2003, 21, 1387-1395, doi:10.1038/nbt896
4. PNAS, 2009 Oct., 106(42), 17741-17746, doi: 10.1073/pnas.0905177106
5. Nature, 1997 Aug., 388, 882-887
6. PNAS, 2006 Dec., 103(51) 19217-19218, doi: 10.1073/pnas.0609223103
7. PNAS, 1967 Aug., 58(2), 719-26. doi:10.1073/pnas.58.2.719
8. Analytical Biochemistry, 1994 Apr., 218(1), 1-13, doi:10.1006/abio.1994.1134
9. Nature Protocols, 2006 Aug., 1, 911–919, doi:10.1038/nprot.2006.122
10. PNAS, 2008 Aug, 105(32), 11176-11181, doi:10.1073/pnas.0801707105
11. Biophysical Journal, 2003 Jan., 84(1), 1-2, doi:10.1016/S0006-3495(03)74827-9
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Ladder Operators are operators that increase or decrease eigenvalue of another operator. There are two types; raising operators and lowering operators. In quantum mechanics the raising operator is called the creation operator because it adds a quantum in the eigenvalue and the annihilation operators removes a quantum from the eigenvalue. For example, in quantum harmonic oscillator, creation operators adds a quantum of energy to the system and annihilation operators removes a quantum of energy from the system. While in potential system quantum field theories, creation operators add a particle to the system and annihilation operators remove a particle from the system.
What are Ladder Operators
Ladder operators increase or decrease eginvaules by a quantum. An example of ladder operators are the a type of angular momentum operators.
Creation operator:
${L_+ Y_{\ell m}}={\hbar}{\sqrt{\ell(\ell+1)-m(m+1)}}{Y_{\ell(m+1)}}$
Annihilation operator:
${L_- Y_{\ell m}}={\hbar}{\sqrt{\ell(\ell+1)-m(m-1)}} {Y_{\ell(m-1)}}$
The creation operators increases the value of m and the annihilation operators decreases the value of m, both without affecting the value of l.
Below is a graphical representation of what ladder operators do when related to energy eigenvalue of the quantum harmonic oscillator. The Creation operators at increases the energy value by a quantum and the annihilation operator decreases the the energy value by a quantum.
Angular Momentum Ladder Operator
In angular momentum, the ladder operators are J+ and J-.
$J_-=J_x-iJ_y$
$J_+=J_x+iJ_y$
Then with the operator Jz
$J_z J_\pm|j m \rangle =(J_\pm J_z +[J_z,J_\pm]) |jm \rangle$
$J_z J_\pm|j m \rangle =(J_\pm J_z \pm \hbar J_\pm)|jm \rangle$
$J_z J_\pm|j m \rangle = \hbar (m \pm 1) J_\pm |jm \rangle$
therefore
$J_z |j m \pm 1 \rangle = \hbar (m \pm 1)|jm \rangle$
so we can see that
Creation Operator:
$J_+|j m \rangle =\alpha|jm+1 \rangle$
Annihilation Operator:
$J_-|j m \rangle =\beta|m-1 \rangle$
Where
$\alpha=\hbar \sqrt{(j-m)(j+m+1)}|jm+1 \rangle$
$\beta=\hbar \sqrt{(j+m)(j-m+1)}|jm-1 \rangle$
This shows why ladder operators are called creation and annihilation operators. The operators relate one eigenvalue to the next one by lower or rising the a quantum number.
Quantum Harmonic Oscillator Ladder Operator
Another example of ladder operators is for the quantum harmonic oscillator. The ladder operators for quantum harmonic oscillator rise or lower the energy of the system by a quantam.
To see where the operators come from, we start with the Schrödinger equation:
$\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega^2x^2\right)\psi(x)=E\psi(x)$
Setting x as
$x=\sqrt{\frac{\hbar}{m\omega}}q$
The Schrödinger equation is now
$\frac{\hbar\omega}{2}\left(-\frac{d^2}{dq^2}+q^2\right) \psi(q)=E\psi(q)$
$\hat H=-\frac{d^2}{dq^2}+q^2$
$\hat H=\left(-\frac{d}{dq}+q\right)\left(\frac{d}{dq}+q\right)+1$
So now the Schrödinger equation becomes
$\hbar\omega \left[\frac{1}{\sqrt2} \left(-\frac{d}{dq}+q\right) \frac{1}{\sqrt2} \left(\frac{d}{dq}+q\right) + \frac{1}{2}\right] \psi(q)=E\psi(q)$
The ladder operators are then
Creation Operator:
$a^t=\frac{1}{\sqrt 2} \left(-\frac{d}{dq}+q \right )$
Annihilation Operator:
$a=\frac{1}{\sqrt 2} \left (\frac{d}{dq}+q \right )$
From this we can get our equation (we won't do it here) for the energy of an our eigenvaules.
$E_n=(n+\frac{1}{2})\hbar\omega$
Conclusion
There are two kinds of ladder operators, creation and annihilation operators. Like the word ladder suggests, these operators move eigenvalues up or down. They are used in angular momentum to rise or lower quantum numbers and quantum harmonic oscillators to move between energy levels.
Contributors
• Kevin Zhang (Fall 2016)
Pressure
Pressure (\(p\)) is the force per unit area applied on a surface, in a direction perpendicular to that surface, i.e. the scalar part of the stress tensor under equilibrium/hydrosatic conditions.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Ladder_Operators_%28Creation_Annihilation_Operators%29.txt
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Quantum tunneling is a phenomenon where particles may "tunnel through" a barrier which they have insufficient kinetic energy to overcome according to classical mechanics. Tunneling is a result of the wavelike nature of quantum particles, and cannot be predicted by any classical system.
Introduction
Tunneling was first directly observed in the early 1900's when researchers were studying the electrical behavior of closely spaced electrodes in gases. An unexplained component of current was observed, even through high vacuum, though no explanation was found. Within a few years, physicists began to find solutions to the new Schrödinger equation. When solving for a double potential well, it was found that the particle could traverse two wells if the barrier separating them was sufficiently small and narrow.
The researchers were observing a phenomenon now called "field emission," where at high potentials electrons may tunnel through the potential barrier at the metal-vacuum interface, where they would then fall through the potential to the counter-electrode. Electron tunneling is exploited in technologies through Esaki diodes, Zener diodes, field emission-based electron guns, and scanning tunneling microscopy. For phenomena more germane to the study of chemistry, tunneling is responsible for the rate of alpha decay, since the alpha particles can escape from the atom even though the reaction that generates them would not produce enough energy to allow them to escape.
Basic description
As previously stated, quantum tunneling is a result of the wave nature of quantum particles. A traveling or standing wave function incident on a non-infinite potential decays in the potential as a function of $A_0 e^{-\alpha x}$, where A0 is the amplitude at the boundary, $\alpha$ is proportional to the potential, and x is the distance into the potential. If a second well exists at infinite distance from the first well, the probability goes to zero, so the probability of a particle existing in the second well is zero. If a second well is brought closer to the first well, the amplitude of the wave function at this boundary is not zero, so the particle may tunnel into that well from the first well. It would appear that the particles are 'leaking' through the barrier; they can travel through it without having to surmount it.
An important point to keep in mind is that tunneling conserves energy. The final sum of the kinetic and potential energy of the system cannot exceed the initial sum. Therefore, the potential on both sides of the barrier does not need to be the same, but the sum of the ground state energy and the potential on the opposite side of the barrier may not be larger than the initial particle energy and potential.
Advanced description
Tunneling is found from the Schrödinger equation, so we will start there. To simplify the calculations, the time-independent, one-dimensional Schrödinger equation is used (Equation \ref{eq1}).
$-{\frac{\hbar^2}{2m}} {\frac{\partial^2 \psi}{\partial x^2}} + V(r) \psi = E \psi \label{eq1}$
To find solutions to a particular system, the potential $V (x)$ must be defined. In this case, we will set the potential to zero for all space, except for the region between 0 and $a$, which we will set as $V_0$. This is represented by the piecewise function in Equation \ref{eq2}.
$V = \begin{cases} 0 & \text{if } -\infty<x\leq 0\[3pt] V_0 & \text{if } 0<x<a\[3pt] 0 & \text{if } a\leq x<\infty \end{cases} \label{eq2}$
To solve this, the equation must be solved separately for each region. However, the boundary conditions at 0 and $a$ for each region must be consistent such that $\psi (x)$ is continuous for all $x$, so that $\psi (x)$ is a valid wavefunction. The general solution for each region, before applying the boundary conditions, is then
$\psi = \begin{cases} A\sin kx + B\cos kx & \text{if } -\infty<x\leq 0\[3pt] Ce^{-\alpha x} + De^{\alpha x} & \text{if } 0<x<a\[3pt] E\sin kx + F\cos kx & \text{if } a\leq x<\infty \end{cases}$
where $k = \frac{\sqrt{2mE}}{\hbar}$ and $\alpha = \frac{\sqrt{2m(V_o -E)}}{\hbar}$. To enforce continuity, the boundaries of each region are set equal. This is expressed as $\psi_1 (0) = \psi_2 (0)$ and $\psi_1 (a) = \psi_2 (a)$.
$A\sin 0 + B\cos 0 = Ce^{0} + De^{0}$
which implies that $A=0$, and $B=C+D$. At the opposite boundary,
$A\sin ka + B\cos ka = Ce^{-\alpha a} + De^{\alpha a}$
It may be observed that, as $a$ goes to infinity, the right hand side of this equation goes to infinity, which does not make physical sense. To reconcile this, $D$ is set to zero. For the final region, $E$ and $F$, present a potentially intractable problem. However, if one realizes that the value at the boundary $a$ is driving the wave in the region $a$ to $\infty$, it may also be realized that the wavefunction could be rewritten as $Ce^{-\alpha a}\cos(k(x-a))$, phase shifting the wavefunction by the value of $a$ and setting the amplitude to the boundary value. The wavefunction is then
$\psi = \begin{cases} B\cos kx & \text{if } -\infty<x\leq 0\[3pt] Be^{-\alpha x} & \text{if } 0<x<a\[3pt] Be^{-\alpha a}\cos(k(x-a)) & \text{if } a\leq x<\infty \end{cases}$
The amplitude of the wavefunction is attenuated by the barrier as $e^{-a\frac{\sqrt{2m(V_o -E)}}{\hbar}}$, where $a$ represents the width of the barrier and $(V_o -E)$ is the difference between the potential energy of the barrier and the current energy of the particle. Since the square of the wavefunction is the probability distribution, the probability of transmission through a barrier is $e^{-2a\frac{\sqrt{2m(V_o -E)}}{\hbar}}$. As the barrier width or height approaches zero, the probability of a particle traveling through the barrier becomes 1. Also of note is that $k$ is unchanged on the other side of the barrier. This implies that the energy of the particles are exactly the same as it was before they tunneled through the barrier, as stated earlier, the only thing that changes is the quantity of particles going in that direction. The rest are reflected off the barrier, and go back the way they came.
• Ryan Bunk
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Quantum_Tunneling.txt
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Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770). In confronting statistical problems we often encounter factorials of very large numbers. The factorial $N!$ is a product $N(N-1)(N-2)...(2)(1)$. Therefore, $\ln \,N!$ is a sum
$\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k. \label{1}$
where we have used the property of logarithms that $\log(abc) =\log(a) + \log(b) +\log(c)$. The sum is shown in figure below.
Using Euler-MacLaurin formula one has
$\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R , \label{2}$
where B1 = −1/2, B2 = 1/6, B3 = 0, B4 = −1/30, B5 = 0, B6 = 1/42, B7 = 0, B8 = −1/30, ... are the Bernoulli numbers, and $R$ is an error term which is normally small for suitable values of $p$.
Then, for large $N$,
$\ln N! \sim \int_1^N \ln x\,dx \approx N \ln N -N . \label{3}$
after some further manipulation one arrives at (apparently Stirling's contribution was the prefactor of $\sqrt{2\pi})$
$N! = \sqrt{2 \pi N} \; N^{N} e^{-N} e^{\lambda_N} \label{4}$
where
$\dfrac{1}{12N+1} < \lambda_N < \frac{1}{12N}. \label{5}$
The sum of the area under the blue rectangles shown below up to $N$ is $\ln N!$. As you can see the rectangles begin to closely approximate the red curve as $m$ gets larger. The area under the curve is given the integral of $\ln x$.
$\ln N! = \sum_{m=1}^N \ln m \approx \int_1^N \ln x\, dx \label{6}$
To solve the integral use integration by parts
$\int u\,dv=uv-\int v\,dy \label{7A}$
Here we let $u = \ln x$ and $dv = dx$. Then $v = x$ and $du = \frac{dx}{x}$.
$\int_0^N \ln x \, dx = x \ln x|_0^N - \int_0^N x \dfrac{dx}{x} \label{7B}$
Notice that $x/x = 1$ in the last integral and $x \ln x$ is 0 when evaluated at zero, so we have
$\int_0^N \ln x \, dx = N \ln N - \int_0^N dx \label{8}$
Which gives us Stirling’s approximation: $\ln N! = N \ln N – N$. As is clear from the figure above Stirling’s approximation gets better as the number N gets larger (Table $1$).
Table $1$: Evaluation of Approximation with absolute values
N N! ln N! N ln N – N Error
10 3.63 x 106 15.1 13.02 13.8%
50 3.04 x 1064 148.4 145.6 1.88%
100 9.33 x 10157 363.7 360.5 0.88%
150 5.71 x 10262 605.0 601.6 0.56%
Calculators often overheat at 200!, which is all right since clearly result are converging. In thermodynamics, we are often dealing very large N (i.e., of the order of Avagadro’s number) and for these values Stirling’s approximation is excellent.
WKB Approximation
The WKB Approximation, named after scientists Wentzel–Kramers–Brillouin, is a method to approximate solutions to a time-independent linear differential equation or in this case, the Schrödinger Equation. Its principal applications are for calculating bound-state energies and tunneling rates through potential barriers. The WKB Approximation is most often applied to 1D problems, but also works for 3D spherically symmetric problems. As a general overview, the wavefunction is assumed to be an exponential function with either amplitude or phase taken to be slowly changing relative to the de Broglie wavelength $λ$. It is then semi-classically expanded.
Solving the Schrödinger Equation
The WKB Approximation states that the wavefunction to the Schrödinger Equation take the form of simple plane waves when at a constant potential $U$ (i.e., acts like a free particle).
$\psi(x)=Ae^{\pm ikx} \nonumber$
where
$k=\dfrac{2\pi}{\lambda}=\sqrt{\dfrac{2m(E-U)}{\hbar^2}}=constant\label{56.1}$
If the potential changes slowly with $x$ $(U \rightarrow U(x))$, the solution of the Schrödinger equation is:
$\psi(x)=Ae^{\pm i\phi(x)}\label{56.2}$
where $\phi (x)=xk(x)$. For a case with constant potential $U$, $\phi (x)=\pm kx$. Thus, phase changes linearly with $x$. For a slowly varying $U$, $\phi (x)$ varies slowly from the linear case $\pm kx$.
The classical turning point is defined as the point at which the potential energy $U$ is approximately equal to total energy $E$ $(U \approx E)$ and the kinetic energy equals zero. This occurs because the mass stops and reverses its velocity is zero. It is an inflection point that marks the boundaries between regions where a classical particle is allowed and where it is not, as well as where two wavefunctions must be properly matched.
If $E > U$, a classical particle has a non-zero kinetic energy and is allowed to move freely. If $U$ is a constant, the solution to the one-dimensional Schrödinger equation is:
$\psi(x)=Ae^{\pm ikx},\label{56.3}$
in which the wavefunction is oscillatory with constant wavelength λ and constant amplitude A. $k(x)$ is defined as:
$k(x)=\sqrt{\dfrac{2m(E-U(x))}{\hbar^2}}\label{56.4}$
If $E < U$, the solution to the Schrödinger equation for a constant $U$ is:
$\psi(x)=Ae^{\pm\kappa x}. \label{56.5}$
In these regions, a classical particle would not be allowed, but there is a finite probability that a particle can pass through a potential energy barrier in quantum mechanics. The quantum particle is described as ‘tunnelling’, which is important in determining the rates of chemical reactions, particularly at lower temperatures. $k(x)$ is defined as:
$k(x)=-i\sqrt{\dfrac{2m(U(x)-E)}{\hbar^2}} = -i\kappa(x).\label{56.6}$
If $U(x)$ is not a constant, but instead varies very slowly on a distance scale of $λ$, then it is reasonable to suppose that ψ remains practically sinusoidal, except that the wavelength and amplitude change slowly with $x$.
WKB Approximation
Substituting in the normalized version of Equation \ref{56.2}, the Schrödinger Equation:
$\dfrac{-\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\psi(x)+U(x)\psi(x)=E\psi(x)\label{56.7}$
becomes
$i\dfrac{\partial^2\phi}{\partial x^2}-\left(\dfrac{\partial\phi}{\partial x}\right)^2+(k(x))^2=0.\label{56.8}$
The WKB Approximation assumes that the potentials, $k(x)$ and $\phi (x)$ are slowly varying.
The 0th order WKB Approximation assumes:
$\dfrac{\partial^2\phi}{\partial x^2} \ge 0.\label{56.9}$
Thus,
$\left(\dfrac{\partial\phi_0}{\partial x}\right)^2=(k(x))^2\label{56.10}$
Solving,
$\phi_0(x)=\pm\int k(x)dx+C_0\label{56.11}$
and substituting $\phi_0(x)$ into Equation \ref{56.2},
$\psi(x)=e^{i(\pm\int k(x)dx+C_0)}.\label{56.12}$
To obtain a more accurate solution, we manipulate Equation \ref{56.8} to solve for $\phi(x)$.
$i\dfrac{\partial^2\phi}{\partial x^2}-\left(\dfrac{\partial\phi}{\partial x}\right)^2+(k(x))^2=0 \nonumber$
$\left(\dfrac{\partial\phi}{\partial x}\right)^2=(k(x))^2+i\dfrac{\partial^2\phi}{\partial x^2} \nonumber$
$\dfrac{\partial\phi}{\partial x}=\sqrt{(k(x))^2+i\dfrac{\partial^2\phi}{\partial x^2}} \nonumber$
So,
$\phi(x)=\pm\int\sqrt{(k(x))^2+i\dfrac{\partial^2\phi}{\partial x^2}}dx+C_1\label{56.13}$
The 1st order WKB Approximation follows the assumption of Equation \ref{56.10} from the 0th order solution. Taking its square root, we find that:
$\dfrac{\partial\phi_0}{\partial x}=\pm k(x).\label{56.14}$
Taking the derivative on both sides with respect to $x$, we find that:
$\dfrac{\partial^2\phi_0}{\partial x^2}=\pm \dfrac{\partial}{\partial x}k(x).\label{56.15}$
Solving,
$\phi_1(x)=\pm\int\sqrt{(k(x))^2\pm i\dfrac{\partial}{\partial x}k(x)}dx+C_1\label{56.16}$
and substituting $\phi_1(x)$ into Equation \ref{56.2},
$\psi(x)=e^{i\left( \pm\int\sqrt{(k(x))^2\pm i\dfrac{\partial}{\partial x}k(x)}dx+C_1\right)}.\label{56.17}$
Example $1$
Determine the tunneling probability $T$ at a finite width potential barrier.
Solution
Given:
$T = \dfrac{\psi^*(L) \psi(L)}{\psi^*(0)\psi(0)} \nonumber$
where
$\psi(x)=\psi(0)e^{i\pm(\int k(x)dx+C_1)} \nonumber$
For tunneling to occur, $E < U$. So,
$k(x)=-i\sqrt{\dfrac{2m(U(x)-E)}{\hbar^2}}$ for $E < U(x)$, assuming $U(x) = U$.
Plugging in $k(x)$ to solve for the wavefunction,
\begin{align*} \psi(x) &=\psi(0)e^{i\left(\pm\int_{0}^{x}-i\sqrt{\dfrac{2m(U(x)-E)}{\hbar^2}}dx\right)} \[4pt] &=\psi(0)e^{\left(-\int_{0}^{x}\sqrt{\dfrac{2m(U-E)}{\hbar^2}}dx\right)} \[4pt] &=\psi(0)e^{\left(-\left(\sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right)x\right)} \end{align*}
Thus solving for the the tunneling probability $T$,
\begin{align*} T &= \dfrac{\psi^*(L) \psi(L)}{\psi^*(0)\psi(0)} \[4pt] &= \dfrac{\psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) L \right)} \psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) L \right)}}{\psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) 0 \right)} \psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) 0 \right)}} \[4pt] &= e^{\left( -2 \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) L \right)} \end{align*}
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Stirlings_Approximation.txt
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Symmetry is very important in chemistry researches and group theory is the tool that is used to determine symmetry. Usually, it is not only the symmetry of molecule but also the symmetries of some local atoms, molecular orbitals, rotations and vibrations of bonds, etc. that are important.
• Group Theory: Theory
Symmetry can help resolve many chemistry problems and usually the first step is to determine the symmetry. If we know how to determine the symmetry of small molecules, we can determine symmetry of other targets which we are interested in. Therefore, this module will introduce basic concepts of group theory and after reading this module, you will know how to determine the symmetries of small molecules.
• Group Theory and its Application to Chemistry
Group Theory is the mathematical application of symmetry to an object to obtain knowledge of its physical properties. What group theory brings to the table, is how the symmetry of a molecule is related to its physical properties and provides a quick simple method to determine the relevant physical information of the molecule.
• Understanding Character Tables of Symmetry Groups
Every molecule has a point group associated with it, which are assigned by a set for rules (explained by Group theory). The character tables takes the point group and represents all of the symmetry that the molecule has.
• Woodward-Hoffmann rules
Electrocyclic reactions. The selection rule of electrocyclization reactions is given in the original statement of the Woodward–Hoffmann rules. If a generalized electrocyclic ring closure occurs in a polyene of 4n π-electrons, then it is conrotatory under thermal conditions and disrotatory under photochemical conditions.
Group Theory
Symmetry can help resolve many chemistry problems and usually the first step is to determine the symmetry. If we know how to determine the symmetry of small molecules, we can determine symmetry of other targets which we are interested in. Therefore, this module will introduce basic concepts of group theory and after reading this module, you will know how to determine the symmetries of small molecules.
Introduction
Symmetry is very important in chemistry researches and group theory is the tool that is used to determine symmetry. Usually, it is not only the symmetry of molecule but also the symmetries of some local atoms, molecular orbitals, rotations and vibrations of bonds, etc. that are important. For example, if the symmetries of molecular orbital wave functions are known, we can find out information about the binding. Also, by the selection rules that are associated with symmetries, we can explain whether the transition is forbidden or not and also we can predict and interpret the bands we can observe in Infrared or Raman spectrum.
Symmetry operations and symmetry elements are two basic and important concepts in group theory. When we perform an operation to a molecule, if we cannot tell any difference before and after we do the operation, we call this operation a symmetry operation. This means that the molecule seems unchanged before and after a symmetry operation. As Cotton defines it in his book, when we do a symmetry operation to a molecule, every points of the molecule will be in an equivalent position.
Symmetry Elements
For different molecules, there are different kinds of symmetry operations we can perform. To finish a symmetry operation, we may rotate a molecule on a line as an axis, reflect it on a mirror plane, or invert it through a point located in the center. These lines, planes, or points are called symmetry elements. There may be more then one symmetry operations associated with a particular symmetry
Identity E
The molecule does not move and all atoms of the molecule stay at the same place when we apply an identity operation, E, on it. All molecules have the identity operation. Identity operation can also be a combination of different operations when the molecule returns to its original position after these operations are performed.1 This will be demonstrated later.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Group_Theory/Group_Theory%3A_Theory.txt
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Group Theory is the mathematical application of symmetry to an object to obtain knowledge of its physical properties. What group theory brings to the table, is how the symmetry of a molecule is related to its physical properties and provides a quick simple method to determine the relevant physical information of the molecule. The symmetry of a molecule provides you with the information of what energy levels the orbitals will be, what the orbitals symmetries are, what transitions can occur between energy levels, even bond order to name a few can be found, all without rigorous calculations. The fact that so many important physical aspects can be derived from symmetry is a very profound statement and this is what makes group theory so powerful.
Introduction
To a fully understand the math behind group theory one needs to take a look at the theory portion of the Group Theory topic or refer to one of the reference text listed at the bottom of the page. Never the less as Chemist the object in question we are examining is usually a molecule. Though we live in the 21st century and much is known about the physical aspects that give rise to molecular and atomic properties. The number of high level calculations that need to be performed can be both time consuming and tedious. To most experimentalist this task is takes away time and is usually not the integral part of their work. When one thinks of group theory applications one doesn't necessarily associated it with everyday life or a simple toy like a Rubik's cube. A Rubik's cube is an a cube that has a $3 \times 3$ array of different colored tiles on each of its six surfaces, for a total of 54 tiles. Since the cube exist in 3D space, the three axis are $x$, $y$, $z$. Since the rubik's cube only allows rotation which are called operations, there are three such operations around each of the $x$, $y$, $z$ axis.
Of course the ultimate challenge of a Rubik's cube is to place all six colors on each of the six faces. By performing a series of such operations on the Rubik's cube one can arrive at a solution (A link of a person solving a Rubik's cube1 in 10.4s with operations performed noted, the operations performed will not translate to chemistry applications but it is a good example of how symmetry operations arrive at a solution). The operations shown in the Rubik's cube case are inherent to the make up of the cube, i.e., the only operations allowed are the rotations along the x, y, z axis. Therefore the Rubik's cube only has x,y,z rotation operations. Similarly the operations that are specific to a molecule are dependent on its symmetry. These operations are given in the top row of the character table.
$C_{3v}$ $E$ $2C_3$ $3\sigma_v$
rotation and translation
quadratic
functions
$A_1$ +1 +1 +1 $z$ $x^2+y^2$, $z^2$
$A_2$ +1 +1 -1 $R_z$ -
$E$ +2 -1 0 ($x$, $y$) ($R_x$, $R_y$) ($x^2-y^2$, $xy$) ($xz$, $yz$)
The character table contains a wealth of information, for a more detailed discussion of the character table can be found in Group Theory Theoretical portion of the chemWiki. All operations in the character table are contained in the first row of the character table, in this case $E$, $C_3$, & $\sigma_v$, these are all of the operations that can be preformed on the molecule that return the original structure. The first column contains the three irreducible representations from now on denoted as $\Gamma_{ir}$, here they are $A_1$, $A_2$ & $E$. The value of the $\Gamma_{ir}$ denotes what the operation does. A value of 1 represents no change, -1 opposite change and 0 is a combination of 1 & -1 (0’s are found in degenerate molecules. The final two columns Rotation and Translation represented by $R_x$,$R_y$, $R_z$ & $x$, $y$, $z$ respectively. Where R's refer to rotation about an axis and the $x$, $y$, $z$ refers to a translation about an axis, the $\Gamma_{ir}$ the each $R_x$, $R_y$, $R_z$ & $x$, $y$, $z$ term is the irreducible symmetry of a rotation or translation operation. Like wise the final column the orbital symmetries relates the orbital wavefunction to a irreducible representation.
Direct Products
This is a quick rule to follow for calculating Direct Products of irreproducible representations, such a calculation will be necessary for working through transition moment integrals. Following the basic rules given by the table given below. One can easily work through symmetry calculations very quickly.
"Symmetric" $\times$ "Symmetric" is "Symmetric" "Symmetric" $\times$ "AntiSymmetric" is "AntiSymmetric" "AntiSymmetric" $\times$ "Symmetric" is "AntiSymmetric" "AntiSymmetric" $\times$ "AntiSymmetric" is "Symmetric"
$g \times g = g$ $g \times u = u$ $u \times g = u$ $u \times u = g$
$' \times ' = '$ $' \times '' = ''$ $'' \times ' = ''$ $'' \times '' = '$
$A \times A= A$ $A \times B= B$ $B \times A= B$ $B \times B= A$
Vibrations
All molecules vibrate. While these vibrations can originate from several events, which will be covered later, the most basic of these occurs when an electron is excited within the electronic state from one eigenstate to another. The Morse potential (electronic state) describes the energy of the eigenstate as a function of the interatomic distance. When an electron is excited form one eigenstate to another within the electronic state there is a change in interatomic distance, this result in a vibration occurring.
Vibrational energies arise from the absorption of polarizing radiation. Each vibrational state is assigned a $\Gamma_{ir}$. A vibration occurs when an electron remains within the electronic state but changes from one eigenstate to another (The vibrations for the moment are only IR active vibrations, there are also Raman vibrations which will be discussed later in electronic spectroscopy), in the case of the Morse diagram above the eigenstates are denoted as $\nu$. As you can see from the diagram the eigenstate is a function of energy versus interatomic distance.
To predicting whether or not a vibrational transition, or for that matter a transition of any kind, will occur we use the transition moment integral.
$\int \Psi_i*\mu \Psi_f d\tau=\langle \Psi_i | \mu| \Psi_f \rangle$
The transition moment integral is written here in standard integral format, but this is equivalent to Bra & Ket format which is standard in most chemistry quantum mechanical text (The $\langle \Psi_i |$ is the Bra portion, $| \Psi_f \rangle$ is the Ket portion). The transition moment operator $\mu$ is the operator the couples the initial state $\Psi_i$ to the final state $\Psi_f$, which is derived from the time independent Schrödinger equation. However using group theory we can ignore the detailed mathematical methods. We can use the $\Gamma_{ir}$ of the vibrational energy levels and the symmetry of the transition moment operator to find out if the transition is allowed by selection rules. The selection rules for vibrations or any transition is that is allowed, for it to by allowed by group theory the answer must contain the totally symmetric $\Gamma_{ir}$, which is always the first $\Gamma_{ir}$ in the character table for the molecule in question.
Symmetry
Let’s work through an example: Ammonia ($NH_3$) with a $C_{3v}$ symmetry. Consequently, all of the properties contained in the $C_{3v}$ character table above are pertinent to the ammonia molecule.
The principle axis is the axis that the highest order rotation can be preformed. In this case the z-axis pass through the lone pairs (pink sphere), which contains a $C_3$ axis. The ?’s or mirror planes ($\sigma_v$ parallel to z-axis & $\sigma_h$ perpendicular to the z-axis). In ammonia there is no $\sigma_h$ only three $\sigma_v$’s. The combination of $C_3$ & $\sigma_v$ leads to $C_{3v}$ point group, which leads to the C3v character table.
The number of transitions is dictated by 3N-6 for non-linear molecules and 3N-5 for linear molecules, where N is the number of atoms. The 6 & the 5 derive from three translations in the x,y,z plan and three rotations also in the x,y,z plan. Where a linear molecule only has two rotations in the x & y plans since the z axis has infinite rotation. This leads to only 5 degrees of freedom in the rotation and translation operations. In the case of Ammonia there will be 3(4)-6=6 vibrational transitions. This can be confirmed by working through the vibrations of the molecule. This work is shown in the table below.
The vibrations that are yielded 2A1 & 2E (where E is doubly degenerate, meaing two vibration modes each) which total 6 vibrations. This calculation was done by using the character table to find out the rotation and translation values and what atoms move during each operation. Using the character table we can characterize the A1 vibration as IR active along the z-axis and raman active as well. The E vibration is IR active along both the x & y axis and is Raman active as well. From the character table the IR symmetries correspond to the x, y & z translations. Where the Raman active vibrations correspond to the symmetries of the d-orbitals.
Vibrational Spectroscopy
Infrared Spectroscopy
Infrared Spectroscopy (IR) measures the vibrations that occur within a single electronic state, such as the one shown above. Because the transition occurs within a single electronic state there is a variation in interatomic distance. The dipole moment is dictate by the equation.
$\vec{\mu} = \alpha\vec{E}$
Where $\vec{\mu}$ is the magnitude of dipole moment; $\alpha$ is the polarizability constant (actually a tensor) & $E$ is the magnitude of the electric field which can be described as the electronegitivity.3 Therefore when a vibration occurs within a single electronic state there is a change in the dipole moment, which is the definition of an active IR transition.
$\left ( \frac{\mathrm{d\mu} }{\mathrm{d} q} \right )_{eq} \neq 0$
In terms of group theory a change in the dipole is a change from one vibrational state to another, as shwon by the equation above. A picture of the vibrational states with respect ot the rotational states and electronic states is given below. In IR spectroscopy the transition occurs only from on vibrational state to another all within the same electronic state, shown below as B.
Electronic Transitions
When an electron is excited from one electronic state to another, this is what is called an electronic transition. A clear example of this is part C in the energy level diagram shown above. Just as in a vibrational transition the selection rules for electronic transitions are dictated by the transition moment integral. However we now must consider both the electronic state symmetries and the vibration state symmetries since the electron will still be coupled between two vibrational states that are between two electronic states. This gives us this modified transition moment integral:
Where you can see that the symmetry of the initial electronic state & vibrational state are in the Bra and the final electronic and vibrational states are in the Ket. Though this appears to be a modified version of the transition moment integral, the same equation holds true for a vibrational transition. The only difference would be the electronic state would be the same in both the initial and final states. Which the dot product of yields the totally symmetric representation, making the electronic state irrelevant for purely vibrational spectroscopy.
Raman
In Resonance Raman spectroscopy transition that occurs is the excitation from one electronic state to another and the selection rules are dictated by the transition moment integral discussed in the electronic spectroscopy segment. However mechanically Raman does produce a vibration like IR, but the selection rules for Raman state there must be a change in the polarization, that is the volume occupied by the molecule must change. But as far as group theory to determine whether or not a transition is allowed one can use the transition moment integral presented in the electronic transition portion. Where one enters the starting electronic state symmetry and vibrational symmetry and final electronic state symmetry and vibrational state, perform the direct product with the different M's or polarizing operators For more information about this topic please explore the Raman spectroscopy portion of the Chemwiki
Fluorescence
For the purposes of Group Theory Raman and Fluorescence are indistinguishable. They can be treated as the same process and in reality they are quantum mechanically but differ only in how Raman photons scatter versus those of fluorescence.
Phosphorescence
Phosphorescence is the same as fluorescence except upon excitation to a singlet state there is an interconversion step that converts the initial singlet state to a triplet state upon relaxation. This process is longer than fluorescence and can last microseconds to several minutes. However despite the singlet to triplet conversion the transition moment integral still holds true and the symmetry of ground state and final state still need to contain the totally symmetric representation.
Molecular Orbital Theory and Symmetry
Molecular Orbitals also follow the symmetry rules and indeed have their own ?ir. Below are the pi molecular orbitals for trans-2-butene and the corresponding symmetry of each molecular orbital.
The ?ir of the molecular orbitals are created by simply preforming the operations of that molecule's character table on that orbital. In the case of trans-2-butene the point group is C2h, the operations are: E, C2, i & ?h. Each operation will result in a change in phase (since were dealing with p-orbitals) or it will result in no change. The first molecular orbital results in the totally symmetric representation, working through all four operations E, C2, i, ?h will only result in 1's meaning there is no change, giving the Ag symmetry state. These molecular orbitals also represent different electronic states and can be arranged energetically. Putting the orbital that has the lowest energy, the orbital with the fewest nodes at the bottom of the energy diagram and like wise working up form lowest energy to highest energy. The highest energy orbital will have the most nodes. Once you've set up your MO diagram and place the four pi electrons in the orbitals you see that the first two orbitals listed (lowest energy) are HOMO orbitals and the bottom two (highest energy) and LUMO orbitals. With this information if you have a transition from the totally symmetric HOMO orbital to the totally symmetric LUMO orbital the transition moment operator would need to have Ag symmetry (using the C2h) to give a result containing the totally symmetric representation.
These four molecular orbitals represent four different electronic states. So transitions from one MO into another would be something that is measured typically with UV-Vis spectrometer.
Problems
1. Follow the links above to Water, Ammonia and Benzene and work out the ?ir of the vibrations. Using the method laid out by the character table. (Follow the example of ammonia for help)
2. From problem 1. work out what possible are the possible transition moment operators for each vibration.
3. Work through the P-orbital molecular orbitals for cis-butadiene. (Note the conservation of "stuff", start by combining four p-orbitals and finish with four molecular orbitals) What is the point group? what are the ?ir of each MO? Finally how many vibrations are there for cis-butadiene and what are their ?ir.
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Every molecule has a point group associated with it, which are assigned by a set for rules (explained by Group theory). The character tables takes the point group and represents all of the symmetry that the molecule has.
Symbols under the first column of the character tables
A (Mulliken Symbol) (singly degenerate or one dimensional) symmetric with respect to rotation of the principle axis
B (Mulliken Symbol) (singly degenerate or one dimensional) anti-symmetric with respect to rotation of the principle axis
E (Mulliken Symbol) (doubly degenerate or two dimensional)
T (Mulliken Symbol) (thirdly degenerate or three dimensional )
Subscript 1 symmetric with respect to the Cnprinciple axis, if no perpendicular axis, then it is with respect to σv
Subscript 2 anti-symmetric with respect to the Cnprinciple axis, if no perpendicular axis, then it is with respect to σv
Subscript g symmetric with respect to the inverse
subscript u
anti-symmetric with respect to the inverse
prime symmetric with respect to $σ_h$ (reflection in horizontal plane)
double prime anti-symmetric with respect to $σ_h$ ( opposite reflection in horizontal plane)
Symbols in the first row of the character tables
E describes the degeneracy of the row (A and B= 1) (E=2) (T=3)
Cn 2pi/n= number of turns in one circle on the main axis without changing the look of a molecule (rotation of the molecule)
Cn' 2π/n= number of turns in one circle perpendicular to the main axis, without changing the structure of the molecule
Cn" 2π/n= number of turns in one circle perpendicular to the Cn' and the main axis, without changing the structure
σ' reflection of the molecule perpendicular to the other sigma
σv (vertical) reflection of the molecule vertically compared to the horizontal highest fold axis.
σh or d (horizontal) reflection of the molecule horizontally compared to the horizontal highest fold axis.
i Inversion of the molecule from the center
Sn rotation of 2π/n and then reflected in a plane perpendicular to rotation axis.
#Cn the # stands for the number of irreducible representation for the Cn
the # stands for the number irreducible representations for the sigmas.
the number in superscript in the same rotation there is another rotation, for instance Oh has 3C2=C42
other useful definitions
(Rx,Ry) the ( , ) means they are the same and can be counted once.
x2+y2, z2 without ( , ) means they are different and can be counted twice.
Looking at a Character Table
D3h E 2C3 3C2 σh 2S3 v IR Raman
A1' 1 1 1 1 1 1 x2+y2, z2
A2' 1 1 -1 1 1 -1 Rz
E' 2 -1 0 2 -1 0 (x,y) (x2-y2, xy)
A1" 1 1 1 -1 -1 -1
A2" 1 1 -1 -1 -1 1 z
E" 2 -1 0 -2 1 0 (Rx, Ry) (xy, yz)
The order is the number in front of the the classes. If there is not number then it is considered to be one. The number of classes is the representation of symmetries.The D3h has six classes and an order of twelve.
Understanding using matrix
The identity does nothing to the matrix.
[1 0 0] [X] [X]
[0 1 0] [Y] = [Y]
[0 0 1] [Z] [Z]
σ(xy) the x and y stay positive, while z turns into a negative.
[1 0 0] [X] [X]
[0 1 0] [Y] = [Y]
[0 0 -1] [Z] [-Z]
Inversion (I) is when all of the matrix turns into a negative.
[-1 0 0] [X] [-X]
[0 -1 0] [Y] = [-Y]
[0 0 -1] [Z] [-Z]
Cnis when one would use cos and sin. for an example C4
[cos (2π/4 -sin (2π/4 0] [X] []
[sin (2π/4) cos (2π/4) 0] [Y] = []
[0 0 1] [Z] []
Practice
There are two columns on the far right. One is ir and the other is Raman. Try moving the molecule around using reflections and rotations. Remember when the positive side of the orbitals goes into the negative side, the number is negative (in the character tables). Also, remember if it is moved or reflected and there is no change then the number is positive (in the character tables). also, molecules are placed on a x, y, and z (three dimensional) graph. The highest C fold rotation is always on the z axis.
IR
$P_x$
$P_y$
$P_z$
$R_x$
$R_y$
$R_z$
Raman
$\sigma$
$d_{xy}$
$d_{yz}$
$d_{xz}$
$d_{z^2}$
$d_{x^2-y^2}$
Woodward-Hoffmann rules
The Woodward-Hoffman rules (R. B. Woodward was one of the U.S.'s most successful and renowned organic chemists; Roald Hoffmann is a theoretical chemist. Hoffmann shared the 1981 Nobel Prize with Kenichi Fukui) allow us to follow the symmetry of the occupied molecular orbitals along a suggested reaction path connecting reactants to products and to then suggest whether a symmetry-imposed additional energy barrier (i.e., above any thermodynamic energy requirement) would be expected. For example, let us consider the disrotatory closing of 1,3-butadiene to produce cyclobutene. The four $p$ orbitals of butadiene, denoted $\pi_1$ through $\pi_4$ in the figure shown below, evolve into the $\sigma$ and $\sigma^{*}$ and $\pi$ and $\pi^{*}$ orbitals of cyclobutene as the closure reaction proceeds. Along the disrotatory reaction path, a plane of symmetry is preserved (i.e., remains a valid symmetry operation) so it can be used to label the symmetries of the four reactant and four product orbitals. Labeling them as odd (o) or even (e) under this plane and then connecting the reactant orbitals to the product orbitals according to their symmetries, produces the orbital correlation diagram shown below.
Figure: Professor Roald Hoffmann (right)
The four $\pi$ electrons of 1,3-butadiene occupy the $\pi_1$ and $\pi_2$ orbitals in the ground electronic state of this molecule. Because these two orbitals do not correlate directly to the two orbitals that are occupied ($\sigma$ and $\pi$) in the ground state of cyclobutene, the Woodward-Hoffmann rules suggest that this reaction, along a disrotatory path, will encounter a symmetry-imposed energy barrier. In contrast, the excited electronic state in which one electron is promoted from $\pi_2$ to $\pi_3$ (as, for example, in a photochemical excitation experiment) does correlate directly to the lowest energy excited state of the cyclobutene product molecule, which has one electron in its p orbital and one in its p* orbital. Therefore, these same rules suggest that photo-exciting butadiene in the particular way in which the excitation promotes one electron from $\pi_2$ to $\pi_3$ will produce a chemical reaction to generate cyclobutene.
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How can an experiment confirm that a reaction is happening in a particular way? What is the mechanism of the reaction? What intermediates are occurring, and in what order do the bond-making and bond-breaking steps take place?
There are many experiments designed to illustrate how reactions happen. One of the methods used is chemical kinetics, in which the rate of a reaction is measured. By making changes in the reaction conditions and measuring the effect of the changes on the rate of reaction, we can infer what is going on at the molecular level.
• Chemical kinetics is the measurement of how quickly reactions occur.
• If changes in conditions affect the speed of reaction, we can learn something about how the reaction happens.
Kinetic studies are important in understanding reactions, and they also have practical implications. For example, in industry, reactions are conducted in reactors in which compounds are mixed together, possibly heated and stirred for a while, and then moved to the next phase of the process. It is important to know how long to hold the reaction at one stage before moving on, to make sure that reaction has finished before starting the next one.
By understanding how a reaction takes place, many processes can be improved. For example, if it is known that a particular intermediate is involved in a reaction, the use of conditions (such as certain solvents) that are incompatible with that intermediate might be avoided. In addition, reagents might be added that would make certain steps in the reaction happen more easily.
Not only are kinetic studies important in industry, but they are also used to understand biological processes, especially enzyme-catalyzed reactions. They also play a role in environmental and atmospheric chemistry, as part of an effort to understand a variety of issues ranging from the fate of prescription pharmaceuticals in wastewater to the cascade of reactions involved in the ozone cycle.
2.1.01: Continuous Flow
Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium. The Reaction Rate for a given chemical reaction is the measure of the change in concentration of the reactants or the change in concentration of the products per unit time.
• 2.1: Experimental Determination of Kinetics
• 2.2: Factors That Affect Reaction Rates
• 2.3: First-Order Reactions
A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.
• 2.4: Half-lives
The half-life of a reaction, t1/2, is the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. Its application is used in chemistry and medicine to predict the concentration of a substance over time.
• 2.5: Reaction Rate
Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium. The Reaction Rate for a given chemical reaction is the measure of the change in concentration of the reactants or the change in concentration of the products per unit time.
• 2.6: Reaction Rates- A Microscopic View
• 2.7: Reaction Rates- Building Intuition
• 2.8: Second-Order Reactions
Many important biological reactions, such as the formation of double-stranded DNA from two complementary strands, can be described using second order kinetics. In a second-order reaction, the sum of the exponents in the rate law is equal to two. The two most common forms of second-order reactions will be discussed in detail in this section.
• 2.9: Third Order Reactions
• 2.10: Zero-Order Reactions
In some reactions, the rate is apparently independent of the reactant concentration. The rates of these zero-order reactions do not vary with increasing nor decreasing reactants concentrations. This means that the rate of the reaction is equal to the rate constant, k, of that reaction.
02: Reaction Rates
• 2.1.1: Continuous Flow
Continuous Flow is a type of assay used to easily measure the progress of a reaction at discrete time points and is commonly used for determining initial rates and inhibition values.
• 2.1.2: Measuring Reaction Rates
The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically. What follows is general guidance and examples of measuring the rates of a reaction.
• 2.1.3: Rate vs. Concentration Proportionalities
• 2.1.4: Relaxation Methods
Many chemical reactions are complete in less than a few seconds, which makes the rate of reaction difficult to determine. In these cases, the relaxation methods can be used to determine the rate of the reaction.
• 2.1.5: Spectrophotometry
Spectrophotometry is a method to measure how much a chemical substance absorbs light by measuring the intensity of light as a beam of light passes through sample solution. The basic principle is that each compound absorbs or transmits light over a certain range of wavelength. This measurement can also be used to measure the amount of a known chemical substance.
• 2.1.6: Stopped Flow
The stopped-flow technique allows for the evaluation of solution-based kinetics on a milliseconds timescale with a very small volume of reactants used.
2.01: Experimental Determination of Kinetics
Continuous Flow is a type of assay used to easily measure the progress of a reaction at discrete time points and is commonly used for determining initial rates and inhibition values.
Theory
In a Continuous Flow experiment, the composition of the reaction is measured continuously, normally by absorbance, while the reactants flow and mix continuously. This is in contrast to Stopped Flow, where the reactants are mixed in a chamber and then measured, or Batch Reactions, where a number of reactions are performed and stopped at various times by the use of a compound which will halt the reaction.
Apparatus
The apparatus used for continuous flow assays will have reactants initially separated in chambers. They are pumped through a mixing chamber and then out a long tube. At various points along this tube, measurements can be taken - generally by spectrophotometers at various points along. Since the flow rate of the reactants is constant, each point on the tube corresponds to a discrete point in time. Reactants which have absorbance values which change during the course of the reaction are therefore necessary to perform a traditional continuous flow assay.
Inhibitors can be added and ratios of reactants can be changed at will to gather data. By taking readings at different points along the output tube, discrete time points can be measured with a high degree of accuracy.
Advantages and Disadvantages
The continuous flow assay offers a high degree of accuracy by allowing what amounts to a large number of readings at each time point, since the reactant mixture at one point is measured multiple times as it passes the sample point. It is very useful for measuring inhibition because reliable data can be obtained and the inhibitor then added upstream while continuously measuring.
Continuous flow assays require a large amount of reagent, since it needs to flow continuously. If the experiment uses expensive or difficult to produce reactants, it is therefore not the best technique.
Problems
1. If your apparatus has a 3 meter long tube for measurement with an internal diameter of 1 centimeter and fluid flows through it 100ml/1min, assuming there is no lag time in the mixing chamber, what time point of the reaction will a measurement taken 1m along the tube represent?
2. If the conditions are the same as the above scenario, assuming that you use two reagents and the reaction volume is 1 part of A:2 parts B, how much of reactant A will you need to run the assay for 30 minutes continuously flowing?
3. Describe the technique (including equations) that you would use to turn a transmittance value from a point on the reaction tube into meaningful data about the progress of the reaction.
4. If your intial conditions are 5m substrate with no product present and you are running an enzyme whose speed is limited only by diffusion, where would you expect to find 50% reaction completion using the apparatus described in problem 1?
5. If the overall flow of reactants from the storage tanks to the mixing chamber is 5 ml/ml MAX for both, the flow through the tube is 1ml/min MAX, the volume of the mixing chamber is 15 ml, and the tube has an internal diameter of 0.25 cm and a length of 10 meters, using two reactants, how long from starting the reaction will it take for the first products to reach the end of the tube?
Resources
1. Atkins, Peter and de Paula, Julio. Physical Chemistry for the Life Sciences. New York, N.Y.: W. H. Freeman Company, 2006. (241).
2. Segel, Irwin. Enzyme Kinetics. Wiley-Interscience Publishing, 1993 (80-89)
3. Kasidas, GP, and GA Rose. "Continuous-flow Assay for Urinary Oxalate Using Immobilised Oxalate Oxidase."Annals of Clinical Biochemistry. 22 (1985): 412-9. Print.
4. Yagminas, AP, and DC Villeneuve. "An Automated Continuous-Flow Assay for Serum Sorbitol Dehydrogenase Activity and Its Use in Experimental Liver Damage." Biochemical Medicine. 18.1 (1977): 117-25. Print.
Contributors
• Tammy Nguyen, Matthew Austin
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The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically. What follows is general guidance and examples of measuring the rates of a reaction.
Introduction
Measuring time change is easy; a stopwatch or any other time device is sufficient. However, determining the change in concentration of the reactants or products involves more complicated processes. The change of concentration in a system can generally be acquired in two ways:
1. By monitoring the depletion of reactant over time, or
2. By monitoring the formation of product over time
Methods for measuring concentration
• For gases, experimenters use a buret to measure the change in volume produced at different times. They then relate these volumes to changes in concentration.
• Chemists can also remove small samples of a reaction mixture at various times and analyze the concentration using titration.
• Additional methods include the use of a spectrophotometer to determine the concentration using beer's law. Or, more advanced techniques for very fast reactions use computers connected to advanced laser technology such as laser magnetic resonance (LMR).
For supplemental information relating to measuring reaction rates, such as the concentration of reactants, the role of catalysts, the characteristics of the rate of a chemical reaction.
Measuring Reagents Versus Product
It does not matter whether an experimenter monitors the reagents or products because there is no effect on the overall reaction. However, since reagents decrease during reaction, and products increase, there is a sign difference between the two rates. Reagent concentration decreases as the reaction proceeds, giving a negative number for the change in concentration. The products, on the other hand, increase concentration with time, giving a positive number. Since the convention is to express the rate of reaction as a positive number, to solve a problem, set the overall rate of the reaction equal to the negative of a reagent's disappearing rate. The overall rate also depends on stoichiometric coefficients.
It is worth noting that the process of measuring the concentration can be greatly simplified by taking advantage of the different physical or chemical properties (ie: phase difference, reduction potential, etc.) of the reagents or products involved in the reaction by using the above methods. We have emphasized the importance of taking the sign of the reaction into account to get a positive reaction rate. Now, we will turn our attention to the importance of stoichiometric coefficients.
Unique Average Rate of Reaction
A reaction rate can be reported quite differently depending on which product or reagent selected to be monitored.
Given a reaction:
$aA+bB \rightarrow cC + dD \nonumber$
rate of reaction = $- \dfrac{1}{a}\dfrac{ \Delta [A]}{ \Delta t} = - \dfrac{1}{b} \dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{c}\dfrac{ \Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{ \Delta [D]}{\Delta t}$
This formula can also be written as:
rate of reaction = $- \dfrac{1}{a}$ (rate of disappearance of A)
= $- \dfrac{1}{b}$ (rate of disappearance of B)
= $\dfrac{1}{c}$ (rate of formation of C)
= $\dfrac{1}{d}$ (rate of formation of D)
Even though the concentrations of A, B, C and D may all change at different rates, there is only one average rate of reaction. To get this unique rate, choose any one rate and divide it by the stoichiometric coefficient. When the reaction has the formula:
$C_{R1}R_1 + \dots + C_{Rn}R_n \rightarrow C_{P1}P_1 + \dots + C_{Pn}P_n \nonumber$
The general case of the unique average rate of reaction has the form:
rate of reaction = $- \dfrac{1}{C_{R1}}\dfrac{\Delta [R_1]}{\Delta t} = \dots = - \dfrac{1}{C_{Rn}}\dfrac{\Delta [R_n]}{\Delta t} = \dfrac{1}{C_{P1}}\dfrac{\Delta [P_1]}{\Delta t} = \dots = \dfrac{1}{C_{Pn}}\dfrac{\Delta [P_n]}{\Delta t}$
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To determine the value of the exponent in a rate equation term, we need to see how the rate varies with the concentration of the substance. For a single-reactant decomposition reaction of the form
A → products
in which the rate is –d[A]/dt, we simply plot [A] as function of time, draw tangents at various intervals, and see how the slopes of these tangents (the instantaneous rates) depend on [A].
• If doubling the concentration of A doubles the rate, then the reaction is first-order in A.
• If doubling the concentration results in a four-fold rate increase, the reaction is second-order in A.
Example 1
Use the tabulated experimental data to determine the order of the reaction
$2 N_2O_5 \rightarrow 4 NO_2 + O_2 \nonumber$
Time (min) p(N2O5) [N2O5] mol L-1
Rate ( mol L-1 min-1)
0 301.6 0.0152
10 224.8 0.0113 3.4 × 10–4
20 166.7 0.0084 2.5
30 123.2 0.0062 1.8
40 92.2 0.0046 1.3
69.1 69.1 0.0035 1.0
Solution
The ideal gas law can be used to convert the partial pressures of $N_2O_5$ to molar concentrations. These are then plotted (left) to follow their decrease with time. The rates are computed from the slopes of the tangents (blue lines) and their values plotted as a function of $[N_2O_5]$ and $[N_2O_5]^2$. It is apparent that the rates are directly proportional to $[N_2O_5]^1$, indicating that this is a first-order reaction.
Initial Rate Method
When there is more than one reactant, the method described above is rarely practical, since the concentrations of the different reactants will generally fall at different rates, depending on the stoichiometry. Instead, we measure only the rate near the beginning of the reaction, before the concentrations have had time to change significantly. The experiment is then repeated with a different starting concentration of the reactant in question, but keeping the concentrations of any others the same. After the order with respect to one component is found, another series of trials is conducted in which the order of another component is found.
The slope of this plot gives the value of the rate constant (Figure 1; right).
rate = (5.2 × 10–3) [N2O5] mol L–1 s–1
Example 2
A study of the gas-phase reduction of nitric oxide by hydrogen
$2 NO + 2 H_2 \rightarrow N_2 + 2 H_2O \nonumber$
yielded the following initial-rate data (all pressures in torr):
experiment P(NO) P(H2)
initial rate (torr s–1)
1
359 300 1.50
2
300 300 1.03
3
152 300 0.25
4
300 289 1.00
5
300 205 0.71
6
200 147 0.51
Find the order of the reaction with respect to each component.
Solution
In looking over this data, take note of the following:
• The six runs recorded here fall into two groups, in which the initial pressures of H2 and of NO, respectively, are held constant.
• All the data are expressed in pressures, rather than in concentrations. We can do this because the reactants are gases, whose concentrations are directly proportional to their partial pressures when T and V are held constant. And since we are only interested in comparing the ratios of pressures and rates, the units cancel out and don't matter. It is far easier experimentally to adjust and measure pressures than concentrations.
Experiments 2 and 3: Reduction of the initial partial pressure of NO by a factor of about 2 (300/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.
Experiments 4 and 6: Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.
The rate law is thus
$\text{rate} = k[NO]^2[H_2]. \nonumber$
Dealing with multiple reactants: the isolation method
It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is
$\ce{A + B + C → products} \nonumber$
and we need to find the order with respect to [B] in the rate law. If we set [B]o to 0.020 M and let [A]o = [C]o = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By "flooding" the reaction mixture with one or more reactants, we are effectively isolating the one in which we are interested.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.01%3A_Experimental_Determination_of_Kinetics/2.1.03%3A_Rate_vs._Concentration_Propo.txt
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Many chemical reactions are complete in less than a few seconds, which makes the rate of reaction difficult to determine. In these cases, the relaxation methods can be used to determine the rate of the reaction.
Introduction
The term relaxation is used to describe a reaction's return to equilibrium. An equilibrium system is subjected to an external perturbation, such a temperature change. When the change is applied suddenly, the lagging time it takes the system to reach the new equilibrium position is related to the $k_f$ and $k_r$ constants and is called relaxation time, $\tau$. To determine the The relaxation method can be utilized by reactions with half-lives of 1 s to 1010 s. There are three different techniques (each are different processes that suddenly disturbs a reaction), that are used to witness the relaxation time. They are:
It can be found that:
$T =\dfrac{1}{k_f([A]_{eq}+[B]_{eq}) + k_r} \label{1}$
where $[A]_{eq}$ and $[B]_{eq}$ are the values at the new equilibrium at the final temperature. These value can be measured in separate experiments. Thus $K$, the equilibrium constant, can also be calculated from
$K =\dfrac{[C]_{eq}}{[A]_{eq}[B]_{eq}} \label{2}$
but
$K = \dfrac{k_f }{k_r} = \dfrac{[C]_{eq}}{[A]_{eq}[B]_{eq}} \label{3}$
Hence, we can find $k_f$ and $k_r$.
The Relaxation time, $\tau$ is a function of the rate constants $k_f$ and $k_r$ of a chemical system at equilibrium with equilibrium concentrations [A]eq and [B]eq
Temperature-jump
$[A]_{eq}$ and $[B]_{eq}$ are the new equilibrium concentrations which resulted after the system was subjected to a temperature-jump while it was under rest at same other equilibrium state.
$T =\dfrac{1}{k_f([A]_{eq}+[B]_{eq}) + k_r} \label{4}$
$A + B \rightleftharpoons C + Q \label{5}$
at new higher temperature.
kf, kr ,[A]eq,[B]eq, [C]eq are the values at the new higher temperature.
The new conditions (higher temperature) can be applied so rapidly that movement of even very responsive systems (very fast moving to new equilibrium) to a new state of equilibrium can be followed. A temperature jump of as much as 10 oC for example can be produced in an aqueous solution by an electric discharge that lasts no more than 1 μs. If the properties of the system can be recorded following such a change, the rates of reaction that produce a new state of equilibrium, even within a few microseconds eg. 35 μs, can be measured.
It has been found that the restoration is always the first order for small displacements from equilibrium, giving:
$X_t = X_oe^{(-t/ T)} \label{6}$
where $X$ is a property like electrical conductance or spectroscopic absorption proportional to the extent of the reaction. $X_t$ and $X_0$ are the values of the linear property at time $t = t$ and $t = 0$, respectively, and $T$ is the relaxation time. When $τ = t$,
$X_t = \dfrac{X_o}{e} = \dfrac{X_o}{2.718} \label{7}$
Thus, the relaxation time can be determined, by measuring the time it takes for X0 to decrease to X0/2.718. In other words, it is the time in which the relaxation process carries the system a fraction 1/e toward the equilibrium position. Thus, τ is a time interval like the half-life t1/2.
Because the property X is proportional to the extent of the reaction eg. to the concentration of the reacting species at time t. We can write the same equation with Xt and Xo representing concentration instead conductance or absorbance.
$X_t = X_oe^{(-t/ T )} \label{8}$
Thus relaxation time is the time it takes for the concentration of C in this case to be decreased to 1/e of its initial value.
Derivation the equation
$\tau =\dfrac{1}{k_f([A]_{eq}+[B]_{eq}) + k_r} \label{9}$
$A + B \rightleftharpoons C + Q \label{10}$
and
$H^+ + OH^- \rightleftharpoons H_2O + Q \label{10a}$
Both new equilibrium in kf and kr at higher temperature with $[A]_{eq}$,$[B]_{eq}$,$[C]_{eq}$. Of course, before the temperature-jump experiment, there was the old equilibrium, kf' and kr' different from kf and kr, with $[A]_o$,$[B]_o$,$[C]_o$ at the initial temperature.
After the temperature-jump, system is at the higher temperature and also the rate constants. The rate equation for C will be:
$\dfrac{d[C]}{dt} = k_f[A]_t[B]_t - k_r[C]_t \label{11}$
From the diagram, we have:
$[A]_t = [A]_{eq} -X_t \label{12a}$
$[B]_t = [B]_{eq} -X_t \label{12b}$
$[C]_t = [C]_{eq} +X_t \label{12c}$
Substituted:
$\dfrac{d([C]_{eq} +X_t)}{dt} = \dfrac{dX_t}{dt} = k_f([A]_{eq} -X_t)([B]_{eq} -X_t) - k_r([C]_{eq} +X_t) \label{13}$
$\dfrac{dX_t}{dt} = k_f[A]_{eq}[B]_{eq} - k_f[A]_{eq}X_t - k_f[B]_{eq}X_t + k_fX_t^2 - k_r[C]_{eq} - k_rX_t \label{14}$
$= k_f[A]_{eq}[B]_{eq} - k_f[A]_{eq}X_t - k_r[C]_{eq} + k_fX_t^2 -X_t( k_f[B]_{eq}X_t + k_f[A]_{eq} + k_r) \label{15}$
Since:
$\dfrac{d[C]}{dt} = 0 = k_f[A]_{eq}[B]_{eq} - k_r[C]_{eq} \label{16}$
kfXt2 can be neglected for a small extent of reaction. Hence,
$\dfrac{dX_t}{dt} = -X_t( k_f[B]_{eq}X_t + k_f[A]_{eq} + k_r) = -L (X_t) \label{17}$
$\dfrac{dX_t}{X_t} = -L (dt) \label{18}$
$\int_{X_o}^{X_t} \dfrac{dX_t}{dt} = -L \int_0^t dt \label{19}$
-> $\ln X_t -\ln X_o = -Lt$ ->
$\dfrac{ln{X_t}}{X_o} = -Lt -> e^{-Lt} = \dfrac{X_t}{X_o} \label{20}$
$X_t = X_o(e^{-Lt}) \label{21}$
From equation 1 & 3,
$X_o(e^{-t/T}) = X_o(e^{-Lt}) \rightarrow \dfrac{-t}{T} = -L \rightarrow T = 1/L \nonumber$
We get:
$T =\dfrac{1}{k_f [A]_{eq}+k_f [B]_{eq}+ k_r} \label{22}$
Overall, the rate constants of a reversible reaction can be determined by using relaxation method.
Contributors and Attributions
• Sarah Navarro
2.1.04: Relaxation Methods
In 1967 the Nobel Prize in Chemistry was awarded to Manfred Eigen, Ronald George, George Porter and Wreyford Norrish for their co-discovery of Flash Photolysis in 1949. Flash Photolysis is used extensively to study reactions that happen extremely quickly, even down to the femtosecond depending on the laser that is used. The technique was born out of cameras developed during and shortly after WWII, which were used to take pictures of fast moving planes, rockets, and missiles. Since then the technology of lasers and optics has progressed allowing faster and faster reactions to be studied.
Flash Photolysis is often used to study reactions that are light dependent such as photosynthesis and reactions in the cones on the retina of the our eye, but the meathod can also be applied to other reactions. The light in the form of a laser excites a molecule into a reactive state, usually in the form of a free radical. From there it is possible to measure the reaction spectroscopically, using the exitory flash as a light source to measure absorbance. The laser pulse must be aproximatly half the length of the reaction, and of sufficient energy to induce the reaction to take place. Further the flash must cover the spectrum of frequencies which are being studied because not only is the flash producing intermediates of the reaction that are usually not observed, it is also producing the source for spectroscopic analysis. Intermediates of most reactions are rarely observed, this techniques isolates even low concentrations of otherwise unobservable portions of reactions allowing research into synthetic, biochemical, and photo-sensitive reactions.
Contributors and Attributions
• Aaron Whiteley
2.1.4.02%
When a reaction is sensitive to changes in pressure, the resulting relaxation can be used to determine the rate constant of reaction. If the reaction is then exposed to a sharp increase in pressure, the reaction will then adjust to a new equilibrium which is controlled by the rate constant(s) for the reaction. Thus, by observing the relaxation of the reaction, the rates can be discovered. This can be utilized for biochemical reactions, such as protein-protein interactions, in order to determine their rate constants.
Contributors and Attributions
• Sarah Navarro
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.01%3A_Experimental_Determination_of_Kinetics/2.1.04%3A_Relaxation_Methods/2.1.4.01%.txt
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Spectrophotometry is a method to measure how much a chemical substance absorbs light by measuring the intensity of light as a beam of light passes through sample solution. The basic principle is that each compound absorbs or transmits light over a certain range of wavelength. This measurement can also be used to measure the amount of a known chemical substance. Spectrophotometry is one of the most useful methods of quantitative analysis in various fields such as chemistry, physics, biochemistry, material and chemical engineering and clinical applications.
Introduction
Every chemical compound absorbs, transmits, or reflects light (electromagnetic radiation) over a certain range of wavelength. Spectrophotometry is a measurement of how much a chemical substance absorbs or transmits. Spectrophotometry is widely used for quantitative analysis in various areas (e.g., chemistry, physics, biology, biochemistry, material and chemical engineering, clinical applications, industrial applications, etc). Any application that deals with chemical substances or materials can use this technique. In biochemistry, for example, it is used to determine enzyme-catalyzed reactions. In clinical applications, it is used to examine blood or tissues for clinical diagnosis. There are also several variations of the spectrophotometry such as atomic absorption spectrophotometry and atomic emission spectrophotometry.
A spectrophotometer is an instrument that measures the amount of photons (the intensity of light) absorbed after it passes through sample solution. With the spectrophotometer, the amount of a known chemical substance (concentrations) can also be determined by measuring the intensity of light detected. Depending on the range of wavelength of light source, it can be classified into two different types:
• UV-visible spectrophotometer: uses light over the ultraviolet range (185 - 400 nm) and visible range (400 - 700 nm) of electromagnetic radiation spectrum.
• IR spectrophotometer: uses light over the infrared range (700 - 15000 nm) of electromagnetic radiation spectrum.
In visible spectrophotometry, the absorption or the transmission of a certain substance can be determined by the observed color. For instance, a solution sample that absorbs light over all visible ranges (i.e., transmits none of visible wavelengths) appears black in theory. On the other hand, if all visible wavelengths are transmitted (i.e., absorbs nothing), the solution sample appears white. If a solution sample absorbs red light (~700 nm), it appears green because green is the complementary color of red. Visible spectrophotometers, in practice, use a prism to narrow down a certain range of wavelength (to filter out other wavelengths) so that the particular beam of light is passed through a solution sample.
Devices and mechanism
Figure 1 illustrates the basic structure of spectrophotometers. It consists of a light source, a collimator, a monochromator, a wavelength selector, a cuvette for sample solution, a photoelectric detector, and a digital display or a meter. Detailed mechanism is described below. Figure 2 shows a sample spectrophotometer (Model: Spectronic 20D).
A spectrophotometer, in general, consists of two devices; a spectrometer and a photometer. A spectrometer is a device that produces, typically disperses and measures light. A photometer indicates the photoelectric detector that measures the intensity of light.
• Spectrometer: It produces a desired range of wavelength of light. First a collimator (lens) transmits a straight beam of light (photons) that passes through a monochromator (prism) to split it into several component wavelengths (spectrum). Then a wavelength selector (slit) transmits only the desired wavelengths, as shown in Figure 1.
• Photometer: After the desired range of wavelength of light passes through the solution of a sample in cuvette, the photometer detects the amount of photons that is absorbed and then sends a signal to a galvanometer or a digital display, as illustrated in Figure 1.
You need a spectrometer to produce a variety of wavelengths because different compounds absorb best at different wavelengths. For example, p-nitrophenol (acid form) has the maximum absorbance at approximately 320 nm and p-nitrophenolate (basic form) absorb best at 400nm, as shown in Figure 3.
Looking at the graph that measures absorbance and wavelength, an isosbestic point can also be observed. An isosbestic point is the wavelength in which the absorbance of two or more species are the same. The appearance of an isosbestic point in a reaction demonstrates that an intermediate is NOT required to form a product from a reactant. Figure 4 shows an example of an isosbestic point.
Referring back to Figure 1 (and Figure 5), the amount of photons that goes through the cuvette and into the detector is dependent on the length of the cuvette and the concentration of the sample. Once you know the intensity of light after it passes through the cuvette, you can relate it to transmittance (T). Transmittance is the fraction of light that passes through the sample. This can be calculated using the equation:
$Transmittance (T) = \dfrac{I_t}{I_o}$
Where It is the light intensity after the beam of light passes through the cuvette and Io is the light intensity before the beam of light passes through the cuvette. Transmittance is related to absorption by the expression:
$Absorbance (A) = - log(T) = - log(\dfrac{I_t}{I_o})$
Where absorbance stands for the amount of photons that is absorbed. With the amount of absorbance known from the above equation, you can determine the unknown concentration of the sample by using Beer-Lambert Law. Figure 5 illustrates transmittance of light through a sample. The length $l$ is used for Beer-Lambert Law described below.
Beer-Lambert Law
Beer-Lambert Law (also known as Beer's Law) states that there is a linear relationship between the absorbance and the concentration of a sample. For this reason, Beer's Law can only be applied when there is a linear relationship. Beer's Law is written as:
$A = \epsilon{lc}$
where
• $A$ is the measure of absorbance (no units),
• $\epsilon$ is the molar extinction coefficient or molar absorptivity (or absorption coefficient),
• $l$ is the path length, and
• $c$ is the concentration.
The molar extinction coefficient is given as a constant and varies for each molecule. Since absorbance does not carry any units, the units for $\epsilon$ must cancel out the units of length and concentration. As a result, $\epsilon$ has the units: L·mol-1·cm-1. The path length is measured in centimeters. Because a standard spectrometer uses a cuvette that is 1 cm in width, $l$ is always assumed to equal 1 cm. Since absorption, $\epsilon$, and path length are known, we can calculate the concentration $c$ of the sample.
Example 1
Guanosine has a maximum absorbance of 275 nm. $\epsilon_{275} = 8400 M^{-1} cm^{-1}$ and the path length is 1 cm. Using a spectrophotometer, you find the that $A_{275}= 0.70$. What is the concentration of guanosine?
Solution
To solve this problem, you must use Beer's Law.
$A = \epsilon lc \nonumber$
0.70 = (8400 M-1 cm-1)(1 cm)($c$)
Next, divide both side by [(8400 M-1 cm-1)(1 cm)]
$c$ = 8.33x10-5 mol/L
Example 2
There is a substance in a solution (4 g/liter). The length of cuvette is 2 cm and only 50% of the certain light beam is transmitted. What is the absorption coefficient?
Solution
Using Beer-Lambert Law, we can compute the absorption coefficient. Thus,
$- \log \left(\dfrac{I_t}{I_o} \right) = - \log(\dfrac{0.5}{1.0}) = A = {8} \epsilon$
Then we obtain that
$\epsilon$ = 0.0376
Example 3
In example 2 above, how much is the beam of light is transmitted when 8 g/liter ?
Solution
Since we know $\epsilon$, we can calculate the transmission using Beer-Lambert Law. Thus,
$\log(1) - \log(I_t) = 0 - \log(I_t)$ = 0.0376 x 8 x 2 = 0.6016
$\log(I_t)$ = -0.6016
Therefore, $I_t$ = 0.2503 = 25%
Example 4
In example 2 above, what is the molar absorption coefficient if the molecular weight is 100?
Solution
It can simply obtained by multiplying the absorption coefficient by the molecular weight. Thus,
$\epsilon$ = 0.0376 x 100 = 3.76 L·mol-1·cm-1
Example 5
The absorption coefficient of a glycogen-iodine complex is 0.20 at light of 450 nm. What is the concentration when the transmission is 40 % in a cuvette of 2 cm?
Solution
It can also be solved using Beer-Lambert Law. Therefore,
$- \log(I_t) = - \log(0.4) = 0.20 \times c \times 2 \nonumber$
Then $c$ = 0.9948
Contributors and Attributions
• Kevin Vo (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.01%3A_Experimental_Determination_of_Kinetics/2.1.05%3A_Spectrophotometry.txt
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The stopped-flow technique allows for the evaluation of solution-based kinetics on a milliseconds timescale with a very small volume of reactants used.
Introduction
Imagine you wanted to compute the initial rate of a reaction, one that, when the reactants combined changed color or fluoresced, however the reaction was far too fast to detect via the human eye alone. How would you go about determining the reaction rate? Could you use the continuous flow technique? But what if you only had a limited volume of the reactants at your disposal? The continuous flow technique requires that you have enough of the reactants to constantly flow through the spectrophotometer at a continuous rate. In cases like this, the Stopped Flow technique is probably more appropriate. Stopped-Flow allows for the rate of a solution-based reaction to be determined in milliseconds, and with a very small volume of reactants.
Mechanism
This technique involves two reactants held in separate reservoirs that are prevented from freely flowing by syringe pumps. The reaction is initiated by depressing the reactant syringes, and thus releasing the reactants into the connecting "mixing chamber" where the solutions are mixed. The reaction is monitored by observing the change in absorbance of the reaction solution as a function of time. As the reaction progresses it fills the “stop syringe” which then expands until it hits a block at the point when the reaction has reached a continuous flow rate, thereby stopping the flow and the reaction, and thus allowing the researcher to calculate the exact initial rate of reaction. See Figure 1 below:
The Stopped Flow technique works because within milliseconds of combining the two reactants the absorbance can be read. In addition, the stop syringe assures for a steady rate of flow pas the spectophotometer so that reactants are being added to solution and forming products at a consistant rate.
Example $1$:
Imagine a reaction that proceeds as follows:
$A + B \rightarrow C + D \nonumber$
Substance C is the product and Substance D is known to fluoresce. The optical path is 1 cm and the e value is 1 L/mol cm-1. Based on the following data gathered by the Stopped Flow method, determine the rate constant and the order of the reaction.
Time (seconds)
Absorbance
.001
.2
.002
.4
.003
.6
.004
.8
.005
1
.006
1.2
.007
1.4
.008
1.6
.009
1.8
(Hint: Make a graph and use that to find the order and the rate. If you need more help, these pages might help: Beer's Law and Reaction Rate)
Solution
This is a zero order reaction with a rate constant of 200 M/S.
We know that the absorbance is proportional to the change of the concentration of D over time because D fluoresces. However, from the reaction above, the concentration of C=D because the stoichiometry is 1:1.
Now lets start with the relationship between absorbance and concentration. From Beer's Law we know that absorbance is directly proportional to concentration, therefore we know, in this case, absorbance is equal to concentration because we divide by 1 and 1 (A = ε l c).
Now for the reaction rate and the reaction order. You can figure the order out by graphing the either concentration by time, ln(concentration) by time, or 1/concentration by time depending on the order of the reaction. If concentration by time is linear, then the reaction is zero order. If ln(concentration) by time is linear, then the reaction is first order. Finally, if 1/concentration by time is linear, then the reaction is 2nd order. Once you have determined that this reaction is zero order, you can find the slope of the line to solve for $k$, and the rate of the reaction.
Resources
1. Drobny, Engel, Philip Reid. Physical Chemistry for the Life Sciences. Pearson Education Inc., 2008. (344-345, 356-357).
2. IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8. doi:10.1351/goldbook.
3. Lehrer, Sam. "Stopped Flow Kinnetics." Boston Biomedical Research Institute. BBRI, n.d. Web. 14 Mar 2011. http://www.bbri.org/index.php/stopped_flow_kin.html.
Contributors and Attributions
• Tammy Nguyen, Rachel Pruitt
2.02: Factors That Affect Reaction Rates
Factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, the solvent, and the presence of a catalyst.
• Anonymous
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.01%3A_Experimental_Determination_of_Kinetics/2.1.06%3A_Stopped_Flow.txt
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A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.
The Differential Representation
Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. The differential equation describing first-order kinetics is given below:
$Rate = - \dfrac{d[A]}{dt} = k[A]^1 = k[A] \label{1}$
The "rate" is the reaction rate (in units of molar/time) and $k$ is the reaction rate coefficient (in units of 1/time). However, the units of $k$ vary for non-first-order reactions. These differential equations are separable, which simplifies the solutions as demonstrated below.
The Integral Representation
First, write the differential form of the rate law.
$Rate = - \dfrac{d[A]}{dt} = k[A] \nonumber$
Rearrange to give:
$\dfrac{d[A]}{[A]} = - k\,dt \nonumber$
Second, integrate both sides of the equation.
\begin{align*} \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]} &= -\int_{t_o}^{t} k\, dt \label{4a} \[4pt] \int_{[A]_{o}}^{[A]} \dfrac{1}{[A]} d[A] &= -\int_{t_o}^{t} k\, dt \label{4b} \end{align*}
Recall from calculus that:
$\int \dfrac{1}{x} = \ln(x) \nonumber$
Upon integration,
$\ln[A] - \ln[A]_o = -kt \nonumber$
Rearrange to solve for [A] to obtain one form of the rate law:
$\ln[A] = \ln[A]_o - kt \nonumber$
This can be rearranged to:
$\ln [A] = -kt + \ln [A]_o \nonumber$
This can further be arranged into y=mx +b form:
$\ln [A] = -kt + \ln [A]_o \nonumber$
The equation is a straight line with slope m:
$mx=-kt \nonumber$
and y-intercept b:
$b=\ln [A]_o \nonumber$
Now, recall from the laws of logarithms that
$\ln {\left(\dfrac{[A]_t}{ [A]_o}\right)}= -kt \nonumber$
where [A] is the concentration at time $t$ and $[A]_o$ is the concentration at time 0, and $k$ is the first-order rate constant.
Because the logarithms of numbers do not have any units, the product $-kt$ also lacks units. This concludes that unit of $k$ in a first order of reaction must be time-1. Examples of time-1 include s-1 or min-1. Thus, the equation of a straight line is applicable:
$\ln [A] = -kt + \ln [A]_o.\label{15}$
To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction.
To create another form of the rate law, raise each side of the previous equation to the exponent, $e$:
$\large e^{\ln[A]} = e^{\ln[A]_o - kt} \label{16}$
Simplifying gives the second form of the rate law:
$[A] = [A]_{o}e^{- kt}\label{17}$
The integrated forms of the rate law can be used to find the population of reactant at any time after the start of the reaction. Plotting $\ln[A]$ with respect to time for a first-order reaction gives a straight line with the slope of the line equal to $-k$. More information can be found in the article on rate laws.
This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law. Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function $y=e^x$ so efficiently describes such changes is that dy/dx = ex; that is, ex is its own derivative, making the rate of change of $y$ identical to its value at any point.
Graphing First-order Reactions
The following graphs represents concentration of reactants versus time for a first-order reaction.
Plotting $\ln[A]$ with respect to time for a first-order reaction gives a straight line with the slope of the line equal to $-k$.
Half-lives of first order reactions
The half-life ($t_{1/2}$) is a timescale on which the initial population is decreased by half of its original value, represented by the following equation.
$[A] = \dfrac{1}{2} [A]_o \nonumber$
After a period of one half-life, $t = t_{1/2}$ and we can write
$\dfrac{[A]_{1/2}}{[A]_o} = \dfrac{1}{2}=e^{-k\,t_{1/2}} \label{18}$
Taking logarithms of both sides (remember that $\ln e^x = x$) yields
$\ln 0.5 = -kt\label{19}$
Solving for the half-life, we obtain the simple relation
$t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{k}\label{20}$
This indicates that the half-life of a first-order reaction is a constant.
Example 1: Estimated Rate Constants
The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant?
Solution
Use Equation 20 that relates half life to rate constant for first order reactions:
$k = \dfrac{0.693}{600 \;s} = 0.00115 \;s^{-1} \nonumber$
As a check, dimensional analysis can be used to confirm that this calculation generates the correct units of inverse time.
Notice that, for first-order reactions, the half-life is independent of the initial concentration of reactant, which is a unique aspect to first-order reactions. The practical implication of this is that it takes as much time for [A] to decrease from 1 M to 0.5 M as it takes for [A] to decrease from 0.1 M to 0.05 M. In addition, the rate constant and the half life of a first-order process are inversely related.
Example $1$: Determining Half life
If 3.0 g of substance $A$ decomposes for 36 minutes the mass of unreacted A remaining is found to be 0.375 g. What is the half life of this reaction if it follows first-order kinetics?
Solution
There are two ways to approach this problem: The "simple inspection approach" and the "brute force approach"
Approach #1: "The simple Inspection Approach"
This approach is used when one can recognize that the final concentration of $A$ is $\frac{1}{8}$ of the initial concentration and hence three half lives $\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)$ have elapsed during this reaction.
$t_{1/2} = \dfrac{36\, \text{min}}{3}= 12 \; \text{min} \nonumber$
This approach works only when the final concentration is $\left(\frac{1}{2}\right)^n$ that of the initial concentration, then $n$ is the number of half lives that have elapsed. If this is not the case, then approach #2 can be used.
Approach #2: "The brute force approach"
This approach involves solving for $k$ from the integral rate law (Equation \ref{17}) and then relating $k$ to the $t_{1/2}$ via Equation \ref{20}. \begin{align*} \dfrac{[A]_t}{[A]_o} &= e^{-k\,t} \[4pt] k &= -\dfrac{\ln \dfrac{[A]_t}{[A]_o}}{t} \[4pt] &= -\dfrac{\ln \dfrac{0.375\, g}{3\, g}}{36\, \text{min}} \[4pt] &= 0.0578 \, \text{min}^{-1} \end{align*} Therefore, via Equation \ref{20} $t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{0.0578 \, \text{min}^{-1}} \approx 12\, \text{min} \nonumber$
The first approach is considerably faster (if the number of half lives evolved is apparent).
Exercise $\PageIndex{2a}$
Calculate the half-life of the reactions below:
1. If 4.00 g A are allowed to decompose for 40 min, the mass of A remaining undecomposed is found to be 0.80 g.
2. If 8.00 g A are allowed to decompose for 34 min, the mass of A remaining undecomposed is found to be 0.70 g.
3. If 9.00 g A are allowed to decompose for 24 min, the mass of A remaining undecomposed is found to be 0.50 g.
Answer
Use the half life reaction that contains initial concentration and final concentration. Plug in the appropriate variables and solve to obtain:
1. 17.2 min
2. 9.67 min
3. 5.75 min
Exercise $\PageIndex{2b}$
Determine the percent $\ce{H2O2}$ that decomposes in the time using $k=6.40 \times 10^{-5} s^{-1}$
1. The time for the concentration to decompose is 600.0 s after the reaction begins.
2. The time for the concentration to decompose is 450 s after the reaction begins.
Answer
1. Rearranging Eq. 17 to solve for the $[H_2O_2]_t/[H_2O_2]_0$ ratio $\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt} \nonumber$ This is a simple plug and play application once you have identified this equation. $\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (600 \, s)} \nonumber$ $\dfrac{[H_2O_2]_{t=600\;s}}{[H_2O_2]_0} = 0.9629 \nonumber$ So 100-96.3=3.71% of the hydrogen peroxide has decayed by 600 s.
2. Rearranging Eq. 17 to solve for the $[H_2O_2]_t/[H_2O_2]_0$ ratio $\dfrac{[H_2O_2]_t}{[H_2O_2]_0} = e^{-kt} \nonumber$ This is a simple plug and play application once you have identified this equation. $\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = e^{-(6.40 \times 10^{-5} s^{-1}) (450 \, s)} \nonumber$ $\dfrac{[H_2O_2]_{t=450\;s}}{[H_2O_2]_0} = 0.9720 \nonumber$ So 100-97.2=2.8% of the hydrogen peroxide has decayed by 450 s.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.03%3A_First-Order_Reactions.txt
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The half-life of a reaction ($t_{1/2}$), is the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. Its application is used in chemistry and medicine to predict the concentration of a substance over time. The concepts of half life plays a key role in the administration of drugs into the target, especially in the elimination phase, where half life is used to determine how quickly a drug decrease in the target after it has been absorbed in units of time (e.g., s, min., day, etc.) or elimination rate constant $k$ in units of 1/time (e.g., min-1, hr-1, day-1, etc.). It is important to note that the half-life is varied between different type of reactions. The following section will go over different type of reaction, as well as how its half-life reaction are derived. The last section will talk about the application of half-life in the elimination phase of pharmacokinetics.
Zero-Order Kinetics
In zero-order kinetics, the rate of a reaction does not depend on the substrate concentration. In other words, saturating the amount of substrate does not speed up the rate of the reaction. Below is a graph of time ($t$) vs. concentration ($[A]$) in a zero order reaction, several observation can be made: the slope of this plot is a straight line with negative slope equal negative $k$, the half-life of zero order reaction decreases as the concentration decreases.
We learn that the zero-order kinetic rate law is as followed, where $[A]$ is the current concentration, $[A]_o$ is the initial concentration, and $k$ is the reaction constant and $t$ is time:
$[A]= [A]_o - kt \label{1}$
We need to isolate $t_{1/2}$ when
$[A]=\dfrac{[A]_o}{2} \nonumber$
Substituting into Equation \ref{1}
\begin{align} \dfrac{[A]_o}{2} &= [A]_o - kt_{1/2} \nonumber \[4pt] kt_{1/2} &= [A]_o - \dfrac{[A]_o}{2} \nonumber \[4pt] t_{1/2} &= \dfrac{[A]_o}{2k} \label{2} \end{align}
Equation \ref{2} show the half-life for a zero-order reaction depends on both the initial concentration and rate constant.
First-Order Kinetics
In First order reactions, the graph represents the half-life is different from zero order reaction in a way that the slope continually decreases as time progresses until it reaches zero. We can also easily see that the length of half-life will be constant, independent of concentration. For example, it takes the same amount of time for the concentration to decrease from one point to another point.
In order to solve the half life of first order reactions, we recall that the rate law of a first order reaction was:
$[A]=[A]_o e^{-kt} \label{4}$
We need to isolate $t_{1/2}$ when
$[A]=\dfrac{[A]_o}{2} \nonumber$
Substituting into Equation \ref{4}
\begin{align} \dfrac{[A]_0}{2} &=[A]_o e^{-kt_{1/2}} \nonumber \[4pt] \dfrac{1}{2} &= e^{-kt_{1/2}} \nonumber \[4pt] \ln \dfrac{1}{2} &= -kt_{1/2} \nonumber \[4pt] t_{1/2} &= \dfrac{\ln 2}{k} \nonumber \[4pt] &\approx \dfrac{0.693}{k} \label{5} \end{align}
Equation \ref{5} shows that for first-order reactions, the half-life depends solely on the reaction rate constant, $k$. We can visually see this on the graph for first order reactions when we note that the amount of time between one half life and the next are the same. Another way to see it is that the half life of a first order reaction is independent of its initial concentration.
Second-Order Kinetics
Half-life of second order reactions shows concentration $[A]$ vs. time ($t$), which is similar to first order plots in that their slopes decrease to zero with time. However, second order reactions decrease at a much faster rate as the graph shows. We can also note that the length of half-life increase while the concentration of substrate constantly decreases, unlike zero and first order reaction.
In order to solve for half life of second order reactions we need to remember that the rate law of a second order reaction is:
$\dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0} \label{6}$
As in zero-order and first-order reactions, we need to isolate $t_{1/2}$ when
$[A]=\dfrac{[A]_o}{2} \nonumber$
Substituting into Equation \ref{6}
\begin{align} \dfrac{2}{[A]_0} &= kt_{1/2} + \dfrac{1}{[A]_0}\nonumber \[4pt] -kt_{1/2} &= \dfrac{1}{[A]_0} - \dfrac{2}{[A]_0} \nonumber \[4pt] t_{1/2} &= \dfrac{1}{k[A]_0} \label{7} \end{align}
Equation \ref{7} shows that for second-order reactions, the half-life depends on both the initial concentration and the rate constant.
Example $1$: Pharmacokinetics
A following example is given below to illustrate the role of half life in pharmacokinetics to determine the drugs dosage interval.
The therapeutic range of drug A is 15-30 mg/L. Its half life in the target in 5 hours. Once the drug is metabolized in the target, its concentration will decrease over time. To ensure its maximal effect of the drug in the target, the administration will be monitored so that the minimum serum concentration will never go lower than 15 mg/L and the maximum serum concentration will never exceed 30 mg/L. As a result, it is important to administer drug A to the target every 5 hours to ensure its effective therapeutic range.
Another important application of half life in pharmacokinetics is that half-life tells how tightly drugs bind to each ligands before it is undergoing decay ($k_s$). The smaller the value of $k_s$, the higher the affinity binding of drug to its target ligand, which is an important aspect of drug design
Exercise $1$
Examine the following graph and answer
1. What is the therapeutic range of drug B?
2. From the graph, estimate the dosage interval of drug B to ensure its maximum effect?
3. The patient forgot to take the drug at the end of the dosage interval, he decided to take double the amount of drug B at the end of the next dosage interval. Will the drug still be in its therapeutic range?
Answer
Looking at the graph, we can see the therapeutic range is the amplitude of the graph, which is 5-15 mg/L
The dosage interval is the half-life of the drug, looking at the graph, the half-life is 10 hours.
Even though it will get in the therapeutic range, such practice is not recommended.
Exercise $2$
A patient is treating with $\ce{^{32}P}$. How long does it takes for the radioactivity to decay by 90%? The half-life of the material is 15 days.
Answer
If we want the product to decay by 90%, that means 10% is left non-decayed, so
$\dfrac{[A]_t}{[A]_o} = 0.1 \nonumber$
From ln([A]t/[A]o) = -kt, plug in value of k and [A]t/[A]o we then have t = 50 days
Exercise $3$
In first order half life, what is the best way to determine the rate constant $k$? Why?
Answer
The best way to determine rate constant $k$ in half-life of first order is to determine half-life by experimental data. The reason is half-life in first order order doesn't depend on initial concentration.
Exercise $4$
In a first order reaction, $\ce{A -> B}$. The half-life is 10 days.
1. Determine its rate constant $k$?
2. How much time required for this reaction to be at least 50% and 60% complete?
Answer
1. This is a direct application of Equation \ref{7}. The rate constant, $k$, will be equal to $k=\dfrac{\ln 2}{t_{1/2}} \nonumber$ so $k = 0.0693 \,day^{-1}$.
2. For the reaction to be 50% complete, that will be exactly the half-life of the reaction at 10 days. For the reaction to be 60% complete, using the similar equation derived from question 4, we have $\dfrac{[A]_t}{[A]_o} = 0.4 \nonumber$ $t = 13.2\, days \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.04%3A_Half-lives.txt
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During the course of the reaction shown below, reactants A and B are consumed while the concentration of product AB increases. The reaction rate can be determined by measuring how fast the concentration of A or B decreases, or by how fast the concentration of AB increases.
$\ A + B \longrightarrow AB \nonumber$
For the stochiometrically complicated Reaction:
$aA + bB \longrightarrow cC + dD \label{1}$
$\text{Rate} = \dfrac{-1}{a}\dfrac{d[A]}{dt} = \dfrac{-1}{b} \dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt} = \dfrac{1}{d}\dfrac{d[D]}{dt} \nonumber$
Looking at Figure $1$ above, we can see that the rate can be measured in terms of either reactant (A or B) or either product (C or D). Not all variables are needed to solve for the rate. Therefore, if you have the value for "A" as well as the value for "a" you can solve for the reaction rate.
You can also notice from Equation \ref{1} that the change in reactants over the change in time must have a negative sign in front of them. The reason for this is because the reactants are decreasing as a function of time, the rate would come out to be negative (because it is the reverse rate). Therefore, putting a negative sign in front of the variable will allow for the solution to be a positive rate.
Chemical reactions vary greatly in the speed at which they occur. Some are ultrafast, while others may take millions of years to reach equilibrium.
Definition of Reaction Rate
The Reaction Rate for a given chemical reaction is the measure of the change in concentration of the reactants or the change in concentration of the products per unit time. The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed:
$\text{rate}=\dfrac{\Delta \text{concentration}}{\Delta \text{time}} \label{2-1}$
For a reaction of the form $A + B \rightarrow C$, the rate can be expressed in terms of the change in concentration of any of its components
$\text{rate}=-\dfrac{\Delta [A]}{\Delta t} \nonumber$
$\text{rate}=-\dfrac{\Delta [B]}{\Delta t} \nonumber$
$\text{rate}=\dfrac{\Delta [C]}{\Delta t} \nonumber$
in which $Δ[A]$ is the difference between the concentration of $A$ over the time interval $t_2 – t_1$:
$\Delta [A] = [A]_2 – [A]_1 \label{2-2}$
Notice the minus signs in the first two examples above. The concentration of a reactant always decreases with time, so $\Delta [A]$ and $\Delta [A]$ are both negative. Since negative rates do not make much sense, rates expressed in terms of a reactant concentration are always preceded by a minus sign to make the rate come out positive.
Consider now a reaction in which the coefficients are different:
$A + 3B \rightarrow 2D \nonumber$
It is clear that $[B]$ decreases three times as rapidly as $[A]$, so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient:
$\text{rate}= -\dfrac{\Delta [A]}{\Delta t} = -\dfrac{\Delta [B]}{3\Delta t} = \dfrac{\Delta [D]}{2\Delta t} \label{2-3}$
Example $1$: Oxidation of Ammonia
For the oxidation of ammonia
$\ce{4 NH3 + 3O2 -> 2 N2 + 6 H2O} \nonumber$
it was found that the rate of formation of N2 was 0.27 mol L–1 s–1.
1. At what rate was water being formed?
2. At what rate was ammonia being consumed?
Solution
a) From the equation stoichiometry, Δ[H2O] = 6/2 Δ[N2], so the rate of formation of H2O is
3 × (0.27 mol L–1 s–1) = 0.81 mol L–1 s–1.
b) 4 moles of NH3 are consumed for every 2 moles of N2 formed, so the rate of disappearance of ammonia is
2 × (0.27 mol L–1 s–1) = 0.54 mol L–1 s–1.
Comment: Because of the way this question is formulated, it would be acceptable to express this last value as a negative number.
Instantaneous rates
Most reactions slow down as the reactants are consumed. Consequently, the rates given by the expressions shown above tend to lose their meaning when measured over longer time intervals Δt. Note: Instantaneous rates are also known as differential rates.
Thus for the reaction whose progress is plotted here, the actual rate (as measured by the increasing concentration of product) varies continuously, being greatest at time zero. The instantaneous rate of a reaction is given by the slope of a tangent to the concentration-vs.-time curve.
An instantaneous rate taken near the beginning of the reaction (t = 0) is known as an initial rate (label (1) here). As we shall soon see, initial rates play an important role in the study of reaction kinetics. If you have studied differential calculus, you will know that these tangent slopes are derivatives whose values can very at each point on the curve, so that these instantaneous rates are really limiting rates defined as
$\text{rate} = \lim_{\Delta t \rightarrow 0} \dfrac{-[A]}{\Delta T} \nonumber$
If you do not know calculus, bear in mind that the larger the time interval Δt, the smaller will be the precision of the instantaneous rate.
Rate Laws and Rate Constants
A rate law is an expression which relates that rate of a reaction to the rate constant and the concentrations of the reactants. A rate constant, $k$, is a proportionality constant for a given reaction. The general rate law is usually expressed as:
$\text{Rate} = k[A]^s[B]^t \label{2}$
As you can see from Equation \ref{2} above, the reaction rate is dependent on the concentration of the reactants as well as the rate constant. However, there are also other factors that can influence the rate of reaction. These factors include temperature and catalysts. When you are able to write a rate law equation for a certain reaction, you can determine the Reaction Order based on the values of s and t.
Reaction Order
The reaction rate for a given reaction is a crucial tool that enables us to calculate the specific order of a reaction. The order of a reaction is important in that it enables us to classify specific chemical reactions easily and efficiently. Knowledge of the reaction order quickly allows us to understand numerous factors within the reaction including the rate law, units of the rate constant, half life, and much more. Reaction order can be calculated from the rate law by adding the exponential values of the reactants in the rate law.
$\text{Rate} = k[A]^s[B]^t \label{4}$
$\text{Reaction Order} = s + t \label{5}$
It is important to note that although the reaction order can be determined from the rate law, there is in general, no relationship between the reaction order and the stoichiometric coefficients in the chemical equation.
NOTE: The rate of reaction must be a non-negative value. It can be zero and does not need to be an integer.
As shown in Equation \ref{5}, the complete reaction order is equal to the sum of "s" and "t." But what does each of these variables mean? Each variable represents the order of the reaction with respect to the reactant it is placed on. In this certain situation, s is the order of the reaction with respect to [A] and t is the order of the reaction with respect to [B].
Here is an example of how you can look at this: If a reaction order with respect to [A] was 2 (s = 2) and [B] was 1 (t = 1), then that basically means that the concentration of reactant A is decreasing by a factor of 2 and the concentration of [B] is decrease by a factor of 1.
So if you have a reaction order of Zero (i.e., $s + t = 0$), this basically means that the concentration of the reactants does not affect the rate of reaction. You could remove or add reactants to the mixture but the rate will not change.
A list of the different reaction rate equations for zero-, first-, and second-order reactions can be seen in Table $1$. This table also includes further equations that can be determine by this equation once the order of the reaction is known (Half life, integrated rate law, etc.)
Table $1$: The table below displays numerous values and equations utilized when observing chemical kinetics for numerous reactions types
Zero-Order First-Order Second-Order
Rate Law
$\ {Rate} = \ {k}$
$\ {Rate} = \ {k[A]}$
$\text{Rate} = \ {k[A]^2}$
Integrated Rate Law
$\ {[A]_t} = \ {-kt + [A]_0}$
$\ {ln[A]_t} = \ {-kt + ln[A]_0}$
$\dfrac{1}{[A]_t} = +kt + \dfrac{1}{[A]_0}$
Units of Rate Constant (k):
$\ {mol L^{-1} s^{-1}}$
$s^{-1}$
$\ {L mol^{-1} s^{-1}}$
Linear Plot to Determine (k):
$[A]$ versus time
$\ln [A]$ versus time
$\dfrac{1}{[A]}$ versus time
Relationship of Rate Constant to the Slope of Straight Line:
$\ {slope} = \ {-k}$
$\ {slope} = \ {-k}$
$\ {slope} = \ {k}$
Half-life:
$\dfrac{[A]_0}{2k}$
$\dfrac{\ln2}{k}$
$\dfrac{1}{k[A]_0}$
Sample Problems
1. Define Reaction Rate
2. TRUE or FALSE: Changes in the temperature or the introduction of a catalyst will affect the rate constant of a reaction
For sample problems 3-6, use Formula 6 to answer the questions
$H_2O \longrightarrow 2H_2+ O_2 \label{6}$
*Assume the reaction occurs at constant temperature
3. For the given reaction above, state the rate law.
4. State the overall order of the reaction.
5. Find the rate, given k = 1.14 x 10-2 and [H2O] = 2.04M
6. Find the half-life of the reaction.
Answers
1. Reaction Rate is the measure of the change in concentration of the disappearance of reactants or the change in concentration of the appearance of products per unit time.
2. FALSE. The rate constant is not dependant on the presence of a catalyst. Catalysts, however, can effect the total rate of a reaction.
3. $\ {Rate} = \ {k[H_2O]}$
4. First - Order
5. 2.33 x 10-2 s-1
6. 60.8 s ((t1/2 = ln 2 / k = ln 2 / 1.14 x 10-2 = 60.8 s).
2.05: Reaction Rate
How long does it take for a chemical reaction to occur under a given set of conditions? As with many "simple" questions, no meaningful answer can be given without being more precise. In this case,
How do we define the point at which the reaction is "completed"?
A reaction is "completed" when it has reached equilibrium — that is, when concentrations of the reactants and products are no longer changing.
If the equilibrium constant is quite large, then the answer reduces to a simpler form: the reaction is completed when the concentration of a reactant falls to zero. In the interest of simplicity, we will assume that this is the case in the remainder of this discussion.
"How long?" may be too long
If the reaction takes place very slowly, the time it takes for every last reactant molecule to disappear may be too long for the answer to be practical. In this case, it might make more sense to define "completed" when a reactant concentration has fallen to some arbitrary fraction of its initial value — 90%, 70%, or even only 20%.
The particular fraction one selects depends on the cost of the reactants in relation to the value of the products, balanced against the cost of operating the process for a longer time or the inconvenience of waiting for more product. This kind of consideration is especially important in industrial processes in which the balances of these costs affect the profitability of the operation.
The half-life of a reaction
Instead of trying to identify the time required for the reaction to become completed, it is far more practical to specify the the time required for the concentration of a reactant to fall to half of its initial value. This is known as the half-life (or half-time) of the reaction.
Contributor
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
2.5.02: The Rate of a Chemical Reaction
The rate of a chemical reaction is the change in concentration over the change in time.
Introduction
The rate of a chemical reaction is the change in concentration over the change in time and is a metric of the "speed" at which a chemical reactions occurs and can be defined in terms of two observables:
1. The Rate of Disappearance of Reactants $-\dfrac{\Delta[Reactants]}{\Delta{t}} \nonumber$ Note this is negative because it measures the rate of disappearance of the reactants.
2. The Rate of Formation of Products $\dfrac{\Delta{[Products]}}{\Delta{t}} \nonumber$ This is the rate at which the products are formed.
They both are linked via the balanced chemical reactions and can both be used to measure the reaction rate.
Problems
1. Consider the reaction $2A + B \longrightarrow C$. The concentration of [A] is 0.54321M and the rate of reaction is $3.45 \times 10^{-6} M/s$. What Concentration will [A] be 3 minutes later?
2. Consider the reaction $A + B \longrightarrow C$. The rate of reaction is 1.23*10-4. [A] will go from a 0.4321 M to a 0.4444 M concentration in what length of time?
3. Write the rate of the chemical reaction with respect to the variables for the given equation. $2A+3B \rightarrow C+2D \nonumber$
4. True or False: The Average Rate and Instantaneous Rate are equal to each other.
5. How is the rate of formation of a product related to the rates of the disappearance of reactants.
Contributors
• Albert Law, Victoria Blanchard, Donald Le
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.05%3A_Reaction_Rate/2.5.01%3A_The_Speed_of_a_Chemical_Reaction.txt
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The Learning Objective of this Module is to determine the individual steps of a simple reaction.
One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism.
In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water:
$2C_8H_{18 \; (l)} + 25O_2(g) \rightarrow 16CO_{2\, (g)} + 18H_2O_{(g)} \tag{14.34}$
For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction, involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction.
Molecularity and the Rate-Determining Step
To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide.
$NO_{2\, (g)} + CO(g) \rightarrow NO_{(g)} + CO_{2\, (g)} \tag{14.35}$
From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO2 with a molecule of CO that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows:
$rate = k[NO_2]^2 \tag{14.36}$
The fact that the reaction is second order in [NO2] and independent of [CO] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be $rate = k[NO_2][CO]$.
The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2:
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\textrm{slow}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{elementary reaction}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\rightarrow\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{elementary reaction}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$ $\textrm{overall reaction}$
According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The $NO_3$ molecule is an intermediate in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step.
Note
The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction.
Using Molecularity to Describe a Rate Law
The molecularity of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular; if there are two reactant molecules, it is bimolecular; and if there are three reactant molecules (a relatively rare situation), it is termolecular. Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!)
Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity (Table 14.7). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is rate = k[A]. For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure 14.19. For a bimolecular elementary reaction of the form A + B → products, the general rate law is rate = k[A][B].
Table 14.7 Common Types of Elementary Reactions and Their Rate Laws
Elementary Reaction Molecularity Rate Law Reaction Order
A → products unimolecular rate = k[A] first
2A → products bimolecular rate = k[A]2 second
A + B → products bimolecular rate = k[A][B] second
2A + B → products termolecular rate = k[A]2[B] third
A + B + C → products termolecular rate = k[A][B][C] third
Figure 14.19 The Basis for Writing Rate Laws of Elementary Reactions. This diagram illustrates how the number of possible collisions per unit time between two reactant species, A and B, depends on the number of A and B particles present. The number of collisions between A and B particles increases as the product of the number of particles, not as the sum. This is why the rate law for an elementary reaction depends on the product of the concentrations of the species that collide in that step.
Identifying the Rate-Determining Step
Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining step, that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions.
Rate-determining step. The phenomenon of a rate-determining step can be compared to a succession of funnels. The smallest-diameter funnel controls the rate at which the bottle is filled, whether it is the first or the last in the series. Pouring liquid into the first funnel faster than it can drain through the smallest results in an overflow.
Look at the rate laws for each elementary reaction in our example as well as for the overall reaction.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{\mathrm{k_1}}\mathrm{NO_3}+\textrm{NO}$ $\textrm{rate}=k_1[\mathrm{NO_2}]^2\textrm{ (predicted)}$
$\textrm{step 2}$ $\underline{\mathrm{NO_3}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO_2}+\mathrm{CO_2}}$ $\textrm{rate}=k_2[\mathrm{NO_3}][\mathrm{CO}]\textrm{ (predicted)}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\xrightarrow{k}\mathrm{NO}+\mathrm{CO_2}$ $\textrm{rate}=k[\mathrm{NO_2}]^2\textrm{ (observed)}$
The experimentally determined rate law for the reaction of $NO_2$ with $CO$ is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so $k$ for the overall reaction must equal $k_1$. That is, NO3 is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2.
Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect.
Example 12
In an alternative mechanism for the reaction of NO2 with CO, N2O4 appears as an intermediate.
$\textrm{step 1}$ $\mathrm{NO_2}+\mathrm{NO_2}\xrightarrow{k_1}\mathrm{N_2O_4}$
$\textrm{step 2}$ $\underline{\mathrm{N_2O_4}+\mathrm{CO}\xrightarrow{k_2}\mathrm{NO}+\mathrm{NO_2}+\mathrm{CO_2}}$
$\textrm{sum}$ $\mathrm{NO_2}+\mathrm{CO}\rightarrow\mathrm{NO}+\mathrm{CO_2}$
Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = k[NO2]2)?
Given: elementary reactions
Asked for: rate law for each elementary reaction and overall rate law
Strategy:
1. Determine the rate law for each elementary reaction in the reaction.
2. Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step.
Solution
A The rate law for step 1 is rate = k1[NO2]2; for step 2, it is rate = k2[N2O4][CO].
B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k1[NO2]2. This is the same as the experimentally determined rate law. Hence this mechanism, with N2O4 as an intermediate, and the one described previously, with NO3 as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO3 and N2O4, directly.
Exercise 12
Iodine monochloride (ICl) reacts with H2 as follows:
$2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s) \nonumber$
The experimentally determined rate law is rate = k[ICl][H2]. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: HI is an intermediate.)
Answer
$\textrm{step 1}$ $\mathrm{ICl}+\mathrm{H_2}\xrightarrow{k_1}\mathrm{HCl}+\mathrm{HI}$ $\mathrm{rate}=k_1[\mathrm{ICl}][\mathrm{H_2}]\,(\textrm{slow})$
$\textrm{step 2}$ $\underline{\mathrm{HI}+\mathrm{ICl}\xrightarrow{k_2}\mathrm{HCl}+\mathrm{I_2}}$ $\mathrm{rate}=k_2[\mathrm{HI}][\mathrm{ICl}]\,(\textrm{fast})$
$\textrm{sum}$ $\mathrm{2ICl}+\mathrm{H_2}\rightarrow\mathrm{2HCl}+\mathrm{I_2}$
This mechanism is consistent with the experimental rate law if the first step is the rate-determining step.
Example 12b: $NO$ with $H_2$
Is the the reaction between $NO$ and $H_2$ occurs via a three-step process:
$\textrm{step 1}$ $\mathrm{NO}+\mathrm{NO}\xrightarrow{k_1}\mathrm{N_2O_2}$ $\textrm{(fast)}$
$\textrm{step 2}$ $\mathrm{N_2O_2}+\mathrm{H_2}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{H_2O}$ $\textrm{(slow)}$
$\textrm{step 3}$ $\mathrm{N_2O}+\mathrm{H_2}\xrightarrow{k_3}\mathrm{N_2}+\mathrm{H_2O}$ $\textrm{(fast)}$
Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: rate = $k[NO]^2[H^2]$?
Answer:
• Step 1: $rate = k_1[NO]^2$;
• Step 2: $rate = k_2[N2O2][H_2]$;
• Step 3: $rate = k_3[N_2O][H_2]$;
• OVer all reaction: $2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g) \nonumber$
• Rate Determining Step : #2
• Yes, because the rate of formation of $[N_2O_2] = k_1[NO]^2$. Substituting $k_1[NO]^2$ for $[N_2O_2]$ in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where $k = k_1k_2$.
Chain Reactions
Many reaction mechanisms, like those discussed so far, consist of only two or three elementary reactions. Many others consist of long series of elementary reactions. The most common mechanisms are chain reactions, in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process. Chain reactions occur in fuel combustion, explosions, the formation of many polymers, and the tissue changes associated with aging. They are also important in the chemistry of the atmosphere.
Chain reactions are described as having three stages. The first is initiation, a step that produces one or more reactive intermediates. Often these intermediates are radicals, species that have an unpaired valence electron. In the second stage, propagation, reactive intermediates are continuously consumed and regenerated while products are formed. Intermediates are also consumed but not regenerated in the final stage of a chain reaction, termination, usually by forming stable products.
Let us look at the reaction of methane with chlorine at elevated temperatures (400°C–450°C), a chain reaction used in industry to manufacture methyl chloride (CH3Cl), dichloromethane (CH2Cl2), chloroform (CHCl3), and carbon tetrachloride (CCl4):
CH4 + Cl2 → CH3Cl + HCl
CH3Cl + Cl2 → CH2Cl2 + HCl
CH2Cl2 + Cl2 → CHCl3 + HCl
CHCl3 + Cl2 → CCl4 + HCl
Direct chlorination generally produces a mixture of all four carbon-containing products, which must then be separated by distillation. In our discussion, we will examine only the chain reactions that lead to the preparation of CH3Cl.
In the initiation stage of this reaction, the relatively weak Cl–Cl bond cleaves at temperatures of about 400°C to produce chlorine atoms (Cl·):
Cl2 → 2Cl·
During propagation, a chlorine atom removes a hydrogen atom from a methane molecule to give HCl and CH3·, the methyl radical:
Cl· + CH4 → CH3· + HCl
The methyl radical then reacts with a chlorine molecule to form methyl chloride and another chlorine atom, Cl·:
CH3· + Cl2 → CH3Cl + Cl·
The sum of the propagation reactions is the same as the overall balanced chemical equation for the reaction:
$\mathrm{Cl\cdot}+\mathrm{CH_4}\rightarrow\mathrm{CH_3\cdot}+\mathrm{HCl}$
$\underline{\mathrm{CH_3\cdot}+\mathrm{Cl_2}\rightarrow\mathrm{CH_3Cl}+\mathrm{Cl\cdot}}$
$\mathrm{Cl_2}+\mathrm{CH_4}\rightarrow\mathrm{CH_3Cl}+\mathrm{HCl}$
Without a chain-terminating reaction, propagation reactions would continue until either the methane or the chlorine was consumed. Because radical species react rapidly with almost anything, however, including each other, they eventually form neutral compounds, thus terminating the chain reaction in any of three ways:
CH3· + Cl· → CH3Cl
CH3· + CH3· → H3CCH3
Cl· + Cl· → Cl2
Here is the overall chain reaction, with the desired product (CH3Cl) in bold:
Initiation: Cl2 → 2Cl·
Propagation: Cl· + CH4 → CH3· + HCl
CH3· + Cl2CH3Cl + Cl·
Termination: CH3· + Cl· → CH3Cl
CH3· + CH3· → H3CCH3
Cl· + Cl· → Cl2
The chain reactions responsible for explosions generally have an additional feature: the existence of one or more chain branching steps, in which one radical reacts to produce two or more radicals, each of which can then go on to start a new chain reaction. Repetition of the branching step has a cascade effect such that a single initiation step generates large numbers of chain reactions. The result is a very rapid reaction or an explosion.
The reaction of H2 and O2, used to propel rockets, is an example of a chain branching reaction:
Initiation: H2 + O2 → HO2· + H·
Propagation: HO2· + H2 → H2O + OH·
OH· + H2 → H2O + H·
Termination: H· + O2 → OH· + ·O·
·O· + H2 → OH· + H·
Termination reactions occur when the extraordinarily reactive H· or OH· radicals react with a third species. The complexity of a chain reaction makes it unfeasible to write a rate law for the overall reaction.
Summary
A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its molecularity, the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. Chain reactions consist of three kinds of reactions: initiation, propagation, and termination. Intermediates in chain reactions are often radicals, species that have an unpaired valence electron.
Key Takeaway
• A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law.
Conceptual Problems
1. How does the term molecularity relate to elementary reactions? How does it relate to the overall balanced chemical equation?
2. What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation?
3. When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction?
4. If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step?
5. Give the rate-determining step for each case.
1. Traffic is backed up on a highway because two lanes merge into one.
2. Gas flows from a pressurized cylinder fitted with a gas regulator and then is bubbled through a solution.
3. A document containing text and graphics is downloaded from the Internet.
1. Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health?
Numerical Problems
1. Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows:
where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction?
1. Above approximately 500 K, the reaction between NO2 and CO to produce CO2 and NO follows the second-order rate law Δ[CO2]/Δt = k[NO2][CO]. At lower temperatures, however, the rate law is Δ[CO2]/Δt = k′[NO2]2, for which it is known that NO3 is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest?
1. Nitramide (O2NNH2) decomposes in aqueous solution to N2O and H2O. What is the experimental rate law (Δ[N2O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows?
$\mathrm{O_2NNH_2}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{O_2NNH^-}+\mathrm{H^+}$ $(\textrm{fast})$
$\mathrm{O_2NNH^-}\xrightarrow{k_2}\mathrm{N_2O}+\mathrm{OH^-}$ $(\textrm{slow})$
$\mathrm{H^+}+\mathrm{OH^-}\xrightarrow{k_3}\mathrm{H_2O}$ $(\textrm{fast})$
Assume that the rates of the forward and reverse reactions in the first equation are equal.
1. The following reactions are given:
$\mathrm{A+B}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{C+D}$
$\mathrm{D+E}\xrightarrow{k_2}\mathrm F$
What is the relationship between the relative magnitudes of k−1 and k2 if these reactions have the rate law Δ[F]/Δt = k[A][B][E]/[C]? How does the magnitude of k1 compare to that of k2? Under what conditions would you expect the rate law to be Δ[F]/Δt =k′[A][B]? Assume that the rates of the forward and reverse reactions in the first equation are equal.
Numerical Answers
1. The k2 step is likely to be rate limiting; the rate cannot proceed any faster than the second step.
$\textrm{rate}=k_2\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}=k\dfrac{[\mathrm{O_2NNH_2}]}{[\mathrm{H^+}]}$
2.07: Reaction Rates- Building Intuition
The rate of a reaction is just the "speed' of the reaction. Just as your speed when driving down the highway can be described in terms of your progress over time (in miles or kilometers per hour), a reaction can be described in terms of the progress of the reaction over time.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.06%3A_Reaction_Rates-_A_Microscopic_View.txt
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Many important biological reactions, such as the formation of double-stranded DNA from two complementary strands, can be described using second order kinetics. In a second-order reaction, the sum of the exponents in the rate law is equal to two. The two most common forms of second-order reactions will be discussed in detail in this section.
To describe how the rate of a second-order reaction changes with concentration of reactants or products, the differential (derivative) rate equation is used as well as the integrated rate equation. The differential rate law can show us how the rate of the reaction changes in time, while the integrated rate equation shows how the concentration of species changes over time. The latter form, when graphed, yields a linear function and is, therefore, more convenient to look at. Nonetheless, both of these equations can be derived from the above expression for the reaction rate. Plotting these equations can also help us determine whether or not a certain reaction is second-order.
Case 1: Identical Reactants
Two of the same reactant ($\ce{A}$) combine in a single elementary step.
\begin{align} \ce{A} + \ce{A} &\ce{->} \ce{P} \label{case1a} \[4pt] \ce{2A} &\ce{->} \ce{P} \label{case1b} \end{align}
The reaction rate for this step can be written as
$\text{Rate} = - \dfrac{1}{2} \dfrac{d[A]}{dt} = + \dfrac{d[P]}{dt} \nonumber$
and the rate of loss of reactant $\ce{A}$
$\dfrac{dA}{dt}= -k[A][A] = -k[A]^2 \label{2ndlaw}$
where $k$ is a second order rate constant with units of $\text{M}^{-1} \text{min}^{-1}$ or $\text{M}^{-1} \text{s}^{-1}$. Therefore, doubling the concentration of reactant $\ce{A}$ will quadruple the rate of the reaction. In this particular case, another reactant ($B$) could be present with $A$; however, its concentration does not affect the rate of the reaction, i.e., the reaction order with respect to B is zero, and we can express the rate law as $v = k[A]^2[B]^0$.
Integration of Equation \ref{2ndlaw} yields
$\dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt \nonumber$
which is easily rearranged into a form of the equation for a straight line and yields plots similar to the one shown below.
The half-life is given by
$t_{1/2}=\dfrac{1}{k[A_o]} \nonumber$
Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to first-order reactions. For this reason, the concept of half-life for a second-order reaction is far less useful. Reaction rates are discussed in more detail here. Reaction orders are defined here. Here are explanations of zero and first order reactions.
For reactions that follow Equation \ref{case1a} or \ref{case1b}, the rate at which $\ce{A}$ decreases can be expressed using the differential rate equation.
$-\dfrac{d[A]}{dt} = k[A]^2 \nonumber$
The equation can then be rearranged:
$\dfrac{d[A]}{[A]^2} = -k\,dt \nonumber$
Since we are interested in the change in concentration of A over a period of time, we integrate between $t = 0$ and $t$, the time of interest.
$\int_{[A]_o}^{[A]_t} \dfrac{d[A]}{[A]^2} = -k \int_0^t dt \nonumber$
To solve this, we use the following rule of integration (power rule):
$\int \dfrac{dx}{x^2} = -\dfrac{1}{x} + constant \nonumber$
We then obtain the integrated rate equation.
$\dfrac{1}{[A]_t} - \dfrac{1}{[A]_o} = kt \nonumber$
Upon rearrangement of the integrated rate equation, we obtain an equation of the line:
$\dfrac{1}{[A]_t} = kt + \dfrac{1}{[A]_o} \nonumber$
The crucial part of this process is not understanding precisely how to derive the integrated rate law equation, rather it is important to understand how the equation directly relates to the graph which provides a linear relationship. In this case, and for all second order reactions, the linear plot of $\dfrac{1}{[A]_t}$ versus time will yield the graph below.
This graph is useful in a variety of ways. If we only know the concentrations at specific times for a reaction, we can attempt to create a graph similar to the one above. If the graph yields a straight line, then the reaction in question must be second order. In addition, with this graph we can find the slope of the line and this slope is $k$, the reaction constant. The slope can be found be finding the "rise" and then dividing it by the "run" of the line. For an example of how to find the slope, please see the example section below. There are alternative graphs that could be drawn.
The plot of $[A]_t$ versus time would result in a straight line if the reaction were zeroth order. It does, however, yield less information for a second order graph. This is because both the graphs of a first or second order reaction would look like exponential decays. The only obvious difference, as seen in the graph below, is that the concentration of reactants approaches zero more slowly in a second-order, compared to that in a first order reaction.
Case 2: Second Order Reaction with Multiple Reactants
Two different reactants ($\ce{A}$ and $\ce{B}$) combine in a single elementary step:
$A + B \longrightarrow P \label{case2}$
The reaction rate for this step can be written as
$\text{Rate} = - \dfrac{d[A]}{dt}= - \dfrac{d[B]}{dt}= + \dfrac{d[P]}{dt} \nonumber$
and the rate of loss of reactant $\ce{A}$
$\dfrac{d[A]}{dt}= - k[A][B] \nonumber$
where the reaction order with respect to each reactant is 1. This means that when the concentration of reactant A is doubled, the rate of the reaction will double, and quadrupling the concentration of reactant in a separate experiment will quadruple the rate. If we double the concentration of $\ce{A}$ and quadruple the concentration of $\ce{B}$ at the same time, then the reaction rate is increased by a factor of 8. This relationship holds true for any varying concentrations of $\ce{A}$ or $\ce{B}$.
As before, the rate at which $A$ decreases can be expressed using the differential rate equation:
$\dfrac{d[A]}{dt} = -k[A][B] \nonumber$
Two situations can be identified.
Situation 2a: $[A]_0 \neq [B]_0$
Situation 2a is the situation that the initial concentration of the two reactants are not equal. Let $x$ be the concentration of each species reacted at time $t$.
Let $[A]_0 =a$ and $[B]_0 =b$, then $[A]= a-x$ ;$[B]= b-x$. The expression of rate law becomes:
$-\dfrac{dx}{dt} = -k([A]_o - x)([B]_o - x)\nonumber$
which can be rearranged to:
$\dfrac{dx}{([A]_o - x)([B]_o - x)} = kdt\nonumber$
We integrate between $t = 0$ (when $x = 0$) and $t$, the time of interest.
$\int_0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = k \int_0^t dt \nonumber$
To solve this integral, we use the method of partial fractions.
$\int_0^x \dfrac{1}{(a - x)(b -x)}dx = \dfrac{1}{b - a}\left(\ln\dfrac{1}{a - x} - \ln\dfrac{1}{b - x}\right)\nonumber$
Evaluating the integral gives us:
$\int_0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = \dfrac{1}{[B]_o - [A]_o}\left(\ln\dfrac{[A]_o}{[A]_o - x} - \ln\dfrac{[B]_o}{[B]_o - x}\right) \nonumber$
Applying the rule of logarithm, the equation simplifies to:
$\int _0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = \dfrac{1}{[B]_o - [A]_o} \ln \dfrac{[B][A]_o}{[A][B]_o} \nonumber$
We then obtain the integrated rate equation (under the condition that [A] and [B] are not equal).
$\dfrac{1}{[B]_o - [A]_o}\ln \dfrac{[B][A]_o}{[A][B]_o} = kt \nonumber$
Upon rearrangement of the integrated rate equation, we obtain:
$\ln\dfrac{[B][A]_o}{[A][B]_o} = k([B]_o - [A]_o)t \nonumber$
Hence, from the last equation, we can see that a linear plot of $\ln\dfrac{[A]_o[B]}{[A][B]_o}$ versus time is characteristic of second-order reactions.
This graph can be used in the same manner as the graph in the section above or written in the other way:
$\ln\dfrac{[A]}{[B]} = k([A]_o - [B]_o)t+\ln\dfrac{[A]_o}{[B]_o}\nonumber$
in form $y = ax + b$ with a slope of $a= k([B]_0-[A]_0)$ and a y-intercept of $b = \ln \dfrac{[A]_0}{[B]_0}$
Situation 2b: $[A]_0 =[B]_0$
Because $A + B \rightarrow P$
Since $A$ and $B$ react with a 1 to 1 stoichiometry, $[A]= [A]_0 -x$ and $[B] = [B]_0 -x$
at any time $t$, $[A] = [B]$ and the rate law will be,
$\text{rate} = k[A][B] = k[A][A] = k[A]^2.\nonumber$
Thus, it is assumed as the first case!!!
Example $1$
The following chemical equation reaction represents the thermal decomposition of gas $E$ into $K$ and $G$ at 200° C
$\ce{ 5E(g) -> 4K(g) + G(g)} \nonumber$
This reaction follows a second order rate law with regards to $\ce{E}$. For this reaction suppose that the rate constant at 200° C is equivalent to $4.0 \times 10^{-2} M^{-1}s^{-1}$ and the initial concentration is $0.050\; M$. What is the initial rate of decomposition of $\ce{E}$.
Solution
Start by defining the reaction rate in terms of the loss of reactants
$\text{Rate (initial)} = - \dfrac{1}{5} \dfrac{d[E]}{dt}\nonumber$
and then use the rate law to define the rate of loss of $\ce{E}$
$\dfrac{d[E]}{dt} = -k [A]_i^2 \nonumber$
We already know $k$ and $[A]_i$ but we need to figure out $x$. To do this look at the units of $k$ and one sees it is M-1s-1 which means the overall reaction is a second order reaction with $x=2$.
\begin{align*} \text{Initial rate} &= (4.0 \times 10^{-2} M^{-1}s^{-1})(0.050\,M)^2 \[4pt] &= 1 \times 10^{-4} \, Ms^{-1}\end{align*} \nonumber
Half-Life
Another characteristic used to determine the order of a reaction from experimental data is the half-life ($t_{1/2}$). By definition, the half life of any reaction is the amount of time it takes to consume half of the starting material. For a second-order reaction, the half-life is inversely related to the initial concentration of the reactant (A). For a second-order reaction each half-life is twice as long as the life span of the one before.
Consider the reaction $2A \rightarrow P$:
We can find an expression for the half-life of a second order reaction by using the previously derived integrated rate equation.
$\dfrac{1}{[A]_t} - \dfrac{1}{[A]_o} = kt \nonumber$
Since,
$[A]_{t_{1/2}} = \dfrac{1}{2}[A]_o \nonumber$
when $t = t_{1/2}$.
Our integrated rate equation becomes:
$\dfrac{1}{\dfrac{1}{2}[A]_o} - \dfrac{1}{[A]_o} = kt_{1/2} \nonumber$
After a series of algebraic steps,
\begin{align*} \dfrac{2}{[A]_o} - \dfrac{1}{[A]_o} &= kt_{1/2} \[4pt] \dfrac{1}{[A]_o} &= kt_{1/2} \end{align*} \nonumber
We obtain the equation for the half-life of a second order reaction:
$t_{1/2} = \dfrac{1}{k[A]_o} \label{2nd halflife}$
This inverse relationship suggests that as the initial concentration of reactant is increased, there is a higher probability of the two reactant molecules interacting to form product. Consequently, the reactant will be consumed in a shorter amount of time, i.e. the reaction will have a shorter half-life. This equation also implies that since the half-life is longer when the concentrations are low, species decaying according to second-order kinetics may exist for a longer amount of time if their initial concentrations are small.
Note that for the second scenario in which $A + B \rightarrow P$, the half-life of the reaction cannot be determined. As stated earlier, $[A]_o$ cannot be equal to $[B]_o$. Hence, the time it takes to consume one-half of A is not the same as the time it takes to consume one-half of B. Because of this, we cannot define a general equation for the half-life of this type of second-order reaction.
Example $2$: Half-Life of a Second-Order Reaciton
If the only reactant is the initial concentration of $A$, and it is equivalent to $[A]_0=4.50 \times 10^{-5}\,M$ and the reaction is a second order with a rate constant $k=0.89 M^{-1}s^{-1}$. What is the half-life of this reaction?
Solution
This is a direct application of Equation \ref{2nd halflife}.
\begin{align*} \dfrac{1}{k[A]_0} &= \dfrac{1}{(4.50 \times 10^{-5} M)(0.89 M^{-1}{s^{-1})}} \[4pt] &= 2.50 \times 10^4 \,s \end{align*} \nonumber
Summary
$2A \rightarrow P$ $A + B \rightarrow P$
Differential Form $-\dfrac{d[A]}{dt} = k[A]^2$ $-\dfrac{d[A]}{dt} = k[A][B]$
Integral Form $\dfrac{1}{[A]_t} = kt + \dfrac{1}{[A]_o}$ $\dfrac{1}{[B]_o - [A]_o}\ln\dfrac{[B][A]_o}{[A][B]_o} = kt$
Half Life $t_{1/2} = \dfrac{1}{k[A]_o}$ Cannot be easily defined; $t_{1/2}$ for A and B are different.
The graph below is the graph that tests if a reaction is second order. The reaction is second order if the graph has a straight line, as is in the example below.
Exercise $3$
Given the following information, determine the order of the reaction and the value of k, the reaction constant.
Concentration (M) Time (s)
1.0 10
0.50 20
0.33 30
*Hint: Begin by graphing
Answer
Make graphs of concentration vs. time (zeroth order), natural log of concentration vs. time (first order), and one over concentration vs. time (second order). Determine which graph results in a straight line. This graph reflects the order of the reaction. For this problem, the straight line should be in the 3rd graph, meaning the reaction is second order. The numbers should have are:
1/Concentration(M-1) Time (s)
1 10
2 20
3 30
The slope can be found by taking the "rise" over the "run". This means taking two points, (10,1) and (20,2). The "rise" is the vertical distance between the points (2-1=1) and the "run" is the horizontal distance (20-10=10). Therefore the slope is 1/10=0.1. The value of k, therefore, is 0.1 M-2s-1.
Exercise $4$
Using the following information, determine the half life of this reaction, assuming there is a single reactant.
Concentration (M) Time (s)
2.0 0
1.3 10
0.9633 20
Answer
Determine the order of the reaction and the reaction constant, $k$, for the reaction using the tactics described in the previous problem. The order of the reaction is second, and the value of k is 0.0269 M-2s-1. Since the reaction order is second, the formula for $t_{1/2} = k^{-1}[A]_0^{-1}$. This means that the half life of the reaction is 0.0259 seconds.
Exercise $5$
Given the information from the previous problem, what is the concentration after 5 minutes?
Answer
Convert the time (5 minutes) to seconds. This means the time is 300 seconds. Use the integrated rate law to find the final concentration. The final concentration is 0.1167 M.
2.08: Second-Order Reactions
Under certain conditions, the 2nd order kinetics can be approximated as first-order kinetics. These pseudo-1st-order reactions greatly simplify quantifying the reaction dynamics.
Introduction
A 2nd-order reaction can be challenging to follow mostly because the two reactants involved must be measured simultaneously. There can be additional complications because certain amounts of each reactant are required to determine the reaction rate, for example, which can make the cost of one's experiment high if one or both of the needed reactants are expensive. To avoid more complicated, expensive experiments and calculations, we can use the pseudo-1st-order reaction, which involves treating a 2nd order reaction like a 1st order reaction.
In second-order reactions with two reactant species,
$\ce{A + B -> products}\nonumber$
the rate of disappearance of $A$ is
$\dfrac{d[A]}{dt}= -k[A][B]\nonumber$
as discussed previously (Case 2a), the integrated rate equation under the condition that [A] and [B] are not equal is
$\dfrac{1}{[B]_o - [A]_o}\ln \dfrac{[B][A]_o}{[A][B]_o} = kt \label{2nd}$
However, when $[B]_0 \gg [A]_0$, then $[B]_0 \approx [B]$ and Equation $\ref{2nd}$ becomes
$\dfrac{1}{[B]_0-[A]_0}\ln \dfrac{[B][A]_0}{[B]_0[A]} \approx \dfrac{1}{[B]}\ln \dfrac{[A]_0}{[A]}=kt \nonumber$
or
$[A] = [A]_0 e^{-[B]kt}\nonumber$
This functional form of the decay kinetics is similar ot the first order kinetics and the system is said to operate under pseudo-first order kinetics. To reach a pseudo-1st-order reaction, we can manipulate the initial concentrations of the reactants. One of the reactants, $\ce{B}$, for example, would have a significantly higher concentration, while the other reactant, $A$, would have a significantly lower concentration. We can then assume that the concentration of reactant $B$ effectively remains constant during the reaction because its consumption is so small that the change in concentration becomes negligible. Because of this assumption, we can multiply the reaction rate, $k$, with the reactant with assumed constant concentration, $B$, to create a new rate constant ($k'=k[B]$) that will be used in the new rate equation,
$\text{Rate}=k'[A]\nonumber$
as the new rate constant so we can treat the 2nd order reaction as a 1st order reaction.
Overloading
One can also do the same by overloading the initial concentration of $A$ so that it effectively remains constant during the course of the reaction For example, if one were to dump a liter of 5 M $\ce{HCl}$ into a 55 M ocean, the concentration of the mixture would be closer or equal to that of the ocean because there is so much water physically compared to the $\ce{HCl}$. Even if the amount of water was one liter this would still be the case because 55 M is relatively large compared to 5 M.
If we have an instance where there are more than two reactants involved in a reaction, all we would have to do to make the reaction pseudo-1st-order is to make the concentrations of all but one of the reactants very large. If there were three reactants, for example, we would make two of the three reactants be in excess (whether in amount or in concentration) and then monitor the dependency of the third reactant.
We can write the pseudost-order reaction equation as:
$[A] = [A]_0 e^{-[B]_{0}kt}\label{eq1}$
or
$[A] = [A]_0 e^{-k^{'}t} \nonumber$
where
• $[A]_o$ is the initial concentration of $A$,
• $[B]_o$ is the initial concentration of $B$,
• $k^{'}$ is the pseudo-1st-order reaction rate constant,
• $k$ is the 2nd order reaction rate constant, and
• $[A]$ is the concentration of A at time $t$.
By using natural log to both sides of the pseudo-1st-order equation we get:
$\ln \left ( \dfrac{A}{A_0} \right ) =-k[B]_0 t\nonumber$
or
$\ln \left ( \dfrac{A}{A_0} \right ) = -k^{'}t \nonumber$
Example $1$
If a 2nd order reaction has the rate equation $\text{Rate} = k[A][B]$, and the rate constant, $k$, is $3.67\, M^{-1}s^{-1}$, $[A]$ is $4.5\, M$ and $[B]$ is $99\, M$, what is the rate constant of its pseudo-1st-order reaction?
Solution
Because $[B]$ is in excess we multiply $99\, M$ with $3.67\, M^{-1}s^{-1}$
$(99\,M)(3.67\, M^{-1}s^{-1}) = 363.33\,s^{-1}\nonumber$
Example $2$
If [A] = 55 M at 39 s, [A]o = 99 M, and [B]o = 1000 M, what is the 2nd order reaction rate constant?
Solution
Use the Equation \ref{eq1}
(55 M) = (99 M)e-k'(1000M)(39s)
k' = 1.507 x 10-5 M-1s-1
Example $3$
What is the concentration of A at time 45 s if [A]o = 1M, [B]o = 45 M, and 2nd order rate constant is 0.6 M-1s-1?
Solution
Use the Equation \ref{eq1}
[A] = (1)e-(0.6 M-1s-1)(45)
[A] = 1.88 x 10-12 M
Example $4$
What is the rate of a reaction if [A]o = 560 M, [B]o = 0.2M, and 2nd order rate constant is 0.1 M-1s-1?
Solution
Use the equation R = k'[A][B]
R = (0.1M-1s-1)(560M)(0.2M)
Rate = 11.2Ms-1
Half-Life in a Pseudo-1st Order reaction
Half-life refers to the time required to decrease the concentration of a reactant by half, so we must solve for $t$. Here, $[B]$ will be the reactant in excess, and its concentration will stay constant. $[A]_o$ is the initial concentration of $A$; thus the half-life concentration of $A$ is $0.5[A]_o$.
The pseudo-1st-order reaction equation can be written as:
$[A] = [A]_o e^{-[B]kt} \;\;or\;\; \dfrac{[A]} {[A]_o}= e^{-k^{'}t}\nonumber$
By taking natural logs on both sides of the pseudo-1st-order equation, we get:
$\ln \left ( \dfrac{[A]} {[A]_o} \right )= k^{'}t \nonumber$
Because the concentration of $A$ for a half-life $t_{1/2}$ is $1/2 [A]_o$ :
$\ln \left ( \dfrac{1/2[A]_o} {[A]_o} \right ) = \ln \left ( \dfrac {1}{2} \right )=- k^{'}t_{1/2} \nonumber$
Recalling that $k^{'}=k [B]$, $[B] \approx [B]_{0}$ and that $-\ln (1/2) = \ln 2$:
$\ln(2)=k[B]_{0}t_{1/2}\nonumber$
or
$t_{1/2}=\dfrac{\ln 2}{k[B]_{0}} \nonumber$
Example $5$
What is the half-life of a reaction with [A]o = 109 M, [B]o = 1 M, k' = 45 M-1s-1 ?
Solution
Because [A] is in excess we can multiply the k' with [A]o to find k
(109M)(45M-1s-1) = 4905s-1
t1/2 = (ln0.5 / -k)
t1/2 = 1.41 X 10-4s
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.08%3A_Second-Order_Reactions/2.8.01%3A_Pseudo-1st-order_reactions.txt
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In some reactions, the rate is apparently independent of the reactant concentration. The rates of these zero-order reactions do not vary with increasing nor decreasing reactants concentrations. This means that the rate of the reaction is equal to the rate constant, $k$, of that reaction. This property differs from both first-order reactions and second-order reactions.
Origin of Zero Order Kinetics
Zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.
There are two general conditions that can give rise to zero-order rates:
1. Only a small fraction of the reactant molecules are in a location or state in which they are able to react, and this fraction is continually replenished from the larger pool.
2. When two or more reactants are involved, the concentrations of some are much greater than those of others
This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface (heterogeneous catalysis) or to an enzyme.
Example 1: Decomposition of Nitrous Oxide
Nitrous oxide will decompose exothermically into nitrogen and oxygen, at a temperature of approximately 575 °C
$\ce{2N_2O ->[\Delta, \,Ni] 2N_2(g) + O_2(g)} \nonumber$
This reaction in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional second order kinetics when carried out entirely in the gas phase.
$\ce{2N_2O -> 2N_2(g) + O_2(g)} \nonumber$
In this case, the $N_2O$ molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site.
Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an enzyme-substrate complex. If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order.
This is most often seen when two or more reactants are involved. Thus if the reaction
$A + B \rightarrow \text{products} \tag{1}$
is first-order in both reactants so that
$\text{rate} = k [A][B] \tag{2}$
If $B$ is present in great excess, then the reaction will appear to be zero order in $B$ (and first order overall). This commonly happens when $B$ is also the solvent that the reaction occurs in.
Differential Form of the Zeroth Order Rate Law
$Rate = - \dfrac{d[A]}{dt} = k[A]^0 = k = constant \tag{3}$
where $Rate$ is the reaction rate and $k$ is the reaction rate coefficient. In this example, the units of $k$ are M/s. The units can vary with other types of reactions. For zero-order reactions, the units of the rate constants are always M/s. In higher order reactions, $k$ will have different units.
Integrated Form of the Zeroth Order Rate Law
Integration of the differential rate law yields the concentration as a function of time. Start with the general rate law equations
$Rate = k[A]^n \tag{4}$
First, write the differential form of the rate law with $n=0$
$Rate = - \dfrac{d[A]^0}{dt} = k \tag{5}$
then rearrange
${d}[A] = -kdt \tag{6}$
Second, integrate both sides of the equation.
$\int_{[A]_{0}}^{[A]} d[A] = - \int_{0}^{t} kdt \tag{7}$
Third, solve for $[A]$. This provides the integrated form of the rate law.
$[A] = [A]_0 -kt \tag{8}$
The integrated form of the rate law allows us to find the population of reactant at any time after the start of the reaction.
Graphing Zero-order Reactions
$[A] = -kt + [A]_0 \tag{9}$
is in the form y = mx+b where slope = m = -k and the y- intercept = b = $[A]_0$
Zero-order reactions are only applicable for a very narrow region of time. Therefore, the linear graph shown below (Figure 2) is only realistic over a limited time range. If we were to extrapolate the line of this graph downward to represent all values of time for a given reaction, it would tell us that as time progresses, the concentration of our reactant becomes negative. We know that concentrations can never be negative, which is why zero-order reaction kinetics is applicable for describing a reaction for only brief window and must eventually transition into kinetics of a different order.
Figure 2: (left) Concentration vs. time of a zero-order reaction. (Right) Concentration vs. time of a zero-order catalyzed reaction.
To understand where the above graph comes from, let us consider a catalyzed reaction. At the beginning of the reaction, and for small values of time, the rate of the reaction is constant; this is indicated by the blue line in Figures 2; right. This situation typically happens when a catalyst is saturated with reactants. With respect to Michaelis-Menton kinetics, this point of catalyst saturation is related to the $V_{max}$. As a reaction progresses through time, however, it is possible that less and less substrate will bind to the catalyst. As this occurs, the reaction slows and we see a tailing off of the graph (Figure 2; right). This portion of the reaction is represented by the dashed black line. In looking at this particular reaction, we can see that reactions are not zero-order under all conditions. They are only zero-order for a limited amount of time.
If we plot rate as a function of time, we obtain the graph below (Figure 3). Again, this only describes a narrow region of time. The slope of the graph is equal to k, the rate constant. Therefore, k is constant with time. In addition, we can see that the reaction rate is completely independent of how much reactant you put in.
Figure 3: Rate vs. time of a zero-order reaction.
Relationship Between Half-life and Zero-order Reactions
The half-life. $t_{1/2}$, is a timescale in which each half-life represents the reduction of the initial population to 50% of its original state. We can represent the relationship by the following equation.
$[A] = \dfrac{1}{2} [A]_o \tag{10}$
Using the integrated form of the rate law, we can develop a relationship between zero-order reactions and the half-life.
$[A] = [A]_o - kt \tag{11}$
Substitute
$\dfrac{1}{2}[A]_o = [A]_o - kt_{\dfrac{1}{2}} \tag{12}$
Solve for $t_{1/2}$
$t_{1/2} = \dfrac{[A]_o}{2k} \tag{13}$
Notice that, for zero-order reactions, the half-life depends on the initial concentration of reactant and the rate constant.
Questions
1. Using the integrated form of the rate law, determine the rate constant k of a zero-order reaction if the initial concentration of substance A is 1.5 M and after 120 seconds the concentration of substance A is 0.75 M.
2. Using the substance from the previous problem, what is the half-life of substance A if its original concentration is 1.2 M?
3. If the original concentration is reduced to 1.0 M in the previous problem, does the half-life decrease, increase, or stay the same? If the half-life changes what is the new half-life?
4. Given are the rate constants k of three different reactions:
• Reaction A: k = 2.3 M-1s-1
• Reaction B: k = 1.8 Ms-1
• Reaction C: k = 0.75 s-1
Which reaction represents a zero-order reaction?
Summary
The kinetics of any reaction depend on the reaction mechanism, or rate law, and the initial conditions. If we assume for the reaction A -> Products that there is an initial concentration of reactant of [A]0 at time t=0, and the rate law is an integral order in A, then we can summarize the kinetics of the zero-order reaction as follows:
Related Topics
Definition of a reaction rate
Contributors and Attributions
• Rachael Curtis, Jessica Martin, David Cao
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.10%3A_Zero-Order_Reactions.txt
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An elementary bimolecular reaction originates from a collision between two reactants. Whether or not a collision results in a chemical reaction is determined by the energy of the reactants and their orientation. The total energy of the two reactants must be in excess of the activation energy, Ea, and the reactants must be in a favorable orientation for the chemical reaction to occur. While there are many different orientations possible for the collisions, usually not all of them will result in a chemical reaction. For most reactions, if the orientation is not correct, the reactants will bounce off of each other without a chemical reaction.
Contributors and Attributions
• Kirsten Deitrick (I. H. Kempner High School, Sugar Land, Texas),
• Mingying Xue (Department of Computer Science, Texas Tech University, Lubbock, Texas)
• William L. Hase (Department of Chemistry and Biochemistry, Texas Tech University, Lubbock, Texas)
3.1.01: Bimolecular Reactions
The following are animations of gas-phase trajectories for reactants with different amounts and types of energies. The reactant energy includes a relative translational energy, Erel, between Cl- + CH3Br, and a temperature, Tvr, for the CH3Br vibrational and rotational energies.
Select Trajectories of the Cl- + CH3Br → ClCH3 + Br- SN2 Reaction
Cl-+ CH3Br → ClCH3 + Br -
Nucleophilic substitution by a direct mechanism; Erel = 50 kcal/mol and Tvr = 300K
Cl- + CH3Br → Cl----CH3Br → ClCH3Br + Br-
Nucleophilic substitution by an indirect mechanism, involving the Cl----CH3 complex; Erel = 1.0 kcal/mol and Tvr = 300K
Cl- + CH3Br → Cl----CH3Br → ClCH3---Br-→ ClCH3+Br-
Nucleophilic substitution by an indirect Reaction mechanism, involving both Cl----CH3Br and ClCH3---Br- complex; Erel = 1.0 kcal/mol and Tvr = 300K
Cl- + CH3Br → Cl- + CH3Br
A non-reactive collision; Erel = 1.0 kcal/mol and Tvr = 300K
Cl-+CH3Br → Cl----CH3Br → Cl-+CH3Br
A non-reactive collision forming the Cl----CH3Br complex, which dissociates back to reactants; Erel = 1.0 kcal/mol and Tvr = 300K.
Cl-+CH3Br → Cl----CH3Br → ClCH3---Br-→ Cl----CH3Br → Cl-+CH3Br
Nucleophilic substitution by an indirect mechanism involving both the Cl----CH3Br and ClCH3---Br - complex and recrossing the transition state separating these complexes; Erel = 1.0 kcal/mol and Tvr = 300K.
References
1. Wang, Yanfei, Hase, William L., and Wang, Haobin, "Trajectory studies of SN2 nucleophilic substitution. IX. Microscopic reaction pathways and kinetics for Cl + CH3Br", Journal of Chemical Physics 118, 2688 (2003)
Contributors
• Prof. William L. Hase (Department of Chemistry and Biochemistry, Texas Tech University, Lubbock, Texas)
3.1.02: Maxwell-Boltzmann Distributions
The Maxwell-Boltzmann equation, which forms the basis of the kinetic theory of gases, defines the distribution of speeds for a gas at a certain temperature. From this distribution function, the most probable speed, the average speed, and the root-mean-square speed can be derived.
Introduction
The kinetic molecular theory is used to determine the motion of a molecule of an ideal gas under a certain set of conditions. However, when looking at a mole of ideal gas, it is impossible to measure the velocity of each molecule at every instant of time. Therefore, the Maxwell-Boltzmann distribution is used to determine how many molecules are moving between velocities $v$ and $v + dv$. Assuming that the one-dimensional distributions are independent of one another, that the velocity in the y and z directions does not affect the x velocity, for example, the Maxwell-Boltzmann distribution is given by
$\dfrac{dN}{N}= \left(\frac{m}{2\pi k_BT} \right)^{1/2}e^{\frac{-mv^2}{2k_BT}}dv \label{1}$
where
• $dN/N$ is the fraction of molecules moving at velocity $v$ to $v + dv$,
• $m$ is the mass of the molecule,
• $k_b$ is the Boltzmann constant, and
• $T$ is the absolute temperature.1
Additionally, the function can be written in terms of the scalar quantity speed c instead of the vector quantity velocity. This form of the function defines the distribution of the gas molecules moving at different speeds, between $c_1$ and $c_2$, thus
$f(c)=4\pi c^2 \left (\frac{m}{2\pi k_BT} \right)^{3/2}e^{\frac{-mc^2}{2k_BT}} \label{2}$
Finally, the Maxwell-Boltzmann distribution can be used to determine the distribution of the kinetic energy of for a set of molecules. The distribution of the kinetic energy is identical to the distribution of the speeds for a certain gas at any temperature.2
Plotting the Maxwell-Boltzmann Distribution Function
Figure 1 shows the Maxwell-Boltzmann distribution of speeds for a certain gas at a certain temperature, such as nitrogen at 298 K. The speed at the top of the curve is called the most probable speed because the largest number of molecules have that speed.
Figure 1: The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures.from OpenStax.
Figure 2 shows how the Maxwell-Boltzmann distribution is affected by temperature. At lower temperatures, the molecules have less energy. Therefore, the speeds of the molecules are lower and the distribution has a smaller range. As the temperature of the molecules increases, the distribution flattens out. Because the molecules have greater energy at higher temperature, the molecules are moving faster.
Figure 2: The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures. from OpenStax.
Figure 3 shows the dependence of the Maxwell-Boltzmann distribution on molecule mass. On average, heavier molecules move more slowly than lighter molecules. Therefore, heavier molecules will have a smaller speed distribution, while lighter molecules will have a speed distribution that is more spread out.
Figure 3: The speed probability density functions of the speeds of a few noble gases at a temperature of 298.15 K (25 °C). The y-axis is in s/m so that the area under any section of the curve (which represents the probability of the speed being in that range) is dimensionless. Figure is used with permission from Wikipedia.
Related Speed Expressions
Three speed expressions can be derived from the Maxwell-Boltzmann distribution: the most probable speed, the average speed, and the root-mean-square speed. The most probable speed is the maximum value on the distribution plot. This is established by finding the velocity when the following derivative is zero
$\dfrac{df(c)}{dc}|_{C_{mp}} = 0 \nonumber$
which is
$C_{mp}=\sqrt {\dfrac {2RT}{M}} \label{3a}$
The average speed is the sum of the speeds of all the molecules divided by the number of molecules.
$C_{avg}=\int_0^{\infty} c f(c) dc = \sqrt {\dfrac{8RT}{\pi M}} \label{3b}$
The root-mean-square speed is square root of the average speed-squared.
$C_{rms}=\sqrt {\dfrac {3RT}{M}} \label{3c}$
where
• $R$ is the gas constant,
• $T$ is the absolute temperature and
• $M$ is the molar mass of the gas.
It always follows that for gases that follow the Maxwell-Boltzmann distribution (if thermallized)
$C_{mp}< C_{avg}< C_{rms} \label{4}$
Problems
1. Using the Maxwell-Boltzman function, calculate the fraction of argon gas molecules with a speed of 305 m/s at 500 K.
2. If the system in problem 1 has 0.46 moles of argon gas, how many molecules have the speed of 305 m/s?
3. Calculate the values of $C_{mp}$, $C_{avg}$, and $C_{rms}$ for xenon gas at 298 K.
4. From the values calculated above, label the Boltzmann distribution plot (Figure 1) with the approximate locations of (C_{mp}\), $C_{avg}$, and $C_{rms}$.
5. What will have a larger speed distribution, helium at 500 K or argon at 300 K? Helium at 300 K or argon at 500 K? Argon at 400 K or argon at 1000 K?
Answers
1. 0.00141
2. $3.92 \times 10^{20}$ argon molecules
3. Cmp = 194.27 m/s
Cavg = 219.21 m/s
Crms = 237.93 m/s
4. As stated above, Cmp is the most probable speed, thus it will be at the top of the distribution curve. To the right of the most probable speed will be the average speed, followed by the root-mean-square speed.
5. Hint: Use the related speed expressions to determine the distribution of the gas molecules: helium at 500 K. helium at at 300 K. argon at 1000 K.
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The mean free path is the average distance traveled by a moving molecule between collisions.
Introduction
Imagine gas leaking out of a pipe. It would take a while for the gas to diffuse and spread into the environment. This is because gas molecules collide with each other, causing them to change in speed and direction. Therefore, they can never move in a straight path without interruptions. Between every two consecutive collisions, a gas molecule travels a straight path. The average distance of all the paths of a molecule is the mean free path.
Analogy
Imagine a ball traveling in a box ; the ball represents a moving molecule. Every time the ball hits a wall, a collision occurs and the direction of the ball changes (Figure 1). The ball hits the wall five times, causing five collisions. Between every two consecutive collisions, the ball travels an individual path. It travels a total of four paths between the five collisions; each path has a specific distance, d. The mean free path, \lambda, of this ball is the average length of all four paths.
Each path traveled by the ball has a distance, denoted dn:
$\lambda = \dfrac{d_1 + d_2 + d_3 + d_4}{4} \nonumber$
Calculations
In reality, the mean free path cannot be calculated by taking the average of all the paths because it is impossible to know the distance of each path traveled by a molecule. However, we can calculate it from the average speed ($\langle c \rangle$) of the molecule divided by the collision frequency ($Z$). The formula for this is:
$\lambda = \dfrac{\langle c \rangle}{Z} \nonumber$
Because $Z$ is equal to $1/ t$, where $t$ is the average time between collisions, the formula can also be:
$\begin{eqnarray} \lambda &=& \dfrac{\langle c \rangle}{\dfrac{1}{t}} \ &=& \langle c \rangle \times t \end{eqnarray} \nonumber$
In addition, because
$\lambda = \sqrt{2} \pi d^2 \langle c \rangle (N/V) \nonumber$
where
• $d$ is the diameter of the molecule and
• ($N/V$) is the density,
The formula can be further modified to:
\begin{align*} \lambda &= \dfrac{\langle c \rangle}{\sqrt{2} \pi d^2 \langle c \rangle \dfrac{N}{V}} \[4pt] &= \dfrac{1}{\sqrt{2} \pi d^2 \dfrac{N}{V}} \end{align*}
Factors affecting mean free path
• Density: As gas density increases, the molecules become closer to each other. Therefore, they are more likely to run into each other, so the mean free path decreases.
• Increasing the number of molecules or decreasing the volume causes density to increase. This decreases the mean free path.
• Radius of molecule: Increasing the radii of the molecules decreases the space between them, causing them to run into each other more often. Therefore, the mean free path decreases.
• Pressure, temperature, and other factors that affect density can indirectly affect mean free path.
Exercise $1$
A gas has an average speed of 10 m/s and a collision frequency of 10 s-1. What is its mean free path?
Answer
l = <c> / = 10 m/s / 10 s^(-1) = 1 m
Exercise $2$
A gas has an average speed of 10 m/s and an average time of 0.1 s between collisions. What is its mean free path?
Answer
l = <c> X (average time between collisions) = 10 m/s X 0.1 s = 1 m
Exercise $3$
1. A gas has an average speed of 10 m/s and an average time of 0.1 s between collisions. What is its mean free path?
2. A gas has a density of 10 particles m-3 and a molecular diameter of 0.1 m. What is its mean free path?
Answer
l = 1 / [$\scriptstyle \sqrt{2}$ (pi) d^2 (N/v)] = 1 / [$\scriptstyle \sqrt{2}$ (pi)(0.1 m)^2 (10 m^(-3))] = 2.25 m
Exercise $4$
A gas in a 1 m3 container has a molecular diameter of 0.1 m. There are 10 molecules. What is its mean free path?
Answer
l = 1 / [$\scriptstyle \sqrt{2}$ (pi) d^2 (N/v)] = 1 / [$\scriptstyle \sqrt{2}$ (pi)(0.1 m)^2 (10 / 1 m^3)] = 2.25 m
Exercise $5$
A gas has a molecular diameter of 0.1 m. It also has a mean free path of 2.25 m. What is its density?
Answer
l = 1 / [$\scriptstyle \sqrt{2}$ (pi) d^2 (N/v)]
2.25 m = 1 / [$\scriptstyle \sqrt{2}$ (pi)(0.1 m)^2 (N/v)]
N/v = 10 m^(-3)
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When studying a chemical reaction, it is important to consider not only the chemical properties of the reactants, but also the conditions under which the reaction occurs, the mechanism by which it takes place, the rate at which it occurs, and the equilibrium toward which it proceeds. According to the law of mass action, the rate of a chemical reaction at a constant temperature depends only on the concentrations of the substances that influence the rate. The substances that influence the rate of reaction are usually one or more of the reactants, but can occasionally be a product. Another influence on the rate of reaction can be a catalyst that does not appear in the balanced overall chemical equation. The rate law can only be experimentally determined and can be used to predict the relationship between the rate of a reaction and the concentrations of reactants.
03: Rate Laws
The study of kinetics, involves very complex and sophisticated reactions that cannot be analyzed without a proposed mechanism. In order to analyze the kinetics of complex reactions, a series of steps in a reaction must be determined. As a result, scientists have developed a proposal that can help in visualizing this process, step by step, in a reaction: the reaction mechanism.
3.02: Reaction Mechanisms
An elementary reaction is a single step reaction with a single transition state and no intermediates.
Introduction
Elementary reactions add up to complex reactions; non-elementary reactions can be described by multiple elementary reaction steps. A set of elementary reactions comprises a reaction mechanism, which predicts the elementary steps involved in a complex reaction. Below are two reaction coordinates of two reactions. One describes an elementary reaction, and the other describes a non-elementary reaction.
Elementary Reaction (one step) Two Step Reaction
$\text{Reactants} \rightarrow \text{Products} \nonumber$
This is a sample reaction coordinate of an elementary reaction. Note that there is one transition state and no intermediate. Elementary steps cannot be broken down into simpler reactions.
$\text{Reactants} \rightarrow \text{Intermediates} \rightarrow \text{Products} \nonumber$
This is a sample reaction coordinate of a complex reaction. Note that it involves an intermediate and multiple transition A complex reaction can be explained in terms of elementary reactions.
Types of Elementary Reactions
The molecularity of a reaction refers to the number of reactant particles involved in the reaction. Because there can only be discrete numbers of particles, the molecularity must take an integer value. Molecularity can be described as unimolecular, bimolecular, or termolecular. There are no known elementary reactions involving four or more molecules.1
Unimolecular Reaction
Bimolecular Reaction
Termolecular Reaction
Unimolecular Reaction
A unimolecular reaction occurs when a molecule rearranges itself to produce one or more products. An example of this is radioactive decay, in which particles are emitted from an atom. Other examples include cis-trans isomerization, thermal decomposition, ring opening, and racemization. The rate at which a substance decomposes is dependent on its concentration. Unimolecular reactions are often first-order reactions as explained by Frederick Alexander Lindemann, which is referred to as the Lindemann mechanism.
Bimolecular Reaction
A bimolecular reaction involves the collision of two particles. Bimolecular reactions are common in organic reactions such as nucleophilic substitution. The rate of reaction depends on the product of the concentrations of both species involved, which makes bimolecular reactions second-order reactions.
Termolecular Reaction
A termolecular reaction requires the collision of three particles at the same place and time. This type of reaction is very uncommon because all three reactants must simultaneously collide with each other, with sufficient energy and correct orientation, to produce a reaction. There are three ways termolecular reactions can react, and all are third order.
Table 1: The three known types of elementary reactions:
Molecularity Elementary Step Rate Law Example
Unimolecular $A \rightarrow Products$ $rate = k[A]$ $N_2O_{4(g)} \rightarrow 2NO_{2(g)}$
Bimolecular $A + A \rightarrow Products$ $rate = k[A]^2$ $2NOCl \rightarrow 2NO_{(g)} + Cl_{2(g)}$
$A + B \rightarrow Products$ $rate = k[A][B]$ $CO_{(g)} + NO_{3(g)} \rightarrow NO_{2(g)} + CO_{2(g)}$
Termolecular $A + A +A \rightarrow Products$ $rate = k[A]^3$
$A + A + B \rightarrow Products$ $rate = k[A]^2[B]$ $2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}$ 2
$A + B + C \rightarrow Products$ $rate = k[A][B][C]$ $H + O_{2(g)} + M \rightarrow HO_{2(g)} + M$ 3
Practice Problems
1. How are non-elementary steps and elementary steps related? 2. Choose the correct statements.
a. An elementary step has 0 intermediates.
b. An elementary step has 1 intermediate.
c. An elementary step has 2 intermediates.
d. An elementary step has 0 transition states.
e. An elementary step has 1 transition state.
f. An elementary step has 2 transition states.
3. Which of the following elementrary reactions is a termolecular reaction?
a. $A + 2B + C \rightarrow D$
b. $A + B + B \rightarrow C$
c. all of the above
d. $2A + 2B + 2C \rightarrow 2D$
e. b and d
f. none of the above
4. Which rate law corresponds to a bimolecular reaction?
a. $rate = k[A][A]^2$
b. $rate = k[A][B]$
c. all of the above
d. $rate = k[A]^2$
e. b and d
f. none of the above
5. Give an example of a reaction with a molecularity of 1/2. 6. True or False: Given species A and B inside a container, instruments detect that three (3) collisions occured before product was formed. That is, we know a reaction occured after detecting three collisions in a box. We can conclude that the reaction is a termolecular reaction (as the reaction could have been produced from A+A+B or A+B+B).
Solutions
1. Non-elementary steps, or complex reactions, are sets of elementary reactions. The addition of elementary steps produces complex, non-elementary reactions.
2. The correct statements are "a" and "e". By definition of elementary reactions they have 0 intermediates because they cannot be broken down. Again by definition of an elementary reaction, a single-step reaction will have 1 transition state. There is no reaction with 0 transition states. Having 2 transition states implies having 1 intermediate, making the reaction non-elementary.
3. "e" is the answer.
"a" is not a termolecular reaction because it involves A + B + B + C, or 4 molecules
"b" is a termolecular reaction because it involves 3 particles: A + B + B
"c" is incorrect because "a" is incorrect
"d" is a termolecular reaction, simplifying to the reaction: $A + B + C \rightarrow D$, which involves 3 particles (A + B + C)
"e" is the correct answer because "b" and "d" are correct
4. "e" is the answer.
"a" is incorrect because the rate law describes a third-order reaction, which is true for termolecular reactions
"b" is a possible rate law for the bimolecular reaction: $A + B \rightarrow Products$
"c" is incorrect because "a" is incorrect
"d" is a possible rate law for the bimolecular reaction: $A + A \rightarrow Products$
"e" is the correct answer because "b" and "d" are correct
5. Impossible. The molecularity of a reaction MUST be an integer because there cannot be a "half particle" producing a reaction.
6. False; nothing can be concluded. Although a termolecular reaction requires the collision of three particles, the reverse logic is not necessarily true. That is, having three collisions is not sufficient for a termolecular reaction.
1. For example, particles A + A + B collide with each other at the same place and time. However, particle B was in the wrong orientation, so no reaction occurred. Instead, the two A particles were in the correct orientation and produced a reaction, which is a bimolecular reaction.
2. Consider a second example: two collisions between particles A + A + B occurred, but there was not enough energy to produce a reaction. Instead, a third collision between A and B had the sufficient energy and correct orientation to produce a reaction. Such a reaction is, again, only bimolecular.
3. A last example: particle A collides twice with a wall, and then once with B to produce a reaction. Such a reaction involving three collisions at different places and different time is only a bimolecular reaction.
Contributors and Attributions
• Tho Nguyen (UCD), Minh Ngo (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.02%3A_Reaction_Mechanisms/3.2.01%3A_Elementary_Reactions.txt
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The pre-equilibrium approximation assumes that the reactants and intermediates of a multi-step reaction exist in dynamic equilibrium.
Introduction
According to the steady state approximation article, reactions involving many steps can be analyzed using approximations. Like the steady state method, the pre-equilibrium approximation method derives an expression for the rate of product formation with approximated concentrations. Unlike the steady state method, the pre-equilibrium approximation does so by assuming that the reactants and intermediate are in equilibrium. Although both methods are used to solve for a rate of reaction, they are used under different conditions. The steady state method can only be used if the first step of a reaction is much slower than the second step, whereas the pre-equilibrium approximation requires the first step to be faster. These opposing conditions prevent the two methods from being interchangeable.
Both the steady state approximation and pre-equilibrium approximation apply to intermediate-forming consecutive reactions, in which the product of the first step of the reaction serves as the reactant for the second step.
As shown in Figure 1, there are three rate constants in the consecutive reaction involving an intermediate. These three rate constants, k1, k-1, and k2, are incredibly important for the pre-equilibrium approximation, as discussed in the next section.
A consecutive reaction found in living systems is the enzyme-substrate reaction. In this type of reaction, an enzyme binds to a substrate to produce an enzyme-substrate intermediate, which then forms the final product. As shown in Figure 2, this reaction follows the basic pattern of the consecutive reaction in Figure 1. The two reactants, E (enzyme) and S (substrate), form the intermediate ES. This enzyme-substrate intermediate forms the product P, usually an essential biomolecule. The enzyme then exits the reaction unchanged and able to catalyze future reactions.
As before, there are three reaction rates in this reaction: k1, k-1, and k2. The pre-equilibrium approximation uses the rate constants to solve for the rate of the reaction, indicating how quickly the reaction is likely to produce the biomolecule.
Derivation of the Rate Law
The enzyme substrate reaction in Figure 2 is used below to demonstrate a rate law derivation. The overall strategy for the pre-equilibrium reaction is as follows:
I. Assume reactants and intermediate are in equilibrium
1. Write out the general reaction
2. Break the overall reaction into elementary steps
3. Assume the reactants and the intermediate are in equilibrium throughout the reaction
II. Solve for the rate of product formation
1. Write out the rate of product formation using the equilibrium constant, K
2. Simplify this rate by incorporating the composite rate constant
3. Simplify to as few variables as possible
I. Assume Reactants and Intermediate are in Equilibrium
1. The general reaction used to derive a rate law is as following:
$E + S \xrightleftharpoons[k_{-1}]{\ k_1\ } ES \xrightarrow{k_2} E + P$
2. Breaking up the overall reaction into elementary steps gives:
$E + S \rightarrow ES \nonumber$ Rate of formation of ES = k1 [E][S]
$ES \rightarrow E + S \nonumber$ Rate of decay of ES = k-1 [ES]
$ES \rightarrow E + P \nonumber$ Rate of formation of P = k2 [ES]
3. In the steady state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. The pre-equilibrium approximation takes a different approach. Because the reversible reaction $E + S \rightarrow ES$ is much faster than the product formation of $ES \rightarrow E + P$, E, S, and ES are considered to be in equilibrium throughout the reaction. This concept greatly simplifies the mathematics leading to the rate law.
II. Solve for the Rate of Product Formation
4. Using the idea that the reactants and intermediate are in equilibrium, the rate of formation of E + P can be written as:
$\frac{d[P]}{dt}=k_2[ES]=k_2K[E][S] \nonumber$
where
$K = \frac{[ES]}{[E][S]} = \frac{k_1}{k_{-1}} \nonumber$
(K is the equilibrium constant)
5. The rate of product formation is simplified through the composite rate constant, k:
$k = k_2 K =\dfrac{k_1 k_2}{k_{-1}} \nonumber$
6. The rate of product formation can now be shortened to:
$\dfrac{d[P]}{dt} = k[E][S] \nonumber$
This equation can be solved for unknown variables in consecutive reactions involving intermediates.
Activation Energies in the Pre-equilibrium Reaction
Figure 4. Potential Energy (PE) vs Rxn Coordinate for two pre-equilibrium reactions.
Left graph
shows a positive overall activation energy (Ea > 0) and right graph shows a
negative overall activation energy (Ea < 0).
Many reactions have at least one activation energy that must be reached in order for the reaction to go forward. There are three activation energies in a pre-equilibrium reaction: two for the reversible steps and one for the final step. The three activation energies are denoted Ea,1 Ea,-1, and Ea,2:
Ea,1
Forward
Ea,-1
Reverse
Ea,2
Product Formation
The overall activation energy is given below:
$E_a= E_a,_1 + E_a,_2 - E_{a, -1} \nonumber$
The overall activation energy is positive if Ea1 + Ea2> Ea-1 The overall activation energy is negative if Ea1 + Ea2 < Ea -1
Practice Problem
For the reaction $A + B \xrightarrow[k_{-1}]{k_1} AB \xrightarrow[]{k_2} C$(see Figure 1), derive a rate law using the pre-equilibrium approximation.
Solution
$\dfrac{d[P]}{dt} = k[A][B]$ where $k =\dfrac{k_1 k_2}{k_{-1}}$
Outside Links
• Discussion of both pre-equilibrium and steady state approximations with Java applet figures: www.ch.cam.ac.uk/magnus
• Supplemental material on pre-equilibrium approximation (zoom in for better view): www.chem.yorku.ca/profs/bohme/notes/II3P11.gif
• An example of the pre-equilibrium approximation being used in current research (see equation 10 in Results and Discussion): http://www.biochemj.org/bj/373/0337/bj3730337.htm
• Tamara Kadir
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The rate determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate determining step can be compared to the neck of a funnel. The rate at which water flows through a funnel is limited/ determined by the width of the neck of the funnel and not by the rate at which the water is poured into the funnel. Like the neck of the funnel, the slow step of a reaction determines the rate of a reaction. Not all reactions have rate determining steps and has one only if one step is significantly slower than the other steps in the reaction.
Introduction
Rate determining step is the slowest step within a chemical reaction. The slowest step determines the rate of chemical reaction.The slowest step of a chemical reaction can be determined by setting up a reaction mechanisms. Many reactions do not occur in a single reaction but they happen in multiple elementary steps.
Consider this reaction:
$\ce{2NO2 +F2 -> 2NO2 F}, \nonumber$
which occurs via this mechanism
elementary step 1: $\ce{NO2 + F2 ->NO2F +F} \tag{slow}$
elementary step 2: $\ce{NO2+F -> NO2F} \tag{fast}$
For elementary step 1 has a rate constant of k1 and for elementary step 2 it has a rate constant of k2. The slowest step in this mechanism is elementary step 1 which is our rate determining step. Looking at this mechanism I see Intermediates. Intermediates are molecules or elements that are found on the product of one step but are also located in the reactant of another step. In this case we have two intermediates $\ce{NO2}$ and $\ce{F}$.
The rate equation is derived by the slowest step in the reaction. When writing a rate equation you set up the equation by writing rate is equal to the rate constant of the slowest step times the concentrations of the reactant or reactants raised to there reaction order. Lets look at elementary step one.
elementary step one: $\ce{NO2 +F2 -> NO2F + F}\nonumber$
Here in this example rate=k1[NO2][F2].
Example $1$
For this reaction
$\ce{2NO + O2 -> 2NO2} \nonumber$
1. What are the intermediates ?
2. What is the rate equation?
elementary step one:
$\ce{NO + NO <=> N2O2 } \tag{fast equilibrium}$
elementary step two:
$\ce{N2O2 + O2 -> 2 NO2 }\tag{slow}$
Solution
1: N2O2 is found on the product side and the reactant side.
2: rate= k2[N2O2][O2]; N2O2 gets canceled out leaving the overall reaction rate.
Example $2$
1. What is the overall reaction?
2. What is the rate equation?
3. Are there intermediates if so what are they?
4. What is the rate determining step?
elementary step one:
$\ce{Br2 + M <=> Br + Br +M}\tag{fast equilibrium}$
elementary step two:
$\ce{Br +H2 -> HBr +H }\tag{slow}$
elementary step three:
$\ce{H + Br2 -> HBR + Br }\tag{fast}$
Solution
1. H2+Br2 -> 2HBr ; Br and H will be canceled out and therefore they won't appear in the overall reaction.
2. rate=k2 [Br][H2]
3. Br, H are the intermediates in the reaction.
4. elementary step 2 is the slowest step in the mechanism.
Contributors and Attributions
• Galaxy Mudda, Pamela Chaha, Florence-Damilola Odufalu, Filmon Tewolde(UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.02%3A_Reaction_Mechanisms/3.2.03%3A_Rate_Determining_Step.txt
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Diffusion in a gas is the random motion of particles involved in the net movement of a substance from an area of high concentration to an area of low concentration. Each particle in a given gas continues to collide with other particles. In regions of the gas where the particle density is the highest, the particles bounce off each other and the boundary of their container at a greater rate than particles in less-dense regions.
Introduction
For a gas, the rate at which diffusion occurs is proportional to the square root of the density of the gas. The density of a gas is equal to the mass of the gas divided by the volume of the gas. If the volume is held constant one gas is compared with another with another,
$\dfrac{R_2}{R_1} = \sqrt{\dfrac{M_1}{M_2}}$
where R is the rate of diffusion in mol/s and M is the molar mass in g/mol. This is known as Graham's law of diffusion.
Fick's First Law of Diffusion
For a volume of solution that does not change:
$J = -D\dfrac{dc}{dx} \nonumber$
• When two different particles end up near each other in solution, they may be trapped as a result of the particles surrounding them, which is known as the cage effect or solvent cage.
Two different particles colliding may be represented as a 2nd order reaction: $A + B \rightarrow AB$
$AB = K_d[A][B] \nonumber$
Notice the use of Kd to denote the diffusion rate constant.
If circumstances change and either of the particles is able to diffuse out of the solvent cage, then the following 1st order reaction $AB \rightarrow A + B$ is possible, then:
$AB = K_d'[AB] \nonumber$
There now exists a reaction for the formation of the AB complex as well as the breakdown of the AB complex into products.
$k = \dfrac{K_aK_d}{K_a + K_d'} \nonumber$
$V = K[A][B]$
$A + B \rightarrow AB \rightarrow Products \nonumber$
The net rate of formation for AB can now be determined:
$\dfrac{d[AB]}{dt} = (A+B \rightarrow AB) - (AB \rightarrow A + B) - (AB \rightarrow Products)$
$\dfrac{d[AB]}{dt} = K_d[A][B] - K_d'[AB] - K_a[AB] \nonumber$
Assuming steady state conditions:
$\dfrac{d[AB]}{dt} = K_d[A][B] - K_d'[AB] - K_a[AB] = 0 \nonumber$
$[AB] = \dfrac{K_d[A][B]}{K_a + K_d'}$
The final rate of product formation taking into account both diffusion and activation:
$V = \dfrac{K_d[A][B]}{K_a + K_d'}$
Diffusion-controlled limit
If the rate at which particle A encounters particle B is much slower than the rate at which AB dissociates, then Kd' is essentially zero.
$V = \dfrac{K_d[A][B]}{K_a}$
Activation-controlled limit
The rate at which particle A encounters and reacts with particle B may exceed the rate at which the AB complex breaks apart into a product by a significant quantity. If the rate of at which AB decomposes is slow enough that Ka in the denominator may be ignored, the following results:
$V = \dfrac{K_d[A][B]}{K_d'}$
The Rate Constant Kd
Viscosity and rate of diffusion may be related by the following formula:
$K_d = \dfrac{8RT}{3n}$
where n is the viscosity of the solution.
Practice Problems
1. Compare the rate of diffusion between fluorine and chlorine gases. Fluorine gas, F2, has a molecular mass of 32 grams. Chlorine gas, Cl2, has a molecular mass of 70.90 grams.
2. Gas A is 0.75 times as fast as Gas B. The mass of Gas B is 32 grams. What is the mass of Gas A?
3. Determine the rate of diffusion (flux) for aspirin dissolving through the stomach lining. C1= 50 mg/L and C2 = 290 mg/L. The diffusivity constant of aspirin is 0.29×10-9 cm2/s and the thickness of the stomach lining is approximately 0.5 cm.
4. How long will it take oxygen to diffuse 0.5 cm below the surface of a still lake if D = 1×10-5cm2/s?
5. If it takes 5 seconds for oxygen to diffuse to the center of a bacterial cell that is 0.02 cm in diameter, determine the diffusivity constant.
Solutions to Practice Problems
1. Using Graham's law of diffusion:
(Rate1/Rate2) = (Mass2/Mass1)1/2
(RateF2/RateCl2) = (70.9g/32g)1/2 = 1.49
Fluorine gas is 1.49 times as fast as chlorine gas.
2. Using Graham's law of diffusion:
(Rate1/Rate2) = (Mass2/Mass1)1/2
(RateA/RateB) = (MassB/MassA)1/2
0.75 = (32g/MassA)1/2
0.752=(32g/MassA)
MassA = (32g/0.5625)
MassA = 56.8888g
3. Using Fick's first law:
J = -D(dc/dx)
Where:
J = unknown flux
D = 0.29×10-9 cm2/s
dc = (C1 - C2) = 50mg/L - 290mg/L = -240mg/L, which is equivalent to -240mg/1000cm3 =-0.24mg/cm3
dx = 0.5cm
J = (0.29×10-9cm2/s)[(-0.24mg/cm3)/(0.5cm)] = 1.39×10-10mg/s×cm2
J = 1.39×10-10mg/s×cm2
4. Using Fick's second law:
T = x2/2D
Where:
T = our unknown (time)
x = 0.5 cm
D = 1×10-5cm2/s
T = (0.5cm)2/[2(1×10-5cm2/s)]
T = 1.25×104 seconds
5. Using Fick's second law:
First, rearrange the equation T = x2/2D to solve for D --> D = x2/2T
Where:
D = our unknown (diffusivity constant)
x = 0.01 cm (distance from the outside to the center of the cell)
T = 5s
D = (0.01cm)2/[2(5s)]
D = 1×10-5cm2/s
Contributors
• Eric Stouffer (UCD), Anne Slisz (UCD)
3.2.05: Reaction Intermediates
A reaction intermediate is transient species within a multi-step reaction mechanism that is produced in the preceding step and consumed in a subsequent step to ultimately generate the final reaction product. Intermediate reactions are common in the biological world; a prime example can be seen in the metabolism of metabolites and nutrients.
Introduction
The lifetime of an intermediate is usually short because it is usually consumed to make the next product of the reaction sequence. In a biochemical pathway, the overall reaction may seem to form only but product but may require multiple smaller steps to achieve that goal. It must be kept “in mind that an intermediate is always formed in an early elementary step and consumed in a later elementary step” (1).
Example $1$:
The general setup of a multi-step equation is:
$CH_4 + Cl_2 \rightarrow X \;\;\; \text{Step 1} \nonumber$
$X \rightarrow CH_3Cl + HCl \;\;\; \text{Step 2} \nonumber$
The first step yields a substance X, which is the intermediate of the chlorination reaction "formed on the pathway between reactants and products" (2).
Example $1$:
An example of a biological process is glycolysis. The overall balanced equation of glycolysis is
D- Glucose + NAD+ + 2 ADP + 2 Pi --> 2 Pyruvate + 2NADH + 2H+ +2ATP + 2H2O
GLC --> PYR
Glucose (GLC) is the beginning reactant and is converted through a series of steps to form the final product, pyruvate (PYR). However, the intermediate “appears in the mechanism of the reaction but is not in the overall balanced equation.” One example of an intermediate in a glycolysis step reaction is the conversion of:
Glucose --> Glucose-6-phosphate
RED = intermediate
The first step of glycolysis yields glucose-6-phosphate from a glucose molecule. However, in the overall balanced reaction of glycolysis, there is no presence of glucose-6-phosphate written because it exists for a short time before it is consumed in the next reaction.
Problems
Identify and circle the intermediates of the glycolysis pathway and explain why they are considered intermediates. NOTE: *- DHAP is isomerized into G-3-P therefore the downstream products should be doubled.
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The steady state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction are constant. This method can only be applied when the first step of the reaction is significantly slower than subsequent step in an intermediate-forming consecutive reaction.
Introduction
Before discussing the steady state approximation, it must be understood that the approximation is derived to simplify the kinetic expression for product concentration, [product]. Consider the following sequential reaction:
$A \xrightarrow[]{k_1} B \xrightarrow {k_2} C \nonumber$
Calculating the [product] depends on all the rate constants in each step. For example, if the kinetic method was used to find the concentration of C, [C], at time t in the above reaction, the expression would be
$[C] = [A]_0 \left(1+ \dfrac{k_2e^{-k_1t}-k_1e^{-k_2t})}{k_1 - k_2}\right) \label{1}$
With a more complicated mechanisms, the kinetic expression becomes harder to derive. To simplify this calculation, we often use one of two approximations for determining the overall reaction rates of consecutive reactions: the steady state approximation and the pre-equilibrium approximation. This article concerns the steady state approximation.
Steady State Approximation
The steady state approximation is applies to a consecutive reaction with a slow first step and a fast second step ($k_1 \ll k_2$). If the first step is very slow in comparison to the second step, there is no accumulation of intermediate product, such as product B in the above example.
$\dfrac{d[B]}{dt} = 0 = k_1[A] - k_2[B] \label{2}$
Thus
$[B] = \dfrac{k_1[A]}{k_2} \label{3}$
From the mechanism:
$\dfrac{d[C]}{dt} = k_2[B] = \dfrac{k_2k_1[A]}{k_2} = k_1[A] \label{4}$
Solving for $[C]$:
$[C] = [A]_0 (1- e^{-k_1t}) \label{5}$
Equation $\ref{5}$ is much simpler to derive than Equation $\ref{1}$, especially with a more complicated multi-step reaction mechanisms.
Example $1$
Consider the reaction:
$A + 2B \xrightarrow[]{} C \nonumber$
1. What is the expected rate law according to the following proposed multi-step mechanism under the steady state approximation with $k_2 \gg k_{-1}$) for the following mechanism: $A + B \ce{<=>[k_1][k_{-1}]} I \tag{Slow}$ $I + B \xrightarrow[]{k_2} C \tag{Fast}$
2. If $x$ is the order of the reaction with respect to $A$, $y$ is the order of the reaction with respect to $B$, and $n$ is the overall reaction order. What are the values of $x$, $y$, and $n$?
Solution
a: First we use the Steady State Approximation for the intermediate (i.e., Equation \ref{2})
$\dfrac{d[I]}{dt} = k_1[A][B] - k_{-1}[I] - k_2[I][B] = 0 \nonumber$
then we solve for the (steady-state) concentration of the intermediate
$[I] = \dfrac{k_1[A][B]}{ k_{-1} +k_2[B]} \nonumber$
Because the second step is much faster than the first step, then $k_2 \gg k_{-1}$ then $k_{-1} \approx 0$ for this approximation and the above equation can be simplified to
$[I] = \dfrac{k_1[A]}{k_2} \nonumber. \nonumber$
The rate law for the production of $[C]$ can be constructed directly from the second step and when the steady-state concentration of $I$ is added, the final rate law expression is derived.
\begin{align*}\dfrac{d[C]}{dt} & = k_2[I][B] \[4pt] &= \dfrac{k_1k_2[A][B]}{k_2} \[4pt] &= k_1[A][B] \end{align*}
b: Direct inspection of the final rate law derived above gives these parameters:
• $x = 1$
• $y = 1$
• $n = 2$
Use of the Steady-State Approximation in Enzyme Kinetics
In 1925, George E. Briggs and John B. S. Haldane applied the steady state approximation method to determine the rate law of the enzyme-catalyzed reaction (Figure 1). The following assumptions were made:
1. The rate constant of the first step must be slower than the rate constant of the second step ($k_1 \ll k_2$), hence $\dfrac{d[ES]}{dt} = 0 \nonumber$
2. Enzyme concentration must be significantly lower than the substrate concentration to keep the first step slower than the second step.
This gives the following:
$\dfrac{d[P]}{dt} = k_2[ES] \label{6}$
where
$\dfrac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES] \label{7}$
Because
$[S] \gg [E] \label{8}$
Using the second assumption and the fact that enzyme concentration equals the initial concentration of enzyme minus the concentration of the enzyme-substrate intermediate,
$[E] = [E]_o - [ES] \label{9}$
The following equation is obtained:
$k_1[E]_o[S] = k_{-1}[ES] + k_2[ES] + k_1[ES][S] \label{10}$
From this equation, the concentration of the ES intermediate can be found:
$[ES] = \dfrac{k_1[E]_o[S]}{(k_{-1} + k_2) + k_1[S]} \label{11}$
Substitute this into Equation $\ref{6}$ gives,
$\dfrac{d[P]}{dt} = \dfrac{k_2[E]_0[S]}{[(k_1+k_2)/k_1]+[S]} = \dfrac{k_2[E]_0[S]}{K_M+[S]} \label{12}$
where
$K_M = \dfrac{k_{-1}+k_{2}}{k_1} \label{13}$
Because in most of the cases, only the initial $d[P]/dt$ is measured to determine the rate of product formation, Equation \ref{12} can be rewritten as:
$v_o = \dfrac{d[P]_0}{dt} = \dfrac{k_2[E]_0[S]}{K_M+[S]} \label{14}$
Because $[E]_o = v_{max}/k_2$. Equation $\ref{14}$ becomes the following:
\begin{align} v_0 &= \dfrac{d[P]_o}{dt} \[4pt] &= \dfrac{(k_2/k_2)v_{max}[S]}{K_M+[S]} \label{15} \ &= \dfrac{v_{max}[S]}{K_M+[S]} \label{16} \end{align}
This equation is a useful tool to in calculating $v_{max}$ and $K_M$ (the Michaelis constant) of an enzyme by using the Lineweaver-Burk plot (1/[S] vs. 1/v0) or the Eadie-Hofstee plot (v0/[S] vs. v0).
Problems
Given the reaction $A \xrightarrow[]{k_1} B \xrightarrow[]{k_2} C$
where k1= 0.2 M-1s-1 , k2 = 2000 s-1
1. Write the reaction rates for A, B, and C.
2. Is this a steady-state reaction?
3. Write the expression for d[C]/dt using the Steady State Approximation
4. Calculate d[C]/dt if [A] = 1M
5. Calculate [C] at t = 3 s and [A]0 = 2M
Solutions
1) d[A]/dt = -k1[A]; d[B]/dt = k1[A] - k2[B]; d[C]/dt = k2[B]
2) Because k1 is much larger than k2, this is a steady state reaction.
3) d[C]/dt = k2[B]
where d[B]/dt = k1[A] - k2[B] = 0
so, [B] = k1[A]/k2
Substitute this into d[C]/dt
d[C]/dt = k1[A]
4) d[C]/dt = 0.2M-1s-1(1M) = 0.2 s-1
5) [C] = [A]0 (1-e-k1t) = 2M(1-e-0.2(3)) = 0.9 M
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In studying a chemical reaction, it is important to consider not only the chemical properties of the reactants, but also the conditions under which the reaction occurs, the mechanism by which it takes place, the rate at which it occurs, and the equilibrium toward which it proceeds. According to the law of mass action, the rate of a chemical reaction at a constant temperature depends only on the concentrations of the substances that influence the rate. The substances that influence the rate of reaction are usually one or more of the reactants, but can occasionally include products. Catalysts, which do not appear in the balanced overall chemical equation, can also influence reaction rate. The rate law is experimentally determined and can be used to predict the relationship between the rate of a reaction and the concentrations of reactants.
• 3.3.1: Order of Reaction Experiments
This is an introduction to some of the experimental methods used in school laboratories to find orders of reaction. There are two fundamentally different approaches to this: investigating what happens to the initial rate of the reaction as concentrations change, and following a particular reaction to completion and processing the results from that single reaction.
• 3.3.2: Rate Laws
Reactions are often monitored by some form of spectroscopy. In spectroscopy, "light" or some other frequency of electromagnetic radiation passes through a reaction sample. The light interacts with the molecules in the sample, which absorb particular frequencies of light. Less light exits the sample than the amount that entered it; the amount that exits is measured by a detector on the other side.
• 3.3.3: Reaction Order
The reaction order is the relationship between the concentrations of species and the rate of a reaction.
3.03: The Rate Law
This is an introduction to some of the experimental methods used in school laboratories to find orders of reaction. There are two fundamentally different approaches to this: investigating what happens to the initial rate of the reaction as concentrations change, and following a particular reaction to completion and processing the results from that single reaction.
Initial rate experiments
The simplest initial rate experiments involve measuring the time taken for some recognizable event to happen early in a reaction. This could be the time required for 5 cm3 of gas to be produced, for a small, measurable amount of precipitate to form, or for a dramatic color change to occur. Examples of these three indicators are discussed below.
The concentration of one of the components of the reaction could be changed, holding everything else constant: the concentrations of other reactants, the total volume of the solution and the temperature. The time required for the event to occur is then measured. This process is repeated for a range of concentrations of the substance of interest. A reasonably wide range of concentrations must be measured.This process could be repeated by altering a different property.
Understanding the results
Consider a simple example of an initial rate experiment in which a gas is produced. This might be a reaction between a metal and an acid, for example, or the catalytic decomposition of hydrogen peroxide. If volume of gas evolved is plotted against time, the first graph below results.
A measure of the rate of the reaction at any point is found by measuring the slope of the graph. The steeper the slope, the faster the rate. Because the initial rate is important, the slope at the beginning is used.
In the second graph, an enlarged image of the very beginning of the first curve, the curve is approximately straight. This is only a reasonable approximation when considering an early stage in the reaction. As the reaction progresses, the curvature of the graph increases. The slope of this linear region is V/t.
Suppose the experiment is repeated with a different (lower) concentration of the reagent. Again, the time it takes for the same volume of gas to evolve is measured, and the initial stage of the reaction is studied:
The initial rates (in terms of volume of gas produced per second) are given below:
experiment 1
experiment 2
Now suppose V is unknown.
Suppose, for example, that instead of measuring the time taken to collect 5 cm3 of gas, the gas is collected up to a mark on the side of a test tube. If simply comparing initial rates, then it the distinction does not matter. As shown in the table above, the initial rate is inversely proportional to the time elapsed:
$\text{initial rate} \propto \frac{1}{t} \nonumber$
In experiments of this type, 1/t is often used as a measure of the initial rate.
The quantity 1/t can be plotted against the varying concentrations of the reactant of interest. If the reaction is first order with respect to that substance, then a straight line results; in a first order reaction, the rate is proportional to the concentration.
If a curve results, the reaction is not first order. It might be second order or a fractional order such as 1.5 or 1.78. The best way to deal with this is to plot what is known as a "log graph." In a reaction involving A, with an order of n with respect to A, the important part of the rate equation is the following:
$\text{rate} \propto [A]^n \nonumber$
Taking the log of each side of the equation gives:
$\log (\text{rate}) \propto n\log [A] \nonumber$
If log(rate) is plotted against log[A], a straight line with slope n is generated. The slope of the line is equal to the order of the reaction. Therefore, all rate values should be converted to log(rate). Then all the values for [A] are converted into log[A], and then the graph is plotted.
Example 1: The catalytic decomposition of hydrogen peroxide
This is an example of measuring the initial rate of a reaction producing a gas.
A simple set-up for this process is given below:
The reason for the weighing bottle containing the catalyst is to avoid introducing errors at the beginning of the experiment. The catalyst must be added to the hydrogen peroxide solution without changing the volume of gas collected. If it is added to the flask using a spatula before replacing the bung, some gas might leak out before the bung is replaced. Alternatively, air might be forced into the measuring cylinder. Either would render results meaningless.
To start the reaction, the flask is shaken until the weighing bottle falls over, and then shaken further to make sure the catalyst mixes evenly with the solution.
Alternatively, a special flask with a divided bottom could be used, with the catalyst in one side and the hydrogen peroxide solution in the other. The two are easily mixed by tipping the flask.
Using a 10 cm3 measuring cylinder, initially full of water, the time taken to collect a small fixed volume of gas can be accurately recorded. A small gas syringe could also be used.
To study the effect of the concentration of hydrogen peroxide on the rate, the concentration of hydrogen peroxide must be changed and everything else held constant —the temperature, the total volume of the solution, and the mass of manganese(IV) oxide. The manganese(IV) oxide must also always come from the same bottle so that its state of division is always the same.
The same apparatus can be used to determine the effects of varying the temperature, catalyst mass, or state of division due to the catalyst
Example 2: The thiosulphate-acid reaction
Mixing dilute hydrochloric acid with sodium thiosulphate solution causes the slow formation of a pale yellow precipitate of sulfur.
$Na_2S_2O_{2(aq)} + 2HCl_{(aq)} \rightarrow 2NaCl_{(aq)} + H_2O_{(l)} + S_{(s)} + SO_{2(g)} \nonumber$
A very simple, but very effective, way of measuring the time taken for a small fixed amount of precipitate to form is to stand the flask on a piece of paper with a cross drawn on it, and then look down through the solution until the cross disappears.
A known volume of sodium thiosulphate solution is placed in a flask. Then a small known volume of dilute hydrochloric acid is added, a timer is started, the flask is swirled to mix the reagents, and the flask is placed on the paper with the cross. The timer is used to determine the time for the cross to disappear. The process is repeated using a smaller volume of sodium thiosulphate, but topped up to the same original volume with water. Everything else is exactly as before.
The actual concentration of the sodium thiosulphate does not need to be known. In each case the relative concentration could be recorded. The solution with 40 cm3 of sodium thiosulphate solution plus 10 cm3 of water has a concentration which is 80% of the original, for example. The one with 10 cm3 of sodium thiosulphate solution plus 40 cm3 of water has a concentration 20% of the original.
The quantity 1/t can again be plotted as a measure of the rate, and the volume of sodium thiosulphate solution as a measure of concentration. Alternatively, relative concentrations could be plotted. In either case, the shape of the graph is the same.
The effect of temperature on this reaction can be measured by warming the sodium thiosulphate solution before adding the acid. The temperature must be measured after adding the acid, because the cold acid cools the solution slightly.This time, the temperature is changed between experiments, keeping everything else constant. To get reasonable times, a diluted version of the sodium thiosulphate solution must be used. Using the full strength, hot solution produces enough precipitate to hide the cross almost instantly.
Example 3: The Iodine Clock Reactions
There are several reactions bearing the name "iodine clock." Each produces iodine as one of the products. This is the simplest of them, because it involves the most familiar reagents.
The reaction below is the oxidation of iodide ions by hydrogen peroxide under acidic conditions:
$H_2O_{2(aq)} + 2I_{(aq)}^- + 2H^+ \rightarrow I_{2(aq)} + 2H_2O_{(l)} \nonumber$
The iodine is formed first as a pale yellow solution, darkening to orange and then dark red before dark gray solid iodine is precipitated.
Iodine reacts with starch solution to give a deep blue solution. If starch solution is added to the reaction above, as soon as the first trace of iodine is formed, the solution turns blue. This gives no useful information. However, iodine also reacts with sodium thiosulphate solution:
$2S_2O^{2-}_{3(aq)} + I_{2(aq)} \rightarrow S_2O_{6(aq)}^{2-} + 2I^-_{(aq)} \nonumber$
If a very small amount of sodium thiosulphate solution is added to the reaction mixture (including the starch solution), it reacts with the iodine that is initially produced, so the iodine does not affect the starch, and there is no blue color. However, when that small amount of sodium thiosulphate is consumed, nothing inhibits further iodine produced from reacting with the starch. The mixture turns blue.
Following the course of a single reaction
Rather than performing a whole set of initial rate experiments, one can gather information about orders of reaction by following a particular reaction from start to finish. There are two different ways this can be accomplished.
1. Samples of the mixture can be collected at intervals and titrated to determine how the concentration of one of the reagents is changing.
2. A physical property of the reaction which changes as the reaction continues can be measured: for example, the volume of gas produced.
These approaches must be considered separately.
Sampling the reaction mixture
Bromoethane reacts with sodium hydroxide solution as follows:
$CH_3CH_2Br + OH^- \rightarrow CH_3CH_2OH + Br^- \nonumber$
During the course of the reaction, both bromoethane and sodium hydroxide are consumed. However, it is relatively easy to measure the concentration of sodium hydroxide at any one time by performing a titration with a standard acid: for example, with hydrochloric acid of a known concentration.
The process starts with known concentrations of sodium hydroxide and bromoethane, and it is often convenient for them to be equal. Because the reaction is 1:1, if the concentrations are equal at the start, they remain equal throughout the reaction. Samples are taken with a pipette at regular intervals during the reaction, and titrated with standard hydrochloric acid in the presence of a suitable indicator. The problem with this approach is that the reaction is still proceeding in the time required for the titration. In addition, only one titration attempt is possible, because by the time another sample is taken, the concentrations have changed.
There are two ways around this problem:
1. The reaction can be slowed by diluting it, adding the sample to a larger volume of cold water before the titration. Then the titration is performed as quickly as possible. This is most effective if the reaction is carried out above room temperature. Cooling it as well as diluting it slows it down even more.
2. If possible (and it is possible in this case) it is better to stop the reaction completely before titrating. In this case, this can be accomplished by adding the sample to a known, excess volume of standard hydrochloric acid. This consumes all the sodium hydroxide in the mixture, stopping the reaction.
At this point the resulting solution is titrated with standard sodium hydroxide solution to determine how much hydrochloric acid is left over in the mixture. This allows one to calculate how much acid was used, and thus how much sodium hydroxide must have been present in the original reaction mixture. This technique is known as a back titration.
Processing the results
This process generates a set of values for concentration of (in this example) sodium hydroxide over time. The concentrations of bromoethane are, of course, the same as those obtained if the same concentrations of each reagent were used. These values are plotted to give a concentration-time graph, such as that below:
The rates of reaction at a number of points on the graph must be calculated; this is done by drawing tangents to the graph and measuring their slopes.
These values are then tabulated. The quickest way to proceed from here is to plot a log graph as described further up the page. All rates are converted to log(rate), and all the concentrations to log(concentration). Then, log(rate) is plotted against log(concentration). The slope of the graph is equal to the order of reaction.
In the example of the reaction between bromoethane and sodium hydroxide solution, the order is calculated to be 2. Notice that this is the overall order of the reaction, not just the order with respect to the reagent whose concentration was measured. The rate of reaction decreases because the concentrations of both of the reactants decrease.
Example 4: Following the course of the reaction
A familiar example is the catalytic decomposition of hydrogen peroxide (used above as an example of an initial rate experiment).
This time, measure the oxygen given off using a gas syringe, recording the volume of oxygen collected at regular intervals.
The practical side of this experiment is straightforward, but the calculation is not. The problem is that the volume of the product is measured, whereas the concentration of the reactants is used to find the reaction order. This means that the concentration of hydrogen peroxide remaining in the solution must be determined for each volume of oxygen recorded. This requires ideal gas law and stoichiometric calculations.
The table of concentrations and times is processed as described above.
Colorimetry
In any reaction involving a colored substance (either reacting or being produced), the reaction progress can be tracked using a colorimeter.
This entire setup can be contained in a small box. The color of the light can be controlled using a colored filter or diffraction grating. The color is chosen to be the frequency of light which is absorbed by the sample.
Taking copper(II) sulfate solution as a familiar example, a red filter is used because copper(II) sulfate solution absorbs red light. The more concentrated the solution is, the more of the red light it absorbs. A commonly quoted example of the use of colorimetry in reaction kinetics is the reaction between propanone and iodine in the presence of an acid catalyst:
$CH_3COCH_3 + I_2 \overset{H^+}{\longrightarrow} CH_3COCH_2I + H^+ + I^- \nonumber$
The solution of iodine in propanone is initially brown, and then fades through orange to yellow to colorless as the iodine is consumed. A colorimeter measures the amount of light absorbed as it passes through a solution, recorded as the absorbance of the solution. It is common to plot a calibration curve for a colorimeter by making up solutions of the colored substance of known concentration and then measuring the absorbance of each under the same conditions. Absorbance is then plotted against concentration to create the calibration curve.
Two other methods
pH measurements: In a reaction in which hydrogen ions are reacting or being produced, changes in their concentration can in principle be tracked using a pH meter. The pH is a measure of hydrogen ion concentration, and an actual hydrogen ion concentration is easily calculated from a pH. However, when measuring pH over a fairly narrow range of hydrogen ion concentrations, the pH does not change much.
For example, a solution containing 0.2 mol dm-3 H+ has a pH of 0.70. By the time that the concentration has decreased to 0.1 mol dm-3, the pH has only increased to 1.00. The feasibility of using a pH meter depends on its accuracy; if the pH meter only records to 0.1 pH units, the results may not be adequate.
Conductivity measurements: The electrical conductivity of a liquid depends on the number of ions present, and the nature of the ions. For example, consider the following reaction:
During the course of the reaction, as hydrogen ions and iodide ions are consumed, the conductivity of the mixture decreases.
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In practice, the easiest way to measure the speed of a reaction is to measure the concentration of either the reactants or the products over time.
The concentration of black dots is higher in the beaker on the right than in the beaker on the left.
Reactions are often monitored by some form of spectroscopy. In spectroscopy, "light" or some other frequency of electromagnetic radiation passes through a reaction sample. The light interacts with the molecules in the sample, which absorb particular frequencies of light. Less light exits the sample than the amount that entered it; the amount that exits is measured by a detector on the other side.
If the concentration of the sample is different, a different amount of light from the spectrometer is absorbed. For instance, suppose the sample is more concentrated. The more molecules there are, the more light is absorbed. Because the beam of light travels through the sample in a straight line, the more concentrated the solution, the more molecules it encounters.
It is simple to calibrate the instrument to determine concentration from the amount of light absorbed. In addition, the light may interact with the reactant molecules and product molecules in different ways. This means that the absorption frequency of the reactants, and not the products, can be monitored, and thus changes in reactant concentration can be measured. This also works for the products..
The rate of reaction is written as:
$\text{rate} = \frac{d[product]}{dt} \nonumber$
In other words, the rate is equal to the change in concentration of product with change in time.
Concentration can be measured in several units. Frequently, the concentration of a solution is measured, and units such as grams per liter or, much more commonly, moles per liter, are used. The change in time is most often measured in seconds.
The rate of the reaction can also be written as follows:
$\text{rate} = -\frac{d[reactant]}{dt} \nonumber$
The rate is the change in concentration of reactant with change in time. The negative sign indicates that the reactant is being consumed over time as it turns into product, so its concentration is decreasing.
Kinetic studies are important in understanding reactions. Not only are they important in industry, but they are also used to understand biological processes, especially enzyme-catalyzed reactions. They also play a role in environmental and atmospheric chemistry, as part of an effort to understand a variety of issues ranging from the fate of prescription pharmaceuticals in wastewater to the cascade of reactions involved in the ozone cycle.
3.3.03: Reaction Order
The reaction order is the relationship between the concentrations of species and the rate of a reaction.
Introduction
The order of a rate law is the sum of the exponents of its concentration terms. Once the rate law of a reaction has been determined, that same law can be used to understand more fully the composition of the reaction mixture. More specifically, the reaction order is the exponent to which the concentration of that species is raised, and it indicates to what extent the concentration of a species affects the rate of a reaction, as well as which species has the greatest effect. For the N2O5 decomposition with a rate law of k[N2O5], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. It can also be said that the reaction is "first order in N2O5". For more complicated rate laws, the overall reaction order and the orders with respect to each component are used. As an example, consider the following reaction,
$A + 3B + 2C \rightarrow \text{products} \nonumber$
whose experimental rate law is given by:
$\text{rate} = k[A][B]^2 \nonumber$
This reaction is third-order overall, first-order in A, second-order in B, and zero-order in C.
Zero-order means that the rate is independent of the concentration of a particular reactant. Of course, enough C must be present to allow the equilibrium mixture to form.
Relation to Rate Law
For the reaction:
$aA + bB \longrightarrow P \nonumber$
The rate law is as follows:
$rate=k[A]^x[B]^y \nonumber$
where
• [A] is the concentration of species A,
• x is the order with respect to species A.
• [B] is the concentration of species B,
• y is the order with respect to species B
• k is the rate constant.
• n is the reaction order for the whole chemical reaction. This can be found by adding the reaction orders with respect to the reactants. In this case, n = x + y.
Simple Rules
The order of a reaction is not necessarily an integer. The following orders are possible:
• Zero: A zero order indicates that the concentration of that species does not affect the rate of a reaction
• Negative integer: A negative order indicates that the concentration of that species INVERSELY affects the rate of a reaction
• Positive integer: A positive order indicates that the concentration of that species DIRECTLY affects the rate of a reaction
• Non-Integer: Non-integer orders, both positive and negative, represent more intricate relationships between concentrations and rate in more complex reactions.
Example 1
The rate of oxidation of bromide ions by bromate in an acidic aqueous solution,
$6H^+ + BrO_3^– + 5Br^– \rightarrow 3 Br_2 + 3 H_2O \nonumber$
is found to follow the following rate law:
$\text{rate} = k[Br^-][BrO_3^-][H^+]^2 \nonumber$
What happens to the rate if, in separate experiments, (a) [BrO3] is doubled;(b) the pH is increased by one unit; (c) the solution is diluted to twice its volume, with the pH held constant using a buffer?
Solution
1. Because the rate is first-order in bromate, doubling its concentration doubles the reaction rate.
2. Increasing the pH by one unit decreases the [H+] by a factor of 10. Because the reaction is second-order in [H+], this decreases the rate by a factor of 100.
3. Dilution reduces the concentrations of both Br2 and BrO3 to half their original values. Doing this to each concentration alone would reduce the rate by a factor of 2, so reducing both concentration reduces the rate by a factor of 4, to (½)×(½) = ¼ of its initial value
Methods to Determining Reaction Order
For chemical reactions that require only one elementary step, the values of x and y are equal to the stoichiometric coefficients of each reactant. For chemical reactions that require more than one elementary step, this is not always the case. However, there are many simple ways of determining the order of a reaction. One very popular method is known as the differential method.
The Differential Method
The differential method, also known as the initial rates method, uses an experimental data table to determine the order of a reaction with respect to the reactants used. Below is an example of a table corresponding with the following chemical reaction:
$A + B \longrightarrow P \nonumber$
Experiment [A] M [B] M Rate M Min-1
1 0.100 0.100 1.0 x 10-3
2 0.200 0.100 1.0 X 10-3
3 0.100 0.200 2.0 x 10-3
When looking at the experiments in the table above, it is important to note factors that change between experiments. In order to determine the reaction order with respect to A, one must note in which experiment A is changing; that is, between experiments 1 and 2. Write a rate law equation based on the chemical reaction above.
This is the rate law:
$\text{rate} = k[A]^x[B]^y \nonumber$
Next, the rate law equation from experiment 2 must be divided by the rate law equation for experiment 1. Notice that the [B]y term cancels out, leaving "x" as the unknown variable. Simple algebra reveals that x = 0.
The same steps must be taken for determining the reaction order with respect to B. However, in this case experiments 1 and 3 are used. After working through the problem and canceling out [A]x from the equation, y = 1.
Finding the reaction order for the whole process is the easy addition of x and y: n = 0 + 1. Therefore, n = 1
After finding the reaction order, several pieces of information can be obtained, such as half-life.
Other methods
Other methods that can be used to solve for reaction order include the integration method, the half-life method, and the isolation method.
Problems
1. Define "reaction order."
Use the following information to solve questions 2 and 3:
Given the rate law equation:
$\text{rate} = k[A]^1[B]^2 \nonumber$
2. Determine: a) the reaction order with respect to A, b) the reaction order with respect to B, and c) the total reaction order for the equation.
3. Assuming the reaction occurs in one elementary step, propose a chemical equation using P as the symbol for your product.
Use the data table below to answer questions 4 and 5:
Experiment [A] M [B] M Rate M Min-1
1 0.100 0.100 1.0 x 10-3
2 0.400 0.100 2.0 X 10-3
3 0.100 0.150 2.0 x 10-3
4. Use the differential method to determine the reaction order with respect to A (x) and B (y). What is the total reaction order (n)?
5. What is the rate constant, k?
Answers
1. The relationship between the concentrations of species and the rate of a reaction
2. a) x = 1, b) y = 2, and c) n = 3
3. $A + 2B \longrightarrow P$
4. x = 0.5 and y = 1.7. n = 2.2
5. k = 0.10 M min-1
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Chemical reactions are studied in terms of reaction rate, reaction orders with respect to reactants, differential and integrated rate laws of chemical reactions, and activaton energy. This type of studies is usually called chemical kinetics.
• 4.1: Activation Energy - Ea
• 4.2: Chain Reactions
Chain reactions usually consist of many repeating elementary steps, each of which has a chain carrier. Once started, chain reactions continue until the reactants are exhausted. Fire and explosions are some of the phenomena associated with chain reactions.
• 4.3: Chain Reactions I
• 4.4: Catalysts and Energy of Activation
• 4.5: Elementary Steps
An elementary process is also called an elementary step or elementary reaction. It expresses how molecules or ions actually react with each other. The equation in an elementary step represents the reaction at the molecular level, not the overall reaction. Based on numbers of molecules involved in the elementary step, there are three kinds of elementary steps: unimolecular step (or process), bimolecular process, and trimolecular process.
• 4.6: Gas Kinetics
• 4.7: Limits of Thermodynamics
• 4.8: Rates and Concentrations
In the introduction to chemical kinetics, we have already defined chemical reaction rates. Rates of chemical reactions depend on the nature of the reactants, the temperature, the presence of a catalyst, and concentration. This page discusses how the concentration affects the chemical reaction rates. Concentration effect is important because chemical reactions are usually carried out in solutions.
• 4.9: Rates and Mechanisms
A reaction mechanism is a collection of elementary processes or steps (also called elementary steps) that explains how the overall reaction proceeds. A mechanism is a proposal from which you can work out a rate law that agrees with the observed rate laws. The fact that a mechanism explains the experimental results is not a proof that the mechanism is correct.
• 4.10: Rate Laws - Differential
This page deals specifically with first- and second-order reaction kinetics, but you should know that other orders such as zeroth-order and 3rd order may also be involved in chemical kinetics.
• 4.11: Rate Laws - Integrated
• 4.12: Steady-State Approximation
The steady-state approximation is a method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated. Its concentration remains the same in a duration of the reaction.
04: Reaction Mechanisms
Learning Objectives
• Explain activation energy.
• Describe how energy varies during a chemical reaction.
• Plot chemical potential energy of the system as a function of the reaction coordinate.
• Explain the model \(k = \ce{A e}^{\Large{E_{\Large\ce a}/RT}}\) for chemical kinetics.
• Plot k versus 1/T when k is measured at several T.
• Calculate Ea when a series of k and T are given.
• Calculate k at certain T.
4.02: Chain Reactions
Learning Objectives
• Explain the mechanisms of chain reactions in terms of elementary steps.
• Define these terms: radical, chain carrier.
• Classify elementary steps as initiation, chain propagation, chain branching, chain inhibition, and chain termination.
Chain reactions usually consist of many repeating elementary steps, each of which has a chain carrier. Once started, chain reactions continue until the reactants are exhausted. Fire and explosions are some of the phenomena associated with chain reactions. The chain carriers are some intermediates that appear in the repeating elementary steps. These are usually free radicals.
Once initiated, repeating elementary steps continue until the reactants are exhausted. When repeating steps generate more chain carriers, they are called chain branching reactions, which leads to explosions. If the repeating elementary steps do not lead to the formation of new product, they are called chain inhibition reactions. Addition of other materials in the reaction mixture can lead to the inhibition reaction to prevent the chain propagation reaction. When chain carriers react with one another forming stable product, the elementary steps are called chain termination reactions.
Explosions, polymerizations, and food spoilage often involve chain reactions. The chain reaction mechanism is involved in nuclear reactors; in this case the chain carriers are neutrons. The mechanisms describing chain reactions are useful models for describing chemical reactions. Most chemical chain reactions have very reactive intermediates called free radicals. The intermediate that maintains the chain reaction is called a chain carrier. These atoms or fragments are usually derived from stable molecules due to photo- or heat-dissociation.
Usually, a free radical is marked by a dot beside the symbol ($\ce{*}$), which represents an odd electron exists on the species. This odd electron makes the intermediate very reactive. For example, the oxygen, chlorine and ethyl radicals are represented by $\ce{O*}$, $\ce{Cl*}$, and $\ce{C2H5*}$, respectively. The $\ce{Cl*}$ radicals can be formed by the homolytic photodissociation reaction:
$\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber$
Mechanism of Chain Reactions
The elementary steps used for mechanisms of chain reactions can be grouped into the following categories:
• initiation step
• chain propagation steps
• chain branching steps
• chain inhibition steps
• chain termination steps
For example, the chlorination of ethane is a chain reaction, and its mechanism is explained in the following way.
If we mix chlorine, $\ce{Cl2}$, and ethane, $\ce{CH3CH3}$, together at room temperature, there is no detectable reaction. However, when the mixture is exposed to light, the reaction suddenly initiates, and explodes. To explain this, the following mechanism is proposed.
Initiation Step
Light ($\ce{h\nu}$) can often be used to initiate chain reactions since they can generate free radical intermediates via a photodissociation reaction. The initiation step can be written as:
$\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber$
Chain Propagation Step
Elementary steps in which the number of free radicals consumed is equal to the number of free radicals generated are called chain propagation steps. Once initiated, the following chain propagation steps repeat indefinitely or until the reactants are exhausted:
$\ce{Cl* +\; H3CCH3 \rightarrow ClH2CCH3 +\; H*} \nonumber$
$\ce{Cl* +\; H3CCH3 \rightarrow H3CCH2* +\; HCl} \nonumber$
$\ce{H* +\; Cl_2 \rightarrow HCl + Cl*} \nonumber$
and many other possibilities.
In each of these steps, a radical is consumed, and another radical is generated. Thus, the chain reactions continue, releasing heat and light. The heat and light cause more radicals to form. Thus, the chain propagation steps cause chain branching reactions.
Chain Branching Steps
Branching reactions are elementary steps that generate more free radicals than they consume. Branching reactions result in an explosion. For example, in the reaction between hydrogen and oxygen, the following reaction may take place:
$\ce{H* +\; O2 \rightarrow HO* + *O*} \nonumber$
where $\ce{*O*}$ is a di-radical, because the $\ce{O}$ atom has an electronic configuration 2s2 2px2 2py1 2pz1. In this elementary step, three radicals are generated, whereas only one is consumed.
The di-radical may react with a $\ce{H2}$ molecule to form two radicals.
$\ce{*O* +\, H2 \rightarrow HO* +\, H*} \nonumber$
Thus, together chain branching reactions increase the number of chain carriers. Branching reactions contribute to the rapid explosion of hydrogen-oxygen mixtures, especially if the mixtures have proper proportions.
Chain Inhibition Steps
The steps not leading to the formation of products are called inhibition reactions or steps. For example, the following steps are inhibition reactions.
$\ce{Cl* +\; ClH2CCH3 \rightarrow H3CCH2* +\; Cl2} \nonumber$
$\ce{Cl* +\; HCl \rightarrow H* +\; Cl2} \nonumber$
$\ce{H* +\; ClH2CCH3 \rightarrow H3CCH3 + Cl*} \nonumber$
Furthermore, sometimes another reactive substance $\ce{*A}$ may be added to the system to reduce the chain carriers to inhibit the chain reactions.
$\ce{Cl* + *A \rightarrow ClA\: (not\: reactive)}$
The species $\ce{A*}$ is often called a radical scavenger. In food industry, radical scavengers are added to prevent spoilage due to oxidation; these are called biological oxidants.
The mechanisms in chain reactions are often quite complicated. When intermediates are detected, a reasonable mechanism can be proposed. Adding radical scavenger to prevent food spoilage is an important application in food chemistry. This application came from the application of the chain reaction model to natural phenomena.
Chain Termination Steps
Chain termination steps are elementary steps that consume radicals. When reactants are exhausted, free radicals combine with one another to give stable molecules (since unpaired electrons become paired). These elementary steps are responsible for the chain reactions' termination:
$\ce{Cl* + *Cl \rightarrow Cl-Cl} \nonumber$
$\ce{H* + *H \rightarrow H-H} \nonumber$
$\ce{H* + *Cl \rightarrow H-Cl} \nonumber$
$\ce{H3CCH2* + *H2CCH3 \rightarrow CH3CH2-CH2CH3\: (forming\: a\: dimer)} \nonumber$
and other possibilities
In chain reactions, many products are produced.
Questions
1. Is argon atom $\ce{Ar}$ a free radical? (yes/no)
2. $\ce{Cl* +\, ClH2CCH3 \rightarrow H3CCH2* +\, Cl2}$
1. initiation step
2. chain propagation step
3. chain branching step
4. chain inhibition reaction
5. chain termination step
Skill -
Identify steps for the names in the multiple choices.
3. Skill -
Predicting the intermediate from the nature of the reactants.
4. Which one of the following is not a chain propagation reaction in the chlorination of ethane?
1. $\ce{Cl* +\; H3CCH3 \rightarrow ClH2CCH3 + H*}$
2. $\ce{Cl* +\; H3CCH3 \rightarrow H3CCH2* +\; HCl}$
3. $\ce{H* +\; Cl2 \rightarrow HCl + Cl*}$
4. $\ce{Cl* +\; HCl \rightarrow H* +\; Cl2}$
Solutions
1. No, argon atoms are monoatomic molecules.
Discussion -
Argon exists as a mono-atomic gas. All noble gases have mono-atomic molecules.
2. d.
3. $\ce{Br*}$
4. d.
Discussion -
The reactant $\ce{HCl}$ in the step is a product in the overall reaction. When $\ce{HCl}$ reacts with $\ce{Cl*}$, the reaction is retarded. $\ce{Cl*}$ attacked one of the product molecule $\ce{HCl}$ causing a reversal of the reaction.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.01%3A_Activation_Energy_-_Ea.txt
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Learning Objectives
• Explain the mechanisms of chain reactions in terms of elementary steps.
• Define these terms: radical, chain carrier.
• Classify elementary steps as initiation, chain propagation, chain branching, chain inhibition, and chain termination.
Chain reactions usually consist of many repeating elementary steps, each of which has a chain carrier. Once started, chain reactions continue until the reactants are exhausted. Fire and explosions are some of the phenomena associated with chain reactions. The chain carriers are some intermediates that appear in the repeating elementary steps. These are usually free radicals.
Once initiated, repeating elementary steps continue until the reactants are exhausted. When repeating steps generate more chain carriers, they are called chain branching reactions, which leads to explosions. If the repeating elementary steps do not lead to the formation of new product, they are called chain inhibition reactions. Addition of other materials in the reaction mixture can lead to the inhibition reaction to prevent the chain propagation reaction. When chain carriers react with one another forming stable product, the elementary steps are called chain termination reactions.
Explosions, polymerizations, and food spoilage often involve chain reactions. The chain reaction mechanism is involved in nuclear reactors; in this case the chain carriers are neutrons. The mechanisms describing chain reactions are useful models for describing chemical reactions. Most chemical chain reactions have very reactive intermediates called free radicals. The intermediate that maintains the chain reaction is called a chain carrier. These atoms or fragments are usually derived from stable molecules due to photo- or heat-dissociation.
Usually, a free radical is marked by a dot beside the symbol ($\ce{*}$), which represents an odd electron exists on the species. This odd electron makes the intermediate very reactive. For example, the oxygen, chlorine and ethyl radicals are represented by $\ce{O*}$, $\ce{Cl*}$, and $\ce{C2H5*}$, respectively. The $\ce{Cl*}$ radicals can be formed by the homolytic photodissociation reaction:
$\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber$
Mechanism of Chain Reactions
The elementary steps used for mechanisms of chain reactions can be grouped into the following categories:
• initiation step
• chain propagation steps
• chain branching steps
• chain inhibition steps
• chain termination steps
For example, the chlorination of ethane is a chain reaction, and its mechanism is explained in the following way.
If we mix chlorine, $\ce{Cl2}$, and ethane, $\ce{CH3CH3}$, together at room temperature, there is no detectable reaction. However, when the mixture is exposed to light, the reaction suddenly initiates, and explodes. To explain this, the following mechanism is proposed.
Initiation Step
Light ($\ce{h\nu}$) can often be used to initiate chain reactions since they can generate free radical intermediates via a photodissociation reaction. The initiation step can be written as:
$\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber$
Chain Propagation Step
Elementary steps in which the number of free radicals consumed is equal to the number of free radicals generated are called chain propagation steps. Once initiated, the following chain propagation steps repeat indefinitely or until the reactants are exhausted:
$\ce{Cl* +\; H3CCH3 \rightarrow ClH2CCH3 +\; H*} \nonumber$
$\ce{Cl* +\; H3CCH3 \rightarrow H3CCH2* +\; HCl} \nonumber$
$\ce{H* +\; Cl_2 \rightarrow HCl + Cl*} \nonumber$
and many other possibilities.
In each of these steps, a radical is consumed, and another radical is generated. Thus, the chain reactions continue, releasing heat and light. The heat and light cause more radicals to form. Thus, the chain propagation steps cause chain branching reactions.
Chain Branching Steps
Branching reactions are elementary steps that generate more free radicals than they consume. Branching reactions result in an explosion. For example, in the reaction between hydrogen and oxygen, the following reaction may take place:
$\ce{H* +\; O2 \rightarrow HO* + *O*} \nonumber$
where $\ce{*O*}$ is a di-radical, because the $\ce{O}$ atom has an electronic configuration 2s2 2px2 2py1 2pz1. In this elementary step, three radicals are generated, whereas only one is consumed.
The di-radical may react with a $\ce{H2}$ molecule to form two radicals.
$\ce{*O* +\, H2 \rightarrow HO* +\, H*} \nonumber$
Thus, together chain branching reactions increase the number of chain carriers. Branching reactions contribute to the rapid explosion of hydrogen-oxygen mixtures, especially if the mixtures have proper proportions.
Chain Inhibition Steps
The steps not leading to the formation of products are called inhibition reactions or steps. For example, the following steps are inhibition reactions.
$\ce{Cl* +\; ClH2CCH3 \rightarrow H3CCH2* +\; Cl2} \nonumber$
$\ce{Cl* +\; HCl \rightarrow H* +\; Cl2} \nonumber$
$\ce{H* +\; ClH2CCH3 \rightarrow H3CCH3 + Cl*} \nonumber$
Furthermore, sometimes another reactive substance $\ce{*A}$ may be added to the system to reduce the chain carriers to inhibit the chain reactions.
$\ce{Cl* + *A \rightarrow ClA\: (not\: reactive)}$
The species $\ce{A*}$ is often called a radical scavenger. In food industry, radical scavengers are added to prevent spoilage due to oxidation; these are called biological oxidants.
The mechanisms in chain reactions are often quite complicated. When intermediates are detected, a reasonable mechanism can be proposed. Adding radical scavenger to prevent food spoilage is an important application in food chemistry. This application came from the application of the chain reaction model to natural phenomena.
Chain Termination Steps
Chain termination steps are elementary steps that consume radicals. When reactants are exhausted, free radicals combine with one another to give stable molecules (since unpaired electrons become paired). These elementary steps are responsible for the chain reactions' termination:
$\ce{Cl* + *Cl \rightarrow Cl-Cl} \nonumber$
$\ce{H* + *H \rightarrow H-H} \nonumber$
$\ce{H* + *Cl \rightarrow H-Cl} \nonumber$
$\ce{H3CCH2* + *H2CCH3 \rightarrow CH3CH2-CH2CH3\: (forming\: a\: dimer)} \nonumber$
and other possibilities
In chain reactions, many products are produced.
Questions
1. Is argon atom $\ce{Ar}$ a free radical? (yes/no)
2. $\ce{Cl* +\, ClH2CCH3 \rightarrow H3CCH2* +\, Cl2}$
1. initiation step
2. chain propagation step
3. chain branching step
4. chain inhibition reaction
5. chain termination step
Skill -
Identify steps for the names in the multiple choices.
3. Skill -
Predicting the intermediate from the nature of the reactants.
4. Which one of the following is not a chain propagation reaction in the chlorination of ethane?
1. $\ce{Cl* +\; H3CCH3 \rightarrow ClH2CCH3 + H*}$
2. $\ce{Cl* +\; H3CCH3 \rightarrow H3CCH2* +\; HCl}$
3. $\ce{H* +\; Cl2 \rightarrow HCl + Cl*}$
4. $\ce{Cl* +\; HCl \rightarrow H* +\; Cl2}$
Solutions
1. No, argon atoms are monoatomic molecules.
Discussion -
Argon exists as a mono-atomic gas. All noble gases have mono-atomic molecules.
2. d.
3. $\ce{Br*}$
4. d.
Discussion -
The reactant $\ce{HCl}$ in the step is a product in the overall reaction. When $\ce{HCl}$ reacts with $\ce{Cl*}$, the reaction is retarded. $\ce{Cl*}$ attacked one of the product molecule $\ce{HCl}$ causing a reversal of the reaction.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.03%3A_Chain_Reactions_I.txt
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Learning Objectives
• Redefine catalyst in terms of energy of activation.
• Calculate Ea when a catalyst is used from rate of reaction.
• If Ea is known, calculate the rate of reaction.
For reactions that follow the Arrhenius rate law a catalyst can be re-defined as a substance that lowers the energy of activation Ea by providing a pathway (reaction mechanism), or transition state. For example, it is well known that iodide ions catalyze the decomposition of hydrogen peroxide $\ce{H2O2}$,
$\ce{2 H2O2 \rightarrow 2 H2O + O2} \nonumber$
Thus, by dissolving solid $\ce{KI}$ in a solution of hydrogen peroxide, the formation of oxygen bubbles is accelerated. Of course, the reaction depends on concentrations of reactants and catalyst, but for a definite (or fixed) concentration, the relative reaction rates can be compared as exemplified by the following examples.
Example 1
At 300 K, the activation energy, Ea, for the decomposition of $\ce{H2O2}$ has been measured to be 75.3 kJ/mol. In the presence of a definite concentration of iodide ions, $\ce{I-}$, the activation energy Ea has been estimated to be 56.5 kJ/mol. How much faster is the decomposition when the same concentration of iodide is present in the reaction?
Solution
If k and k' represent the reaction rate constants of decomposition in the absence of iodide and in the presence of iodide ion (at a definite concentration) respectively, then
$rate = k\ce{f([H2O2])}$
$rate\,' = k\,'\ce{f([H2O2])}$
where $\ce{f([H2O2])}$ is a function of the concentration of the reactant. Note that rate and rate' are rates of the reactions in the absence of and in the presence of iodide ions.
\begin{align*} k &= \ce{A e}^{-75300/(8.3145\times300)}\ \ k\,' &= \ce{A e}^{-56500/(8.3145\times300)}\ \ \dfrac{k}{k\,'} &= \dfrac{\ce e^{-75300/(8.3145\times300)}}{\ce e^{-56500/(8.3145\times300)}}\ \ &= \ce e^{-(75300 - 56500)/(8.3145\times300)}\ \ &= \ce e^{-18800/(8.3145\times300)}\ \ &= 1877 \end{align*}
Thus, rate' = 1877 times rate, because the rate constant has increased 1877 times.
DISCUSSION
Note that the presence of a catalyst allows the reaction to proceed at the same low temperature, but achieve a much faster rate of reaction.
Example 2
The activation energy of a reaction is 19.0 kJ/mol. When a catalyst is used at a particular concentration, the rate increases ten fold. Calculate the energy of activation when the catalyst is present.
Solution
This example differs from example 1 in that we know how much faster the reaction is and want to evaluate the activation energy. Let the activation energy in the presence of the catalyst be Ea, then
\begin{align*} \dfrac{\ce e^{-E\ce a/(8.3145\times300)}}{\ce e^{-19000/(8.3145\times300)}} &= \dfrac{k_{\ce{catalyst}}}{k}\ \ &= 10\ \ \ce e^{-E\ce a/(8.3145\times300)} &= 10 \times \ce e^{-19000/(8.3145\times300)}\ \ &= 0.00492\ \ \dfrac{-E\ce a}{(8.3145\times300)} &= \ln0.00492\ \ -E\ce a &= (8.3145\times300)\times(-5.315)\ \ E\ce a &= \mathrm{13257\: J/mol}\ \ &= \mathrm{13.3\: kJ/mol} \end{align*}
DISCUSSION
The details of the calculation are given to illustrate the mathematic skills involved. Check out the units throughout the calculation please.
Questions
1. At 300 K, the activation energy, Ea for the decomposition of $\ce{H2O2}$ has been measured to be 75.3 kJ/mol. In the presence of iodide ion, $\ce{I-}$, the activation energy Ea has been estimated to be 56.5 kJ/mol. If 11 s is required to collect 0.1 mL of oxygen when iodide is present, how many seconds are required to collect 0.1 mL if iodide is absent?
Skill -
Calculate the time required to accomplish a certain task when the rate is different.
2. At 298 K, the activation energy, Ea for the decomposition of $\ce{H2O2}$ has been measured to be 75.3 kJ/mol. In the presence of iodide ion, $\ce{I-}$, the activation energy Ea has been estimated to be 56.5 kJ/mol. When catalyzed by the enzyme catalase, the activation energy is 8.4 kJ/mol. If 1 s is required to collect 100 mL of oxygen when catalase is present, how many seconds are required to collect 100 mL if iodide is used as the catalyst?
3. The activation energy of a reaction is 19.0 kJ/mol. When a catalyst is used at 300 K, the rate increases one hundred fold. Calculate the energy of activation when the catalyst is present.
Skill -
Calculate energy of activation when the rate of increase is known.
Solutions
1. 20647 s = 5.74 hours
2. 2.7E8 s = 3215 days or close to 10 years
Discussion -
Calculate the time required to collect when no catalyst is used.
3. 7.51 kJ/mol
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.04%3A_Ea_and_Catalysts.txt
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Learning Objectives
• Explain elementary steps.
• Write the expression for elementary steps.
Reaction Mechanism - Elementary Process
A mechanism for a reaction is a collection of elementary processes (also called elementary steps or elementary reactions) that explains how the overall reaction proceeds. A mechanism is a proposal from which you can work out a rate law that agrees with the observed rate laws. The fact that a mechanism explains the experimental results is not a proof that the mechanism is correct. A mechanism is our rationalization of a chemical reaction, and devising mechanisms is an excellent academic exercise.
Elementary Processes or Steps
An elementary process is also called an elementary step or elementary reaction. It expresses how molecules or ions actually react with each other. The equation in an elementary step represents the reaction at the molecular level, not the overall reaction. Based on numbers of molecules involved in the elementary step, there are three kinds of elementary steps: unimolecular step (or process), bimolecular process, and trimolecular process.
An elementary step is proposed to give the reaction rate expression. The rate of an elementary step is always written according to the proposed equation. This practice is very different from the derivation of rate laws for an overall reaction.
When a molecule or ion decomposes by itself, such an elementary step is called a unimolecular step (or process). A unimolecular step is always a first order reaction. The following examples are given to illustrate this point:
$\ce{O3 \rightarrow O2 + O}$, $Rate = k \ce{[O3]}$
or in general
$\ce{A \rightarrow B + C + D}$, $Rate = k \ce{[A]}$
$\mathrm{A^* \rightarrow X + Y}$, $Rate = k \mathrm{[A^*]}$
$\mathrm{A^*}$ represents an excited molecule.
A bimolecular process involves two reacting molecules or ions. The rates for these steps are 2nd order, and some examples are given to illustrate how you should give the rate expression. The simulation illustrates a bimolecular process.
$\ce{NO + O3 \rightarrow NO2 +O2}$, $Rate = k \ce{[NO] [O3]}$
$\ce{Cl + CH4 \rightarrow HCl + CH3}$, $Rate = k \ce{[Cl] [CH4]}$
$\mathrm{Ar + O_3 \rightarrow Ar + O_3^*}$, $Rate = k \ce{[Ar] [O3]}$
$\ce{A + A \rightarrow B + C}$, $Rate = k \ce{[A]^2}$
$\ce{A + B \rightarrow X + Y}$, $Rate = k \ce{[A] [B]}$
A trimolecular process involves the collision of three molecules. For example:
$\ce{O + O2 + N2 \rightarrow O3 + N2}$, $Rate = k \ce{[O] [O2] [N2]}$
$\ce{O + NO + N2 \rightarrow NO2 + N2}$, $Rate = k \ce{[O] [NO] [N2]}$
The $\ce{N2}$ molecules in the above trimolecular elementary steps are involved with energy transfer. They can not be canceled. They are written in the equation to give an expression for the Rates. In general, trimolecular steps may be,
$\ce{A + A + A \rightarrow products}$, $Rate = k \ce{[A]^3}$
$\ce{A + A + B \rightarrow products}$, $Rate = k \ce{[A]^2 [B]}$
$\ce{A + B + C \rightarrow products}$, $Rate = k \ce{[A] [B] [C]}$
Three molecules colliding at an instant is rare, but occasionally these are some of the ways reactions take place.
Elementary processes are written to show how a chemical reaction progresses leading to an overall reaction. Such a collection is called a reaction mechanism. In a mechanism, elementary steps proceed at various speeds. The slowest step is the rate-determining step. The order for that elementary process is the order of a reaction, but the concentrations of reactants in that step must be expressed in terms of the concentrations of the reactants.
Deriving Rate Laws From Reaction Mechanisms
The following example illustrates how elementary steps are used to represent a reaction mechanism. In particular, a slow step in a mechanism determines the rate of a reaction.
Example 1
If the reaction
$\ce{2 NO2 + F2 \rightarrow 2 NO2F}$
follows the mechanism
$\mathrm{i.}\:\; \ce{NO2 + F2 \rightarrow NO2F + F\: (slow)}\ \mathrm{ii.}\: \ce{NO2 + F \rightarrow NO2F\: (fast)}$
Work out the rate law.
Solution
Since step i. is the rate-determining step, the rate law is
$-\dfrac{1}{2} \ce{\dfrac{d[NO2]}{dt}} = k \ce{[NO2] [F2]}$
Addition of i. and ii. gives the overall reaction.
DISCUSSION
This example illustrates that the overall reaction equation has nothing to do with the order of the reaction. The elementary process in the rate-determining step determines the order.
Other possible elementary steps in this reaction are:
$\ce{F + F \rightarrow F2}$
$\ce{F + F2 \rightarrow F2 + F}$
$\ce{NO2F + F \rightarrow F + NO2F}$
but they do not lead to the formation of products.
To propose a mechanism requires the knowledge of chemistry to give plausible elementary processes. A freshman in chemistry will not be asked to propose mechanisms, but you will be asked to give the rate laws from a given mechanism.
Summary
The number of particles involved in an elementary step is called the molecularity, and in general, we consider only the molecularity of 1, 2, and 3. Types of elementary steps are summarized below. In the table, $\ce{A}$, $\ce{B}$, and $\ce{C}$ represent reactants, intermediates, or products in the elementary process.
Molecularity Elementary step Rate law
1 $\ce{A \rightarrow products}$ $rate = k \ce{[A]}$
2 $\ce{A + A \rightarrow products}$
$\ce{A + B \rightarrow products}$
$rate = k \ce{[A]^2}$
$rate = k \ce{[A] [B]}$
3 $\ce{A + A + A \rightarrow products}$
$\ce{A + 2 B \rightarrow products}$
$\ce{A + B + C \rightarrow products}$
$rate = k \ce{[A]^3}$
$rate = k \ce{[A] [B]^2}$
$rate = k \ce{[A] [B] [C]}$
Questions
1. Which one of the following is a bimolecular process?
1. $\ce{A \rightarrow products}$
2. $\ce{A + A \rightarrow products}$
3. $\ce{A + A + A \rightarrow products}$
Hint: b. $\ce{A + A \rightarrow products}$
Skill -
Recognize and name all three elementary steps.
2. What is the order of an trimolecular process (or step)?
Hint: Third order
Skill -
Give the order of any elementary step.
3. $\mathrm{Hg^* \rightarrow Hg + light}$
$\mathrm{Hg^* + Ar \rightarrow Hg + Ar^*}$
What is the order of light emission?
Hint: First order for the light emission step.
Skill -
Emission of light is first order, but use a number.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.05%3A_Elementary_Steps.txt
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Learning Objectives
• Explain the distribution of molecular speed of a gas.
Molecular Speed of Gases
The quantitative relationship of temperature effect on chemical reaction rates is discussed in form of activation energy, depicted as Ea in the diagram. For reactions having large Ea values, high temperatures are required to have a measurable reaction rate. The reason behind this is due to the small number of molecules having sufficient energy to overcome Ea, the energy barrier of forming an activated reaction intermediate.
At a certain temperature, not all gas molecules are moving with the same speed, some fast and some slow. The kinetic energies of molecules are not all equal. The fast moving ones have high kinetic energy, and they may have enough energy to overcome the reaction energy barrier Ea.
A plot of number of gas molecules at certain speed versus speed gives a distribution curve. Careful studies of gases show that the distribution is not a bell or normal distribution, but a Maxwell- Boltzmann distribution.
A sketch of a Maxwell distribution is given here. The peaks are not symmetrical. There are more molecules at higher speed than at lower speeds. When the temperature increases, the peak shifts to the right. A more carefully plotted diagram is shown below.
You will learn the theory and the statistics at a higher year. For the moment, you may notice how the distribution curve shifts as temperature increases.
Questions
1. Which curve has a bell shape?
1. Normal distribution
2. Maxwell-Boltzmann distribution
3. Chi-square distribution
Skill -
Name the various distributions.
2. Which of the following gases has the lowest most-probable speed: $\ce{H2}$, $\ce{He}$, $\ce{Ne}$, $\ce{N2}$, $\ce{O2}$, $\ce{F2}$, $\ce{Ar}$, or $\ce{Xe}$?
Skill -
Explain how molecular masses affect the average speed.
3. For $\ce{N2}$ at 300 K, which of the following speeds is the highest?
1. Most probable speed
2. Average speed
3. Root-mean-square speed
4. For $\ce{N2}$ at 300 K, which of the following speeds is the lowest?
1. Most probable speed
2. Average speed
3. Root-mean-square speed
5. For the gas $\ce{N2}$, which temperature gives the highest most probable speed: 500 C, 600 K, or 700 F?
Solutions
1. a. Normal distributions have bell shape.
2. $\ce{Xe}$, because it is the heaviest one here.
Discussion -
The gas with the heaviest molecular mass has the lowest average speed. $\ce{UF6}$ is a gas used for enrichment of the isotope $\ce{U-235}$.
3. c.
Discussion -
The root-mean-square speed is $\sqrt{\dfrac{3 R T}{M}}$, where R = gas constant, T = temperature, M = molecular mass.
4. a.
Discussion -
The formulas for the three speeds mentioned here can be derived by mathematical techniques. The most probable speed is $\sqrt{\dfrac{2 R T}{M}}$, where R = gas constant, T = temperature, M = molecular mass.
5. 500 C
Discussion -
It is easy to calculate the most probable speed if you know the formula to use. Note that 500 C is higher than 700 F. Do you know what the temperature scales measure?
4.07: Limits of Thermodynamics
Thermodynamics is a powerful approach toward understanding chemical reactions, but only provides part of the picture. Specifically:
1. Thermodynamics only points the way
2. Thermodynamics says nothing about how long it takes to get there
3. The stoichiometric equation for the reaction says nothing about its mechanism
Thermodynamics only points the way: Chemical change is driven by the tendency of atoms and molecules to rearrange themselves in a way that results in the maximum possible dispersion of thermal energy into the world. The observable quantity that measures this spreading and sharing of energy is the free energy of the system. As a chemical change takes place, the quantities of reactants and products change in a way that leads to a more negative free energy. When the free energy reaches its minimum possible value, there is no more net change and the system is said to be in equilibrium.
The beauty of thermodynamics is that it enables us to unfailingly predict the net direction of a reaction and the composition of the equilibrium state even without conducting the experiment; the standard free energies of the reactants and products, which can be independently measured or obtained from tables, are all we need.
Note
When the free energy reaches its minimum possible value, there is no more net change and the system is said to be in equilibrium
Thermodynamics says nothing about how long it takes to get there: It is worth noting that the concept of "time" plays no role whatsoever in thermodynamics. But kinetics is all about time. The "speed" of a reaction — how long it takes to reach equilibrium — bears no relation at all to how spontaneous it is (as given by the sign and value of ΔG°) or whether it is exothermic or endothermic (given by the sign of ΔH°). Moreover, there is no way that reaction rates can be predicted in advance; each reaction must be studied individually.
Note
The concept of "time" plays no role whatsoever in thermodynamics.
The stoichiometric equation for the reaction says nothing about its mechanism: The term "mechanism" refers to, "who does what to whom". Think of a reaction mechanism as something that goes on in a "black box" that joins reactants to products. The inner workings of the black box are ordinarily hidden from researchers, are highly unpredictable, and can only be inferred by indirect means.
Note
The stoichiometric equation for the reaction says nothing about its mechanism!
Three reactions that look alike, but are different
Consider, for example, the gas-phase formation reactions of the hydrogen halides from the elements. The thermodynamics of these reactions are all similar (they are all highly exothermic), but their dynamics (their kinetics and mechanisms) could not be more different.
$\ce{H2(g) + I2(g) -> 2 HI(g)} \nonumber$
Careful experiments, carried out over many years, are consistent with the simplest imaginable mechanism: a collision between the two reactant molecules results in a rearrangement of the bonds.
$\ce{H2(g) + Br2(g) -> 2 HBr(g)} \nonumber$
One might be tempted to suppose that this would proceed in a similar way, but experiments reveal that the mechanism of this reaction is far more complex. The reaction takes place in a succession of steps, some of which involve atomic H and Br.
$\ce{H2(g) + Cl2(g) -> 2 HCl(g)} \nonumber$
The mechanism of this reaction is different again. Although the first two reactions reach equilibrium in minutes to an hour or so at temperatures of 300 to 600 K, a mixture of hydrogen and chlorine will not react at all in the dark, but if you shine a light on the mixture, it goes off with a bang as the instantaneous reaction releases heat and expands the gas explosively.
What is particularly noteworthy is that these striking differences cannot be reliably predicted from theory; they were revealed only by experimentation.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.06%3A_Gas_Kinetics.txt
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Learning Objectives
• Explain chemical reaction rates.
• Explain the concentration effects on reaction rate.
• Define the order of a reaction with respect to a reactant.
• Define the overall order of a chemical reaction.
• Define the rate constant of a chemical reaction.
• Determine the order of a reaction by experiment.
Concentration and Chemical Reaction Rate
In the introduction to chemical kinetics, we have already defined chemical reaction rates. Rates of chemical reactions depend on the nature of the reactants, the temperature, the presence of a catalyst, and concentration. This page discusses how the concentration affects the chemical reaction rates. Concentration effect is important because chemical reactions are usually carried out in solutions.
Chemical Reaction Rates
The reaction rates of chemical reactions are the amounts of a reactant reacted or the amount of a product formed per unit time (moles per second). Often, the amount can be expressed in terms of concentrations.
$Rate = \dfrac{amount\: reacted\: or\: produced}{time\: interval}\: \textrm{units: }\mathrm{g/s,\: mol/s,\: or\: \%/s}$
At certain conditions, the rates are functions of concentrations. Depending on the time interval between measurements, the rates are called
• average rate: rate measured between long time interval
• instantaneous rate: rate measured between very short interval
• initial rate: instantaneous rate at the beginning of an experiment
However, a more realistic representation for a reaction rate is the change in concentration per unit time, either the decrease of concentration per unit time of a reactant or the increase of concentration per unit time of a product. In this case, the rate is expressed in mol/(L sec).
$Rate = \dfrac{concentration\: change\: of\: a\: reactant\: or\: product}{time\: interval}\: \textrm{units: g/(M s), M/s, ppm/s etc.}$
Measuring Reaction Rate
To measure a reaction rate, we usually monitor either a product or a reactant for its change. Any physical characteristic related to the quantity or concentration of a product or reactant can be monitored. Some of the characteristics to be monitored are:
• change in pressure,
• change in color (spectroscopic measurement),
• temperature for exothermic or endothermic reaction, and
• presence of certain key substance
The change can be plotted on a graph, and from the graph, we can get the average rate or the instantaneous rate by either graphical methods or using a computer for the data analysis.
Rate Constants and the Orders
Usually, the rate of a reaction is a function of the concentrations of reactants. For example, the rate of the reaction
$\ce{2 NO + O2 \rightarrow 2 NO2}$
has the form:
$Rate = k \ce{[O2] [NO]^2}$
The rate is proportional to the concentration of $\ce{O2}$, usually written as $\ce{[O2]}$ and is proportional to the square of $\ce{[NO]}$, or $\ce{[NO]^2}$. The orders of 1 and 2 for $\ce{[O2]}$ and $\ce{[NO]}$ respectively have been determined by experiment, NOT from the chemical equation. The total order of this reaction is 3 (=2+1).
Note the rates and order in the following example reactions:
$\ce{H2 + I2 \rightarrow 2 HI}$,
$Rate = k \ce{[H2] [I2]}$,
Total order 2.
$\ce{H2 + Br2 \rightarrow 2 HBr}$,
$Rate = k \ce{[H2] [Br2]}^{1/2}$,
Total order 1.5.
In particular, note that orders are NOT determined from the stoichiometry of the reaction equation.
Rates as Functions of Reactant Concentrations
The order with respect to (wrt) a reactant is determined experimentally by keeping the concentration of other reactants constant, but varying the concentration of one of the reactants, say $\ce{A}$ in a general reaction
$a \,\textrm A + b \,\textrm B + c \,\textrm C = products$
If concentrations of $\ce{B}$ and $\ce{C}$ are kept constant, you can measure the reaction rate of $\ce{A}$ at various concentrations. You can then plot the rate as a function of $\ce{[A]}$. For a zeroth order reaction, you will get a horizontal line, because
$rate = k$ (a horizontal line)
rate
| /
| /rate = k [A]
| /
|- -/- - - - - rate = k
| /
| /
|/_________________
0 1 2 3 4 [A]
For a first order reaction, the plot is a straight line (linear), as shown above, because
$rate = k \ce{[A]}$ (a straight line)
Note that $rate = k$ when $\mathrm{[A] = 1}$.
For a second order reaction, the plot is a branch of a parabola, because
$rate = k \ce{[A]}^2$
rate
| .
| rate = k [A]2
| . (a branch of
| a parabola)
| .
| - . - - - - - -
| .
|._________________
0 1 2 3 4 [A]
For a reaction with an infinite order, the plot is a step function. The rate is small, almost zero, when $\ce{[A]}$ is less than 1. When $\ce{[A]}$ is greater than or equal to 1, then the reaction rate is very large. This model applies to nuclear explosion, except that $\mathrm{[A] = 1}$ is actually the critical mass of the fission material.
$rate = k \ce{[A]}^{\infty}$
rate
| (order = infinity)
| | rate = k [A]00
| | (a vertical line)
| |
| |
| |
| |
|...|_________________
0 1 2 3 4 [A]
Is there a chemical process like this? Well, we all know that one of the key conditions in an atomic bomb is to have a critical mass of the fission material, $\ce{^235U}$ or $\ce{^239Pu}$. When such a mass is put together, the reaction rate increases dramatically, leading to an explosion. Thus, this model seems to apply; however, the mechanism for the fission reaction is not described by the order of the fission material.
Variation of Rate, Rate Constant, and Order of a Reaction
If only $\ce{[A]}$ is varied in experiments, and the order wrt $\ce{[A]}$ is n, then the rate has the general expression,
$rate = k \ce{[A]}^n$
In this expression, k is the specific rate constant, or the rate when $\mathrm{[A] = 1.00}$. Again, the order n is not necessarily an integer, but its most common values are 0.5 (1/2), 1, 2, or 3. Cases in which n is a negative number are rare.
Mathematical models for the effect of concentration on rates are interesting. In general, the rate of a reaction of order n with respect to $\ce{A}$ can be represented by the equation:
$y = k x^n$, (n = various values including 0.5, 1, 2, 3, ...)
Plots of equations for various values of n illustrate the dependence of rate on concentration for various orders.
Evaluation of Order by Experiments
For a chemical reaction, we often determine the order with respect to a reagent by determining the initial rate. When more than one reactants are involved, we vary the concentrations in a systematic way so that the effect of concentration of one of the reactants can be measured.
For example, in a reaction involving three reactants, $\ce{A}$, $\ce{B}$, and $\ce{C}$, we vary $\ce{[A]}$ from experiment 1 to experiment 2 and find out how the rate varies. Similarly, we vary concentrations of $\ce{B}$ or $\ce{C}$ in other experiments, keeping others constant, and investigate its effect. The example below illustrates the strategy for such an approach.
Example
Derive the rate law for the reaction $\mathrm{A + B + C \rightarrow products}$ from the following data, where rate is measured as soon as the reactants are mixed.
Experiment 1 2 3 4
[A]o 0.100 0.200 0.200 0.100
[B]o 0.100 0.100 0.300 0.100
[C]o 0.100 0.100 0.100 0.400
rate 0.100 0.800 7.200 0.400
Solution
Assuming the orders to be x, y, and z respectively for $\ce{A}$, $\ce{B}$, and $\ce{C}$, we have
$rate = k \ce{[A]}^{\Large x} \ce{[B]}^{\Large y} \ce{[C]}^{\Large z}$
From experiment 1 and 2, we have:
$\dfrac{0.800}{0.100}=\dfrac{k\: 0.2^{\Large x}\: 0.1^{\Large y}\: 0.1^{\Large z}}{k\: 0.1^{\Large x}\: 0.1^{\Large y}\: 0.1^{\Large z}}$
Thus, $8 = 2^{\Large x}$; and $x = \dfrac{\ln8}{\ln2} = 3$
By similar procedures, we get $y = 2$ and $z = 1$. Thus, the rate law is:
$rate = k \ce{[A]}^3 \ce{[B]}^2 \ce{[C]}$
Discussion
Note the following relationships:
$x = y^{\Large z}$
$\ln x = z \ln y$
$z = \dfrac{\ln x}{\ln y}$
Summary
The variation of reaction rates as functions of order and concentrations are summarized in the form of a Table below.
Differential
rate law
Plot of rate vs
[A]
0th
order
- d[A]/dt = k horizontal
line
first
order
- d[A]/dt = k[A] straight line with
slope = k
second
order
- d[A]/dt = [A]2 a branch of
parabola
order
= infinity
- d[A]/dt = k [A]oo rate = 0 when [A] < 1
rate = infinite when [A] > 1
a vertical line at [A] = 1
Confidence Building Questions
1. Hint: First order wrt $\ce{A}$
Skill -
Recognize the order when rate is linear dependent on $\ce{[A]}$. Only when n = 1 does the rate depend linearly on the concentration.
2. Hint: Zeroth order wrt $\ce{A}$.
Skill -
Recognize the order when $rate = k$ when $n = 0$.
3. For a reaction that is second order with respect to a reactant $\ce{A}$, how many times does the rate increase as $\ce{[A]}$ increases by a factor of 2?
Hint: The rate increases four times.
Skill -
Predict rate increases as the concentration doubles for reactions of various orders.
4. What is the reaction rate when the concentrations of all reactants are unity ($\mathrm{[A] = [B] = 1}$) for a second order reaction?
Hint: Rate = rate constant k.
Discussion -
$Rate = k$, when $\mathrm{[A] = [B] = 1}$ regardless of the order of the reaction. Note the difference between rate and rate constant, k.
5. For a reaction that is 2nd order with respect to $\ce{A}$, you can keep concentrations of other reactants constant, and vary the concentration of $\ce{A}$. You can measure the initial rate in the experiments, and then plot rate as a function of $\ce{[A]}$. What type of curve or line is such a plot?
Hint: A branch of a parabola.
6. In general, the rates increase as the concentrations increase for any reaction with a positive order, true or false?
Hint: True
Skill -
What if n is a negative value?
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.08%3A_Rates_and_Concentrations.txt
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Learning Objectives
• Explain reaction mechanism.
• Derive a rate law from a given mechanism.
Reaction Mechanisms - Derive Rate Laws
A reaction mechanism is a collection of elementary processes or steps (also called elementary steps) that explains how the overall reaction proceeds. A mechanism is a proposal from which you can work out a rate law that agrees with the observed rate laws. The fact that a mechanism explains the experimental results is not a proof that the mechanism is correct. As a mechanism, no proof is required. Proposing a mechanism is an interesting academic exercise for a mature chemist. Students in general chemistry will not be required to propose a mechanism, but they are required to derive the rate law from a proposed mechanism.
Elementary Processes or Steps
A summary of the elementary process or steps is given in a table form here. This previous link provides details about these steps. In the table below, $\ce{A}$, $\ce{B}$, and $\ce{C}$ represent reactants, intermediates, or products in the elementary process.
Molecularity Elementary step(s) Rate law
1 $\ce{A \rightarrow products}$ $rate = k \ce{[A]}$
2 $\ce{A + A \rightarrow products}$
$\ce{A + B \rightarrow products}$
$rate = k \ce{[A]^2}$
$rate = k \ce{[A] [B]}$
3 $\ce{A + A + A \rightarrow products}$
$\ce{A + 2 B \rightarrow products}$
$\ce{A + B + C \rightarrow products}$
$rate = k \ce{[A]^3}$
$rate = k \ce{[A] [B]^2}$
$rate = k \ce{[A] [B] [C]}$
Deriving Rate Laws from Mechanisms
You all have experienced that an accident or road construction on a free way slows all the traffic on the road, because limitations imposed by the accident and constructions apply to all cars on that road. The narrow passing stretch limits the speed of the traffic.
If several steps are involved in an overall chemical reaction, the slowest step limits the rate of the reaction. Thus, a slow step is called a rate determining step. The following examples illustrate the method of deriving rate laws from the proposed mechanism. Please learn the technique.
Example 1
If the reaction
$\ce{2 NO2 + F2 \rightarrow 2 NO2F}$
follows the mechanism,
1. $\ce{NO2 + F2 \rightarrow NO2F + F\: (slow)}$
2. $\ce{NO2 + F \rightarrow NO2F\: (fast)}$
what is the rate law?
Solution
Since step i. is the rate-determining step, the rate law is
$\mathrm{- \dfrac{1}{2}\dfrac{d[NO_2]}{dt} = \mathit{k} [NO_2] [F_2]}$
Since both $\ce{NO2}$ and $\ce{F2}$ are reactants, this is the rate law for the reaction.
DISCUSSION
Addition of i. and ii. gives the overall reaction, but step ii. does not affect the rate law. Note that the rate law is not derived from the overall equation either.
Example 2
For the reaction
$\ce{H2 + Br2 \rightarrow 2 HBr}$,
the following mechanism has been proposed
1. $\ce{Br2 \underset{\Large{\mathit{k}_{-1}}}{\overset{\Large{\mathit{k}_1}}{\rightleftharpoons}} 2 Br}\textrm{ (both directions are fast)}$
2. $\mathrm{Br + H_2 \xrightarrow{\Large{\mathit{k}_2}}HBr + H\: (slow)}$
3. $\mathrm{H + Br_2 \xrightarrow{\Large{\mathit{k}_3}}HBr + Br\: (fast)}$
Derive the rate law that is consistent with this mechanism.
Solution
For a problem of this type, you should give the rate law according to the rate-determining (slow) elementary process. In this case, step ii. is the rate-determining step, and the rate law is
$\mathrm{\dfrac{1}{2}\dfrac{d[HBr]}{dt} = \mathit{k}_2 [H_2] [Br]}$
The factor 1/2 results from the $\ce{2 HBr}$ formed every time, one in step ii. and one in step iii. Since $\ce{[Br]}$ is not one of the reactants, its relationship with the concentration of the reactants must be sought. The rapid reaction in both directions of step i. implies the following relationship:
$k_1 [\ce{Br2}] = k_{-1} [\ce{Br}]^2$
or
$\ce{[Br]} = \left(\dfrac{k_1}{k_{-1}} [\ce{Br2}]\right)^{1/2}$
Substituting this in the rate expression results in
$\ce{rate} = k_2 \left(\dfrac{k_1}{k_{-1}}\right)^{1/2} \ce{[H2] [Br2]}^{1/2}$
The overall reaction order is 3/2, 1 with respect to $\ce{[H2]}$ and 1/2 with respect to $\ce{[Br2]}$.
DISCUSSION
The important point in this example is that the rapid equilibrium in step i. allows you to express the concentration of an intermediate ($\ce{[Br]}$) in terms of concentrations of reactants ($\ce{[Br2]}$) so that the rate law can be expressed by concentrations of the reactants.
The ratio k1 / k-1 is often written as K, and it is called the equilibrium constant for the reversible elementary steps.
Example 3
Derive the rate law that is consistent with the proposed mechanism in the formation of phosgene from $\ce{Cl2}$ and $\ce{CO}$. ($K_1 = \dfrac{k_1}{k_{-1}}$ and $K_2 = \dfrac{k_2}{k_{-2}}$ may be considered as equilibrium constants of the elementary processes, and $\ce{M}$ is any inert molecule.)
1. $\ce{Cl2 + M \rightleftharpoons 2 Cl + M\:\: (fast\: equilibrium,\: \mathit{K}_1)}$
2. $\ce{Cl + CO + M \rightleftharpoons ClCO + M\:\: (fast\: equilibrium,\: \mathit{K}_2)}$
3. $\ce{ClCO + Cl2 \rightarrow Cl2CO + Cl\:\: (slow,\: \mathit{k}_3)}$
The overall reaction is
$\ce{Cl2 + CO \rightarrow Cl2CO}$
Solution
From the rate-determining (slow) step,
$\mathrm{\dfrac{d[Cl_2CO]}{dt}} = k_3 \mathrm{[ClCO] [Cl_2]}\tag{1}$
You should express $\ce{[ClCO]}$ in terms of concentrations of $\ce{Cl2}$ and $\ce{CO}$. This is done by considering step ii.
$\ce{[ClCO]} = K_2 \ce{[Cl] [CO]}\tag{2}$
You should express $\ce{[Cl]}$ in terms of $\ce{[Cl2]}$. For this, you may use step i.
$\ce{[Cl]} = K_1^{1/2} \ce{[Cl2]}^{1/2} \tag{3}$
Substituting (3) in (2) and then in (1) gives the Rate,
\begin{align*} \ce{Rate} &= k_3 K_1^{1/2} K_2 \ce{[CO] [Cl2]}^{3/2}\ &= k \ce{[CO] [Cl2]}^{3/2} \end{align*}
where $k = k_1 K_1^{1/2} K_2$, the observed rate constant. The overall order of the reaction is 5/2: strange, but that is the observed rate law.
DISCUSSION
This example shows how the concentrations of intermediates are related to those of the reactants in a two-step equilibrium.
If the third step is
1. $\ce{ClCO + Cl \rightarrow Cl2CO\:\: (slow,\: \mathit{k}_3)}$
the rate law will be different from the result derived above.
Exercise
Derive the rate law using the alternate step from the Discussion in Example 3.
Example 4
In an acid solution, the mechanism for the reaction
$\ce{NH4+ + HNO2 \rightarrow N2 + 2 H2O + H+ }$
is:
1. $\ce{HNO2 + H+ \rightleftharpoons H2O + NO+ \:\: (equilibrium,\: \mathit{K}_1)}$
2. $\ce{NH4+ \rightleftharpoons NH3 + H+ \:\: (equilibrium,\: \mathit{K}_2)}$
3. $\ce{NO+ + NH3 \rightarrow NH3NO+ \:\: (slow,\: \mathit{k}_3)}$
4. $\ce{NH3NO+ \rightarrow H2O + H+ + N2 \:\: (fast,\: \mathit{k}_4)}$
Derive the rate law.
Solution
From the rate-determining step, you have
$\mathrm{\dfrac{d[NH_3NO^+ ]}{dt}} = k_3 \mathrm{[NO^+] [NH_3]} \tag{4}$
Neither $\ce{NO+}$ nor $\ce{NH3}$ is a reactant. You must express their concentrations in terms of $\ce{[NH4+]}$ and $\ce{[HNO2]}$ from elementary processes i. and ii.
From i, $\ce{[NO+]} = K_1 \ce{\dfrac{[HNO2] [H+]}{[H2O]}} \tag{5}$
From ii, $\ce{[NH3]} = K_2 \ce{\dfrac{[NH4+]}{[H+]}} \tag{6}$
Substituting (6) and (5) in (4) gives,
\begin{align*} \ce{Rate} &= k_3 K_1 K_2 \ce{\dfrac{[HNO2] [NH4+]}{[H2O]}}\ \ &= k \ce{[HNO2] [NH4+ ]} \end{align*}
where $k = \dfrac{k_3 K_1 K_2}{[\ce{H2O}]}$ is the overall rate constant.
Questions
1. $\ce{2 A + B2 \rightarrow 2 AB}$ follows the mechanism,
1. $\ce{A + B2 \rightarrow AB + B\: (slow)}$
2. $\ce{A + B \rightarrow AB\: (fast)}$
What is the order with respect to $\ce{[B2]}$?
Hint: First order
Skill -
Figure out the order from a given mechanism.
What is the overall order?
2. $\ce{A2 + B2 \rightarrow 2 AB}$ suggested the mechanism as follows:
1. $\ce{A2 \rightleftharpoons 2 A\: (fast\: equilibrium)}$
2. $\ce{A + B2 \rightarrow AB + B\: (slow)}$
3. $\ce{B + A2 \rightarrow AB + A\: (fast)}$
What is the order of the reaction with respect to $\ce{[A2]}$?
Hint: The order is half (1/2).
Discussion -
If you got $\ce{rate} = k \mathrm{[A_2]^{1/2} [B_2]}$, congratulations.
3. $\ce{2 NO + O2 \rightarrow 2 NO2}$ is
1. $\ce{NO + NO \rightleftharpoons N2O2\: (fast\: equilibrium)}$
2. $\ce{N2O2 + O2 \rightarrow 2 NO2\: (slow)}$
what is the power of $\ce{[NO]}$ in the differential rate law?
Hint: The order is 2 with respect to $\ce{[NO]}$.
Skill -
Figure out the order from a given mechanism.
You should get: $rate = k \ce{[NO]^2 [O2]}$
4. If the reaction mechanism consists of these elementary processes,
1. $\ce{A \rightleftharpoons 2 B\: (fast,\: equilibrium)}$
2. $\ce{B + 2 C \rightarrow E\: (slow)}$
3. $\ce{E \rightarrow F\: (fast)}$
choose the correct differential rate law for the reaction
$\ce{A + 4 C \rightarrow 2 F}$
1. $\mathrm{\dfrac{1}{2}\dfrac{d[F]}{dt} = \mathit{k}[A] [C]^4}$
2. $\mathrm{rate = \mathit{k}[A] [C]^2}$
3. $\mathrm{-\dfrac{d[A]}{dt} = \mathit{k}[A]^{1/2} [C]}$
4. $\mathrm{\dfrac{d[F]}{dt} = \mathit{k}[A] [C]}$
5. $\mathrm{\dfrac{d[F]}{dt} = \mathit{k}[A]^{1/2}[C]^2}$
Hint: e.
The method -
The rate determining step is ii. Express $\ce{[B]}$ in terms of $\ce{[A]}$ from i.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.09%3A_Rates_and_Mechanisms.txt
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Learning Objectives
• Derive the integrated rate laws from differential rate laws.
• Describe the variation of concentration vs. time for 1st order reactions.
• Describe the variation of concentration vs. time for 2nd order reactions.
• Figure out order of reaction from concentration vs. time plots.
Rate and Order of Reactions
The rate of a chemical reaction is the amount of substance reacted or produced per unit time. The rate law is an expression indicating how the rate depends on the concentrations of the reactants and catalysts. The power of the concentration in the rate law expression is called the order with respect to the reactant or catalyst.
This page deals specifically with first- and second-order reaction kinetics, but you should know that other orders such as zeroth-order and 3rd order may also be involved in chemical kinetics.
Reaction Rates and Stoichiometry
In acidic solutions, hydrogen peroxide and iodide ion react according to the equation:
$\ce{H2O2 + 2H+ + 3I- \rightarrow 2 H2O + I3-}$.
In this reaction, the reaction Rate can be expressed as
• decreasing Rate of $\ce{H2O2}$, $\ce{- \dfrac{d[H2O2]}{dt}}$
• decreasing Rate of $\ce{H+}$, $\ce{- \dfrac{d[H+]}{dt}}$
• decreasing Rate of $\ce{I-}$, $\ce{- \dfrac{d[I- ]}{dt}}$
• increasing Rate of $\ce{H2O}$, $\ce{+ \dfrac{d[H2O]}{dt}}$
• increasing Rate of $\ce{I3-}$, $\ce{\dfrac{d[I3- ]}{dt}}$
However, from the stoichiometry, you can easily see the following relationship:
$\ce{-\dfrac{d[H2O2]}{dt}}=\ce{-\dfrac{1}{2}\dfrac{d[H+]}{dt}}=\ce{-\dfrac{1}{3}\dfrac{d[I- ]}{dt}}=\ce{\dfrac{1}{2}\dfrac{d[H2O]}{dt}}=\ce{\dfrac{d[I3- ]}{dt}}$
In reality, if the concentration of $\ce{H2O2}$ is low, the changes in concentrations of $\ce{H+}$ and $\ce{H2O}$ are very difficult to detect because their quantities are so large in the solution. The merit in this equation is to show you that the rates of decreasing of reactant concentrations are governed by the stoichiometry. So are the rates of increasing of product concentrations.
There are many ways to express the reaction Rates, for example
$\ce{-\dfrac{d[H2O2]}{dt}}= \ce{-\dfrac{1}{2}\dfrac{d[H+]}{dt}} = \ce{-\dfrac{1}{3}\dfrac{d[I- ]}{dt}}=\ce{\dfrac{1}{2}\dfrac{d[H2O]}{dt}}=\ce{\dfrac{d[I3- ]}{dt}}= Rate$
or
$\ce{-\dfrac{d[I- ]}{dt}}= Rate\,'$
Obviously,
$Rate = 3 Rate\,'$
But both Rate and Rate' are reasonable expressions.
To generalize it, let the chemical reaction be represented by
$a\, \ce A + b\, \ce B \rightarrow c\, \ce C + d\, \ce D$
then the rate is represented by any one of the following
$rate = -\dfrac{1}{a}\ce{\dfrac{d[A]}{dt}}= -\dfrac{1}{b}\ce{\dfrac{d[B]}{dt}}=\dfrac{1}{c}\ce{\dfrac{d[C]}{dt}}=\dfrac{1}{d}\ce{\dfrac{d[D]}{dt}}$
Unless the rate expression is specified.
Differential Rate Laws and Integrated Rate Laws
For simplicity, let us consider the reactions:
$\mathrm{A + other\: reactants} \xrightarrow{\:\:\large{k}\:} \mathrm{products}$
where $\ce{A}$ is one of the reactants, and k is the rate constant.
For most experiments in chemical kinetics, the concentration of one reactant or product is monitored. If the concentrations of other reactants are high, they are not greatly changed. Thus, we have a pseudo decomposition reaction.
For the decomposition of $\ce{A}$ with a rate constant k,
$\mathrm{A \rightarrow products}$
The concentration $\ce{[A]}$ can be monitored. Let the order of the reaction be n, then the expression
$\mathrm{-\dfrac{d[A]}{dt}= \mathit k [A]^n}$
is called the differential rate law.
The differential expressions can be integrated to give an explicit relation of $\ce{[A]}$ with respect to time t. These explicit relations are called integrated rate laws.
Depending on the value of n, the integrated equations are different. If the reaction is first order with respect to $\ce{[A]}$, integration with respect to time, t, gives:
$\mathrm{[A] = [A]_o}\, e^{\large{-k\, t}} \tag{1}$
where $\mathrm{[A]_o}$ is the concentration of $\ce{A}$ at t = 0, and $\ce{[A]}$ is the concentration at time t.
For a second order reaction, the integrated rate law is:
$\mathrm{\dfrac{1}{[A]}=\dfrac{1}{[A]_o}}+ k t \tag{2}$
Derive the above equations yourself.
Determination of Rate Constants Using the Integrated Rate Laws
The usual approach to calculate the rate constant k makes use of the differential rate law. A series of experiments are performed with various initial concentrations, and their rates measured. The rate constants are calculated from the initial concentration and time of measurement. Often, you should construct a graph for their evaluation. However, results so obtained contain errors due to the approximation whereas values for k calculated using the integrated rate laws (1) and (2) are more accurate. Today, calculations can easily be performed with the aid of calculators and computers.
Thus, we emphasize that you use the integrated rate laws whenever possible. Whether you use the differential rate laws or the integrated laws, you have to evaluate the order first.
Equation (1) may be rewritten as
$k =\mathrm{\dfrac{\ln [A]_o - \ln [A]}{t}}\tag{1'}$
In a real experiment, you should plot $\ce{- \ln [A]}$ vs. t and find a line to best fit your data. The slope of the line is k.
Similarly, equation (2) may be rewritten as:
$k =\mathrm{\dfrac{\dfrac{1}{[A]}-\dfrac{1}{[A]_o}}{t}}\tag{2'}$
Thus, plot of $\ce{\dfrac{1}{[A]}}$ vs. t should yield a straight line, and the slope is k.
Half Lives of First and Second Order Reactions
The half-life ($t_{\frac{1}{2}}$) of a reaction is the time period required to reduce the reactant to half of its original value. The half life of a first order reaction is a constant, independent of the initial concentration. The rate constant and half-life have the relationship:
$t_{\frac{1}{2}} = \dfrac{\ln(2)}{k}$
$t_{\frac{1}{2}} \times k = \ln(2)$
For 2nd order reactions, the half life depends on the initial concentration, $\mathrm{[A]_o}$, and the rate constant k:
$t_{\frac{1}{2}} = \dfrac{1}{k \mathrm{[A]_o}}$
$t_{\frac{1}{2}} \times k \mathrm{[A]_o} = 1$
Since the concentration is reduced to half of its original value at the end of its first half-life, the second half-life is twice as long as the first half life. Thus, a plot of $\ce{[A]}$ vs. t easily reveals the order of the reaction by tracking its half life.
Either plotting $\ce{[A]}$ vs. t or plotting the appropriate linear relationship will reveal the order of the reaction.
A summary of the relationships for first and second order reactions is given below:
Differential
rate law
Integrated
rate law
Linear plot Half life
first
order
$\mathrm{-\dfrac{d[A]}{dt} = \mathit k [A]}$ $\mathrm{[A] = [A]_o} e^{-\large{k\, t}}$
$\mathrm{\ln [A]\: vs}\: t$
$\textrm{slope} = - k$
$t_{\frac{1}{2}}=\dfrac{\ln(2)}{k}$
second
order
$\mathrm{- \dfrac{d[A]}{dt} = \mathit k [A]^2}$
$\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t$
$\mathrm{\dfrac{1}{[A]}\: vs}\: t$
$\textrm{slope} = k$
$t_{\frac{1}{2}}=\dfrac{1}{k\mathrm{[A]_o}}$
Further applications of rate laws will be discussed in Integrated Rate Laws. Students must analyze the problems carefully, and appropriately apply the differential or integrated rate laws to derive the desirable results.
Confidence Building Questions
1. If the reaction is zeroth (0th) order with respect to $\ce{[X]}$, which of the following quantities when plotted vs. t should be a straight line: $\ce{\ln [X]}$, $\ce{\dfrac{1}{[X]}}$, $\ce{[X]}$, or $\ce{[X]^2}$?
Hint: $\ce{[X]}$
Discussion -
Since: $\mathrm{- \dfrac{d[X]}{dt} = k}$; => $\mathrm{[X] - [X]_o = - \mathit k t}$; The plot of $\ce{[X]}$ and t has a linear relationship.
2. The half-lives for all first order reactions are independent of the initial concentration, true or false?
Hint: true
Discussion -
All radioactive decays follow first order kinetics.
3. For a first order reaction, the product of rate constant and half-life is a constant. What is the constant? (give value)
Hint: ln(2) = 0.693
Discussion -
$k \times t_{\frac{1}{2}} = \ln(2) = 0.693$
4. Tritium (T) decays to helium ($\ce{He}$ of mass 3) with a half-life of 12.5 years. How long will it take 1.0 g of T to reduce to 0.25 g of T (and 0.75 g of $\ce{He}$)?
Hint: 25 years
Discussion -
The first 12.5 y reduced 1.0 g of T to 1/2 g. Another 12.5 y reduced 0.5 g of T to 1/4 g.
Calculate the time for 1.0 g T to reduce to 0.9 g. Calculation of half-life from experiment is also a common exercise.
5. $\ce{2 HI \rightarrow H2 + I_{2\large{(s)}}}$ In an experiment starting at 8:00 hr, the partial pressure of $\ce{HI}$ decreases from 30.0 Pa to 15.0 Pa at 8:45 hr (at a constant temperature). What shall the time be for the partial pressure of $\ce{HI}$ to reduce to 3.75 Pa?
Hint: 13:15 hr
Discussion -
Amount left: 30, 15, 7.5, 3.75
Time: 8:00 8:45 10:15 13:15
t'increment: 45 90 180 (min)
Calculate the time when 10 percent of the $\ce{HI}$ has decomposed.
6. The dust explosion is a
1. homogeneous reaction
2. heterogeneous reaction
3. catalytic reaction
4. chain reaction
Hint: b. heterogeneous reaction
Discussion -
Chain reactions may lead to explosion, but dust explosion is not a chain reaction. Chemists explained the causes of a grain elevator explosion.
7. $\ce{2 NO + O2 \rightarrow 2 NO2}$, the following results were obtained:
run [NO] [O2] Rate
1 0.12 0.05 0.12
2 0.12 0.10 0.24
3 0.24 0.05 0.48
What is the order of the reaction with respect to $\ce{[NO]}$?
Hint: 2
Discussion -
What is the order with respect to $\ce{[O2]}$?
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This is a continuation of Rate and Order of Reactions.
Learning Objectives
• Apply either the differential rate laws or integrated rate laws to solve chemical kinetic problems.
In this module, more examples are given to show applications of integrated rate laws in problem solving.
Integrated Rate Laws
The differential rate laws and integrated rate laws are summarized in the table below to give you an overall view of reactions of these types.
Differential
rate law
Integrated
rate law
Linear plot Half life
first
order
$\mathrm{-\dfrac{d[A]}{dt} = \mathit k [A]}$ $\mathrm{[A] = [A]_o} e^{-\large{k\, t}}$
$\mathrm{\ln [A]\: vs}\: t$
$\textrm{slope} = - k$
$t_{\frac{1}{2}}=\dfrac{\ln(2)}{k}$
second
order
$\mathrm{- \dfrac{d[A]}{dt} = \mathit k [A]^2}$
$\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t$
$\mathrm{\dfrac{1}{[A]}\: vs}\: t$
$\textrm{slope} = k$
$t_{\frac{1}{2}}=\dfrac{1}{k\mathrm{[A]_o}}$
Example 1
The decomposition of $\ce{A}$ is first order, and $\ce{[A]}$ is monitored. The following data are recorded:
t / min 0 1 2 3
[A]/[M] 0.100 0.0905 0.0819 0.670
1. Calculate k. (What is the rate constant?)
2. Calculate the half life. (What is the half life?)
3. Calculate $\ce{[A]}$ when t = 5 min.
4. Calculate t when $\mathrm{[A] = 0.0100}$ (i.e., estimate the time required for 90% of $\ce{A}$ to decompose.)
Solution
1. We can calculate k from any two data points.
The integrated rate law for 1st order is
$\mathrm{A = A_o}\, e^{- \large{k\, t}}$
Using the the first two points,
$0.0905 = 0.100\, e^{(- \large k \times 1)}$
\begin{align*} - k &= \ln \left(\dfrac{0.0905}{0.100}\right)\ &= \mathrm{\ln(0.905) = -.0998\: min^{-1}} \end{align*}
Using the point when t = 2
$0.100 = 0.0819\, e^{(-\large k \times 2)}$
\begin{align*} - k\, 2 &= \ln \left(\dfrac{0.0819}{0.100}\right)\ &= \ln (0.819)\ &= -0.200\ k &= \mathrm{0.100\: min^{-1}} \end{align*}
Using the point when t = 4
$0.0670 = 0.100\, e^{(- \large k \times 4)}$
\begin{align*} 4\, k &= \ln 1.49\ &= 0.400\ k &= \mathrm{0.100\: min^{-1}} \end{align*}
2. Two methods to evaluate half life are:
1. $t_{\frac{1}{2}} \times k = \ln 2$
$= 0.693$
$t_{\frac{1}{2}} = \dfrac{0.693}{0.1} = \mathrm{6.93\: min}$
(note the calculation of units)
2. Calculate the time t when $\mathrm{[A] = 0.0500}$
$0.0500 = 0.100\, e^{(\large{-0.100\: t})}$
$0.100 \times t = \ln \left(\dfrac{0.100}{0.0500}\right)$
gives the same result.
3. When t = 5 min
\begin{align*} \mathrm{[A]} &= 0.100\, e^{(\large{-0.100 \times5})}\ &= 0.100 \times 0.6065\ &= 0.0607 \end{align*}
4. When $\ce{[A]}$ is reduced by 90%, we have
$0.01 = 0.1\, e^{(\large{-0.100 \times t})}$
$0.100 \times t = \ln (10)$
$\mathrm{t = \dfrac{2.303}{0.100} = 23.03\: min}$
Check:
\begin{align*} \mathrm{[A]} &= 0.1\, e^{(\large{-0.1 \times 23.03})}\ &= 0.010 \end{align*}
Example 2
The dimerization reaction of butadiene is second order process:
$\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}} \nonumber$
The rate constant at some temperature is 0.100 /min with an initial concentration of butadiene ([B]) of 1 M. Calculate the concentration of butadiene at 1, 2, 5, 10, 20, 30, and 70 minutes.
Solution
Many of the following values can be evaluated without using a calculator.
t / min 0 1 2 5 10 20 30 70
[B]/[M] 1.0 0.909 0.833 0.667 0.50 0.33 0.25 0.125
Discussion:
This is an applicaiton of this equation:
$\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t \nonumber$
Exercise
From the data and results in Example 2:
1. Calculate k. (What is the rate constant?)
2. Calculate the half life for [B] = 1.0. (What is the half life for an initial concentration of 1.0?)
3. Calculate [B] when t = 40 min.
4. Calculate t when [B] = 0.100. (Estimate the time required for 90% of B to polymerize.)
5. How does the total pressure change with time?
Questions
1. $\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}}$ the rate of decreasing of partial pressure of butadiene is 2 kPa/min. What is the rate of increasing of partial pressure of (octadiene)?
Hint: 1 kPa/min
Discussion -
Consider the relationship
$-\ce{\dfrac{d[C4H6]}{dt}} = \ce{2 \dfrac{d[C8H12]}{dt}}$
2. $\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}}$
Hint: 6 kPa
Discussion -
The drop in pressure is 101-95 = 6 kPa. Consider the reaction:
\begin{alignat}{3} \ce{2 &C4H_{6\large{(g)}} \rightarrow\: &&C8H_{12\large{(g)}}}\ \ce{&12\: kPa &&\:\:6\: kPa} \end{alignat}
Since two moles of butadiene combine to give one mole octadiene, the difference in total pressure is the partial pressure of octadiene, 12 kPa of butadiene converted to 6 kPa of octadiene. What is the partial pressure of butadiene?
3. At some temperature, the total pressure of a butadiene cylinder drops from 101 kPa to 95 kPa in 10 min. The polymerization is known to be second order. What is the rate constant?
Hint: 0.000133 kPa/min
Discussion -
The reaction is:
$\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}}$
Since the total pressure drops from 101 to 95 kPa after 10 min, the partial pressure of $\ce{C4H6}$ goes from 101 to [101 - 2*(101-95)] = 89 kPa. Using the integrated rate law:
$\dfrac{1}{89} - \dfrac{1}{101} = k \times 10$
$k =\mathrm{1.33\,e^{-4}}$
What is the total pressure when the reaction is completed?
4. Radioactive decay always follows first order kinetics. Carbon-11 is a radioactive isotope of carbon, and its half life is 20.3 min. What is the decay (or rate) constant?
Hint: 0.0341 /m
Skill -
Evaluate half-life according to conditions. Apply the equation,
$k \times t_{\frac{1}{2}} = \ln 2$
in this case.
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Learning Objectives
• Explain steady state and steady-state approximation.
• Derive a rate law when a mechanism is given but the rate determining step is not identified.
• Derive a general expression of the rate law using the steady-state approximation.
• Make appropriate assumptions so that the derived rate law agrees with the observed rate law.
• Give expressions for the producing rate of an intermediate.
• Give expressions for the consuming rate of an intermediate.
• Express concentration of intermediate in terms of concentration of reactants.
• Eliminate concentrations of intermediates using concentrations of reactants.
• Derive a rate law from the many elementary steps.
• Discuss the derived rate law.
The Steady-State Approximation
When a reaction mechanism has several steps of comparable rates, the rate-determining step is often not obvious. However, there is an intermediate in some of the steps. An intermediate is a species that is neither one of the reactants, nor one of the products. The steady-state approximation is a method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated. Its concentration remains the same in a duration of the reaction.
Definition: Intermediates
An intermediate is a species that is neither one of the reactants, nor one of the products. It transiently exists during the course of the reaction.
When a reaction involves one or more intermediates, the concentration of one of the intermediates remains constant at some stage of the reaction. Thus, the system has reached a steady-state. The concentration of one of the intermediates, $[Int]$, varies with time as shown in Figure $1$. At the start and end of the reaction, [Int] does vary with time.
$\dfrac{d[Int]}{dt}= 0 \nonumber$
When a reaction mechanism has several steps with comparable rates, the rate-determining step is not obvious. However, there is an intermediate in some of the steps. The steady-state approximation implies that you select an intermediate in the reaction mechanism, and calculate its concentration by assuming that it is consumed as quickly as it is generated. In the following, an example is given to show how the steady-state approximation method works.
Example $1$
Use the steady-state approximation to derive the rate law for this reaction
$\ce{2 N2O5 \rightarrow 4 NO2 + O2}\nonumber$
assuming it follows the following three-step mechanism:
\begin{align} \ce{N_2O_5} &\underset{\Large{k_{\textrm b}}}{\overset{\Large{k_{\textrm f}}}\rightleftharpoons} \ce{NO_2 + NO_3} \tag{step 1} \[4pt] \ce{NO3 + NO2} &\ce{->[\large{k_2}] NO + NO2 + O2} \tag{step 2} \[4pt] \ce{NO3 + NO} & \ce{->[\Large{k_3}] 2 NO2} \tag{step 3} \end{align}
Solution
In these steps, $\ce{NO}$ and $\ce{NO3}$ are intermediates. You have
$\ce{production\: rate\: of\: NO} = k_2 \ce{[NO3] [NO2]}$
$\ce{consumption\: rate\: of\: NO} = k_3 \ce{[NO3] [NO]}$
A steady-state approach makes use of the assumption that the rate of production of an intermediate is equal to the rate of its consumption. Thus, we have
$k_2 \ce{[NO3] [NO2]} = k_3 \ce{[NO3] [NO]}$
and solving for $\ce{[NO]}$ gives the result,
$\ce{[NO]} = \dfrac{k_2 \ce{[NO3] [NO2]}}{k_3 \ce{[NO3]}} \tag{1}$
For the other intermediate $\ce{NO3}$,
$\ce{production\: rate\: of\: NO3} = k_{\ce f} \ce{[N2O5]}$
$\ce{consumption\: rate\: of\: NO3} = k_2\ce{[NO3] [NO2]} + k_3\ce{[NO3] [NO]} + k_{\ce b}\ce{[NO3] [NO2]}$
Applying the steady-state assumption gives:
$k_{\ce f} \ce{[N2O5]} = k_2\ce{[NO3] [NO2]} + k_3\ce{[NO3] [NO]} + k_{\ce b}\ce{[NO3] [NO2]}$
Thus,
$\ce{[NO3]} = \dfrac{k_{\ce f} \ce{[N2O5]}}{k_2\ce{[NO2]} + k_3\ce{[NO]} + k_{\ce b}\ce{[NO2]}}\tag{2}$
Let's review the three equations (steps) in the mechanism:
Step i. is at equilibrium and thus can not give a rate expression.
Step ii. leads to the production of some products, and the active species $\ce{NO}$ causes further reaction in step iii. This consideration led to a rate expression from step ii. as:
$\ce{\dfrac{d[O2]}{dt}} = k_2 \ce{[NO3] [NO2]} \tag{3}$
Substituting (1) in (2) and then in (3) gives
$\ce{\dfrac{d[O2]}{dt}}= \dfrac{k_{\ce f} k_2 \ce{[N2O5]}}{k_{\ce b} + 2 k_2} = \ce{k [N2O5]}$
where $\ce{k} = \dfrac{k_{\ce f} k_2}{k_{\ce b} + 2 k_2}$.
This is the differential rate law, and it agrees with the experimental results. Carry out the above manipulation yourself on a piece of paper. Simply reading the above will not lead to solid learning yet.
This page gives another example to illustrate the technique of deriving rate laws using the steady-state approximation. The reaction considered here is between $\ce{H2}$ and $\ce{I2}$ gases.
Example $1$
For the reaction:
$\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightarrow 2 HI_{\large{(g)}}}$
what mechanisms might be appropriate? Derive a rate law from the proposed mechanism.
Solution
Well, this question does not have a simple answer, and there is no way to prove one over another for its validity. Beginning chemistry students will not be asked to propose a mechanism, but you will be asked to derive the rate law from the proposed mechanism.
First of all, you should be able to express the rate of reaction in terms of the concentration changes,
$rate = - \ce{\dfrac{d[H2]}{dt}} = - \ce{\dfrac{ d[I2]}{dt}} = \ce{\dfrac{1}{2}\dfrac{d[HI]}{dt}}$
Look at the overall reaction equation again to see its relationship and the rate expressions.
Proposing a mechanism
In order to propose a mechanism, we apply the following reasoning. Since the bonding between $\ce{I-I}$ is weak, we expect $\ce{I2}$ to dissociate into atoms or radicals. These radicals are active, and they react with $\ce{H2}$ to produce the products. Thus we propose the three-step mechanism:
1. $\ce{I_{2\large{(g)}}} \xrightarrow{\Large{k_1}} \ce{2 I_{\large{(g)}}}$
2. $\ce{2 I_{\large{(g)}}} \xrightarrow{\Large{k_2}} \ce{I_{2\large{(g)}}}$
3. $\ce{H_{2\large{(g)}} + 2 I_{\large{(g)}}} \xrightarrow{\Large{k_3}} \ce{2 HI_{\large{(g)}}}$
Which step would you use to write the differential rate law?
Since only step iii. gives the real products, we expect you to recognize that step iii. hints the rate law to be:
$rate = k_3 \ce{[H2] [I]^2}$
However, this is not a proper rate law, because $\ce{I}$ is an intermediate, not a reactant. So, you have to express $\ce{[I]}$ or $\ce{[I]^2}$ in terms of the concentration of reactants. To do this, we use the steady-state approximation and write out the following relationships:
$\textrm{rate of producing I} = 2 k_1 \ce{[I2]}$
$\textrm{rate of consuming I} = 2 k_2 \ce{[I]^2} + 2 k_3 \ce{[H2] [I]^2}$
$\textrm{producing rate of I} = \textrm{consuming rate of I}$
Thus,
$\ce{[I]^2} = \dfrac{k_1 \ce{[I2]}}{k_2 + k_3 \ce{[H2]}}$
Substituting this for $\ce{[I]^2}$ into the rate expression, you have
\begin{align} rate &= k_3 \ce{[H2]} \dfrac{k_1 \ce{[I2]}}{k_2 + k_3 \ce{[H2]}}\ &= \dfrac{k_1 k_3 \ce{[H2] [I2]}}{k_2 + k_3 \ce{[H2]}} \end{align}
Discussion
If step iii. is slow, then $k_3$ and $k_2 >> k_3 \ce{[H2]}$. The rate law is reduced to
$\ce{rate} = \ce{k [H2] [I2]}$,
where $\ce{k} = \dfrac{k_1 k_3}{k_2}$. (Work this out on paper yourself; reading the above derivation does not lead to learning.)
Since the rate law is first order with respect to both reactants, one may argue that the rate law also supports a one-step mechanism,
$\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightarrow 2 HI}$
This elementary step is the same as the overall reaction.
Suppose we use a large quantity of $\ce{H2}$ compared to $\ce{I2}$, then the change in $\ce{[H2]}$ is insignificant. For example, if $\mathrm{[H_2] = 10}$, and $\mathrm{[I_2] = 0.1}$ initially, $\ce{[H2]}$ remains essentially 10 (9.9 with only one significant figure). In other words, $\ce{[H2]}$ hardly changed when the reaction ended. Thus,
$k_3 \ce{[H2]} >> k_3$
and the rate law becomes:
$rate = k_1 \ce{[I2]}$.
Thus, the reaction is a pseudo first order reaction, due to the large quantity of one reactant. The results suggest iii. a fast step (due to large quantity of $\ce{H2}$), and i. the rate determining step.
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The basic method of obtaining the information needed to determine rate constants and reaction orders is to bring the reactants together and then measure successive changes in concentration of one of the components as a function of time. Two important requirements are:
• The time required to take a measurement must be very short compared to the time the reaction takes to run to completion;
• The temperature must be held constant — something than can pose a problem if the reaction is highly exothermic.
Measuring the concentration of a reactant or product directly — that is by chemical analysis—is awkward and seldom necessary. When it cannot be avoided, the reaction sample must usually be quenched in some way in order to stop any further change until its composition can be analyzed. This may be accomplished in various ways, depending on the particular reaction. For reactions carried out in solution, especially enzyme-catalyzed ones, it is sometimes practical to add a known quantity of acid or base to change the pH, or to add some other inhibitory agent.
More commonly, however, the preferred approach is to observe some physical property whose magnitude is proportional to the extent of the reaction.
Contributors and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
05: Experimental Methods
Kinetics are used to study of the rate of the reaction and how factors such as temperature, concentration of the reactants, and catalysts affect it. The reaction order shows how the concentration of reactants affects the reaction rate. Determining the reaction rate also helps determine the reaction mechanism. For the reaction $\ce{A \rightarrow B}$ the rate is given by
$\text{rate}=[\text{A}]^x[\text{B}]^y \nonumber$
The overall rate is the sum of these superscripts.
Introduction
Although there is an infinite variety of rates laws possible, three simple reaction orders commonly taught: zeroth, first, and second order.
In zero order reactions, the disappearance of reactants is
$\dfrac{-d[\text{A}]}{dt}= k[\text{A}]^0= k \nonumber$
Its integrated form is
$[\text{A}]=-kt+[\text{A}]_0 \nonumber$
In first order reactions, the disappearance of reactants is
$\dfrac{-d[\text{A}]}{dt}=k[\text{A}]^1 \nonumber$
The integrated form is
$[\text{A}]=[\text{A}]_0 e^{-kt} \label{1st}$
In second order reactions with two reactant species, the rate of disappearance of $\text{A}$ is
$\dfrac{-d[\text{A}]}{dt}= k[\text{A}][\text{B}] \nonumber$
The integrated form is
$\dfrac{1}{[\text{B}]_0-[\text{A}]_0}\ln \dfrac{[\text{B}][\text{A}]_0}{[\text{B}]_0[\text{A}]}=kt \label{2nd}$
when $[\text{B}]_0>[\text{A}]_0$.
When $[\text{B}]_0>>[\text{A}]_0$, then $[\text{B}] \approx [\text{B}]_0$ and Equation $\ref{2nd}$ becomes
$\dfrac{1}{[\text{B}]_0-[\text{A}]_0}\ln \dfrac{[\text{B}][\text{A}]_0}{[\text{B}]_0[\text{A}]} \approx \dfrac{1}{[\text{B}]_0}\ln \dfrac{[\text{A}]_0}{[\text{A}]}=kt \nonumber$
or
$[\text{A}] = [\text{A}]_0 e^{-[\text{B}]kt} \nonumber$
This functional form of the decay kinetics is similar ot the first order kinetics in Equation $\ref{1st}$ and the system is said to operate under pseudo-first order kinetics.
In second order reactions with one reactant, the disappearance of reactants is
$\dfrac{-d[\text{A}]}{dt}= k[\text{A}]^2 \nonumber$
The integrated form is
$\dfrac{1}{[\text{A}]}= kt+\dfrac{1}{[\text{A}]_0} \nonumber$
Graphical Method
The graphical method makes use of the concentrations of reactants. It is most useful when one reactant is isolated by having the others in large excess. A series of standards is used to make a Beer's Law plot, the best-fit equation of which is used to determine the concentration of the isolated reactant. To test for the order of reaction with regard to that reactant, three plots are made. The first is concentration of the isolated reactant versus time. The second is of inverse concentration versus time, while the third is of the natural log of concentration versus time. These graphs, respectively, show zero, first, and second order dependence on the specific reactant. The graph that is most linear shows the order of the reaction with regard to that reactant.
$\ce{ CH_3COCH_3 + Br_2 + H^+ \rightarrow CH_3COCH_2Br + HBr} \nonumber$
For example, in the acid-catalyzed bromination of acetone, the concentration of bromine is isolated and measured, while the concentrations of the hydrogen ion and acetone are not. The three graphs from a run are below. The graph of concentration versus time is the one that is most linear, so the order of the bromination reaction with respect to bromine is zero. The slope of the best-fit line gives $-k_{obs}$. In theory, the other reactants could be isolated like bromine was, but the data from bromine can be used to determine the reaction order with regard to them as well. It is found by comparing two reactions where only the concentration of the reactant in question is changed. If p is the reaction order with regard to acetone,
$p = \dfrac{ \log \dfrac{k_{obsII}}{k_{obsI}}}{ \log u } \nonumber$
The observed rate constant at the higher concentration of acetone is $k_{obsII}$, while $k_{obsI}$ is the observed rate constant at the lower concentration of acetone. The ratio of the higher concentration to the lower concentration is given by u. The process is repeated for the hydrogen ion.
Mathematical Method
The mathematical method is useful when the means to graph are not available. It is essentially determining the slope of the plot that "linearizes the data". This requires plotting concentration versus time data. As in the graphical method, the inverse and natural log of the concentration must be calculated. A linear plot has a constant slope, so the slopes calculated from two pairs of adjacent points should be the same.Take three consecutive points from the concentration versus time data. Calculate $\dfrac{\Delta y}{\Delta x}$ for the first and second points. The concentration is the $y$ value, while time is the $x$ value. Do the same for the second and third point. If the reaction is zero order with regard to the reactant, the numbers will be the same. If not, then calculate the slope for the inverse concentration versus time data or natural log of the concentration versus time data.
Table 1: Bromine vs. time data
[Bromine] (M) Time (min)
0.00349 4.25
0.00344 4.42
0.00339 4.58
The slope of the first two points is
$m= \dfrac{0.00344\,\text{M} - 0.00349\,\text{M}}{4.42\, \text{min}-4.25\, \text{min}} \nonumber$
or
$m= -2.9 \times 10^{-4}\dfrac{\text{M}}{\text{min}} \nonumber$
The slope of the second two points is
$\dfrac{0.00476\, \text{M}-0.00479\,\text{M}}{4.58 \,\text{min}-4.42\, \text{min}} \nonumber$
or
$-3.1 \times 10^{-4}\dfrac{\text{M}}{\text{min}} \nonumber$
These are approximately the same, so the bromine depletion follows zero order kinetics.
Another method uses half lives. For a zero-order reaction,
$t_{1/2}=\dfrac{[\text{A}]_0}{2k} \nonumber$
For a first-order reaction,
$t_{1/2}=\dfrac{\ln 2}{k} \nonumber$
For a second-order reaction,
$t_{1/2}=\dfrac{1}{[\text{A}]_0k} \nonumber$
If an increase in reactant increases the half life, the reaction has zero-order kinetics. If it has no effect, it has first-order kinetics. If the increase in reactant decreases the half life, the reaction has second-order kinetics.
Problems
1. Use the data in the table to find the reaction order with respect to the reactant being studied.
Time (minutes) $\ln [\text{A}]$
0.83 -5.320
0.25 -5.341
0.42 -5.347
0.58 -5.352
0.75 -5.362
1. Suppose another reactant, B, is involved in the reaction. The reaction order, x, with regard to B is 2. $\dfrac{\ k_{obsII}}{\ k_{obsI}}$ is 0.433. The larger concentration of B is 2.21 M. What is the smaller concentration? 0.067 M
2. What is the overall order of the reaction between A and B? 2
3. If the concentration of A were doubled, what would happen to the reaction rate? What would happen to the reaction rate if the concentration of B were doubled? Nothing; it would quadruple
Contributors and Attributions
• Lydia Rau (Hope College)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/05%3A_Experimental_Methods/5.01%3A_Determining_Reaction_Order.txt
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Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, the exponents in the rate law are the positive integers: 1 and 2 or even 0. Thus the reactions are zeroth, first, or second order in each reactant. The common patterns used to identify the reaction order are described in this section, where we focus on characteristic types of differential and integrated rate laws and how to determine the reaction order from experimental data. The learning objective of this Module is to know how to determine the reaction order from experimental data.
Zeroth-Order Reactions
A zeroth-order reaction is one whose rate is independent of concentration; its differential rate law is rate = k. We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \tag{14.15}$
Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of −k. The value of k is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of k, a positive value.
The graph of a zeroth-order reaction. The change in concentration of reactant and product with time produces a straight line.
The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form
$[A] = [A]_0 − kt \tag{14.16}$
where [A]0 is the initial concentration of reactant A. (Equation 14.16 has the form of the algebraic equation for a straight line, y = mx + b, with y = [A], mx = −kt, and b = [A]0.) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second.
Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N2O on a platinum (Pt) surface to produce N2 and O2, which occurs at temperatures ranging from 200°C to 400°C:
$\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \tag{14.17}$
Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N2O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N2O to react with the entire Pt surface, doubling or quadrupling the N2O concentration will have no effect on the reaction rate. At very low concentrations of N2O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N2O concentration. The reaction rate is as follows:
$\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \tag{14.18}$
Thus the rate at which N2O is consumed and the rates at which N2 and O2 are produced are independent of concentration. As shown in Figure 14.8, the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N2O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally.
A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the enzyme alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is
where NAD+ (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero (part (a) in Figure 14.10). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in Figure 14.10).
These examples illustrate two important points:
1. In a zeroth-order reaction, the reaction rate does not depend on the reactant concentration.
2. A linear change in concentration with time is a clear indication of a zeroth-order reaction.
First-Order Reactions
In a first-order reaction, the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \tag{14.19}$
If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s−1).
The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$[A] = [A]_0e^{−kt} \tag{14.20}$
where [A]0 is the initial concentration of reactant A at t = 0; k is the rate constant; and e is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. Equation 14.20 predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of Equation 14.20 and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and t:
$\ln[A] = \ln[A]_0 − kt \tag{14.21}$
Because Equation 14.21 has the form of the algebraic equation for a straight line, y = mx + b, with y = ln[A] and b = ln[A]0, a plot of ln[A] versus t for a first-order reaction should give a straight line with a slope of −k and an intercept of ln[A]0. Either the differential rate law (Equation 14.19) or the integrated rate law (Equation 14.21) can be used to determine whether a particular reaction is first order.
Graphs of a first-order reaction. The expected shapes of the curves for plots of reactant concentration versus time (top) and the natural logarithm of reactant concentration versus time (bottom) for a first-order reaction.
First-order reactions are very common. In this chapter, we have already encountered two examples of first-order reactions: the hydrolysis of aspirin (Figure 14.6) and the reaction of t-butyl bromide with water to give t-butanol (Equation 14.10). Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin.
Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows:
Figure 14.11
Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in Figure 14.11 is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in Figure 14.11 have been studied extensively to find ways of maximizing the concentration of the active species.
Note
If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order.
The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in Table 14.2. The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in Table 14.2 shows that the reaction rate doubles [(1.8 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10−5 M/min) ÷ (9.0 × 10−6 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = k[cisplatin]1. Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table 14.2. For example, substituting the values for Experiment 3 into Equation 14.19,
3.6 × 10−5 M/min = k(0.024 M)
1.5 × 10−3 min−1 = k
Table 14.2 Rates of Hydrolysis of Cisplatin as a Function of Concentration at pH 7.0 and 25°C
Experiment [Cisplatin]0 (M) Initial Rate (M/min)
1 0.0060 9.0 × 10−6
2 0.012 1.8 × 10−5
3 0.024 3.6 × 10−5
4 0.030 4.5 × 10−5
Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug.
Example 4
At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction:
$\mathrm{CH_3CH_2Cl(g)}\xrightarrow{\Delta}\mathrm{HCl(g)}+\mathrm{C_2H_4(g)} \nonumber$
Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction.
Experiment [CH3CH2Cl]0 (M) Initial Rate (M/s)
1 0.010 1.6 × 10−8
2 0.015 2.4 × 10−8
3 0.030 4.8 × 10−8
4 0.040 6.4 × 10−8
Given: balanced chemical equation, initial concentrations of reactant, and initial rates of reaction
Asked for: reaction order and rate constant
Strategy:
1. Compare the data from two experiments to determine the effect on the reaction rate of changing the concentration of a species.
2. Compare the observed effect with behaviors characteristic of zeroth- and first-order reactions to determine the reaction order. Write the rate law for the reaction.
C Use measured concentrations and rate data from any of the experiments to find the rate constant.
Solution
The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate.
A Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH3CH2Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH3CH2Cl].
B This behavior is characteristic of a first-order reaction, for which the rate law is rate = k[CH3CH2Cl].
C We can calculate the rate constant (k) using any row in the table. Selecting Experiment 1 gives the following:
1.60 × 10−8 M/s = k(0.010 M)
1.6 × 10−6 s−1 = k
Exercise
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
SO2Cl2(g) → SO2(g) + Cl2(g)
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Exercise 1: I see you
Sulfuryl chloride (SO2Cl2) decomposes to SO2 and Cl2 by the following reaction:
SO2Cl2(g) → SO2(g) + Cl2(g)
Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction.
Experiment [SO2Cl2]0 (M) Initial Rate (M/s)
1 0.0050 1.10 × 10−7
2 0.0075 1.65 × 10−7
3 0.0100 2.20 × 10−7
4 0.0125 2.75 × 10−7
Answer first order; k = 2.2 × 10−5 s−1
We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Part (a) in Figure 14.12 shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in Figure 14.12. The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in Figure 14.12 for t = 100 min ([cisplatin] = 0.0086 M) and t = 1000 min ([cisplatin] = 0.0022 M),
\begin{align}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}} \-k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}} \k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align}
The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min−1 because the times plotted on the horizontal axes in parts (a) and (b) in Figure 14.12 are in minutes rather than seconds.
The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions.
Example 5
Refer back to Example 4. If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M? (Recall that we calculated the rate constant for this reaction in Example 4.)
Given: initial concentration, rate constant, and time interval
Asked for: concentration at specified time and time required to obtain particular concentration
Strategy:
1. Substitute values for the initial concentration ([A]0) and the calculated rate constant for the reaction (k) into the integrated rate law for a first-order reaction. Calculate the concentration ([A]) at the given time t.
2. Given a concentration [A], solve the integrated rate law for time t.
Solution
The exponential form of the integrated rate law for a first-order reaction (Equation 14.20) is [A] = [A]0ekt.
A Having been given the initial concentration of ethyl chloride ([A]0) and having calculated the rate constant in Example 4 (k = 1.6 × 10−6 s−1), we can use the rate law to calculate the concentration of the reactant at a given time t. Substituting the known values into the integrated rate law,
\begin{align}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt} \&=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}}) \&=0.0189\textrm{ M}\end{align}
We could also have used the logarithmic form of the integrated rate law (Equation 14.21):
\begin{align}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt \ &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})] \ &=-3.912-0.0576=-3.970 \ [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \ &=0.0189\textrm{ M}\end{align}
B To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for t. Equation 14.21 gives the following:
\begin{align}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt \kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \ t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right) \ &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h}\end{align}
Exercise 5
In the exercise in Example 4, you found that the decomposition of sulfuryl chloride (SO2Cl2) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO2Cl2 that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO2Cl2 to decompose?
Answer 0.0252 M; 29 h
Second-Order Reactions
The simplest kind of second-order reaction is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer).
The differential rate law for the simplest second-order reaction in which 2A → products is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \tag{14.22}$
Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M−1·s−1). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s).
For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time:
$\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \tag{14.23}$
Because Equation 14.23 has the form of an algebraic equation for a straight line, y = mx + b, with y = 1/[A] and b = 1/[A]0, a plot of 1/[A] versus t for a simple second-order reaction is a straight line with a slope of k and an intercept of 1/[A]0.
Note
Second-order reactions generally have the form 2A → products or A + B → products.
Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO2 to NO and O2 and the decomposition of HI to I2 and H2. Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture.
Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows:
Figure 14.13
For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law (Equation 14.22) or the integrated rate law (Equation 14.23).
To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in Table 14.3. From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7:
$\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7$
Table 14.3 Rates of Reaction as a Function of Monomer Concentration for an Initial Monomer Concentration of 0.0054 M
Time (min) [Monomer] (M) Instantaneous Rate (M/min)
10 0.0044 8.0 × 10−5
26 0.0034 5.0 × 10−5
44 0.0027 3.1 × 10−5
70 0.0020 1.8 × 10−5
120 0.0014 8.0 × 10−6
Because (1.7)2 = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration.
rate ∝ [monomer]2
This means that the reaction is second order in the monomer. Using Equation 14.22 and the data from any row in Table 14.3, we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following:
\begin{align}\textrm{rate}&=k[\textrm A]^2 \8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align}
We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in Figure 14.14. The measurements show that the concentration of the monomer (initially 5.4 × 10−3 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus t should be a straight line, as shown in part (b) in Figure 14.14. Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, k = 4.1 M−1·min−1, which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally.
Figure 14.14 Dimerization of a Monomeric Compound, a Second-Order Reaction. These plots correspond to dimerization of the monomer in Figure 14.13 as (a) the experimentally determined concentration of monomer versus time and (b) 1/[monomer] versus time. The straight line in (b) is expected for a simple second-order reaction.
For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. The differential and integrated rate laws for zeroth-, first-, and second-order reactions and their corresponding graphs are shown in Figure 14.16.
Example 6
At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen.
$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber$
Experimental data for the reaction at 300°C and four initial concentrations of NO2 are listed in the following table:
Experiment [NO2]0 (M) Initial Rate (M/s)
1 0.015 1.22 × 10−4
2 0.010 5.40 × 10−5
3 0.0080 3.46 × 10−5
4 0.0050 1.35 × 10−5
Determine the reaction order and the rate constant.
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: reaction order and rate constant
Strategy:
1. From the experiments, compare the changes in the initial reaction rates with the corresponding changes in the initial concentrations. Determine whether the changes are characteristic of zeroth-, first-, or second-order reactions.
2. Determine the appropriate rate law. Using this rate law and data from any experiment, solve for the rate constant (k).
Solution
A We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO2 concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10−5) ÷ (1.35 × 10−5) = 4.0], which means that the reaction rate is proportional to [NO2]2. Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO2]2. This behavior is characteristic of a second-order reaction.
B We have rate = k[NO2]2. We can calculate the rate constant (k) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following:
\begin{align}\textrm{rate}&=k[\mathrm{NO_2}]^2 \5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2 \0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align}
Exercise
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
$2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber$
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Exercise
When the highly reactive species HO2 forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows:
$2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber$
The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table:
Experiment [HO2]0 (M) Initial Rate (M/s)
1 1.1 × 10−8 1.7 × 10−7
2 2.5 × 10−8 8.8 × 10−7
3 3.4 × 10−8 1.6 × 10−6
4 5.0 × 10−8 3.5 × 10−6
as
Determine the reaction order and the rate constant.
Answer second order in HO2; k = 1.4 × 109 M−1·s−1
Note
If a plot of reactant concentration versus time is not linear but a plot of 1/reaction concentration versus time is linear, then the reaction is second order.
Example 7
If a flask that initially contains 0.056 M NO2 is heated at 300°C, what will be the concentration of NO2 after 1.0 h? How long will it take for the concentration of NO2 to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction (Equation 14.23) and the rate constant calculated in Example 6.
Given: balanced chemical equation, rate constant, time interval, and initial concentration
Asked for: final concentration and time required to reach specified concentration
Strategy:
1. Given k, t, and [A]0, use the integrated rate law for a second-order reaction to calculate [A].
2. Setting [A] equal to 1/10 of [A]0, use the same equation to solve for t.
Solution
A We know k and [NO2]0, and we are asked to determine [NO2] at t = 1 h (3600 s). Substituting the appropriate values into Equation 14.23,
\begin{align}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \&=2.0\times10^3\textrm{ M}^{-1}\end{align}
Thus [NO2]3600 = 5.1 × 10−4 M.
B In this case, we know k and [NO2]0, and we are asked to calculate at what time [NO2] = 0.1[NO2]0 = 0.1(0.056 M) = 0.0056 M. To do this, we solve Equation 14.23 for t, using the concentrations given.
$t=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k}=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}}=3.0\times10^2\textrm{ s}=5.0\textrm{ min}$
NO2 decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min.
Exercise 7
In the exercise in Example 6, you calculated the rate constant for the decomposition of HO2 as k = 1.4 × 109 M−1·s−1. This high rate constant means that HO2 decomposes rapidly under the reaction conditions given in the problem. In fact, the HO2 molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO2, calculate the concentration of HO2 that remains after 1.0 h at 25°C. How long will it take for 90% of the HO2 to decompose? Use the integrated rate law for a second-order reaction (Equation 14.23) and the rate constant calculated in the exercise in Example 6.
Answer 2.0 × 10−13 M; 6.4 × 10−6 s
In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form $A + B \rightarrow products$, in which the reaction is first order in $A$ and first order in $B$. The differential rate law for this reaction is as follows:
$\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A][\textrm B] \tag{14.24}$
Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant.
Determining the Rate Law of a Reaction
The number of fundamentally different mechanisms (sets of steps in a reaction) is actually rather small compared to the large number of chemical reactions that can occur. Thus understanding reaction mechanisms can simplify what might seem to be a confusing variety of chemical reactions. The first step in discovering the reaction mechanism is to determine the reaction’s rate law. This can be done by designing experiments that measure the concentration(s) of one or more reactants or products as a function of time. For the reaction $A + B \rightarrow products$, for example, we need to determine k and the exponents m and n in the following equation:
$\text{rate} = k[A]^m[B]^n \tag{14.25}$
To do this, we might keep the initial concentration of B constant while varying the initial concentration of A and calculating the initial reaction rate. This information would permit us to deduce the reaction order with respect to A. Similarly, we could determine the reaction order with respect to B by studying the initial reaction rate when the initial concentration of A is kept constant while the initial concentration of B is varied. In earlier examples, we determined the reaction order with respect to a given reactant by comparing the different rates obtained when only the concentration of the reactant in question was changed. An alternative way of determining reaction orders is to set up a proportion using the rate laws for two different experiments. Rate data for a hypothetical reaction of the type $A + B \rightarrow products$ are given in Table 14.4.
Table 14.4 Rate Data for a Hypothetical Reaction of the Form $A + B \rightarrow products$
Experiment [A] (M) [B] (M) Initial Rate (M/min)
1 0.50 0.50 8.5 × 10−3
2 0.75 0.50 19 × 10−3
3 1.00 0.50 34 × 10−3
4 0.50 0.75 8.5 × 10−3
5 0.50 1.00 8.5 × 10−3
The general rate law for the reaction is given in Equation 14.25. We can obtain m or n directly by using a proportion of the rate laws for two experiments in which the concentration of one reactant is the same, such as Experiments 1 and 3 in Table 14.4.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_3}}=\dfrac{k[\textrm A_1]^m[\textrm B_1]^n}{k[\textrm A_3]^m[\textrm B_3]^n} \nonumber$
Inserting the appropriate values from Table 14.4,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{34\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{1.00 M}]^m[\textrm{0.50 M}]^n} \nonumber$
Because 1.00 to any power is 1, [1.00 M]m = 1.00 M. We can cancel like terms to give 0.25 = [0.50]m, which can also be written as 1/4 = [1/2]m. Thus we can conclude that m = 2 and that the reaction is second order in A. By selecting two experiments in which the concentration of B is the same, we were able to solve for m.
Conversely, by selecting two experiments in which the concentration of A is the same (e.g., Experiments 5 and 1), we can solve for n.
$\dfrac{\mathrm{rate_1}}{\mathrm{rate_5}}=\dfrac{k[\mathrm{A_1}]^m[\mathrm{B_1}]^n}{k[\mathrm{A_5}]^m[\mathrm{B_5}]^n}$
Substituting the appropriate values from Table 14.4,
$\dfrac{8.5\times10^{-3}\textrm{ M/min}}{8.5\times10^{-3}\textrm{ M/min}}=\dfrac{k[\textrm{0.50 M}]^m[\textrm{0.50 M}]^n}{k[\textrm{0.50 M}]^m[\textrm{1.00 M}]^n} \nonumber$
Canceling leaves 1.0 = [0.50]n, which gives $n = 0$; that is, the reaction is zeroth order in $B$. The experimentally determined rate law is therefore
rate = k[A]2[B]0 = k[A]2
We can now calculate the rate constant by inserting the data from any row of Table 14.4 into the experimentally determined rate law and solving for $k$. Using Experiment 2, we obtain
19 × 10−3 M/min = k(0.75 M)2
3.4 × 10−2 M−1·min−1 = k
You should verify that using data from any other row of Table 14.4 gives the same rate constant. This must be true as long as the experimental conditions, such as temperature and solvent, are the same.
Example 8
Nitric oxide is produced in the body by several different enzymes and acts as a signal that controls blood pressure, long-term memory, and other critical functions. The major route for removing NO from biological fluids is via reaction with $O_2$ to give $NO_2$, which then reacts rapidly with water to give nitrous acid and nitric acid:
These reactions are important in maintaining steady levels of NO. The following table lists kinetics data for the reaction of NO with O2 at 25°C:
$2NO(g) + O_2(g) \rightarrow 2NO_2(g) \nonumber$
Determine the rate law for the reaction and calculate the rate constant.
Experiment [NO]0 (M) [O2]0 (M) Initial Rate (M/s)
1 0.0235 0.0125 7.98 × 10−3
2 0.0235 0.0250 15.9 × 10−3
3 0.0470 0.0125 32.0 × 10−3
4 0.0470 0.0250 63.5 × 10−3
Given: balanced chemical equation, initial concentrations, and initial rates
Asked for: rate law and rate constant
Strategy:
1. Compare the changes in initial concentrations with the corresponding changes in rates of reaction to determine the reaction order for each species. Write the rate law for the reaction.
2. Using data from any experiment, substitute appropriate values into the rate law. Solve the rate equation for k.
Solution
A Comparing Experiments 1 and 2 shows that as [O2] is doubled at a constant value of [NO2], the reaction rate approximately doubles. Thus the reaction rate is proportional to [O2]1, so the reaction is first order in O2. Comparing Experiments 1 and 3 shows that the reaction rate essentially quadruples when [NO] is doubled and [O2] is held constant. That is, the reaction rate is proportional to [NO]2, which indicates that the reaction is second order in NO. Using these relationships, we can write the rate law for the reaction:
rate = k[NO]2[O2]
B The data in any row can be used to calculate the rate constant. Using Experiment 1, for example, gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{7.98\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0125\textrm{ M})}=1.16\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
Alternatively, using Experiment 2 gives
$k=\dfrac{\textrm{rate}}{[\mathrm{NO}]^2[\mathrm{O_2}]}=\dfrac{15.9\times10^{-3}\textrm{ M/s}}{(0.0235\textrm{ M})^2(0.0250\textrm{ M})}=1.15\times10^3\;\mathrm{ M^{-2}\cdot s^{-1}} \nonumber$
The difference is minor and associated with significant digits and likely experimental error in making the table.
The overall reaction order $(m + n) = 3$, so this is a third-order reaction whose rate is determined by three reactants. The units of the rate constant become more complex as the overall reaction order increases.
Exercise 8
The peroxydisulfate ion (S2O82−) is a potent oxidizing agent that reacts rapidly with iodide ion in water:
S2O82−(aq) + 3I(aq) → 2SO42−(aq) + I3(aq)
The following table lists kinetics data for this reaction at 25°C. Determine the rate law and calculate the rate constant.
Experiment [S2O82−]0 (M) [I]0 (M) Initial Rate (M/s)
1 0.27 0.38 2.05
2 0.40 0.38 3.06
3 0.40 0.22 1.76
Answer rate = k[S2O82−][I]; k = 20 M−1·s−1
Summary
The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism.
Note the Pattern
If a plot of reactant concentration versus time is linear, then the reaction is zeroth order in that reactant.
Key Equations
zeroth-order reaction
Equation 14.15: $\textrm{rate}=-\frac{\Delta[\textrm A]}{\Delta t}=k$
Equation 14.16: [A] = [A]0kt
first-order reaction
Equation 14.19: $\textrm{rate}=-\frac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]$
Equation 14.20: [A] = [A]0e−kt
Equation 14.21: ln[A] = ln[A]0kt
second-order reaction
Equation 14.22: $\textrm{rate}=-\frac{\Delta[\textrm A]}{\Delta t}=k[\textrm A]^2$
Equation 14.23: $\frac{1}{[\textrm A]}=\frac{1}{[\textrm A]_0}+kt$
Conceptual Problems
1. What are the characteristics of a zeroth-order reaction? Experimentally, how would you determine whether a reaction is zeroth order?
2. Predict whether the following reactions are zeroth order and explain your reasoning.
1. a substitution reaction of an alcohol with HCl to form an alkyl halide and water
2. catalytic hydrogenation of an alkene
3. hydrolysis of an alkyl halide to an alcohol
4. enzymatic conversion of nitrate to nitrite in a soil bacterium
1. In a first-order reaction, what is the advantage of using the integrated rate law expressed in natural logarithms over the rate law expressed in exponential form?
1. If the reaction rate is directly proportional to the concentration of a reactant, what does this tell you about (a) the reaction order with respect to the reactant and (b) the overall reaction order?
1. The reaction of NO with O2 is found to be second order with respect to NO and first order with respect to O2. What is the overall reaction order? What is the effect of doubling the concentration of each reagent on the reaction rate?
Numerical Problems
1. Iodide reduces Fe(III) according to the following reaction:
2Fe3+(soln) + 2I(soln) → 2Fe2+(soln) + I2(soln)
Experimentally, it was found that doubling the concentration of Fe(III) doubled the reaction rate, and doubling the iodide concentration increased the reaction rate by a factor of 4. What is the reaction order with respect to each species? What is the overall rate law? What is the overall reaction order?
1. Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:
Experiment [Benzoyl Peroxide]0 (M) Initial Rate (M/s)
1 1.00 2.22 × 10−4
2 0.70 1.64 × 10−4
3 0.50 1.12 × 10−4
4 0.25 0.59 × 10−4
What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?
1. 1-Bromopropane is a colorless liquid that reacts with S2O32− according to the following reaction:
C3H7Br + S2O32− → C3H7S2O3 + Br
The reaction is first order in 1-bromopropane and first order in S2O32−, with a rate constant of 8.05 × 10−4 M−1·s−1. If you began a reaction with 40 mmol/100 mL of C3H7Br and an equivalent concentration of S2O32−, what would the initial reaction rate be? If you were to decrease the concentration of each reactant to 20 mmol/100 mL, what would the initial reaction rate be?
1. The experimental rate law for the reaction 3A + 2B → C + D was found to be Δ[C]/Δt = k[A]2[B] for an overall reaction that is third order. Because graphical analysis is difficult beyond second-order reactions, explain the procedure for determining the rate law experimentally.
Numerical Answers
1. First order in Fe3+; second order in I; third order overall; rate = k[Fe3+][I]2.
1. 1.29 × 10−4 M/s; 3.22 × 10−5 M/s
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/05%3A_Experimental_Methods/5.02%3A_Methods_of_Determining_Reaction_Order.txt
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To investigate reactions that are complete in less than a millisecond, one can start with a pre-mixed sample in which one of active reactants is generated in situ. Alternatively, a rapid change in pressure or temperature can alter the composition of a reaction that has already achieved equilibrium.
Flash photolysis
Many reactions are known which do not take place without light of wavelength sufficiently short to supply the activation energy needed to break a bond, often leading to the creation of a highly reactive radical. A good example is the combination of gaseous Cl2 with H2, which proceeds explosively when the system is illuminated with visible light. In flash photolysis, a short pulse of light is used to initiate a reaction whose progress can be observed by optical or other means.
Photolysis refers to the use of light to decompose a molecule into simpler units, often ions or free radicals. In contrast to thermolysis (decomposition induced by high temperature), photolysis is able to inject energy into a molecule almost instantaneously and can be much "cleaner," meaning that there are fewer side reactions that often lead to complex mixtures of products. Photolysis can also be highly specific; the wavelength of the light that triggers the reaction can often be adjusted to activate one particular kind of molecule without affecting others that might be present.
1. All this had been known for a very long time, but until the mid-1940's there was no practical way of studying the kinetics of the reactions involving the highly reactive species produced by photolysis.
In 1945, Ronald Norrish of Cambridge University and his graduate student George Porter conceived the idea of using a short-duration flash lamp to generate gas-phase CH2 radicals, and then following the progress of the reaction of these radicals with other species by means of absorption spectroscopy.
In a flash photolysis experiment, recording of the absorbance of the sample cell contents is timed to follow the flash by an interval that can be varied in order to capture the effects produced by the product or intermediate as it is formed or decays.
Norrish and Porter shared the 1967 Nobel Prize in Chemistry for this work.
Nanosecond flash photolysis setup
Flash durations of around 1 millisecond permitted one to follow processes having lifetimes in the microsecond range, but the advent of fast lasers gradually extended this to picoseconds and femtoseconds. Flash photolysis revolutionized the study of organic photochemistry, especially that relating to the chemistry of free radicals and other reactive species that cannot be isolated or stored, but which can easily be produced by photolysis of a suitable precursor. It has proven invaluable for understanding the complicated kinetics relating to atmospheric chemistry and smog formation. More recently, flash photolysis has become an important tool in biochemistry and cellular physiology.
Perturbation-relaxation methods
Many reactions, especially those that take place in solution, occur too rapidly to follow by flow techniques, and can therefore only be observed when they are already at equilibrium. The classical examples of such reactions are two of the fastest ones ever observed, the dissociation of water
$2 H_2O \rightarrow H_3O^+ + OH^- \nonumber$
and the formation of the triiodide ion in aqueous solution
$I^– + I_2 \rightarrow I_3^– \nonumber$
Reactions of these kinds could not be studied until the mid-1950s when techniques were developed to shift the equilibrium by imposing an abrupt physical change on the system.
For example, if the reaction A B is endothermic, then according to the Le Chatelier principle, subjecting the system to a rapid jump in temperature will shift the equilibrium state to one in which the product B has a higher concentration. The composition of the system will than begin to shift toward the new equilibrium composition at a rate determined by the kinetics of the process.
For the general case illustrated here, the quantity "x" being plotted is a measurable quantity such as light absorption or electrical conductivity that varies linearly with the composition of the system. In a first-order process, x will vary with time according to
$x_t = x_o e^{-kt} \nonumber$
After the abrupt perturbation at time to, the relaxation time t* is defined as the half-time for the return to equilibrium — that is, as the time required for xo to decrease by Δx/e = Δx/2.718. The derivation of t* and the relations highlighted in yellow can be found in most standard kinetics textbooks. Temperature jumps are likely most commonly used.
Manfred Eigen
This is the method that Manfred Eigen (Germany, 1927-) pioneered when, in the early 1960's, he measured the rate constant of what was then the fastest reaction ever observed:
$H^+ + OH^– \rightarrow H_2O \;\;\;\; k = 1.3 \times 10^{11} \; M^{–1}\; sec^{–1} \nonumber$
BioLogic rapid-mixing T-Jump System
• high-voltage electric discharge: A capacitor, charged to 5-10 kV, is discharged through a solution to which an electrolyte has been added to provide a conductive path.
• laser irradiation: The sample is irradiated with a laser whose wavelength corresponds to an absorption peak in the sample. Infrared lasers are often used for this purpose.
• mixing of two pre-equilibrated solutions: Two solutions, otherwise identical but at different temperatures, are rapidly mixed in a stopped-flow type of apparatus. Although this method is not as fast, it has the advantage of allowing both negative and positive T-jumps. The device shown here uses 0.1-mL samples and provides jumps of up to ±40 C° over a few microseconds. Observation times, however, are limited to 1-2 milliseconds owing to thermal dissipation.
Pressure jumps
According to the Le Chatelier principle, a change in the applied pressure will shift the equilibrium state of any reaction which involves a change in the volume of a system. Aside from the obvious examples associated with changes in the number of moles of gases, there are many more subtle cases involving formation of complexes, hydration shells and surface adsorption, and phase changes. One area of considerable interest is the study of protein folding, which has implications in diseases such as Parkinson's and Alzheimer's.
The pressure-jump is applied to the cell through a flexible membrane that is activated by a high-pressure gas supply, or through an electrically-actuated piezoelectric crystal. The latter method is employed in the device shown here, which can produce P-jumps of around 1 GPa over sub-millisecond time intervals.
Shock tubes: extreme jumping
When a change in pressure propagates through a gas at a rate greater than the ordinary compressions and rarefactions associated with the travel of sound, a moving front (a shock wave) of very high pressure forms. This in turn generates an almost instantaneous rise in the temperature that can approach several thousand degrees in magnitude.
A shock tube is an apparatus in which shock waves can be generated and used to study the kinetics of gas-phase reactions that are otherwise inaccessible to kinetic measurements. Since all molecules tend to dissociate at high temperatures, shock tubes are widely used to study dissociation processes and the chemistry of the resulting fragments. For example, the shock-induced decomposition of carbon suboxide provides an efficient means of investigating carbon atom reactions:
$C_3O_2 \rightarrow C + CO \nonumber$
Shock tube techniques are also useful for studying combustion reactions, including those that proceed explosively.
The shock tube itself consists of two sections separated by a breakable diaphragm of metal or plastic. One section is filled with a "driver" gas at a very high pressure, commonly helium, but often mixed with other inert gases to adjust the properties of the shock. The other, longer section of the tube contains the "driven" gas—the reactants—at a low pressure, usually less than 1 atmosphere.
The reaction is initiated by causing the diaphragm to rupture, either by means of a mechanical plunger or by raising the pressure beyond its bursting point. The kinetics of the reaction are monitored by means of an absorption or other optical monitoring device that is positioned at a location along the reaction tube that is appropriate to the time course of the reaction.
Contributors and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/05%3A_Experimental_Methods/5.04%3A_Resolving_Kinetics-_Faster_Methods.txt
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Rapid mixing
The traditional experimental methods described above all assume the possibility of following the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can be done if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion?
For reactions that take place in milliseconds, the standard approach since the 1950s has been to employ a flow technique of some kind. An early example was used to study fast gas-phase reactions in which one of the reactants is a free radical such as OH that can be produced by an intense microwave discharge acting on a suitable source gas mixture. This gas, along with the other reactant being investigated, is made to flow through a narrow tube at a known velocity.
If the distance between the point at which the reaction is initiated and the product detector is known, then the time interval can be found from the flow rate. By varying this distance, the time required to obtain the maximum yield can then be determined.
Although this method is very simple in principle, it can be complicated in practice, as the illustration shows. Owing to the rather large volumes required, his method is more practical for the study of gas-phase reactions than for solutions, for which the stopped-flow method described below is generally preferred.
Stopped-flow and Quenched-flow methods
These are by far the most common means of studying fast solution-phase reactions over time intervals of down to a fraction of a millisecond. The use of reasonably simple devices is now practical even in student laboratory experiments. These techniques make it possible to follow not only changes in the concentrations of reactants and products, but also the buildup and decay of reaction intermediates.
The basic stopped-flow apparatus consists of two or more coupled syringes that rapidly inject the reactants into a small mixing chamber and then through an observation cell that can be coupled to instruments that measure absorption, fluorescence, light scattering, or other optical or electrical properties of the solution. As the solution flows through the cell, it empties into a stopping syringe that, when filled, strikes a backstop that abruptly stops the flow. The volume that the stopping syringe can accept is adjusted so that the mixture in the cell has just become uniform and has reached a steady state; at this point, recording of the cell measurement begins and its change is followed.
Of course, there are many reactions that cannot be followed by changes in light absorption or other physical properties that are conveniently monitored. In such cases, it is often practical to quench (stop) the reaction after a desired interval by adding an appropriate quenching agent. For example, an enzyme-catalyzed reaction can be stopped by adding an acid, base, or salt solution that denatures (destroys the activity of) the protein enzyme. Once the reaction has been stopped, the mixture is withdrawn and analyzed in an appropriate manner.
The quenched-flow technique works something like the stopped-flow method described above, with a slightly altered plumbing arrangement.
The reactants A and B are mixed and fed directly through the diverter valve to the measuring cell, which is not shown in this diagram. After a set interval that can vary from a few milliseconds to 200 sec or more, the controller activates the quenching syringe and diverter valve, flooding the cell with the quenching solution.
Contributors and Attributions
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/05%3A_Experimental_Methods/5.05%3A_Resolving_Kinetics-_Fast_Methods.txt
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These methods are applicable to reactions that are not excessively fast, typically requiring a few minutes or hours to run to completion. They were about the only methods available before 1950. Most first-year laboratory courses students will include at least one experiment based on one of these methods.
Optical methods
Light absorption is perhaps the most widely-used technique. If either a reactant or a product is colored, the reaction is easily followed by recording the change in transmission of an appropriate wavelength after the reaction is started. When a beam of light passes through a solution containing a colored substance, the fraction that is absorbed is directly proportional to the concentration of that substance and to the length of the light's path through the solution. The latter can be controlled by employing a cell or cuvette having a fixed path length.
If I0 is the intensity of the light incident on the cell and I is the intensity that emerges on the other side, then the percent absorption is just 100 × I / I0. Because a limited 1-100 scale of light absorption is often inadequate to express the many orders of magnitude frequently encountered, a logarithmic term optical density is often employed. The relation between this, the cell path length, concentration, and innate absorption ability of the colored substance is expressed by Beer's Law.
The simplest absorbance measuring device is a colorimeter in which a beam of white light from an incandescent lamp is passed through a cell (often just a test tube) and onto a photodetector whose electrical output is directly proportional to the light intensity. Before beginning the experiment, a zeroing control sets the meter to zero when the light path is blocked off, and a sensitivity control sets it to 100 when a cell containing an uncolored solution (a "blank") is inserted.
The sensitivity and selectivity of such an arrangement is greatly enhanced by adjusting the wavelength of the light to match the absorption spectrum of the substance being measured. Thus if the substance has a yellow color, it is because blue light is being absorbed, so a blue color filter is placed in the light path.
More sophisticated absorption spectrophotometers employ two cells, one for the sample and another reference cell for the blank. They also allow one to select the particular wavelength range that is most strongly absorbed by the substance under investigation, which can often extend into the near-ultraviolet region.
Light scattering measurements (made with a nephthelometer) can be useful for reactions that lead to the formation a fine precipitate. A very simple student laboratory experiment of this kind can be carried out by placing a conical flask containing the reaction mixture on top of a marked piece of paper. The effects of changing the temperature or reactant concentrations can be made by observing how long it takes for precipitate formation to obscure the mark on the paper.
Other optical methods such a fluorescence and polarimetry (measurement of the degree to which a solution rotates the plane of polarized light) are also employed when applicable.
Other methods
• Measurement of gases: The classical method of following reactions that produce changes in the number of moles of gases is to observe changes in pressure or volume. An alternative method for following the loss of a gas is to place the reaction container on an electronic balance and monitor the loss in weight. This is generally less accurate than pressure measurements, but is sometimes used in student experiments.
• pH measurements: Many reactions yield or consume hydrogen ions and are conveniently followed by means of a pH meter.
• Electrical Conductance: Reactions that yield or consume ionic substances are often studied by measuring the electrical conductance of the solution. Conductimetry is usually carried out by balancing the solution conductance with a known resistance in a bridge arrangement. An audio-frequency alternating current is used in order to avoid electrolysis.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/05%3A_Experimental_Methods/5.06%3A_Resolving_Kinetics-_Slow_Methods.txt
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Learning Objectives
• To use graphs to analyze the kinetics of a reaction.
You learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order. We will illustrate the use of these graphs by considering the thermal decomposition of NO2 gas at elevated temperatures, which occurs according to the following reaction:
$\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \label{14.26}$
Experimental data for this reaction at 330°C are listed in Table $1$; they are provided as [NO2], ln[NO2], and 1/[NO2] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively.
Table $1$: Concentration of NO2 as a Function of Time at 330°C
Time (s) [NO2] (M) ln[NO2] 1/[NO2] (M−1)
0 1.00 × 10−2 −4.605 100
60 6.83 × 10−3 −4.986 146
120 5.18 × 10−3 −5.263 193
180 4.18 × 10−3 −5.477 239
240 3.50 × 10−3 −5.655 286
300 3.01 × 10−3 −5.806 332
360 2.64 × 10−3 −5.937 379
The actual concentrations of NO2 are plotted versus time in part (a) in Figure $1$. Because the plot of [NO2] versus t is not a straight line, we know the reaction is not zeroth order in NO2. A plot of ln[NO2] versus t (part (b) in Figure $1$) shows us that the reaction is not first order in NO2 because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO2] versus t (part (c) in Figure $1$). This plot is a straight line, indicating that the reaction is second order in NO2.
We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure $2$, the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates required multiple experiments at different NO2 concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions.
Example $1$
Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction:
$2N_2O_5(soln) → 4NO_2(soln) + O_2(g) \nonumber$
This reaction is carried out in a CCl4 solution at 45°C. The concentrations of N2O5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N2O5 concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant.
Time (s) [N2O5] (M) ln[N2O5] 1/[N2O5] (M−1)
0 0.0365 −3.310 27.4
600 0.0274 −3.597 36.5
1200 0.0206 −3.882 48.5
1800 0.0157 −4.154 63.7
2400 0.0117 −4.448 85.5
3000 0.00860 −4.756 116
3600 0.00640 −5.051 156
Given: balanced chemical equation, reaction times, and concentrations
Asked for: graph of data, rate law, and rate constant
Strategy:
A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure $1$ to determine the reaction order.
B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction.
Solution
A Here are plots of [N2O5] versus t, ln[N2O5] versus t, and 1/[N2O5] versus t:
The plot of ln[N2O5] versus t gives a straight line, whereas the plots of [N2O5] versus t and 1/[N2O5] versus t do not. This means that the decomposition of N2O5 is first order in [N2O5].
B The rate law for the reaction is therefore
rate = k[N2O5]
Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s,
\begin{align*} \textrm{slope} &=\dfrac{\ln[\mathrm{N_2O_5}]_{3000}-\ln[\mathrm{N_2O_5}]_0}{3000\textrm{ s}-0\textrm{ s}} \[4pt] &=\dfrac{(-4.756)-(-3.310)}{3000\textrm{ s}} \[4pt] &=-4.820\times10^{-4}\textrm{ s}^{-1} \end{align*}
Thus $k = 4.820 \times 10^{−4}\, s^{−1}$.
Exercise $1$
1,3-Butadiene (CH2=CH—CH=CH2; C4H6) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C4H6 as a function of time at 326°C are listed in the following table along with ln[C4H6] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C4H6, the rate law, and the rate constant for the reaction.
Time (s) [C4H6] (M) ln[C4H6] 1/[C4H6] (M−1)
0 1.72 × 10−2 −4.063 58.1
900 1.43 × 10−2 −4.247 69.9
1800 1.23 × 10−2 −4.398 81.3
3600 9.52 × 10−3 −4.654 105
6000 7.30 × 10−3 −4.920 137
Answer
second order in C4H6; rate = k[C4H6]2; k = 1.3 × 10−2 M−1·s−1
Summary
For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/05%3A_Experimental_Methods/5.07%3A_Using_Graphs_to_Determine_Integrated_Rate_Laws.txt
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The collision theory explains that gas-phase chemical reactions occur when molecules collide with sufficient kinetic energy. The collision theory is based on the kinetic theory of gases; therefore only dealing with gas-phase chemical reactions are dealt with. Ideal gas assumptions are applied. Furthermore, we also are assuming:
1. All molecules are traveling through space in a straight line.
2. All molecules are rigid spheres.
3. The reactions concerned are between only two molecules.
4. The molecules need to collide.
Ultimately, the collision theory of gases gives the rate constant for bimolecular gas-phase reactions; it is equal to the rate of successful collisions. The rate of successful collisions is proportional to the fraction of successful collisions multiplied by the overall collision frequency.
6.01: Collision Theory
The collisional cross section is an "effective area" that quantifies the likelihood of a scattering event when an incident species strikes a target species. In a a hard object approximation, the cross section is the area of the conventional geometric cross section. The collisional cross sections typically denoted σ and measured in units of area.
Introduction
Atoms and molecules can move around in space and bump into each other. If certain conditions of the collision are met, a chemical reaction occurs and a product forms. Sometimes, however, particles may get extremely close to each other but do not strike. We can use the collisional cross section to find how large the distance between two particles must be in order for a collision to occur.
A few assumptions must be made:
• All particles travel through space linearly
• All particles are hard spheres
• Only two particles are involved in the collision
The collisional cross section is defined as the area around a particle in which the center of another particle must be in order for a collision to occur. In the image below, the area within the black circle is molecule A's collisional cross section. That area can change depending on the size of the two particles involved.
The collisional cross section $\sigma_{AA}$ between molecule $A$ and molecule $A$ can be calculated using the following equation:
$\sigma_{AB} = \pi{(r_A + r_B)^2} \nonumber$
Collision occurs when the distance between the center of the two reactant molecules is less than the sum of the radii of these molecules, as shown in Figure $2$. The collisional cross section describes the area around a single reactant. For a collisional reaction to occur, the center of one reactant must be within the collisional cross section of a corresponding reactant.
It is first assumed that all particles, whether it be an atom or a molecule, are hard spheres. Particle A must come in contact with Particle B in order for a collision to occur. Particle B can approach particle A from any direction; thus, consider a circle with radius $r$:
$Area \; of \; a \; Circle = \pi{r}^2 \nonumber$
Assume that the two particles involved in the collision are the same in size and have the same radius. The furthest distance the two centers can be and still have a collision is $2r$. Substitute $r$ with $2r$:
$\sigma_{AA}= \pi{(2r)}^2 \label{SameEq}$
We will define this area as the collisional cross section. Anytime the center of another particle is within this area, there will be parts of the two particles that will overlap, touch, and cause a collision.
Example $1$: Hydrogen/Hydrogen Collisions
Consider the following reaction: $\ce{H + H -> H_2}$ The radius of hydrogen is $5.3 \times 10^{-11}\; m$. What is the collisional cross section for this reaction?
Solution
Use Equation \ref{SameEq} with the given atomic radius for hydrogen atoms:
$\text{Collisional Cross Section}= \pi{(2r)}^2= \pi{[(2)(5.3 \times 10^{-11}\,m)]}^2= 3.53 \times 10^{-20} m^2 \nonumber$
Although the collisional cross section of a particle can be calculated, it is usually not used on its own (Table $1$). Instead, it is a component of more complex theories such as collision frequency and collision theory.
Figure $1$: Table of Cross-Sections for Common Gases
Molecule Cross-Section (nm2)
Ar 0.36
C2H4 0.64
C6H6 0.88
CH4 0.46
Cl2 0.93
CO2 0.52
H2 0.27
He 0.21
N2 0.43
Ne 0.24
O2 0.40
SO2 0.58
Now what if the particles were of different size and different radii? The $2r$ term in Equation $\ref{SameEq}$ is really the sum of the radius of each molecule (i.e., $2r= r + r$). However, if the colliding molecules have differing sizes (e.g., $r_A$ and $r_B$) for the radius of particle A and B, respectively, then $2r$ in Equation $\ref{SameEq}$ is substituted with $r_A+r_B$:
$\sigma_{AB} = \pi{(r_A + r_B)}^2 \nonumber$
Example $1$: Hydrogen/Fluorine Collisions
What is the collisional cross section for this reaction?
$H + F \rightarrow HF \nonumber$
The radius of fluorine atom is 4.2 x 10-11 m.
Solution
$\text{ Collisional Cross-Section}= \pi{(r_A+r_B)}^2= \pi{[(5.3 \times 10^{-11})+(4.3 \times 10^{-11})]}^2= 2.90 \times 10^{-20} \,m^2 \nonumber$
Exercise $\PageIndex{1A}$
$H_2 + O_2 \rightarrow H_2O$. The radius of oxygen is 4.8 x 10-11 m. What is the collisional cross section for this reaction? (Hint: Assume that the radius of a molecule is just the sum of its atoms.)
Answer
The radius of H2 is 2(rH)=1.06 x 10-10 m. The radius of O2 is 2(rO)=9.6 x 10-11 m. $Collisional \; Cross \; Section= \pi{(r_A+r_B)}^2= \pi{[(1.06 \times 10^{-10})+(9.6 \times 10^{-11})]}^2= 1.28 \times 10^{-19} \nonumber$
Exercise $\PageIndex{1B}$
$N_2 + O_2 \rightarrow N_2O$. The radius of nitrogen is 5.6 x 10-11 m. If the distance between the two centers is 2.00 x 10-19 m, is there a collision between the two molecules?
Answer
Yes, there is a collision. The radius of N2 is 2(rN)=1.12 x 10-10 m. The radius of O2 is 2(rO)=9.6 x 10-11 m. $Collisional \; Cross \; Section= \pi{(r_A+r_B)}^2= \pi{[(1.12 \times 10^{-10})+(9.6 \times 10^{-11})]}^2= 1.36 \times 10^{-19} \nonumber$ 1.36 x 10-11 m is the furthest the two molecules can be and still get a collision. Since 2.00 x 10-11 m is larger than that distance, the center of one molecule is not in the collisional cross section of the other molecule. Therefore, no collision occurs. The molecules are too far apart.
Exercise $\PageIndex{1C}$
$F + F \rightarrow F_2$ The center of the two atoms are 3.50 x 10-20 m apart from each other. How much closer do the centers have to be in order for a collision to occur?
Answer
$Collisional \; Cross \; Section= \pi{(2r)}^2= \pi{[(2)(4.2 \times 10^{-11})]}^2= 2.22 \times 10^{-20} \nonumber$ Because the molecules are 3.50 x 10-20 m apart, the center of one F is not in the Collisional Cross Section of the other F. They must be closer. $(3.50 \times 10^{-20})-(2.22 \times 10^{-20})=1.28 \times 10^{-20} \nonumber$ The molecules must be at least 1.28 x 10-20 m closer.
Contributors and Attributions
• Lily Feng, Keith Dunaway (UC Davis)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.01%3A_Collision_Theory/6.1.01%3A_Collisional_Cross_Section.txt
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For two molecules to react, they must first come into contact with each other. This contact can be considered a "collision." The more mobile the molecules are, the more likely they are to collide. In addition, the closer the molecules are together, the more likely they are to collide.
In the following drawings, the molecules are closer together in the picture on the right than they are in the picture on the left. The molecules are more likely to collide and react in the picture on the right.
The two figures above might be described in terms of population density. Both drawings appear to offer the same amount of space, but are inhabited by different amounts of molecules. The difference is much like the difference between human population densities in various locations around the world. Some places, such as Mexico City or Tokyo, are very crowded; they have high population densities. Some places, such as the Australian Outback or the Canadian Arctic, have low population densities.
6.1.03: Collisions and Phase
Reaction rates depend on the energy required (the activation barrier) and the energy available. They also may involve collisions between molecules.
If two molecules need to collide in order for a reaction to take place, then factors that influence the ease of collisions will be important. The more energy there is available to the molecules, the faster they will move around, and the more likely they are to bump into each other. Higher temperatures ought to lead to more collisions and a greater frequency of reactions between molecules. In the drawing below, the cold, sluggish molecules on the left are not likely to collide, but the energetic molecules on the right are due to collide at any time.
These principles are summarized as follows:
• Temperature affects molecule mobility
• The higher the temperature, the more mobile the molecules will be, and the more likely they are to collide and react.
Phase also has a pronounced effect on the mobility of molecules. Molecules in the gas phase are free to move around, and they do so quickly. On the other hand, they are pretty well spread out. Nevertheless, collisions in the gas phase happen quite easily, which make gas-phase reactions happen more readily.
At the opposite extreme, molecules in the solid phase are not very mobile at all (reactions may involve atoms or ions, rather than molecules, but the same arguments apply). Not many collisions happen. As a result, reactions often happen extremely slowly in the solid state. Reactions are mostly limited to the grain boundaries: the surfaces of the grains, where they are in contact with each other. Nothing happens in the middle of a lump of solid, which remains unreacted.
If a solid is heated, the molecules can move around a little more. They may even leave their crystal lattice (if the solid is crystalline) and diffuse very slowly through the solid. Many solid state reactions are run at elevated temperature.
There are also many solid state reactions that are conducted in combination with gas-phase reactants. The solid reactants are often heated in a furnace while gas-phase reactants flow over them.
Many reactions are performed in the liquid state, either because the reactants are already liquid of because the solid reactants are heated past their melting point. In the liquid phase, molecules are much more mobile and collisions are much more frequent than in the solid phase.
It is also very common to run reactions in solution. In solution, a compound that is meant to undergo reaction is dissolved in a solvent. The solvent must be a liquid at the temperature at which the reaction will be run, so that molecules will be very mobile, but will still be close together, so collisions are favored.
There are many advantages to running reactions in solution. The reactant molecules are very mobile and pretty close together, so that collisions are facilitated. If the reaction is exothermic and gives of a lot of heat, the excess heat can be absorbed by the solvent molecules and carried away. That can be important in controlling reactions and avoiding decomposition. Also, we will see that the rate of collisions can be controlled by adding more solvent or less, in order to slow the reaction down or speed it up. In this way, the reaction rate can be controlled to some extent.
There are many liquids that are commonly used as solvents. Dichloromethane, toluene, dimethylformamide, tetrahydrofuran and acetonitrile are some common "organic" solvents, so called because they are based on carbon, which forms the basis of molecules in organisms. These different solvents offer a range of polarities, offering an ability to dissolve a variety of different reactants.
Water may be the most common solvent on the planet, and it is non-toxic, so it is very appealing for use in large-scale, industrial reactions. However, it is not very good at dissolving non-polar reactants.
The reactant compound could be a liquid or a solid. It only has to have strong enough intermolecular attractions with the solvent molecules so that it can be dissolved. Individual molecules of the reactant become lost among the solvent molecules move freely.
The disadvantage of using a solvent is that the solvent must be removed at the end of the reaction so that the desired product can be isolated and used. That means the solvent may eventually be thrown away as waste. That practice is less efficient and less environmentally friendly, although the solvent could possibly be recycled.
One method of dissolving reactants that is potentially greener is the use of supercritical fluids. In this approach, gases such as carbon dioxide are pressurized until they turn into liquids. In this form, carbon dioxide is a pretty good solvent, and reactions can be run when reactants are dissolved in it. At the end of the reaction, a valve is opened, releasing the pressure, and the carbon dioxide turns back into a gas. It can be stored and re-pressurized for another reaction.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.01%3A_Collision_Theory/6.1.02%3A_Collisions_and_Concentration.txt
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Collisional Frequency is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system.
Background and Overview
To fully understand how the collisional frequency equation is derived, consider a simple system (a jar full of helium) and add each new concept in a step-by-step fashion. Before continuing with this topic, it is suggested that the articles on collision theory and collisional cross section are reviewed, as these topics are essential to understanding collisional frequency.
In determining the collisional frequency for a single molecule, $Z_i$, picture a jar filled with helium atoms. These atoms collide by hitting other helium atoms within the jar. If every atom except one is frozen and the number of collisions in one minute is counted, the collisional frequency (per minute) of a single atom of helium within the container could be determined. This is the basis for the equation.
$Z_i = \dfrac{(\text{Volume of Collisional Cylinder}) (\text{Density})}{\text{Time}} \nonumber$
While the helium atom is moving through space, it sweeps out a collisional cylinder, as shown above. If the center of another helium atom is present within the cylinder, a collision occurs. The length of the cylinder is the helium atom's mean relative speed, $\sqrt{2}\langle c \rangle$, multiplied by change in time, $\Delta{t}$. The mean relative speed is used instead of average speed because, in reality, the other atoms are moving and this factor accounts for some of that. The area of the cylinder is the helium atom's collisional cross section.
Although collision will most likely change the direction an atom moves, it does not affect the volume of the collisional cylinder, which is due to density being uniform throughout the system. Therefore, an atom has the same chance of colliding with another atom regardless of direction as long as the distance traveled is the same.
$\text{Volume of Collisional Cylinder} = \sqrt{2}\pi{d^2}\langle c \rangle\Delta{t} \nonumber$
Density
Next, account must be taken of the other atoms that are moving that helium can hit; which is simply the density $\rho$ of helium within the system. The density component can be expanded in terms of $N$ and $V$. $N$ is the number of atoms in the system, and $V$ is the volume of the system. Alternatively, the density in terms of pressure (relating pressure to volume using the perfect gas law equation, $PV = nRT$:
$\rho = \left(\dfrac{N}{V}\right) = \left(\dfrac{\rho{N_A}}{V}\right) = \left(\dfrac{\rho{N_A}}{RT}\right) = \left(\dfrac{\rho}{kT}\right) \nonumber$
The Full Equation
When you substitute in the values for $Z_i$, the following equation results:
${Z_{i} = \dfrac{\sqrt{2}\pi d^{2} \left \langle c \right \rangle\Delta{t}\left(\dfrac{N}{V}\right)}{\Delta{t}}} \nonumber$
Cancel Δt:
$Z_{i} = \sqrt{2}\pi d^{2} \left \langle c \right \rangle\left(\dfrac{N}{V}\right) \nonumber$
All Molecules Moving System: $Z_{ii}$
Now imagine that all of the helium atoms in the jar are moving again. When all of the collisions for every atom of helium moving within the jar in a minute are counted, $Z_{ii}$ results. The relation is thus:
$Z_{ii} = \dfrac{1}{2}Z_{i}\left(\dfrac{N}{V}\right) \nonumber$
This expands to:
$Z_{ii} = \dfrac{\sqrt{2}}{2}\pi d^{2}\left \langle c \right \rangle\left(\dfrac{N}{V}\right)^2 \nonumber$
System With Collisions Between Different Types of Molecules: $Z_{AB}$
Consider a system of hydrogen in a jar:
$\ce{H_A + H_{BC} <=> H_{AB} + H_{C}} \nonumber$
In considering hydrogen in a jar instead of helium, there are several problems. First, the HA ions have a smaller radius than the HBC molecules. This is easily solved by accounting for the different radii which changes $d^{2}$ to $\left(r_A + r_B\right)^2$.
The second problem is that the number of HA ions could be much different than the number of HBC molecules. So we expand $\dfrac{\sqrt{2}}{2}\left(\dfrac{N}{V}\right)^2$ to account for the number of both reacting molecules to get $N_AN_B$. Because two reactants are considered, Zii becomes ZAB, and the two changes are combined to give the following equation:
$Z_{AB} = N_{A}N_{B}\pi\left(r_{A} + r_{B}\right)^2 \left \langle c \right \rangle \nonumber$
Mean speed, $\left \langle c \right \rangle$, can be expanded:
$\left \langle c \right \rangle = \sqrt{\dfrac{8k_BT}{\pi m}} \nonumber$
This leads to the final change to the collisional frequency equation. Because two different molecules must be taken into account, the equation must accommodate molecules of different masses ($m$). So, mass ($m$) must be converted to reduced mass, $\mu_{AB}$, converting a two bodied system to a one bodied system. Now we substitute $\left \langle c \right \rangle$ in the ZAB equation to obtain:
$Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\pi\sqrt{ \dfrac{8k_{B}T}{\pi\mu_{AB}}} \nonumber$
Cancel $\pi$:
$Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{\dfrac{8\pi{k_{B}T}}{\mu_{AB}}} \nonumber$
with
• $N_A$ is the number of A molecules in the system
• $N_B$ is the number of B molecules in the system
• $r_a$ is the radius of molecule A
• $r_b$ is the radius of molecule B
• $k_B$ is the Boltzmann constant $k_B$ =1.380 x 10-23 Joules Kelvin
• $T$ is the temperature in Kelvin
• $\mu_{AB}$ is the reduced mass found by using the equation $\mu_{AB} = \dfrac{m_Am_B}{m_A + m_B}$
The following assumptions are used when deriving and calculating the collisional frequency:
1. All molecules travel through space in straight lines.
2. All molecules are hard, solid spheres.
3. The reaction of interest is between only two molecules.
4. Collisions are hit or miss only. They occur when distance between the center of the two reactants is less than or equal to the sum of their respective radii. Even if the two molecules barely miss each other, it is still considered a complete miss (in reality, their electron clouds would interact).
Variables that affect Collisional Frequency
• Temperature: As is evident from the collisional frequency equation, when temperature increases, the collisional frequency increases.
• Density: From a conceptual point, if the density is increased, the number of molecules per volume is also increased. If everything else remains constant, a single reactant comes in contact with more atoms in a denser system. Thus if density is increased, the collisional frequency must also increase.
• Size of Reactants: Increasing the size of the reactants increases the collisional frequency. This is directly due to increasing the radius of the reactants as this increases the collisional cross section. This in turn increases the collisional cylinder. Because radius term is squared, if the radius of one of the reactants is doubled, the collisional frequency is quadrupled. If the radii of both reactants are doubled, the collisional frequency is increased by a factor of 16.
Exercise $1$
If the temperature of the system was increased, how would the collisional frequency be affected?
Exercise $2$
If the masses of both the reactants were increased, how would the collisional frequency be affected?
Exercise $3$
0.4 moles of N2 gas (molecular diameter= 3.8x10-10 m and mass= 28 g/mol) occupies a 1-liter (0.001m3) container at 1 atm of pressure and at room temperature (298 K)
1. Calculate the number of collisions a single molecule makes in one second.(hint use Zi)
2. Calculate the binary collision frequency. (hint use $Z_{ii}$)
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Collision Frequency
The rate at which molecules collide which is the frequency of collisions is called the collision frequency, Z, which has units of collisions per unit of time. Given a box of molecules A and B, the collision frequency between molecules A and B is defined by the following equation:
$Z=N_{A}N_{B}\sigma_{AB}\sqrt{\dfrac{8k_{B}T}{\pi\mu_{AB}}} \nonumber$
where:
• NA and NB are the numbers of molecules A and B, and is directly related to the concentrations of A and B.
• $\sqrt{\dfrac{8k_{B}T}{\pi\mu_{AB}}}$ is the mean speed of molecules.
• $\sigma_{AB}$ is the averaged sum of the collision cross sections of molecules A and B. The collision cross section represents the collision region presented by one molecule to another.
Successful Collisions
In order for a successful collision to occur, the reactant molecules must collide with enough kinetic energy to break original bonds and form new bonds to become the product molecules. This energy is called the activation energy for the reaction; it is also often referred to as the energy barrier.
The fraction of collisions with enough energy to overcome the activation barrier is given by:
$f = e^{\frac{-E_a}{RT}}$
where:
• f is the fraction of collisions with enough energy
• Ea is the activation energy
The fraction of successful collisions is directly proportional to the temperature and inversely proportional to the activation energy.
Putting it all Together
The rate constant of the gas-phase reaction is proportional to the product of the collision frequency and the fraction of successful reactions. As stated above, sufficient kinetic energy is required for a successful reaction; however, they must also collide properly (see below). Compare the following equation to the Arrhenius equation:
$k = Z\rho{e^{\frac{-E_a}{RT}}}$
where
• k is the rate constant for the reaction
• ρ is the steric factor. It is the probability of the reactant molecules colliding with the right orientation and positioning to achieve a product with the desirable geometry and stereospecificity. The steric factor is very difficult to assess on paper; it is determined experimentally.
• is the pre-exponential factor, A, of the Arrhenius equation. In theory, it is the frequency of total collisions that collide with the right orientation. In practice, it is the pre-exponential factor that is directly determined by experiment and then used to calculate the steric factor.
• Ea is activation energy
• T is absolute temperature
• R is gas constant.
Collision Frequency Equation
The collision frequency equation can thus be given as follows:
$Z = N_A\rho_{AB} \sqrt{\frac{8k_BT}{\pi\mu_{AB}}} \nonumber$
where:
• NA is number of molecules per unit volume
• kB is Boltzmann's constant
Examples of gas-phase reactions and their steric factors
Reactions with more complex reactants and greater needs for collision orientation specificity, such as that below, have smaller steric factors (and therefore lower chances of success).
$H_2 + C_2H_4 \rightarrow C_2H_6 \nonumber$
The opposite holds true for simpler reactions; they have a relatively larger steric factors:
$H^. + H^. \rightarrow H_2 \nonumber$
Applications
The collision theory is used to predict the reaction rates for gas-phase reactions. It is a rough approximation due to the complications of the steric factor; furthermore, some of the assumptions do not hold in practical scenarios. For example, in real life molecules are not perfect spheres. Finally, the concepts of collision frequency can be applied in the laboratory:
• The temperature of the environment affects the average speed of molecules. Thus, reactions are heated to increase the reaction rate.
• The initial concentration of reactants is directly proportional to the collision frequency; increasing the initial concentration will speed up the reaction.
Although the collision theory deals with gas-phase reactions, its concepts can also be applied to reactions that take place in solvents; however, the properties of the solvents (for example: solvent cage) will affect the rate of reactions. Ultimately, collision theory illustrates how reactions occur; it can be used to approximate the rate constants of reactions, and its concepts can be directly applied in the laboratory.
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Collision theory explains why different reactions occur at different rates, and suggests ways to change the rate of a reaction. Collision theory states that for a chemical reaction to occur, the reacting particles must collide with one another. The rate of the reaction depends on the frequency of collisions. The theory also tells us that reacting particles often collide without reacting. For collisions to be successful, reacting particles must (1) collide with (2) sufficient energy, and (3) with the proper orientation.
Molecules must collide before they can react
This rule is fundamental to any analysis of an ordinary reaction mechanism. It explains why termolecular processes are so uncommon. The kinetic theory of gases states that for every 1000 binary collisions, there will be only one event in which three molecules simultaneously come together. Four-way collisions are so improbable that this process has never been demonstrated in an elementary reaction.
Consider a simple bimolecular step:
$A + B \rightarrow Products \nonumber$
If the two molecules A and B are to react, they must approach closely enough to disrupt some of their existing bonds and to permit the creation of any new ones that are needed in the products. Such an encounter is called a collision.
The frequency of collisions between A and B in a gas is proportional to the concentration of each; if [A] is doubled, the frequency of A-B collisions will double, and doubling [B] will have the same effect. If all collisions lead to products, then the rate of a bimolecular process is first-order in A and in B, or second-order overall:
$rate = k[A][B] \nonumber$
Not all Collisions are Equal
For a gas at room temperature and normal atmospheric pressure, there are about 1033 collisions in each cubic centimeter of space every second. If every collision between two reactant molecules yielded products, all reactions would be complete in a fraction of a second. For example, when two billiard balls collide, they simply bounce off of each other. This is the most likely outcome if the reaction between A and B requires a significant disruption or rearrangement of the bonds between their atoms. In order to effectively initiate a reaction, collisions must be sufficiently energetic (or have sufficient kinetic energy) to bring about this bond disruption. This is further discussed below.
There is often one additional requirement. In many reactions, especially those involving more complex molecules, the reacting species must be oriented in a manner that is appropriate for the particular process. For example, in the gas-phase reaction of dinitrogen oxide with nitric oxide, the oxygen end of N2O must hit the nitrogen end of NO; altering the orientation of either molecule prevents the reaction. Owing to the extensive randomization of molecular motions in a gas or liquid, there are always enough correctly-oriented molecules for some of the molecules to react. However, the more critical this orientational requirement is, the fewer collisions will be effective.
Energetic collisions between molecules cause interatomic bonds to stretch and bend, temporarily weakening them so that they become more susceptible to cleavage. Distortion of the bonds can expose their associated electron clouds to interactions with other reactants that might lead to the formation of new bonds.
Chemical bonds have some of the properties of mechanical springs: their potential energies depend on the extent to which they are stretched or compressed. Each atom-to-atom bond can be described by a potential energy diagram that shows how its energy changes with its length. When the bond absorbs energy (either from heating or through a collision), it is elevated to a higher quantized vibrational state (indicated by the horizontal lines) that weakens the bond as its length oscillates between the extended limits corresponding to the curve.
A particular collision will typically excite a number of bonds in this way. Within about 10–13 seconds, this excitation is distributed among the other bonds in the molecule in complex and unpredictable ways that can concentrate the added energy at a particularly vulnerable point. The affected bond can stretch and bend farther, making it more susceptible to cleavage. Even if the bond does not break by pure stretching, it can become distorted or twisted so as to expose nearby electron clouds to interactions with other reactants that might encourage a reaction.
Unimolecular processes also begin with a collision
Until about 1921, chemists did not understand the role of collisions in unimolecular processes. It turns out that the mechanisms of such reactions are actually quite complicated, and that at very low pressures they do follow second-order kinetics. Such reactions are more properly described as pseudounimolecular. The cyclopropane isomerization described in Example 1 is typical of many decomposition reactions found to follow first-order kinetics, implying that the process is unimolecular.
Example 1: Isomerization of Cyclopropane
Consider, for example, the isomerization of cyclopropane to propene, which takes place at fairly high temperatures in the gas phase.
The collision-to-product sequence can be conceptualized in the following [grossly oversimplified] way:
Note that
• For simplicity, the hydrogen atoms are not shown here. This is reasonable because C–C bonds are weaker then C–H bonds, which are less likely to be affected.
• The collision at is most likely with another cyclopropane molecule, but because no part of the colliding molecule gets incorporated into the product, it can in principle be a noble gas or some other non-reacting species;
• Although the C–C bonds in cyclopropane are all identical, the instantaneous localization of the collisional energy can distort the molecule in various ways (), leading to a configuration sufficiently unstable to initiate the rearrangement to the product.
Note: In the Classroom
Here are some simple little "experiments" to help illustrate this concepts. These are "thought experiments," to be reasoned out rather than actually performed.
In an empty classroom, blindfold a group of 10 or so students and instruct them to walk slowly around the room. Occasionally, a pair of students will bump into one another. If they are moving slowly enough, nothing much will happen (that is, no pain upon contact). But if these blindfolded students begin running around the room (more speed means more energy), then a collision is likely to be successful (painfully so). If some students move about quickly, while others stroll about at a more leisurely pace, successful reactions will still occur, but not as often as if all students are running.
Next, have the students move at a brisk pace, but without running. This time, however, consider that collisions that occur shoulder-to-shoulder are not successful—shoulders act as sufficient bumpers and collisions are not painful. But if one student steps on another's toes, then a successful collision occurs. Let the movement begin. Of the frequent collisions that result, only a few will involve one student stepping on another's toes. This experiment illustrates that collisions must occur with the proper orientation. Again, collisions between students will occur, but only some will be successful.
Factors that increase the rate of a reaction must influence at least one of the following:
• how often collisions occur. More frequent collisions will mean a faster rate.
• more effective collisions in terms of collisions occurring with sufficient energy
• more effective collisions in terms of collisions occurring with the proper orientation.
Two more topics must be examined before these can be discussed in depth: reaction mechanisms and the concept of threshold energy.
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This page describes the collision theory of reaction rates, concentrating on the key factors that determine whether a particular collision will result in a reaction—in particular, the energy of the collision, and the orientation of the collision. Reactions in which a single species falls apart are simpler because the orientation of the molecule is unimportant. Reactions involving collisions between more than two species are extremely uncommon (see below).
Two species can only react with each other if they come into contact with each other and then they may react. However, it is not sufficient for the two species to just collide:
1. they must collide the right way,
2. and they must collide with enough energy for their chemical bonds to break.
The chances of this happening for more than two particles at a time are remote. All three (or more) particles would have to arrive at exactly the same point in space at the same time, with everything lined up exactly right, and with enough energy to react. Such circumstances are rarely observed in nature.
Requirement 1: The Orientation of Collision
Consider a simple reaction involving a collision between two molecules: for example, ethene, $\ce{CH2=CH2}$, and hydrogen chloride, $\ce{HCl}$. These react to give chloroethane as shown:
$\ce{H_2C=CH_2 + HCl \rightarrow CH_3CH_2Cl} \nonumber$
As a result of the collision between the two molecules, the double bond in ethene is converted into a single bond. A hydrogen atom is now attached to one of the carbons and a chlorine atom to the other. The reaction can only happen if the hydrogen end of the $\ce{H-Cl}$ bond approaches the carbon-carbon double bond. No other collision between the two molecules produces the same effect. The two simply bounce off each other.
Of the collisions shown in Figure $2$ only collision 1 may possibly lead on to a reaction. With no knowledge of the reaction mechanism, one might wonder why collision 2 would be unsuccessful. The double bond has a high concentration of negative charge around it due to the electrons in the bonds. The approaching chlorine atom is also partially negative due to dipole created by the electronegativity difference between it and hydrogen. The repulsion simply causes the molecules to bounce off each other. In any collision involving unsymmetrical species, the way they hit each other is important in determining whether a reaction occurs.
Requirement 2: Activation Energy
Even if the species are orientated properly, a reaction will not take place unless the particles collide with a certain minimum energy called the activation energy of the reaction. Activation energy is the minimum energy required to make a reaction occur. This can be illustrated on an energy profile for the reaction. An energy profile for a simple exothermic reaction is given in Figure $2$.
If the particles collide with less energy than the activation energy, nothing interesting happens. They bounce apart. The activation energy can be thought of as a barrier to the reaction. Only those collisions with energies equal to or greater than the activation energy result in a reaction.
Any chemical reaction results in the breaking of some bonds (which requires energy) and the formation of new ones (which releases energy). Some bonds must be broken before new ones can be formed. Activation energy is involved in breaking some of the original bonds. If a collision is relatively gentle, there is insufficient energy available to initiate the bond-breaking process, and thus the particles do not react.
The Maxwell-Boltzmann Distribution
Because of the key role of activation energy in deciding whether a collision will result in a reaction, it is useful to know the proportion of the particles present with high enough energies to react when they collide. In any system, the particles present will have a very wide range of energies. For gases, this can be shown in Figure $3$ and is called the Maxwell-Boltzmann distribution, a plot showing the number of particles with each particular energy.
The area under the curve measures of the total number of particles present. Remember that for a reaction to occur, particles must collide with energies equal to or greater than the activation energy for the reaction. The activation energy is marked on the Maxwell-Boltzmann distribution with a green line. Notice that the large majority of the particles have insufficient energy to react when they collide. To enable them to react, either the shape of the curve must be altered, or the activation energy shifted further to the left. This is described on other pages.
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It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days.
6.02: Temperature Dependence of Reaction Rates
Determination of Activation Parameters
It seems intuitive that a reaction goes faster as the temperature is raised, as more reactant molecules have the energy needed to overcome the activation barrier to the reaction. The Arrhenius equation relates the reaction rate constant (k) and temperature. One of the forms of the Arrhenius equation is given below:
$\ln k = -{E_a \over RT} + \ln A \nonumber$
where Ea is the activation energy for the reaction, T is the absolute temperature (in Kelvin) at which a corresponding k is determined, R is the gas constant, and A is a pre-exponential factor. The activation energy may then be extracted from a plot of ln k vs. 1/T, which should be linear. This plot is called an "Arrhenius plot."
Example ARK1.1.
Recall that y = mx + b.
1. In an “Arrhenius plot” plot, what is the slope?
2. What is the intercept?
Example ARK1.2.
Using the following data, construct an Arrhenius plot and determine the activation energy (in both kcal/mol and kJ/mol) and the pre-exponential factor.
1/T (K-1)
ln k (unitless)
0.00152 3.7
0.00157 3.2
0.00160 2.9
0.00165 2.2
0.00170 1.6
Example ARK1.3.
Using the following data, construct an Arrhenius plot and determine the activation energy (in both kcal/mol and kJ/mol) and the pre-exponential factor.
T (°C) k (mol L-1 s-1)
40 1.3 x 10-4
50 2.2 x 10-4
60 4.0 x 10-4
70 7.5 x 10-4
80 1.4 x 10-3
In practice, activation energies are not often cited in the current literature. Instead, a similar but more useful equation called the Eyring equation is used. The Eyring equation is:
$\ln \left ( {k \over T} \right ) = - {\Delta H^{\ddagger} \over RT} + \ln \left ( {k_B \over h}\right ) + {\Delta S^{\ddagger} \over R} \nonumber$
where k, T and R are the same as in the Arrhenius equation, kB is Boltzmann’s constant, h is Planck’s constant and ΔH and ΔS are the enthalpy and entropy of activation, respectively.
Example ARK1.4.
1. What information is required to make an Eyring plot?
2. What is the slope?
3. What is the intercept?
Note that the activation parameters (ΔH and ΔS ) are not the same as the entropy and enthalpy of the reaction, which can usually be calculated from tables of values. Since they depend on how the reaction proceeds, not just the initial and final states of the reaction, they must be determined experimentally. Once that has been done, interpretation of the numerical values provides insight into the mechanism of the reaction.
6.2.02: Changing React
The vast majority of reactions depend on thermal activation, so the major factor to consider is the fraction of the molecules that possess enough kinetic energy to react at a given temperature. According to kinetic molecular theory, a population of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law.
The two distribution plots shown here are for a lower temperature T1 and a higher temperature T2. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of Ea that are shown.
It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures.
Temperature is considered a major factor that affects the rate of a chemical reaction. It is considered a source of energy in order to have a chemical reaction occur. Svante Arrhenius, a Swedish chemist, believed that the reactants in a chemical reaction needed to gain a small amount of energy in order to become products. He called this type of energy the activation energy. The amount of energy used in the reaction is known to be greater than the activation energy in the reaction. Arrhenius came up with an equation that demonstrated that rate constants of different kinds of chemical reactions varied with temperature. This equation indicates a rate constant that has a proportional relationship with temperature. For example, as the rate constant increases, the temperature of the chemical reaction generally also increases. The result is given below:
$\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \nonumber$
This equation is known as Arrhenius' equation. T1and T2are temperature variables expressed in Kelvin. T1 can be expressed as the initial or lower temperature of the reaction, while T2 is the final or higher temperature of the reaction. Rate constants, k1and k2, are values at T1 and T2. Ea is the activation energy expressed in (Joules/mole)=(J/mol). R is the gas constant expressed as 8.3145 (Joules/mole × Kelvin)=(J/mol×K)
Some may ask how the temperature actually affects the chemical reaction rate. The answer to this is that this phenomenon is related to the collision theory. Molecules only react if they have sufficient energy for a reaction to take place. When the temperature of a solution increases, the molecular energy levels also increase, causing the reaction to proceedfaster.
The graph of ln K vs. 1/T is linear, allowing the calculation of the activation energy needed for the reaction. An alternate form of the Arrhenius equation is given below:
$k = A_e^{-\frac{E_a}{RT}} \nonumber$
Some interesting examples:
1. Salt or food coloring is added to cold water, room temperature water, and hot water. When the substance mixes with the hot water, the high temperatures allow it become a homogeneous mixture. This is because due to water molecules moving faster when the temperature is higher and speeding up the dissolution reaction.
2. Another form of energy is light. One example of the effect of temperature on chemical reaction rates is the use of lightsticks or glowsticks. The lightstick undergoes a chemical reaction that is called chemiluminescence; but this reaction does not require or produce heat. Its rate, however, is influenced by temperature. If the lightstick is put in a cold environment, the chemical reaction slows down, allowing it to give off light longer. If the lightstick is in a hot environment, the reaction speeds up causing the light to wear out faster. (This example is from "How Things Work—Lightsticks" from chemistry.about.com)
Contributors and Attributions
• Andrea B.
Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook
6.2.03: The Arrhenius
It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days.
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This page examines simple energy profiles for reactions, and shows how they are slightly different for reactions involving an intermediate, as opposed to just a transition state.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
6.3.02: Basics of Reaction Profiles
Most reactions involving neutral molecules cannot take place at all until they have acquired the energy needed to stretch, bend, or otherwise distort one or more bonds. This critical energy is known as the activation energy of the reaction. Activation energy diagrams of the kind shown below plot the total energy input to a reaction system as it proceeds from reactants to products.
In examining such diagrams, take special note of the following:
• The "reaction coordinate" plotted along the abscissa represents the changes in atomic coordinates as the system progresses from reactants to products. In the very simplest elementary reactions it might correspond to the stretching or twisting of a particular bond, and be shown to a scale. In general, however, the reaction coordinate is a rather abstract concept that cannot be tied to any single measurable and scalable quantity.
• The activated complex (also known as the transition state) represents the structure of the system as it exists at the peak of the activation energy curve. It does not correspond to an identifiable intermediate structure (which would more properly be considered the product of a separate elementary process), but rather to whatever configuration of atoms exists during the collision, which lasts for only about 0.1 picosecond.
• Activation energy diagrams always incorporate the energetics (ΔG or ΔH) of the net reaction, but it is important to understand that the latter quantities depend solely on the thermodynamics of the process which are always independent of the reaction pathway. This means that the same reaction can exhibit different activation energies if it can follow alternative pathways.
• With a few exceptions for very simple processes, activation energy diagrams are largely conceptual constructs based on the standard collision model for chemical reactions. It is unwise to read too much into them.
Contributors and Attributions
• Yuliani Sanjaya
6.3.03: RK3. Activation Barriers
Why do reactions take place at different rates? Why do some happen quickly, and others proceed very slowly? Why might the same reaction proceed at different rates under different conditions? There are a number of factors that influence reaction rates, but this article focuses on the activation barrier.
An activation barrier is a sort of energetic hurdle that a reaction must bypass. Some reactions have higher hurdles and some have lower hurdles. It is easier to overcome lower hurdles, so reactions with low activation barriers can proceed more quickly than ones with higher activation barriers:
• A low activation barrier allows a reaction to happen quickly.
• A high activation barrier makes a reaction proceed more slowly.
A reaction can be exergonic overall, but still have an activation barrier at the beginning. Even if the system decreases in energy by the end of the reaction, it generally experiences an initial increase in energy.
• Even if a reaction gives off energy overall, energy must be added initially to get the reaction started.
This situation is similar to investing in a business. A business generally requires a financial investment to get started. If the business is successful, it will eventually make products and pay money back to the investors. If the business is unable to make back its initial investment, it may fail.
Reactions require an initial investment of energy. This energy may come from surrounding molecules or the environment in general. If the reaction is successful, it will proceed to make products and it will emit energy back to its surroundings.
• It always "costs" a molecule energy to enter into a reaction; it "borrows" that energy from its environment.
• That initial investment of energy may be "paid back" as the reaction proceeds.
All reactions must overcome activation barriers in order to occur. The activation barrier is the sum of the energy that must be expended to get the reaction going. An activation barrier is often pictured as a hill the reactants must climb over during the reaction. Once, there, it can slide down the other side of the hill to become products. At the top of the hill, the molecule exists in what is called the "transition state." At the transition state, the structure is somewhere between its original form and the structure of the products.
The type of diagram shown above is sometimes called a "reaction progress diagram." It shows energy changes in the system as a reaction proceeds. One or more activation barriers may exist along the reaction pathways, due to various elementary steps in the reaction.
In order to understand more concretely the terms "reaction progress" and "transition state," consider a real reaction. Suppose a nucleophile, such as an acetylide ion, donates its electrons to an electrophilic carbonyl. The π bond breaks and an alkoxide ion is formed.
"Reaction progress" refers to how far the reaction has proceeded. The transition state refers specifically to the highest energy point on the pathway from reactants to products. It refers to the structure at that point, and the energy associated with that structure.
In the following diagram, the term "reaction progress" has been replaced by an illustration that matches the status of the reaction with the corresponding point in the energy curve. The structure in the square brackets is the transition state, corresponding to the maximum of the curve. The "double dagger" symbol indicates a transition state structure.
The transition state is not a true chemical structure. It does not necessarily obey the rules of Lewis structures, because some new bonds have started to form and some old bonds have started to break; partial bonds have no place in a Lewis structure.
Physically, the transition state structure cannot be isolated. Because it sits at the top of an energy curve, the transition state tends to convert into something else. A change in either direction will lower its energy. The tendency is to proceed to lowest energy if possible. As soon as the transition state forms, it will either slide back into the original starting materials or slip forward into the final products.
• The transition state is inherently a high-energy, unstable structure with a very short lifetime. As soon as it is formed, it disappears.
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Transition State Theory provides a more accurate alternative to the previously used Arrhenius equation and the Collision theory. The Transition State Theory attempts to provide a better understanding of activation energy, Ea and the thermodynamic properties involving the transition state.
6.04: Transition State Theory
The Eyring Equation, developed by Henry Eyring in 1935, is based on transition state theory and is used to describe the relationship between reaction rate and temperature. It is similar to the Arrhenius Equation, which also describes the temperature dependence of reaction rates. However, whereas Arrhenius Equation can be applied only to gas-phase kinetics, the Eyring Equation is useful in the study of gas, condensed, and mixed-phase reactions that have no relevance to the collision model.
Introduction
The Eyring Equation gives a more accurate calculation of rate constants and provides insight into how a reaction progresses at the molecular level. The Equation is given below:
$k = \dfrac{k_BT}{h}e^{-\left ( \frac{\bigtriangleup H^\ddagger}{RT} \right )}e^{ \left ( \frac{\bigtriangleup S^\ddagger}{R} \right )} \label{1}$
Consider a bimolecular reaction:
$A~+B~\rightarrow~C \label{2}$
$K = \dfrac{[C]}{[A][B]} \label{3}$
where $K$ is the equilibrium constant. In the transition state model, the activated complex AB is formed:
$A~+~B~\rightleftharpoons ~AB^\ddagger~\rightarrow ~C \label{4}$
$K^\ddagger=\dfrac{[AB]^\ddagger}{[A][B]} \label{5}$
There is an energy barrier, called activation energy, in the reaction pathway. A certain amount of energy is required for the reaction to occur. The transition state, $AB^\ddagger$, is formed at maximum energy. This high-energy complex represents an unstable intermediate. Once the energy barrier is overcome, the reaction is able to proceed and product formation occurs.
The rate of a reaction is equal to the number of activated complexes decomposing to form products. Hence, it is the concentration of the high-energy complex multiplied by the frequency of it surmounting the barrier.
$\begin{eqnarray} rate~&=&~v[AB^\ddagger] \label{6} \ &=&~v[A][B]K^\ddagger \label{7} \end{eqnarray}$
The rate can be rewritten:
$rate~=~k[A][B] \label{8}$
Combining Equations $\ref{8}$ and $\ref{7}$ gives:
$\begin{eqnarray} k[A][B]~&=&~v[A][B]K^\ddagger \label{9} \ k~&=&~vK^\ddagger \label{10} \end{eqnarray}$
where
• $v$ is the frequency of vibration,
• $k$ is the rate constant and
• $K ^\ddagger$ is the thermodynamic equilibrium constant.
The frequency of vibration is given by:
$v~=~\dfrac{k_BT}{h} \label{11}$
where
• $k_B$ is the Boltzmann's constant (1.381 x 10-23 J/K),
• $T$ is the absolute temperature in Kelvin (K) and
• $h$ is Planck's constant (6.626 x 10-34 Js).
Substituting Equation $\ref{11}$ into Equation $\ref{10}$ :
$k~=~\dfrac{k_BT}{h}K^\ddagger \label{12}$
Equation ${ref}$ is often tagged with another term $(M^{1-m})$ that makes the units equal with $M$ is the molarity and $m$ is the molecularly of the reaction.
$k~=~\dfrac{k_BT}{h}K^\ddagger (M^{1-m}) \label{E12}$
The following thermodynamic equations further describe the equilibrium constant:
$\begin{eqnarray} \Delta G^\ddagger~&=&~-RT\ln{K^\ddagger}\label{13} \ \Delta G^\ddagger~&=&~\Delta H^\ddagger~-~T\Delta S^\ddagger \label{14} \end{eqnarray}$
where $\Delta G^\ddagger$ is the Gibbs energy of activation, $\Delta H^\ddagger$ is the enthalpy of activation and $\Delta S^\ddagger$ is the entropy of activation. Combining Equations $\ref{10}$ and $\ref{11}$ to solve for $\ln K ^\ddagger$ gives:
$\ln{K}^\ddagger~=~-\dfrac{\Delta H^\ddagger}{RT}~+~\dfrac{\Delta S^\ddagger}{R} \label{15}$
The Eyring Equation is finally given by substituting Equation $\ref{15}$ into Equation $\ref{12}$:
$k~=~\dfrac{k_BT}{h}e^{-\frac{\Delta H^\ddagger}{RT}}e^\frac{\Delta S^\ddagger}{R} \label{16}$
Application of the Eyring Equation
The linear form of the Eyring Equation is given below:
$\ln{\dfrac{k}{T}}~=~\dfrac{-\Delta H^\dagger}{R}\dfrac{1}{T}~+~\ln{\dfrac{k_B}{h}}~+~\dfrac{\Delta S^\ddagger}{R} \label{17}$
The values for $\Delta H^\ddagger$ and $\Delta S^\ddagger$ can be determined from kinetic data obtained from a $\ln{\dfrac{k}{T}}$ vs. $\dfrac{1}{T}$ plot. The Equation is a straight line with negative slope, $\dfrac{-\Delta H^\ddagger}{R}$, and a y-intercept, $\dfrac{\Delta S^\ddagger}{R}+\ln{\dfrac{k_B}{h}}$.
• Kelly Louie
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• 7.1: Catalytic Converters
A catalytic converter is a device used to reduce the emissions from an internal combustion engine (used in most modern day automobiles and vehicles). Not enough oxygen is available to oxidize the carbon fuel in these engines completely into carbon dioxide and water; thus toxic by-products are produced. Catalytic converters are used in exhaust systems to provide a site for the oxidation and reduction of toxic by-products.
• 7.2: Chymotrypsin
Chymotrypsin is a digestive enzyme belonging to a super family of enzymes called serine proteases. It uses an active serine residue to perform hydrolysis on the C-terminus of the aromatic amino acids of other proteins. Chymotrypsin is a protease enzyme that cleaves on the C-terminal phenylalanine (F), tryptophan (W), and tyrosine (Y) on peptide chains. It shows specificity for aromatic amino acids because of its hydrophobic pocket.
• 7.3: Depletion of the Ozone Layer
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine.
• 7.4: Smog
Smog is a common form of air pollution found mainly in urban areas and large population centers. The term refers to any type of atmospheric pollution—regardless of source, composition, or concentration—that creates a significant reduction in atmospheric visibility. Smog encompasses a broad category of air pollutants created through a multitude of processes that relate specifically to the atmospheric conditions of the formation region.
07: Case Studies- Kinetics
A catalytic converter is a device used to reduce the emissions from an internal combustion engine (used in most modern day automobiles and vehicles). Not enough oxygen is available to oxidize the carbon fuel in these engines completely into carbon dioxide and water; thus toxic by-products are produced. Catalytic converters are used in exhaust systems to provide a site for the oxidation and reduction of toxic by-products (like nitrogen oxides, carbon monoxide, and hydrocarbons) of fuel into less hazardous substances such as carbon dioxide, water vapor, and nitrogen gas.
Introduction
Catalytic converters were first widely introduced in American production cars in 1975 due to EPA regulations on toxic emissions reductions. The United States Clean Air Act required a 75% decrease in emissions in all new model vehicles after 1975, a decrease to be carried out with the use of catalytic converters. Without catalytic converters, vehicles release hydrocarbons, carbon monoxide, and nitrogen oxide. These gases are the largest source of ground level ozone, which causes smog and is harmful to plant life. Catalytic converters can also be found in generators, buses, trucks, and trainsalmost everything with an internal combustion engine has a form of catalytic converter attached to its exhaust system.
A catalytic converter is a simple device that uses basic redox reactions to reduce the pollutants a car makes. It converts around 98% of the harmful fumes produced by a car engine into less harmful gases. It is composed of a metal housing with a ceramic honeycomb-like interior with insulating layers. This honeycomb interior has thin wall channels that are coated with a "washcoat" of aluminum oxide. This coating is porous and increases the surface area, allowing more reactions to take place and containing precious metals such as platinum, rhodium, and palladium. No more than 4-9 grams of these precious metals are used in a single converter.
The converter uses simple oxidation and reduction reactions to convert the unwanted fumes. Recall that oxidation is the loss of electrons and that reduction is the gaining of electrons. The precious metals mentioned earlier promote the transfer of electrons and, in turn, the conversion of toxic fumes. The last section of the converter controls the fuel-injection system. This control system is aided by an oxygen sensor that monitors how much oxygen is in the exhaust stream, and in turn tells the engine computer to adjust the air-to-fuel ratio, keeping the catalytic converter running at the stoichiometric point and near 100% efficiency.
Catalytic Converter Theft
Due to the precious metals in the coating of the inner ceramic structure, many catalytic converters have been targeted for theft. The converter is the most easily-accessible component because it lies on the outside and under the car. A thief could easily slide under the car, saw the connecting tubes on each end, and leave with the catalytic converter. Depending on the type and amount of precious metals inside, a catalytic converter can be easily sold for \$200 apiece.
Functions
A three-way catalytic converter has three simultaneous functions:
1. Reduction of nitrogen oxides into elemental nitrogen and oxygen: $\ce{NO_{x} \rightarrow N_{x} + O_{x}} \nonumber$
2. Oxidation of carbon monoxide to carbon dioxide: $\ce{CO + O_2 \rightarrow CO_2} \nonumber$
3. Oxidation of hydrocarbons into carbon dioxide and water: $\ce{C_{x}H_{4x} + 2xO_2 \rightarrow xCO_2 + 2xH_2O} \nonumber$
There are two types of "systems" running in a catalytic converter, "lean" and "rich." When the system is running "lean," there is more oxygen than required, and the reactions therefore favor the oxidation of carbon monoxide and hydrocarbons (at the expense of the reduction of nitrogen oxides). On the contrary, when the system is running "rich," there is more fuel than needed, and the reactions favor the reduction of nitrogen oxides into elemental nitrogen and oxygen (at the expense of the two oxidation reactions). With a constant imbalance of the reactions, the system never achieves 100% efficiency.
Note: converters can store "extra" oxygen in the exhaust stream for later use. This storage usually occurs when the system is running lean; the gas is released when there is not enough oxygen in the exhaust stream. The released oxygen compensates for the lack of oxygen derived from $\ce{NO_{x}}$ reduction, or when there is hard acceleration and the air-to-fuel ratio system becomes rich faster than the catalytic converter can adapt to it. In addition, the release of the stored oxygen stimulates the the oxidation processes of $\ce{CO}$ and $\ce{C_{x}H_{4x}}$.
Dangers of Pollutants
Without the redox process to filter and convert the nitrogen oxides, carbon monoxides, and hydrocarbons, the air quality (especially in large cities) becomes harmful to the human being.
Nitrogen oxides ($\ce{NO_{x}}$) are compounds are of the same family as nitrogen dioxide, nitric acid, nitrous oxide, nitrates, and nitric oxide. When $\ce{NO_{x}}$ is released into the air, it reacts, stimulated by sunlight, with organic compounds in the air; the result is smog. Smog is a pollutant and has adverse effects on children's lungs. $\ce{NO_{x}}$ reacting with sulfur dioxide produces acid rain, which is highly destructive to everything it lands on. Acid rain corrodes cars, plants, buildings, national monuments and pollutes lakes and streams to an acidity unsuitable for fish. $\ce{NO_{x}}$ can also bind with ozone to create biological mutations (such as smog), and reduce the transmission of light.
Carbon monoxide ($\ce{CO}$) is a harmful variant of a naturally occurring gas, $\ce{CO2}$. Odorless and colorless, this gas does not have many useful functions in everyday processes.
Hydrocarbons: Inhaling hydrocarbons from gasoline, household cleaners, propellants, kerosene and other fuels can be fatal to children. Further complications include central nervous system impairments and cardiovascular problems.
Catalytic Inhibition and Destruction
The catalytic converter is a sensitive device with precious metals coating the inside and without these metals, the desired redox reactions cannot occur. There are several substances and chemicals that inhibit the catalytic converter.
1. Lead: Most vehicles now run on unleaded gasoline with no lead compounds added to the fuel. However, if lead were added (typically as tetraethyllead) to the fuel and is burned, it leaves a residue that coats the catalytic metals ($\ce{Pt}$, $\ce{Rh}$, $\ce{Pd}$, and $\ce{Au}$) and prevents contact with the exhaust fumes, which is necessary in performing the necessary redox reactions.
2. Manganese and silicon: Manganese is primarily found in the organometallic compound MMT (methylcyclopentadienyl manganese tricarbonyl). MMT is a compound used in in the 1990's to increase fuel's octane rating (a higher octane rating indicates that the gas is less likely to combust outside of normal operation). This is important since higher performing engines have a high compression ratio, which would need a higher octane gas to complement the amount of compression in the engine), and has now been banned from commercial sale due to the EPA's regulations. Silicon can leak from the combustion chamber into the exhaust stream from the coolant inside the engine.
These contaminants prevent the catalytic converter from functioning properly. However, this process could be reversed by running the engine at a high temperature to increase the hot exhaust flow through the converter, melting or liquefying some of the contaminants and removing them from the exhaust pipe. This process does not work if the metal is coated with lead, because lead has a high boiling point. If the lead poisoning is severe enough, the whole converter is rendered useless and must be replaced.
Thermodynamics of Catalytic Converters
Recall that thermodynamics predicts whether or not a reaction or process is spontaneous under certain conditions, but not the rate of the process. The redox reactions below occur slowly without a catalyst; even if the processes are thermodynamically favorable, they cannot occur without proper energy. This energy is the activation energy ($E_a$), which is required to overcome the initial energy barrier preventing the reaction. A catalyst accelerates the reaction by lowering the activation energy although the catalyst itself does not produce a product, but it does affect the amount and the speed at which the products are formed.
1. Reduction of nitrogen oxides into elemental nitrogen and oxygen: $\ce{NO_{x} \rightarrow N_{x} + O_{x}} \nonumber$
2. Oxidation of carbon monoxide to carbon dioxide. $\ce{CO + O_2 \rightarrow CO_2 } \nonumber$
3. Oxidation of hydrocarbons into carbon dioxide and water. $\ce{C_{x}H_{4x} + 2xO_2 \rightarrow {x}CO_2 + {2x}H_2O} \nonumber$
Global Warming
Although the catalytic converter has helped reduce toxic emissions from car engines, it also has detrimental environmental effects. In the conversion of hydrocarbons and carbon monoxide, carbon dioxide is produced. Carbon dioxide is one of the most common greenhouse gases and contributes significantly to global warming. Along with carbon dioxide, the converters sometimes rearrange the nitrogen-oxygen compounds to form nitrous oxide. This is the same compound used in laughing gas and as a speed enhancer in cars. As a greenhouse gas, nitrous oxide is a 300 times more potent than carbon dioxide, and contributes proportionally to global warming.
Problems
1. What are the potential hazards of the toxic substances emitted by a car without a catalytic converter?
2. Which three redox reactions occur in a three-way catalytic converter?
3. Does the catalytic converter run at 100% efficiency? Why or why not?
4. How can catalytic converters be damaged or misused?
5. Why are catalytic converters targeted for theft?
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/07%3A_Case_Studies-_Kinetics/7.01%3A_Catalytic_Converters.txt
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Chymotrypsin is a digestive enzyme belonging to a super family of enzymes called serine proteases. It uses an active serine residue to perform hydrolysis on the C-terminus of the aromatic amino acids of other proteins. Chymotrypsin is a protease enzyme that cleaves on the C-terminal phenylalanine (F), tryptophan (W), and tyrosine (Y) on peptide chains. It shows specificity for aromatic amino acids because of its hydrophobic pocket.
Introduction
Chymotrypsin is one of the most studied enzymes due to its two phase kinetics: pre-steady-state and steady state. The study of these two kinetic states gives evidence of the "Ping-Pong" mechanism, the formation of covalent complexes leading to covalent hydrolysis reactions, and the rate of the catalyzed reactions. Synthesis of chymotrypsin occurs primarily in the pancreas. Instead of the active form, however, it is produced as an inactive zymogen called chymotrypsinogen to prevent its protease activity from digesting the pancreas. Upon secretion into the lumen of the small intestine, it is converted to its active form by another enzyme called trypsin. This dependence of a different enzyme for the activation of a protease is a common way for the body to prevent the digestion of organs and other harmful enzymatic side-effects.
Chymotrypsin operates through a general mechanism known as the ping-pong mechanism (Figure $1$) whereby the enzyme reacts with a substrate to form an enzyme intermediate. This intermediate has different properties than the initial enzyme, so to regenerate the initial enzymatic activity, it must react with a secondary substrate. This process is illustrated below:
More specifically, chymotrypsin operates through a particular type of ping-pong mechanism called covalent hydrolysis. This means that the enzyme first forms a covalent bond with the target substrate, displacing the more stable moiety into solution. This enzyme-substrate complex is called the enzyme intermediate. The intermediate then reacts with water, which displaces the remaining part of the initial substrate and reforms the initial enzyme.
Chymotrypsin, like most enzymes, is specific in the types of substrates with which it reacts. As a protease, it cleaves polypeptides, and its inherent specificity allows it to act only on the carboxy-terminal of aromatic residues. It is a somewhat complicated mechanism, and is best explained in a series of steps.
Step 1: The target enters the active site of chymotrypsin, and it is held there by hydrophobic interactions between exposed non-polar groups of enzyme residues and the non-polar aromatic side-chain of the substrate. It is important to note the hydrogen bond between the Schiff nitrogen on histidine-57 and the oxygen side-chain of serine-195.
Step 2: Aided by the histidine-serine hydrogen bonding, the hydroxyl group on serine-195 performs a nucleophilic attack on the carbonyl carbon of an aromatic amino acid while simultaneously transferring the hydroxyl hydrogen to the histidine Schiff nitrogen. This attack pushes the pi carbonyl electrons onto the carbonyl oxygen, forming a short-lived intermediate consisting of a c-terminal carbon with four single bonds: an oxygen anion, the beta-carbon of the aromatic amino acid, the n-terminus of the subsequent amino acid of the substrate protein, and the serine-195 side-chain oxygen.
Step 3: This intermediate is short-lived, as the oxyanion electrons reform the pi bond with the c-terminus of the aromatic amino acid. The bond between the carboxy-terminus of the aromatic amino acid and the n-terminus of the subsequent residue is cleaved, and its electrons are used to extract the hydrogen of the protonated Schiff nitrogen on histidine-57. The bonds between the carbonyl carbon and the serine-195 oxygen remain in an ester configuration. This is called the acyl-enzyme intermediate. The c-terminal side of the polypeptide is now free to dissociate from the active site of the enzyme.
Step 4: Water molecules are now able to enter and bind to active site through hydrogen bonding between the hydrogen atoms of water and the histidine-57 Schiff nitrogen.
Step 5: The water oxygen now makes a nucleophilic attack on the carbonyl carbon of the acyl-enzyme intermediate, pushing the carbonyl’s pi electrons onto the carbonyl carbon as histidine-57 extracts one proton from water. This forms another quaternary carbon covalently bonded with serine, a hydroxyl, an oxyanion, and the aromatic amino acid. The proton on the recently protonated histidine-57 is now able to make a hydrogen bond with the serine oxygen.
Step 6: The oxyanion electrons reform the carbonyl pi bond, cleaving the bond between the carbonyl carbon and the serine hydroxyl. The electrons in this bond are used by the serine oxygen to deprotonate the histidine Schiff nitrogen and reform the original enzyme. The substrate no longer has affinity for the active site, and it soon dissociates from the complex.
Kinetics
Experiments were conducted in 1953 by B.S. Hartley and B.A. Kilby to investigate the kinetics of chymotrypsin-catalyzed hydrolysis. Instead of using a poly-peptide chain as a substrate, they used a nitro-phenyl ester, p-nitrophenyl acetate, that resembles an aromatic amino acid. Hydrolysis of this compound by chymotrypsin at the carbonyl group yields acetate and nitrophenolate, the latter of which absorbs near 400 nm light and its concentration can thus be measured by spectrophotometry (Figure $2$).
Spectrophotometric analysis of chymotrypsin acting on nitrophenylacetate showed that nitrophenolate was produced at a rate independent of substrate concentration, proving that the only factor contributing to the rate of product formation is the concentration of enzyme; this is typical for enzyme-substrate kinetics. However, when the slope of the 0-order absorbance plot was traced back to the starting point (time = 0), it was found that the initial concentration of nitrophenolate was not 0. In fact, it showed a 1:1 stoichiometric ratio with the amount of chymotrypsin used in the assay. This can only be explained by the fact that hydrolysis by chymotrypsin is biphasic in nature, meaning that it proceeds in two distinct steps.
1. The first step, which describes the initial burst of nitrophenolate seen in Hartley and Kilby’s absorbance plot, is the fastest. The attack of the nitrophenyl acetate substrate by chymotrypsin immediately cleaves the nitrophenolate moiety and leaves the acetate group attached to chymotrypsin, rendering the enzyme inactive.
2. The second step has been deduced to involve the hydrolysis of the acetate group from the inactivated chymotrypsin to regenerate the original enzyme.
To analytically determine the rate of catalysis, all substrates, products, and intermediates must be defined. Refer to the figure below:
Using these abbreviations, kinetic analysis becomes less cumbersome.
1. The initial amount of enzyme can be represented as the sum of the free enzyme, the bound enzyme, and the inactive intermediate.
$[E]_o = [ES] + [^*ES] + [E] \nonumber$
2. Assuming the initial step is the faster than the subsequent steps, the rate of nitrophenolate production can be described as:
$\dfrac{d[P_1]}{dt}=k_2[ES] \nonumber$
3. Likewise, the rate of acetate formation can be represented by the equation:
$\dfrac{d[P_2]}{dt}=k_3[^*ES] \nonumber$
4. Therefore, the net change in concentration of the inactive intermediate can be deduced:
$\dfrac{d[*ES]}{dt}=k_2[ES]-k_3[^*ES] \nonumber$
5. The last inference that can be made from analysis of the measured kinetics data (Figure $2$) is that the first step of the reaction equilibrates rapidly, and thus the change in bound substrate can be described in the following equation. This is a principal tenet in analyzing the kinetics of chymotrypsin and is a ubiquitous mechanism in biological enzyme catalysis.
$\dfrac{d[ES]}{dt}=k_1[E][S]-k_{-1}[ES]=0 \nonumber$
6. Where:
$\dfrac{k_{-1}}{k_1} = K_s = \dfrac{[E][S]}{[ES]} \nonumber$
7. Combining all of these quantities, we can deduce the catalytic rate constant as:
$k_{cat} = \dfrac{k_2k_3}{k_2+k_3} \nonumber$
8. In ester hydrolysis, $k_3 >> k_2$, so the resultant catalytic rate constant simplifies to:
$k_{cat}=k_2 \nonumber$
which is in agreement with the observed zeroth-order kinetics of Figure $2$.
Table $1$: Kinetic Constants of the Chymotrypsin-Catalyzed Hydrolysis of p-Nitrophenyl Trimethylacetate at pH 8.2. 0.01 M tris-HCL buffer, ionic strength 0.06, 25.6 ± 0.1 °C, 1.8% (v/v) acetonitrile-water. From M.L. Bender, F.J. Kezdy and F.C. Wedler, J. Chem. Educ. 44, 84 (1967
k2 0.37 ± 0.11 s-1
k3 (1.3 ± 0.02) × 10-4 s-1
Ks (1.6 ± 0.5) × 10-3
kcat 1.3 × 10-4 s-1
KM 5.6 × 10-7
Exercise $1$
Speculate on how the catalytic rate constant can be determined from the spectrophotogram.
Answer
The catalytic rate constant can be deduced from the graph by simply determining the slope of the line where the reaction demonstrates 0-order kinetics (the linear part)
Exercise $2$
How can product be consistently produced if the rate of change of the ES complex is 0?
Answer
This is pre-equilibrium kinetics in action. The ES complex is formed from E and S at a faster rate than any other step in the reaction. As soon as ES is converted to *ES, another mole of ES is produced from an infinite supply of E + S. This means that the amount of ES and E + S is constantly at equilibrium, and thus the change of either with respect to time is 0.
Exercise $3$
How would the rate of product formation change if:
1. the substrate concentration was doubled.
2. the enzyme concentration was doubled.
3. The reaction was carried out in mono-deuterated water instead of H2O (comment qualitatively).
Answer
1. No change.
2. Two fold increase.
3. Since water is involved in the final, slowest step of the mechanism, deuterating the water would decrease the rate of the overall reaction from 5 to 30-fold.
Exercise $4$
Explain the role of hydrogen bonding in protein hydrolysis catalyzed by chymotrypsin.
Answer
Initially, hydrogen bonding between the enzymes histidine and serine side chains weakens the bond of serine’s O-H. This allows for a facilitated nucleophilic attack of the hydroxyl Oxygen on the substrates carbonyl group. Conversely, in the final step of the reaction, the bound serine oxygen forms a hydrogen bond with a protonated histidine, which allows for easier cleavage from the substrate.
Exercise $5$
What would the spectrophotogram look like if the reaction proceeded via a steady-state mechanism instead of pre-equilibrium.
Answer
The graph would show similar 0-order kinetics, but the line would intercept the Y-axis at an absorbance of 0 instead of the 1:1 mole ratio of nitrophenolate to enzyme.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/07%3A_Case_Studies-_Kinetics/7.02%3A_Chymotrypsin.txt
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The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine.
Introduction
The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion.
Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone ($O_3$) to produce $O_2$ molecules; however, $NO$ molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for $NO$ to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with $O_3$ and destroy the molecule. This process is called the Rowland-Molina theory of $O_3$ depletion.
The Chapman Cycle
The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O2) to create ozone (O3) and atomic oxygen (O). This process is called the Chapman cycle.
Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals:
$h\nu + O_2 \rightarrow 2O^. \nonumber$
Step 2: Oxygen radicals then react with molecular oxygen to produce ozone:
$O_2 + O^. \rightarrow O_3 \nonumber$
Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen:
$O_3 + O^. \rightarrow 2O_2 \nonumber$
Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon:
$O_3 + h\nu \rightarrow O_2 + O^. \nonumber$
It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O2 is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself.
Chemistry of Ozone Depletion
CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:
$Cl + O_3 \rightarrow ClO + O_2 \tag{step 1}$
$ClO + O^. \rightarrow Cl + O_2 \tag{step 2}$
$O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction}$
Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface.
From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's.
The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction.
• Because Antarctica is surrounded by water, winds over the continent blow in a unique clockwise direction creating a so called "polar vortex" that effectively contains a single static air mass over the continent. As a result, air over Antarctica does not mix with air in the rest of the earth's atmosphere.
• Antarctica has the coldest winter temperatures on earth, often reaching -110 F. These chilling temperatures result in the formation of polar stratospheric clouds (PSC's) which are a conglomeration of frozen H2O and HNO3. Due to their extremely cold temperatures, PSC's form an electrostatic attraction with CFC molecules as well as other halogenated compounds
As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O3. Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process.
Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink
Ozone Depleting Substances
The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act:
Table $1$: Ozone Depleting Substances And Their Ozone-Depletion Potential. Taken directly from the Clean Air Act, as of June 2010.
Substance Ozone- depletion potential
chlorofluorocarbon-11 (CFC–11) 1.0
chlorofluorocarbon-12 (CFC–12) 1.0
chlorofluorocarbon-13 (CFC–13) 1.0
chlorofluorocarbon-111 (CFC–111) 1.0
chlorofluorocarbon-112 (CFC–112) 1.0
chlorofluorocarbon-113 (CFC–113) 0.8
chlorofluorocarbon-114 (CFC–114) 1.0
chlorofluorocarbon-115 (CFC–115) 0.6
chlorofluorocarbon-211 (CFC–211) 1.0
chlorofluorocarbon-212 (CFC–212) 1.0
chlorofluorocarbon-213 (CFC–213) 1.0
chlorofluorocarbon-214 (CFC–214) 1.0
chlorofluorocarbon-215 (CFC–215) 1.0
chlorofluorocarbon-216 (CFC–216) 1.0
chlorofluorocarbon-217 (CFC–217) 1.0
halon-1211 3.0
halon-1301 10.0
halon-2402 6.0
carbon tetrachloride 1.1
methyl chloroform 0.1
hydrochlorofluorocarbon-22 (HCFC–22) 0.05
hydrochlorofluorocarbon-123 (HCFC–123) 0.02
hydrochlorofluorocarbon-124 (HCFC–124) 0.02
hydrochlorofluorocarbon-141(b) (HCFC–141(b)) 0.1
hydrochlorofluorocarbon-142(b) (HCFC–142(b)) 0.06
General Questions
• What are the causes of the depletion of our ozone layer?
• the release of free radicals, the use of CFC's, the excessive burning of fossil fuels
• What is the chemical reaction that displays how ozone is created?
• UV + O2 -> 2O + heat, O2 + O -> O3, O3 + O -> 2O2
• Which reactions demonstrate the destruction of the ozone layer?
• Cl + O3 ------> ClO + O2 and ClO + O ------> Cl + O
• How do CFC's destroy the ozone layer?
• the atomic chlorine freed from CFC reacts in a catalytic manner with ozone and atomic oxygen to make more oxygen molecules
• Why should regulations be enforced now in regards to pollution and harmful chemicals?
• without regulation, the production and use of chemicals will run out of hand and do irreversible damage to the stratosphere
• What type of atom in the CFC molecule is most destructive to the ozone?
• chlorine
• In which layer of the atmosphere does the ozone layer?
• the stratosphere, the second closest to the Earth's surface
• What cycle is responsible for ozone in the stratosphere?
• the Chapman cycle
• What factor is responsible for breaking up stable molecules?
• ultraviolet rays from the sun
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/07%3A_Case_Studies-_Kinetics/7.03%3A_Depletion_of_the_Ozone_Layer.txt
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Smog is a common form of air pollution found mainly in urban areas and large population centers. The term refers to any type of atmospheric pollution—regardless of source, composition, or concentration—that creates a significant reduction in atmospheric visibility. Smog encompasses a broad category of air pollutants created through a multitude of processes that relate specifically to the atmospheric conditions of the formation region. In the early 1900s, London was plagued by a particular type of smog that resulted from a combination of dense fog and soot from coal combustion. In modern times, the Los Angeles Basin is often associated with dense photochemical smog, produced through a combination of vehicle exhaust and sunlight. These are two of many examples of pollution classified as smog, but they are in no way chemically related. Smog refers to a diverse category of air pollutants with varying chemical composition; however, all types of smog form a visible haze that reduces atmospheric visibility.
Introduction
The term smog was first coined in 1905 in a paper by Dr. Henry Antoine Des Voeux to describe the combination of smoke and fog that had been plaguing London during that time. London has since enacted strict air pollution regulations which have drastically reduced incidents of smog in that region; however, London-type smog is still a major problem in areas of the world that burn large quantities of coal for heat. In the United States, smog is most typically associated with the Los Angeles Basin of Southern California and its photochemical smog. In Los Angeles, a combination of orographic features, ample sunlight, and a dense population combine to form some of the worst air quality in the United States.
Photochemical Smog
Photochemical smog, as commonly seen in the Los Angeles Basin, is mainly composed of ozone and nitrogen dioxide. During the formation of ozone, nitrogen dioxide from vehicle exhaust is photolyzed by incoming solar radiation to produce nitrogen oxide and an unpaired oxygen atom. The lone oxygen atom then combines with an oxygen molecule to produce ozone. Under normal conditions, the majority of ozone molecules oxidize nitrogen oxide back into nitrogen dioxide, creating a virtual cycle that leads to only a very slight build up of ozone near ground level. However, when volatile organic compounds (VOCs) are present in the atmosphere, the equation changes entirely. Highly reactive VOCs oxidize nitrogen oxide into nitrogen dioxide without breaking down any ozone molecules in the process. This leads to a proliferation of ozone near ground level and dense smog formation. Although photochemical smog in the United States is mainly associated with the Los Angeles Basin, photochemical smog episodes have been reported in Denver, Philadelphia and New York.
Composition of Photochemical Smog
The following substances are identified in photochemical smog:
1. Nitrogen Dioxide ($NO_2$) from vehicle exhaust, is photolyzed by ultraviolet (UV) radiation ($h\nu$) from the sun and decomposes into Nitrogen Oxide ($NO$ and an oxygen radical:
$NO_2 + h\nu \rightarrow NO + O^. \label{1}$
2. The oxygen radical then reacts with an atmospheric oxygen molecule to create ozone, O3:
$O^. + O_2 \rightarrow O_3 \label{2}$
3. Under normal conditions, O3 reacts with NO, to produce NO2 and an oxygen molecule:
$O_3 + NO \rightarrow O_2 + NO_2 \label{3}$
This is a continual cycle that leads only to a temporary increase in net ozone production. To create photochemical smog on the scale observed in Los Angeles, the process must include Volatile organic compounds (VOC's).
4. VOC's react with hydroxide in the atmosphere to create water and a reactive VOC molecule:
$RH + OH^. \rightarrow R^. + H_2O \label{4}$
5. The reactive VOC can then bind with an oxygen molecule to create an oxidized VOC:
$R^. + O_2 \rightarrow RO_2 \label{5}$
6. The oxidized VOC can now bond with the nitrogen oxide produced in the earlier set of equations to form nitrogen dioxide and a reactive VOC molecule:
$RO_2+ NO \rightarrow RO-. + NO_2 \label{6}$
In the second set of equations, it is apparent that nitrogen oxide, produced in equation 1, is oxidized in equation 6 without the destruction of any ozone. This means that in the presence of VOCs, equation 3 is essentially eliminated, leading to a large and rapid build up in the photochemical smog in the lower atmosphere.
Figure 1, courtesy of the EPA, depicts concentrations and constituents of photochemical smog throughout the course of an average work day. In the morning, NO and VOC concentrations are high, as people fill their cars with gas and drive to work. By midmorning , VOC's begin to oxidize NO into NO2, thus reducing their respective concentrations. At midday, NO2 concentrations peak just before solar radiation becomes intense enough to photolyze the NO2 bond, releasing an oxygen atom that quickly gets converted into O3. By late afternoon, peak concentrations of photochemical smog are present.
Controlling Photochemical Smog
Every new vehicle sold in the United States must include a catalytic converter to reduce photochemical emissions. Catalytic converters force CO and incompletely combusted hydrocarbons to react with a metal catalyst, typically platinum, to produce CO2 and H2O. Additionally, catalytic converters reduce nitrogen oxides from exhaust gases into O2 and N2, eliminating the cycle of ozone formation. Many scientists have suggested that pumping gas at night could reduce photochemical ozone formation by limiting the amount of exposure VOCs have with sunlight.
London Smog
London-type smog is mainly a product of burning large amounts of high sulfur coal. Clean air laws passed in 1956 have greatly reduced smog formation in the United Kingdom; however, in other parts of the world London-type smog is still very prevalent. The main constituent of London-type smog is soot; however, these smogs also contain large quantities of fly ash, sulfur dioxide, sodium chloride and calcium sulfate particles. If concentrations are high enough, sulfur dioxide can react with atmospheric hydroxide to produce sulfuric acid, which will precipitate as acid rain.
$SO_2 + OH^. \rightarrow HOSO_2 \label{1}$
$HOSO_2 + O_2 \rightarrow HO_2 + SO_3 \label{2}$
$SO_3 + H_2O \rightarrow H_2SO_4 \label{3}$
Health Hazards
Because ozone is highly reactive, it has the ability to oxidize and destroy lung tissue. Short term exposures to elevated levels of ozone (above .75 ppm) have been linked to a host of respiratory irritations including coughing, wheezing, substernal soreness, pharyngitis, and dyspnea. Prolonged exposure to the molecule has been proven to cause a permanent reduction in lung function, as well as elevate the risk of developing asthma. Sulfur dioxide is a common component of London smog. Epidemiological studies have linked short term sulfur dioxide exposure to respiratory irritations including coughing, wheezing, and pharyngitis.
Problems
1. True or False: Reducing NO emissions will always reduce the amount of photochemical smog that can be produced ?
• False; from equation 3, listed above, NO reacts with O3to produce NO2 and O2. Therefore, reducing NO emissions can actually increase the amount of photochemical smog in the atmosphere
2. List three factors that make the Los Angeles Basin an ideal place for photochemical smog formation.
• Los Angeles is located in an arid environment so sunlight is almost always intense enough to photolyze NO2 into NO + O.
• Los Angeles sits in a geographic bowl, surrounded by mountain ranges so photochemical smog has a tendency to be trapped directly over the city
• Los Angeles has a very dense population with millions of cars on the road every day emitting vast quantities of NOx and VOC's
3. How can pumping your gas at night reduce photochemical ozone formation?
• Pumping gas releases lots of VOC's into the atmosphere, If these VOC's are released at night, NO concentrations are low because there is no sunlight to photolyze the NO2 bond. By morning, the VOC's will likely have broken down into a less reactive compound.
4. True or False: Photochemical smog is created from a combination of soot, fly ash, and sulfur dioxide.
• False, London smog is created from a combination of soot, fly ash, and sulfur dioxide.
5. Why is it not a good idea to breath ozone?
• Ozone is a highly oxidative chemical; when ozone comes in contact with lung tissue it can oxidize and destroy it.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/07%3A_Case_Studies-_Kinetics/7.04%3A_Smog.txt
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A reaction’s equilibrium position defines the extent to which the reaction can occur. For example, we expect a reaction with a large equilibrium constant, such as the dissociation of HCl in water
$\ce{HCl}(aq) + \ce{H2O}(l) ⇋ \ce{H3O+}(aq) + \ce{Cl-}(aq) \nonumber$
to proceed nearly to completion. A large equilibrium constant, however, does not guarantee that a reaction will reach its equilibrium position. Many reactions with large equilibrium constants, such as the reduction of MnO4 by H2O
$\ce{4MnO4-}(aq) + \ce{2H2O}(l) ⇋ \ce{4MnO2}(s) + \ce{3O2}(g) + \ce{4OH-}(aq) \nonumber$
do not occur to an appreciable extent. The study of the rate at which a chemical reaction approaches its equilibrium position is called kinetics.
A17.1 Chemical Reaction Rates
A study of a reaction’s kinetics begins with the measurement of its reaction rate. Consider, for example, the general reaction shown below, involving the aqueous solutes A, B, C, and D, with stoichiometries of a, b, c, and d.
$a\ce{A} + b\ce{B} ⇋ c\ce{C} + d\ce{D}\tag{A17.1}$
The rate, or velocity, at which this reaction approaches its equilibrium position is determined by following the change in concentration of one reactant or one product as a function of time. For example, if we monitor the concentration of reactant A, we express the rate as
$R = -\dfrac{d[\ce A]}{dt}\tag{A17.2}$
where R is the measured rate expressed as a change in concentration of A as a function of time. Because a reactant’s concentration decreases with time, we include a negative sign so that the rate has a positive value.
We also can determine the rate by following the change in concentration of a product as a function of time, which we express as
$R´ = −\dfrac{d[\ce C]}{dt}\tag{A17.3}$
Rates determined by monitoring different species do not necessarily have the same value. The rate R in equation A17.2 and the rate R´ in equation A17.3 have the same value only if the stoichiometric coefficients of A and C in reaction A17.1 are identical. In general, the relationship between the rates R and R´ is
$R = \dfrac{a}{c} × R´ \nonumber$
A17.2 The Rate Law
A rate law describes how a reaction’s rate is affected by the concentration of each species in the reaction mixture. The rate law for reaction A17.1 takes the general form of
$R = k\mathrm{[A]^α[B]^β[C]^γ[D]^δ[E]^ε...}\tag{A17.4}$
where k is the rate constant, and α, β, γ, δ, and ε are the reaction orders of the reaction for each species present in the reaction.
There are several important points about the rate law in equation A17.4. First, a reaction’s rate may depend on the concentrations of both reactants and products, as well as the concentration of a species that does not appear in the reaction’s overall stoichiometry. Species E in equation A17.4, for example, may be a catalyst that does not appear in the reaction’s overall stoichiometry, but which increases the reaction’s rate. Second, the reaction order for a given species is not necessarily the same as its stoichiometry in the chemical reaction. Reaction orders may be positive, negative, or zero, and may take integer or non-integer values. Finally, the reaction’s overall reaction order is the sum of the individual reaction orders for each species. Thus, the overall reaction order for equation A17.4 is α + β + γ + δ + ε.
A17.3 Kinetic Analysis of Selected Reactions
In this section we review the application of kinetics to several simple chemical reactions, focusing on how we can use the integrated form of the rate law to determine reaction orders. In addition, we consider how we can determine the rate law for a more complex system.
First-Order Reactions
The simplest case we can treat is a first-order reaction in which the reaction’s rate depends on the concentration of only one species. The best example of a first-order reaction is an irreversible thermal decomposition of a single reactant, which we represent as
$\mathrm{A → Products}\tag{A17.5}$
with a rate law of
$R = -\dfrac{d[\ce A]}{dt} = k[\ce A]\tag{A17.6}$
The simplest way to demonstrate that a reaction is first-order in A, is to double the concentration of A and note the effect on the reaction’s rate. If the observed rate doubles, then the reaction must be first-order in A. Alternatively, we can derive a relationship between the concentration of A and time by rearranging equation A17.6 and integrating.
$\dfrac{d[\ce A]}{[\ce A]} = -k dt \nonumber$
$\int_{[\ce A]_0}^{[\ce A]_t} \dfrac{d[\ce A]}{[\ce A]} = -k \int_{0}^{t}dt\tag{A17.7}$
Evaluating the integrals in equation A17.7 and rearranging
$\ln\dfrac{[\ce A]_t}{[\ce A]_0}= -kt\tag{A17.8}$
$\ln[\ce A]_t = -kt+ \ln[\ce A]_0\tag{A17.9}$
shows that for a first-order reaction, a plot of ln[A]t versus time is linear with a slope of –k and a y-intercept of ln[A]0. Equation A17.8 and equation A17.9 are known as integrated forms of the rate law.
Reaction A17.5 is not the only possible form of a first-order reaction. For example, the reaction
$\mathrm{A + B → Products}\tag{A17.10}$
will follow first-order kinetics if the reaction is first-order in A and if the concentration of B does not affect the reaction’s rate. This may happen if the reaction’s mechanism involves at least two steps. Imagine that in the first step, A slowly converts to an intermediate species, C, which rapidly reacts with the remaining reactant, B, in one or more steps, to form the products.
$\mathrm{A → B \hspace{20px}(slow)} \nonumber$
$\mathrm{B + C → Products \hspace{20px} (fast)} \nonumber$
Because a reaction’s rate depends only on those species in the slowest step—usually called the rate-determining step—and any preceding steps, species B will not appear in the rate law.
Second-Order Reactions
The simplest reaction demonstrating second-order behavior is
$\mathrm{2A → Products} \nonumber$
for which the rate law is
$R = -\dfrac{d[\ce A]}{dt}= k[\ce A]^2 \nonumber$
Proceeding as we did earlier for a first-order reaction, we can easily derive the integrated form of the rate law.
$\dfrac{d[\ce A]}{[\ce A]^2}= -k\, dt \nonumber$
$\int_{[\ce A]_0}^{[\ce A]_t} \dfrac{d[\ce A]}{[\ce A]^2} = -k \int_{0}^{t}dt \nonumber$
$\dfrac{1}{[\ce A]_t} = kt + \dfrac{1}{[\ce A]_0} \nonumber$
For a second-order reaction, therefore, a plot of [A]t–1 versus t is linear with a slope of k and a y-intercept of [A]0–1. Alternatively, we can show that a reaction is second-order in A by observing the effect on the rate when we change the concentration of A. In this case, doubling the concentration of A produces a four-fold increase in the reaction’s rate.
Example A17.1
The following data were obtained during a kinetic study of the hydration of p-methoxyphenylacetylene by measuring the relative amounts of reactants and products by NMR.1
time (min)
% p-methyoxyphenylacetylene
67
85.9
161
70.0
241
57.6
381
40.7
479
32.4
545
27.7
604
24
Determine whether this reaction is first-order or second-order in p-methoxyphenylacetylene.
Solution
To determine the reaction’s order we plot ln(%pmethoxyphenylacetylene) versus time for a first-order reaction, and (%p-methoxyphenylacetylene)–1 versus time for a second-order reaction (see Figure A17.1). Because a straight-line for the first-order plot fits the data nicely, we conclude that the reaction is first-order in p-methoxyphenylacetylene. Note that when we plot the data using the equation for a second-order reaction, the data show curvature that does not fit the straight-line model.
Pseudo-Order Reactions and the Method of Initial Rates
Unfortunately, most reactions of importance in analytical chemistry do not follow the simple first-order or second-order rate laws discussed above. We are more likely to encounter the second-order rate law given in equation A17.11 than that in equation A17.10.
$R = k\mathrm{[A][B]}\tag{A17.11}$
Demonstrating that a reaction obeys the rate law in equation A17.11 is complicated by the lack of a simple integrated form of the rate law. Often we can simplify the kinetics by carrying out the analysis under conditions where the concentrations of all species but one are so large that their concentrations remain effectively constant during the reaction. For example, if the concentration of B is selected such that [B] >> [A], then equation A17.11 simplifies to
$R = k´[\ce A] \nonumber$
where the rate constant k´ is equal to k[B]. Under these conditions, the reaction appears to follow first-order kinetics in A and, for this reason we identify the reaction as pseudo-first-order in A. We can verify the reaction order for A using either the integrated rate law or by observing the effect on the reaction’s rate of changing the concentration of A. To find the reaction order for B, we repeat the process under conditions where [A] >> [B].
A variation on the use of pseudo-ordered reactions is the initial rate method. In this approach we run a series of experiments in which we change one at a time the concentration of those species expected to affect the reaction’s rate and measure the resulting initial rate. Comparing the reaction’s initial rate for two experiments in which only the concentration of one species is different allows us to determine the reaction order for that species. The application of this method is outlined in the following example.
Example A17.2
The following data was collected during a kinetic study of the iodation of acetone by measuring the concentration of unreacted I2 in solution.2
experiment number
[C3H6O] (M)
[H3O+] (M)
[I2] (M)
Rate (M s–1)
1
1.33
0.0404
6.65×10–3
1.78×10–6
2
1.33
0.0809
6.65×10–3
3.89×10–6
3
1.33
0.162
6.65×10–3
8.11×10–6
4
1.33
0.323
6.65×10–3
1.66×10–5
5
0.167
0.323
6.65×10–3
1.64×10–6
6
0.333
0.323
6.65×10–3
3.76×10–6
7
0.667
0.323
6.65×10–3
7.55×10–6
8
0.333
0.323
3.32×10–3
3.57×10–6
Solution
The order of the rate law with respect to the three reactants is determined by comparing the rates of two experiments in which there is a change in concentration for only one of the reactants. For example, in experiment 2 the [H3O+] and the rate are approximately twice as large as in experiment 1, indicating that the reaction is first-order in [H3O+]. Working in the same manner, experiments 6 and 7 show that the reaction is also first order with respect to [C3H6O], and experiments 6 and 8 show that the rate of the reaction is independent of the [I2]. Thus, the rate law is
$R = k\ce{[C3H6O][H3O+]} \nonumber$
To determine the value of the rate constant, we substitute the rate, the [C3H6O], and the [H3O+] for each experiment into the rate law and solve for k. Using the data from experiment 1, for example, gives a rate constant of 3.31×10–5 M–1 sec–1. The average rate constant for the eight experiments is 3.49×10–5 M–1 sec–1.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/08%3A_Review_of_Chemical_Kinetics.txt
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Diffusion can be described as the random movement of particles through space, usually due to a concentration gradient. Diffusion is a spontaneous process and is a result of the random thermal motions between two particles. The diffusion coefficient ($D$) can be solved for with Fick’s laws of diffusion, which are broken up into two laws.
Fick’s First Law of Diffusion
Fick's first law of diffusion is given by the following equation:
$J = -D \dfrac{dc}{dx} \label{1}$
where
• $J$ is the flux and is defined by the number or particles that are moving past a given region divided by the area of that region multiplied by the time interval.
• The units of $J$ are mol m-2 s-1.
• The letter $D$ represents the diffusion coefficient with units m2 s and
• $c$ is the concentration of the gradient with units molecules m-3.
Equation \ref{1} indicates that if the flux and the change in the concentration over time are known, then the diffusion coefficient can be calculated. The negative sign indicates that the concentration gradient is negative. The first law can only be applied to systems in which the conditions remain the same— in other words, if the flux coming into the system equals the flux going out. Fick’s second law is more applicable to physical science and other systems that are changing. This second law is applied to systems in which the condition are not steady, or the solution in not equal throughout.
Fick’s Second Law of Diffusion
$\dfrac{dc}{dt} = D \dfrac{d^2c}{dx^2} \label{2}$
Where $\frac{dc}{dt}$ represents the rate of change of concentration in a certain area and $\frac{d^2c}{dx^2}$ represents the changes that the change in concentration can take; this would not be a smooth curve. This term accounts for a varying concentration in the system. If the concentration does not vary, then
$\dfrac{d^2c}{dx^2} = 0 \label{3}$
Diffusion can be thought of as a series of random steps that the particle takes as it moves from where it started. These steps can either lead the particle away from where it started, or lead back to where it started. This means that the overall path that a particle takes can look like line overlapping itself multiple times. A third way to calculate the diffusion concentration is through the Einstein-Smoluchowski equation:
$D = \dfrac{\lambda^2}{2\tau} \label{4}$
where $λ$ is the length that each step takes, and $τ$ is the time that each step takes. In this particular model, each step is the same distance.
The diffusion coefficient is useful because it can tell you something about the system. For example, different substances have different diffusion coefficients, so knowing this can give you an idea of the substance. Ions at room temperature usually have a diffusion coefficient of 0.6×10-9 to 2×10-9 m2/s, and biological molecules fall in the range 10-11 to 10-10 m2/s. The diffusion coefficient changes as the properties of the system change. For example, at higher temperatures, the diffusion coefficient is greater because the molecules have more thermal motion. The diffusion coefficient is also related to the viscosity of the solution. The greater the diffusion coefficient, the lower the viscosity. Because the rate of diffusion depends on the temperature of the system, the Arrhenius equation can be applied. Applying this equation gives:
$D = [D]_{o}e^{-Ea/RT} \label{5}$
The dependence of the diffusion coefficient on the viscosity can be modeled by the Stokes-Einstein relation:
$D = \dfrac{kT}{6\pi\eta{a}} \label{6}$
where a is the radius of the molecule and η is the coefficient of viscosity and is defined by:
$\eta = \eta_o{e^{Ea/RT}} \label{7}$
This equation demonstrates the dependence of viscosity on temperature.
Contributors and Attributions
• Laura Dickson (UC Davis)
10: Using logarithms - Log vs. Ln
A common question exists regarding the use of logarithm base 10 ($\log$ or $\log_{10}$) vs. logarithm base $e$ ($\ln$). The logarithm base $e$ is called the natural logarithm since it arises from the integral:
$\ln (a) = \int_1^a \dfrac{dx}{x}\nonumber$
Of course, one can convert from $\ln$ to $\log$ with a constant multiplier.
$\ln (10^{\log a}) = \log (a) \ln(10) \approx 2.3025 \log (a)\nonumber$
but $10^{\log a}=a$ so
$\ln a \approx 2.4025 \log a\nonumber$
The analysis of the reaction order and rate constant using the method of initial rates is performed using the $\log_{10}$ function. This could have been done using the $\ln$ function just as well. The initial rate is given by
$r_o=k'[A]_0^a\nonumber$
The analysis can proceed by taking the logarithm base 10 of each side of the equation
$\log r_o = \log k' + a\log [A]_0\nonumber$
or the $\ln$ of each side of the equation
$\ln r_o = \ln k' + a\ln [A]_0\nonumber$
as long as one is consistent.
Once can think of the $\log$ or the $\ln$ as a way to 'linearize data' that has some kind of power law dependence. The only difference between these two functions is a scaling factor ($\ln 10 \approx 2.3025$) in the slope. Obviously, if you multiply both sides of the equation by the same number the relative values of the constants remains the same on both sides.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/09%3A_Diffusion.txt
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Nuclear chemistry has many applications in agriculture, medicine, industry and research. They greatly improve the day to day quality of our lives.
• Nuclear Reactors
A nuclear reactor is a device in which nuclear reactions are generated, and the chain reaction is controlled to release large amount of steady heat, thereby producing energy.
• Nuclear Reactors: Chernobyl
The Chernobyl disaster was a nuclear accident that occurred at the Chernobyl Nuclear Power Plant in on April 26, 1986. It is considered the worst nuclear power plant disaster in history. A nuclear meltdown in one of the reactors caused a fire that sent a plume of radioactive fallout that eventually spread all over Europe.
• Nuclear Reactors: Nuclear Waste
Nuclear waste is radioactive waste, meaning that it spontaneously emits radiation. It usually originates from the by-products of nuclear reactions in applications such as medicine and research. Radioactive waste degrades with time, releasing alpha, beta, and gamma radiation that pose many health risks to the environment and most organisms, including humans. Due to the harmful nature of nuclear waste, there is strict government regulation on the safe disposal of it.
• Nuclear Weapons
A nuclear weapon is commonly defined as a device, which uses a nuclear reaction for destructive means.
• Radiation in Biology and Medicine
What comes to mind when you think of radiation? If you ask the average American, radiation would conjure images of deformed humans and malignant diseases, namely cancer. Despite radiation's notorious history, radiation has revolutionized the medical field, enhancing the ability of medical professionals to treat and diagnose diseases.
• Radiation in Biology and Medicine: Positron Emission Tomography
Positron emission tomography (PET) is one of the beneficial real-life uses of nuclear chemistry. Simply, it is a handy instrument that physicians use to take images of an individual's body to determine if a person is at risk for a certain disease or carries one. This module will focus on the procedures of a patient receiving a PET scan, the nuclear reactions associated with the images that are produced, commonly used tracer molecules, and useful applications of PET scans in clinical diagnosis.
• Radiocarbon Dating
Radiocarbon Dating is the process of determining the age of a sample by examining the amount of C-14 remaining against the known half-life, 5,730 years. The reason this process works is because when organisms are alive they are constantly replenishing their C-14 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 C-14 isotope begins to decay.
• Radiocarbon Dating: The Shroud of Turin
The Shroud of Turin is a linen wrapping cloth that appears to possess the image of Jesus Christ. Some people believe this to be the cloth that he was wrapped in following his crucifixion. In 1988, several groups of scientists were allowed samples of the shroud to subject these samples to 14C dating. The carbon-14 to carbon-12 ratio was found to be 92% of that in living organisms.
• Radiology
Applications of Nuclear Chemistry
Applications of Nuclear Technology
There are many applications of nuclear technology. These are discussed in separate topics as follows:
• X-ray imaging in Varian
• Radioactive dating is an application of radioactive decay kinetics to determine the age of certain things since they became inactive, dead or being isolated.
• Apollo 15 Gamma-ray Spectrometer
The X-ray Century
Teaching radiology on the Internet
The following will be separated later in other webpages.
Neutron Activation Analysis
Neutron Activation Analysis
Nuclear Medicine
The practice of nuclear medicine involves injecting a liquid radioactive pharmaceutical (radiopharmaceutical) into a patient. Signals from the radiopharmaceutical are then detected and processed into a useful image by sophisticated instruments in order to diagnose or treat a disease state. A radiopharmaceutical is formed by combining radioactive atoms radioisotopes) with chemical or biological material formulated to collect temporarily in the part of the body to be imaged...
A nuclear pharmaceutical is a physiologically active carrier to which a radioisotope is attached. It is possible to manufacture chemical or biological carriers which migrate to a particular part of the human body. Calcium, for example, is a bone 'seeker', and iodine concentrates in the thyroid gland. The radioisotope attached to these compounds emits radiation so that the relevant organ and its functioning can be 'observed'. Radiation is easy to detect. Even radiation which is many times weaker than natural background radiation can be measured. The location in the patient's body which emits the radiation can thus be very accurately pinpointed.
So we see that radioactive preparations can be very useful for diagnostic examination if we choose them in such a way that they emit sufficient radiation to be easily detectable in the body, but only for a long enough time to enable completion of the examination. Nuclear pharmaceuticals for diagnosis must therefore have a rather short half-life, preferably no longer than a few hours. Useful radioisotopes for diagnostic purposes are technetium-99, gallium-67, indium-111, iodine-123, iodine-131, thallium-201, krypton-81m.
Therapeutics
Tissue dies rather quickly after receiving a large dose of radiation. This aspect of radiation can be utilized for treating tumours. The goal of therapy in nuclear medicine is to use radiation to destroy diseased or cancerous tissue while sparing adjacent healthy tissue. For certain types of cancer, this is achieved by using an external radioactive beam directed at the cancerous tumor. It is also possible to insert a small radioactive source through body openings, via the bloodstream or by means of surgery into a tumour and leave it there for a period lasting from days to weeks until sufficient dose has been given off. With the exception of radioactive iodine, which is used to treat cancer of the thyroid (see picture), few radioactive therapeutic agents are injected or swallowed. A much higher dose of radioactivity is administered in a therapeutic situation than in a diagnostic one; thus the therapeutic radiopharmaceutical must have a high affinity for the diseased tissue relative to the healthy tissue. Nuclear pharmaceuticals which are used for therapy must have a rather longer half-life.
Useful radioisotopes for therapeutical purposes are iodine-131 (in NaI or in metaiodobenzylguanidine, MIBG), phosphorus-32, iridium-192, gold-198. Radioactive sources which are placed in the body near the tumour for local irradiation of the tumour are also called nuclear pharmaceuticals. Iridium-192 sources are normally used for this purpose.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Applications_of_Nuclear_Chemistry/Applications.txt
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The Chernobyl disaster was a nuclear accident that occurred at the Chernobyl Nuclear Power Plant in on April 26, 1986. It is considered the worst nuclear power plant disaster in history. A nuclear meltdown in one of the reactors caused a fire that sent a plume of radioactive fallout that eventually spread all over Europe.
Introduction
On April 26, 1986, a test was scheduled at the Chernobyl Nuclear Power Plant to test a method of keeping the reactors properly cooled in the event of a power grid failure. If the test had gone as planned, the risk to the plant was very small. When things did go wrong, though, the potential for disaster was miscalculated and the test was continued even as serious problems arose. Meltdown occurred at 1:23 AM, starting a fire that dispersed large quantities of radioactive materials into the atmosphere. The amount of radioactive material released was 400 times more than the amount the atomic bombing of Hiroshima released. The fallout would be detected in almost all parts of Europe.
Before the accident
Nuclear reactors require active cooling in order to remove the heat generated by radioactive decay. Even when not generating power, reactors still generate some heat, which must be removed in order to prevent damage to the reactor core. Cooling is usually accomplished through fluid flow, water in Chernobyl's case.
The problem at the Chernobyl plant was that following an emergency shutdown of all power, diesel generators were needed to run the cooling pumps. These generators took about a minute to attain full speed, which was deemed an unacceptably long time for the reactor to be without cooling. It was suggested that the rotational momentum of the winding down steam turbine be used to power the pumps in the time between shutdown and the generators being ready. A test was devised to test this method in 1982, but the turbine did not prove to be successful in providing the required voltage as it spooled down. Two more tests would be conducted in the following years, but would also be unsuccessful. The fourth test was scheduled to be run on April 25, 1986.
The experiment was devised in such a way that if it had gone as planned, the disruption and danger to the plant would be very minimal. First, the reactors would be brought down to low power, between 700 and 800 megawatts. Then the steam turbine would be run up to full speed and then turned off. The power generated by the winding down generators would then be measured to determine if it was sufficient to power the cooling pumps in the time before the diesel generators got up to full speed.
By 1986, the plant had been running for two years without the implementation of a method to keep the cooling pumps running continuously following an emergency shutdown. This was an important safety measure that the plant was lacking, which presumably gave the plant managers a considerable amount of urgency in completing another test.
The experiment
Preparing for the experiment
The experiment was scheduled to run during the day shift of 1985, while the night shift would only have to maintain cooling of the radioactive decay in the shut-down plant. However, another power generator nearby unexpectedly shut down, necessitating the need for the Chernobyl plant to delay the test and continue producing power. The experiment would be resumed at 11:04 PM, by which time the day shift had departed and the evening shift was about to leave. This meant that the experiment would be conducted in the middle of two shifts, leaving very little time for the night shift employees to be briefed about the experiment and told what to do.
The power reduction of reactor 4 to 700 MW was accomplished at 00:05 AM on the 26th of April. However, the natural production of a neutrino absorber, Xenon-135, led to a further decrease in power. When the power dropped to about 500 MW, the night shift operator committed an error and inserted the reactor control rods too far. This caused the reactor to go into a near-shutdown state, dropping power output to around 30 MW.
Since this was too low for the test, it was decided to restore power by extracting the control rods. Power would eventually rise and stabilize at around 200 MW.
The operation of the reactor at such a low power level would lead to unstable temperature and flow. Numerous alarms and warnings were recorded regarding emergency measures taken to keep the reactor stable. In the time between 0:35 and 0:45 AM, alarm signals regarding thermal-hydraulic parameters were ignored in order to preserve the reactor's power level.
The test continued, and at 1:05 AM extra water pumps were activated in order to increase the water flow. The increased coolant flow rate led to an increase of the coolant temperature in the core, reducing the safety margin. The extra water flow also led to a decrease in the core's temperature and increased the neutron absorption rate, decreasing the reactor's power output. Operators removed the manual control rods in order to maintain power.
All these actions led to the reactor being in an unstable state that was clearly outside safe operation protocol. Almost all the control rods had been removed, which reduced the effectiveness of inserting safety rods in an emergency shutdown. The water was very close to boiling, which meant that any power increase would cause it to boil. If it started boiling, it would be less effective at absorbing neutrons, further increasing the reactor's power output.
Conducting the experiment
The experiment was started at 1:23:04 AM. The steam to the turbines was shut off, causing the turbines to start spooling down. Four of the eight cooling pumps were also shut down. The diesel generator was started and began powering the cooling pumps after at 1:23:43. Between this time, the four pumps were powered by the slowing steam turbines. As the turbines slowed down, their power output decreased, slowing the cooling pumps. This lead to increased formation of steam voids in the core, reducing the ability of the cooling water to absorb neutrons. This increased the power output of the reactor, which caused more water to boil into steam, further increasing the reactor's power. However, during this time the automatic control system was successful in limiting power increase through the insertion of control rods.
At 1:23:40, a button was pressed that initiated the emergency shutdown of the reactor and the insertion of all control rods. It is believed that this was done as a routine method to shut down the reactor to conclude the experiment and not as an emergency measure.
The process of inserting the control rods was initiated, but it took about 20 seconds for the rods to be completely inserted. A flawed design in the graphite-tip control rod meant that coolant was displaced before the neutron absorbing material could be fully inserted and slow down the reaction. This meant that the process of inserting the control rods actually increased the reaction rate in the lower half of the core.
A massive power spike occurred, causing the core to overheat. Some of the fuel rods fractured, causing the control rods to become stuck before they were fully inserted. Within three seconds the core's power output rose to above 500 MW. According to simulation, it is estimated that power output then rose to 30 GW, ten times the normal power output. This was caused by the rising power output causing massive steam buildup, which destroyed fuel elements and ruptured their channels.
It is not possible to know precisely what sequence of events led to the destruction of the reactor. It is believed that the steam buildup entered the reactor's inner structure and lifted the 2000 ton upper plate. This steam explosion further ruptured fuel channels, resulting in more coolant turning into steam and leaving the reactor core. This loss of coolant further increased the reactor's power. A nuclear excursion (an increasing nuclear chain reaction) caused a second, even more powerful explosion.
The explosion destroyed the core and scattered its contents in the surrounding area, igniting the red-hot graphite blocks. Against safety regulations, a flammable material, bitumen, had been used in roof of the reactor. When this was ignited and scattered into the surrounding area, it started several fires on reactor 3. Those working there were not aware of the damage that had been done and continued running the reactor until it was shut down at 5:00 AM.
Crisis Management
Radiation Levels
In the worst-hit parts of the reactor building, radiation levels were high enough to cause fatal doses in a matter of minutes. However, all dosimeters available to the workers did not have the ability to read radiation levels so high and thus read "off scale." Thus, the crew did not know exactly how much radiation they were being exposed to. It was assumed that radiation levels were much lower than they actually were, leading to the reactor crew chief to believe that the reactor was still intact. He and his crew would try to pump water into the reactor for several hours, causing most of them to receive fatal doses of radiation.
Containment
Fire crew were called in to protect the remaining buildings from catching fire and to extinguish the still burning reactor 4. While some firefighters were not aware of the harmful doses of radiation they were receiving and had assumed it to be a simple electrical fire, others knew that they would probably receive fatal doses of radiation. However, their heroic efforts were necessary in order to try to contain the large amounts of radiation being released into the atmosphere. The fires in the surrounding buildings were extinguished by 5:00 AM, but it would take firefighters until May 10 before they could fully extinguish reactor 4.
In order to prevent a steam explosion from occurring, volunteers were needed to swim through radioactive water and drain a pool of water under the reactor core. While they were successful, they would later succumb to the high doses of radiation that they had received.
The worst of the radioactive debris was shoveled back into the reactor by crew wearing heavy protective gear.
In total, 600,000 people worked in the cleanup, about 250,000 of which received their lifetimes' limit of radiation. It is estimated that over 10,000 eventually died from the radiation.
By December, a concrete sarcophagus had been completed that sealed off the reactor. This was never meant to be a completely permanent solution, however, and is now in danger of collapsing. A collapse could cause a large amount of radioactive material to once again be released and spread around the world. This is why it is necessary that a new structure be constructed to contain reactor 4.
Evacuation of Pripyat
Pripyat, a city nearby the power plant, was not immediately evacuated. At first, the government denied that the reactor had exploded and insisted that it was only a small accident. By April 27, though, investigators were forced to acknowledge that the reactor had exploded and ordered Pripyat to be immediately evacuated.
Effects
400 times more radiation was released by the disaster than had been by the atomic bombing of Hiroshima. The radiation would later be detected in almost all parts of Europe. Over one million people could have been adversely affected by the radiation. The radiation would cause numerous problems, including Down's Syndrome, chromosomal aberrations, mutations, leukemia, thyroid cancer, and birth defects.
The radiation would affect all parts of the environment surrounding the plant, killing plants and animals and infecting the soil and groundwater. Life has returned to the area and seems to be flourishing, possibly due to the lack of human intrusion. Remarkably, numerous species have been reported to have adapted to their environment and have developed increased tolerance of radiation, making it possible for them to live with the radiation that is still prevalent in the soil and plants around the plant. It has even been reported that radiotrophic fungi have been growing on the walls of reactor 4.
Today, radiation levels are still higher than normal in the areas surrounding the plant, but have dropped considerably from the levels that they were at twenty years ago. It is now considered safe to visit the areas immediately surrounding the plant for short periods of time. However, it is estimated that it will take 20,000 years for reactor 4's core to be completely safe.
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Nuclear waste is radioactive waste, meaning that it spontaneously emits radiation. It usually originates from the by-products of nuclear reactions in applications such as medicine and research. Radioactive waste degrades with time, releasing alpha, beta, and gamma radiation that pose many health risks to the environment and most organisms, including humans. Due to the harmful nature of nuclear waste, there is strict government regulation on the safe disposal of it. There are several types of waste, and as such, many different ways of discarding it.
Introduction
The use of nuclear energy provides cheaper and more powerful energy through fission reactions. Fission is the process by which a heavy nucleus splits into two smaller, separate nuclei. For example, the splitting of a large atom such as Uranium creates neutrons and a substantial amount of energy. This energy is created due to the difference in mass between the two nuclei products and the large nucleus product. However, the remains of the nuclear reactor becomes nuclear waste, which is extremely lethal due to its radioactivity. As a response, researchers have found several ways to shield and isolate radioactive waste till it degrades completely.
Importance of Nuclear Waste
The controversy behind nuclear technology is due to the radioactive waste it creates. Some elements used in nuclear reactors have extremely long half-lives and must be shielded from humans and the environment for thousands of years. For example, plutonium-239, an isotope used in the production of nuclear weapons, has a half-life of 24,200 years while uranium-235 Hiroshima has a half-life of 700 million years. These elements emit large quantities of radioactivity that is extremely dangerous. Too much exposure can be followed by Acute Radiation Syndrome (ARS), which includes skin burns, nausea, vomiting, and eventually death within days if the exposure and dosage of radiation is high.
Sources of Nuclear Waste
There are currently a number of nuclear productions that result in radioactive waste.
1. Nuclear Weapons: From the weapons to the tools and machinery used in its production, proportional amounts of radioactivity can be found in all of them. After their use, these contaminated items must be disposed of while the radioactivity slowly degrades.
2. Medicine and Research Applications: X-rays and other disease detecting technology in the medical field also consist of radiation albeit in a less harmful amount. For example, technetium-99m is an isomer that can be consumed to allow doctors to take images of the body’s process. However, even the syringes contribute to the problem of nuclear waste
3. Nuclear Power: Most radioactive waste comes from the nuclear power plants situated around the world. There is a 20-30 ton waste that comes from each nuclear reactor every month it is in use.
4. Agriculture: Nuclear power is also used in eliminating bacteria through the disruption of their genetic structure. This insures that they can no longer proliferate and grow in the food.
Classification of Nuclear Waste
There are four different types of radioactive waste that result from nuclear power:
1. Low Level Waste (LLW) – waste that is usually results from medicine or other industrial uses such as tools, rags, medical tubes, protective clothing, and others. There are three facilities in the U.S. that handle low-level waste by land disposal.
2. Intermediate Level Waste (ILW) – waste from processing-plants and reactors that require some shielding. The waste is usually mixed with cement and buried.
3. High Level Waste (HLW) – waste that is produced from the nuclear reactor cores that are highly radioactive. Because of the level of contamination, the waste must be discarded by some sort of geological repository. It is not uncommon for spent nuclear fuel to be stored underwater.
4. Uranium Mill Tailings – this category of waste comes from remains after extracting uranium from its natural ore. Large amounts are currently left out in the open of abandoned mining sites. However, if it remains uncovered and not disposed of correctly, the waste can mix with the sand and travel to water sources, polluting the environment to a great extent.
Management of Nuclear Waste
1. Deep Geological Repository: This method is currently still being tested and facilities are being built. The idea is to seal the radioactive waste into special casks and to deposit them hundreds meters deep into a geographically stable area to allow it to decay. There is currently only one facility in the U.S. but a few more located around the globe.
2. Transmutation: at the moment researchers are searching for a method to transmute the dangerous material into something less harmful, making it easier to dispose.
3. Re-use: Currently, there is a Canadian patent to filter the waste and reuse whatever material that has not been altered. However, this practice requires a substantial amount of energy and is still in the very early stages of development.
4. Space disposal: This has the best alternative to geological repositories because the waste can no longer harm the environment on earth. Of course, there are significant consequences of shooting radioactive waste into space such as damaging space shuttles and pollution. At the same time, space disposal would have to be agreed upon internationally.
Problems
1. Many citizens of Nevada are against geological repositories in the Yucca Mountains, which is situated far from Vegas but still in the state. Why do you think they protest nuclear waste duping in that location despite the distance from the human environment?
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A nuclear reactor is a device in which nuclear reactions are generated, and the chain reaction is controlled to release large amount of steady heat, thereby producing energy.
Introduction
Nuclear fission is the process in which the nucleus of an atom is split, forming nuclei of lighter atoms and neutrons. The mass of these products is less than the original mass. According to Einstein's equation \(E=mc^2\), the small amount of missing mass is converted into a large amount of energy.
A chain reaction occurs when the neutrons released in fission collide with at least one other nuclei, causing the fission of another nuclei. The process then repeats. In today's nuclear reactors, Uranium-235 is commonly used. During each U235/92 fission, 2.5 neutrons are released on average. Note here; Uranium 235 is used because it has a fairly large nucleus which facilitates the process of fission. An explosion could only occur if the reaction becomes uncontrolled. When one mass of U-235 exceeds the mass of U-235 that is large enough to hold down a chain reaction, also known as critical mass, an explosion occurs. A great example of this phenomenon would be a nuclear bomb. In uncontrolled reactions, neutrons escape too quickly to maintain a chain reaction. This rapid release of nuclear energy causes an explosion. However, in a nuclear reactor, energy is being produced at a controlled, constant rate; a nuclear explosion is unlikely to occur.
The opposite of a nuclear explosion, nuclear reactors are the controlled release of fission energy. They serve the purpose of converting “nuclear energy” to heat. To produce energy, a nuclear reactor contains several major components: fuel elements (or rods), control rods, and coolant/moderator, besides the vessel itself containing everything. The fuel elements contains the fissile material, typically uranium or plutonium, which is used as the fuel to undergo fission and provide the nuclear energy. The fissile material is encased in a solid cladding, made of Zircalloy (alloy of zirconium, having low capacity to absorb neutrons), to contain both the fuel and the resulting fission products and keep then from escaping into the moderator, coolant, or anywhere outside the cladding. The moderator and coolant flows between the fuel elements (or rods) moderating the neutrons and carrying away the heat. The region inside the nuclear reactor where the fuel elements undergo fission to generate heat is called the nuclear reactor core. The control rods, usually made of cadmium metal, absorb neutrons in order to control the rate of fission. By raising or lowering the control rods in the reactor, the concentration of neutrons, called the neutron flux, in the core increases or decreases respectively. The other component is the coolant. Since the process of fission produces large amounts of heat, the coolant is used to carry away the heat. The heat carried away causes incoming cooler water to turn into steam. This steam spins a turbine, which then powers an electric generator. Generally speaking, water is used, however; helium gas and liquid sodium can be used as substitutes. A pressurized water reactor (PWR) is a common design of a nuclear reactor. In a PWR, water functions as a coolant and as a moderator. A moderator slows down the neutrons, because slower moving neutrons are better at causing fission to occur. A moderator is usually water; however, graphite and heavy water can also be used. A boiling water reactor (BWR) is another common design of current nuclear reactor plants in commercial use. In a BWR, water also functions as both a coolant and moderator. Another type of reactor used in civilian power plants in the former Soviet Union called an RBMK reactor used graphite (carbon) as a moderator, but is considered insufficiently safe.
Most of the power from nuclear fission reactors comes directly from the fission, which can be rapidly stopped by a shutdown of the reactor. However, about 7% comes from heat of decay of highly radioactive fission products, which cannot be stopped by shutting down the reactor, and must continue to be removed from the reactor to prevent overheating and damage to the reactor core. Therefore, coolant pumps must continue to be run for many hours after the reactor is shutdown to remove the decay heat, which over the course of hours eventually decreases. If the coolant pumps do not work, emergency cooling methods must be used to remove decay heat to prevent damage to the core, possible meltdown or release of highly radioactive fission products to where they should not be.
Nuclear Fusion
Nuclear Fusion is the process by which two elements collide to form a new element, releasing a tremendous amount of energy much greater than that of a fission reaction. Similar to nuclear fission, the mass of the resulting element does not exactly match the combined masses of the two smaller elements, but is converted to energy. Stars throughout the universe, including our Sun, release energy through the fusion of two hydrogen atoms into a single helium atom (Figure 3). While electrostatic forces of repulsion would normally drive the positively-charged regions of two hydrogen nuclei away from each other and prevent such a reaction, the extreme heat (15,000,000 °C in the Sun) and density of a star overcomes these forces and allows fusion to occur. With so much potential as a source of energy, the prospect of a fusion reactor on Earth has become a highly sought after technological advancement, even though the challenges of creating such a reactor are immense.
The most promising reaction being attempted by scientists today to produce fusion power is the collision of two hydrogen isotopes, deuterium (2H) and tritium (3H), to produce a 4He atom, a neutron, and a large quantity of energy. Just as in a star, however, the nuclei of the hydrogen isotopes repel each other and must be brought together for fusion at extremely high thermal energies reaching 150,000,000 °C (ten times greater than that required in the Sun). At temperatures this high, the reactant gases transform into a plasma- a hot, fully ionized gas consisting of atomic nuclei and electrons. A problem of central importance to the as of yet theoretical fusion reactor is containing that plasma so that it does not lose thermal energy by touching surrounding materials. A collaborative effort funded by multiple nations known as the International Thermonuclear Experimental Reactor (ITER) aims to solve this problem by confining the plasma in a magnetic field created by powerful superconducting magnets. Such a design is known as a tokamak reactor (See Figure 4).
While the feasibility of a controlled fusion reaction occurring on Earth has yet to be adequately verified, the potential benefits of fusion as opposed to fission may be immense. Deuterium may be extracted from water and lithium, the tritium source for the fusion reaction, is estimated to exist on earth in quantities that will last for one million years. Additionally, there is far less nuclear waste that decays much faster compared to that produced by fission.
Nuclear Safety
There have been three major accidents involving full-scale civilian nuclear power plants.
• The first occurred in 1979 at Three Mile Island Unit 2 in Pennsylvania. Due mechanical failure, the main water pumps stopped running, leading to a partial meltdown of the fuel rods. Excessive heat caused a fracture in one of the reactors, allowing a small amount of radioactive steam into the atmosphere. Fortunately, no one was killed or even injured. This incident also lead to heightened regulation and safety precautions of nuclear reactors in the United States.
• On April 26, 1986, the worst accident in nuclear history occurred in Chernobyl, Ukraine. During a routine test, an uncontrollable power surge burned the control rods, and massive amounts of radioactive smoke were released. 237 people suffered from acute radiation sickness, and 31 died within the first three months of the accident. Other effects of the radiation included an increase in down's syndrome, chromosomal aberrations, neural tube defects, and thyroid cancer. Perhaps the most important effect was psychological as the accident caused severe anxiety for the survivors and a general lack of trust in the government.
• Due to a severe earthquake and tsunami in Japan in March 11, 2011, several BWR (Boiling Water Reactor) nuclear reactors at the Fukushima power plant lost electrical power for cooling, underwent explosions, and suffered reactor core damage from post-shutdown decay heat coming from highly radioactive fission products. Workers eventually pumped seawater into the reactors to cool them down and limit any further damage.
Problems
1. What is the function of control rods in a nuclear reactor?
2. True or False? The accident at Three Mile Island led to the radiation poisoning of possibly hundreds of thousands of people.
3. How does nuclear fission lead to a chain reaction?
4. What two functions can water serve in a nuclear reactor?
5. What is critical mass? What will occur if the mass of a reaction surpasses its critical mass?
Answers
1. In a fission reactor, control rods absorb neutrons to control the rate of a reaction. Lowering the rods into the reactor decreases the rate of fission and removing them increases the rate.
2. False. The accident at Three Mile Island was only minor and no one was killed or injured.
3. When a neutron strikes a fissile material breaking it into smaller fragments, more neutrons are released (2.5 on average). These neutrons then collide with other fissile atoms producing even more neutrons, which continue the chain reaction either in a stable manner (reactor) or violently (atomic bomb).
4. In a fission reactor, water serves as a moderator (slows down highly energetic neutrons to the appropriate thermal energy for a reaction) and produces steam by coming into contact with the hot water near the reactor and transferring heat. This steam can then power turbines to generate electricity.
5. The critical mass of a fission reactor is the mass of fissile material required to maintain a chain reaction. If the mass of a reaction surpasses its critical mass, the result is an uncontrolled chain reaction that culminates in a large explosion (this is how an atomic bomb works).
External Links
For more information on the subject of nuclear reactor, the following links may be helpful:
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A nuclear weapon is commonly defined as a device, which uses a nuclear reaction for destructive means. The first nuclear weapon was successfully detonated on July 16, 1945. The nuclear weapon, code named “Trinity”, yielded an explosion which was equivalent to 20 kilotons of Trinitrotoluene (TNT). This reaction unexpectedly had a shock wave which could be felt 100 miles away. Compared to chemical reactions, the amount of energy that can be released from nuclear reactions can be up to a million times greater. Before we can fully understand the chemical complexity and appreciate the engineering elegance of a nuclear weapon, however, it is important to grasp basic nuclear chemistry concepts.
Nuclear Fission
This type of nuclear reaction is caused by nuclear decay of an unstable atom that has been hit by a neutron. As a result of the instability of the atom, the nucleus splits into two fission fragments also yielding free neutrons and exorbitant amounts of energy (both in the form of electromagnetic radiation and kinetic energy). A neutron carries no electric charge but the nucleus of an atom does - a positive charge. Like with magnets, like charges repel each other, so particles that carry positive electric charges like alpha particles are repelled by the nucleus of an atom and thus do not stick to the nucleus. However, a neutrally charged neutron can combine with the nucleus of an atom which then causes the nucleus to become unstable and split into 2 nuclei. According to Einstein's formula, $E=mc^2$, energy is released from this reaction along with other neutrons that have the same effect on nearby nuclei and a chain reaction occurs which yields an incredible amount of energy.
Nuclear Fusion
Fusion is almost completely complimentary to fission. It is the process in which a nuclear reaction where two nuclei are joined together to form a heavier nucleus. The reaction also yields free neutrons and exorbitant amounts of energy from binding energy. Our sun is a "nuclear furnace" in the sense that it is a place where nuclear fusion occurs. Although a nuclear fusion reactor would undoubtedly be better for our environment, such reactions are not yet possible because no known material can withstand the incredible high temperatures needed for such reactions to occur.
Example of Fusion Reaction
$\ce{^2D+^3T → ^4_2He +^1_0N} + 17.6\, MeV$
• Fissile / Fissionable An atom is fissionable if it is capable of undergoing fission. If an atom is Fissile it is not only able to undergo fission, but it is also capable of sustaining a nuclear chain reaction.
• Nuclear Chain Reaction: As stated earlier in the text, a nuclear fission reaction yields free neutrons. In a nuclear chain reaction, the free neutrons from a nuclear fission reaction bombard nearby Fissile isotopes resulting in multiple fission reactions. These reactions result in colossal amounts of kinetic energy and gamma radiation.
• Critical Mass: The amount of fissile isotope required to successfully assist a nuclear chain reaction. If the fissile material is at a subcritical mass it cannot sustain a nuclear chain reaction. On the other hand, if the fissile material is at a supercritical mass, it will undergo a chain reaction at a faster rate.
Nuclear Weaponry
A nuclear weapon can either undergo a nuclear fission reaction (atomic bomb) or a nuclear fusion reaction (H bomb or thermonuclear bomb). The first nuclear weapons built underwent pure nuclear fission. Uranium-235 and Plutonium-239 were the most common fissile isotopes used. (Uranium-235 is less than 1% naturally abundant. It requires complex methods of enrichment to be a fissile isotope). There are 2 basic nuclear fission weapon designs:
1. Gun assembly- An uranium-235 bullet is fired through a barrel at a fissile Uranium-235 target. The collision of the two isotopes initiates a chain reaction. The gun assembly design only proves practical with the uranium-235 isotope.
2. Implosion- A critical mass of a fissile material (U-235 or Pu-239) is surrounded by highly explosive material. When detonated, the high explosives compress the fissile material causing it to assume a state of supercritical mass. The supercritical mass instantaneously begins to undergo a chain reaction.
Weapons Utilizing Fusion Reactions
The first nuclear fusion weapons (also known as thermonuclear weapons) were designed to initiate a fission-based chain reaction. The fusion reaction between tritium and deuterium would result in the free neutrons necessary to bombard a fissile isotope and start a nuclear chain reaction. The first thermonuclear weapon was detonated in November of 1952. Similar to the nuclear implosion design, thermonuclear weapons use the heat and radiation from a fission reaction to make the fissile material assume a state of supercritical mass. The supercritical mass then instantaneously undergoes a fusion chain reaction yielding exponentially more energy than a fission chain reaction.
Exercise
1. Given the following nuclear fusion reaction, what particle corresponds to $\ce{^{?}_{?}X}$?
$\ce{^2_1H + ^2_1H \rightarrow ^{?}_{?}X + \rm{energy}}$
1. $\ce{^4_2He}$
2. $\ce{^{175}_{87}Fr}$
3. $\ce{^{14}_6C}$
4. $\ce{^0_{-1}e}$
5. none of the above
2. Which of the following best describes a Nuclear Fusion Reaction?
1. A nuclear reaction in which 2 small nuclei combine to form one large nucleus.
2. A reaction in which a large nucleus splits into two smaller and fast moving particles.
3. The combination of a nucleus and an electron.
4. The combination of a nucleus and a positron.
5. The combination of a nucleus and an alpha particle?
3. Which of the following facts describes a Nuclear Fission Reaction?
1. A nuclear reaction in which 2 small nuclei combine to form one large nucleus.
4. The following nuclear reaction is carried out with the corresponding masses of the reactants:
31 H + 11 H yields 42 He + Energy
7.556g 2.143g Xg
If 4.56 X1011 Joules of energy was released, what mass in grams of 42 He is produced?
1. A reaction in which a large nucleus splits into two smaller and fast moving particles.
2. The combination of a nucleus and an electron.
3. The combination of a nucleus and a positron.
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Nuclear chemistry is the study of changes associated with the nuclei of atoms. Positron emission tomography (PET) is one of the beneficial real-life uses of nuclear chemistry. Simply, it is a handy instrument that physicians use to take images of an individual's body to determine if a person is at risk for a certain disease or carries one. This module will focus on the procedures of a patient receiving a PET scan, the nuclear reactions associated with the images that are produced, commonly used tracer molecules, and useful applications of PET scans in clinical diagnosis.
History
The first PET camera was completed in January 1974 by Michael Phelps's and the first whole-body system for human and animal studies was finished in December of that year. The first commercial system, also designed by Phelps, was delivered to UCLA in December 1976. The research that went into creating the first PET scanner dates several decades back to the early 1950s. It was suggested by William Sweet, Chief at the Massachusetts General Hospital, that the radiation from positron emission has the potential to increase the resolution of nuclear images taken in detecting any brain damages. Soon after this proposal, many patients under the suspicion of brain tumors were imaged with a simple positron scanner and the image results were much clearer, as was predicted by Sweet. The simple positron imaging device uses two-dimensional arrays and only a few radiation sensors to trace the radioactive substances, so the images were very low quality.
In 1952, the first clinical positron scanner device was invented. This device was essentially the same as the first simple positron scanner from 1950 but with a few upgrades. It not only produced coincidence scans, but unbalance scans as well. The unbalance scan was a huge refinement because it produced a lower resolution image that allowed for easier determination of a brain tumor to the right or left of the midline of a patient's brain.
Then in 1962, the first multiple detector positron imaging device was invented. A few unique features on this new scanner include higher sensitivity as well as being able to obtain another three-dimensional image by focusing on different planes of the two detector arrays. This allowed for the imaging of lesions in two dimensions while the exact position of the tumor growth can be determined in three dimension. Refinement of the PET scanner focuses on increasing the sensitivity and adding features to find more ways to have the scanners produce images in higher definition.
In 1970, the PET scanner was formally introduced for use in the medical field. A few years following the introduction of the PET scanner, the radiopharmaceutical fluorine-18-2-fluoro-2-deoxyglucose (FDG) was made by chemists as a tracer so studying of the images were easier in locating brain tumors. Technological advances in the 1980s allowed for the invention of PET scanners with higher resolution than ever before that produced very precise, clear images.
Nuclear Reactions
The procedure of PET scans are made possible by nuclear reactions involving the emission of the positive beta particle known as the positron. The positron is a mass-less particle but has a positive charge of one. In other words, it is very similar to an electron (they have the same mass), but instead of a negative charge, it carries a positive charge. In nuclear reactions, it is commonly represented as any of the symbols $\ce{^0_{+1}e}$, $\ce{e^+}$, or $\ce{_{+1}\beta}$. The emission of a positron is represented by:
$\ce{^1_1p \rightarrow ^1_0n + ^0_{+1}e}$
This shows that the positron (represented here by $\ce{^0_{+1}e}$) speeds out of the nucleus while the neutron stays inside the nucleus. Consider the following nuclear reaction that is common in PET scans of the brain where carbon-11 is used as the tracer molecule.
$\ce{^{11}_6C \rightarrow ^{11}_{5}B + ^0_{+1}e}$
Notice that in this example of positron emission, the nuclide changes into a different element and as it gives off a positron particle, the atomic number is lowered by one, but the mass of the new element stays the same as the carbon that has decayed.
General Procedure
The first step of the process is to inject the patient with a solution commonly used by the body to produce energy, such as glucose. It must be a positron-emitting substance because when the scanner with its array of detectors detect the collisions of the positrons with electrons, the two species disappear, and in the process, create two gamma photons to move apart in opposite directions. The scanner's electronics record these detected gamma rays and map an image of the area where the active glucose is located. PET allows the physician to locate the exact areas where metabolic activities are occurring. For example, the PET scan shows exactly where the injected glucose is being used in the brain, the heart muscle, or a growing tumor. In a cancer patient, the cancer cells show up as denser areas on a PET scan due to the cells' high metabolic rates in comparison with normal cells.
Computed Axial Tomography (CAT) and Magnetic Resonance Imaging (MRI)
Computed Axial Tomography (CAT) scanners are very similar to PET scanners. The major difference is that patients are scanned with x-rays in a CAT scan so they are at risk of developing cancer if exposure to the ionizing radiation is too long or too often. The Magnetic Resonance Imaging (MRI), which was invented after the CAT scanner, is an expensive test and uses radio waves and magnets that show information on a computer screen. Patients are not at risk with MRIs because the test does not use ionizing radiation of any kind. The reason CAT scans are still used is most likely because it is much more affordable than MRIs.
Compared to CAT and MRI scanners, PET is much more efficient in detecting the differences between benign and malignant tumors because it focuses on the metabolic activity of radiopharmaceutical chemicals (glucose).
Radionuclides
Radionuclides used in PET scanning are typically isotopes with short half-lives that stay active just long enough to trace its activities in the body effectively and produce corresponding images. Below is a list of some of the more common isotopes used as tracers in the body and what each of their primary uses are:
• Carbon-11: PET brain scan
• Copper-64: Lung and liver disease diagnosis
• Fluorine-18: Bone scanning and studying the cerebral sugar metabolism
• Krypton-79: Assessing cardiovascular function
• Nitrogen-13: Brain, heart, and liver imaging
• Oxygen-15: Lung function test
Figure 3: Graphic representation of a patient's PET scan shows abnormal activity in the abdominal region. Image courtesy of Wikipedia Commons.
Applications of PET Scans
PET scanning is useful in the medical field as well as a research tool. In the medical field, it is heavily used in the clinical diagnosis of certain diseases and disorders because it is effective in targeting radionuclides (tracers) used in particular bodily functions. It is also useful in researching certain brain diseases and checking whether cardiovascular functions in an individual are normal because the produced images map this out very clearly as shown in the graphic representation below.
1. Oncology
PET scanning using the tracer radionuclide fluorine-18 (commonly known as FDG) is used because glucose-using cells take this molecule up and the images show where the metabolism of glucose is abnormal. The FDG molecule is trapped inside of the cell that takes it up until it decays completely, so a relatively short half-life is important. The use of FDG in oncology PET scans make up the majority of all PET scans in current practice.
PET scanning is especially effective in the detection of cancer because the activity of the injected glucose is especially active and is easily traced in the produced images.
Listed below are a few reasons why PET scanning is effective in detecting cancer:
1. Early and immediate detection: PET scanners effectively produce images of the immediate activity of glucose, accurately distinguishing a tumor growth due to its significantly active form, as is shown on the images in higher resolution. However, a PET scan images the entire body, so if it catches the activity of secondary tumor growths first, this can change the treatment that a patient will undergo.
2. Stages of cancer: PET is very sensitive in determining the full extent of disease, especially in lymphoma, malignant melanoma, breast, lung, colon and cervical cancers. Confirmation of metastatic disease allows the physician and patient to more accurately decide how to proceed with a certain patient's treatment.
3. Checking for recurrences: PET scanning is considered to be one of the most accurate procedures in differentiating tumor recurrences. By being able to scan a patient without the risk of over-radiation, this particular procedure allows for the development of a treatment plan that works most favorable to the patient, while also incorporating all functioning data of the tracer chemical.
4. Effectiveness of chemotherapy: The levels of tumor metabolism are compared on PET scans before and after a chemotherapy cycle. A successful chemotherapy treatment is visible by a PET scan by comparing the images from before and after the treatment and noting whether there are any noticeable changes in the activity of glucose throughout the body.
2. Neurology/Neuropsychology
PET imaging of the brain is based on the assumption that brain activity is associated with high radioactivity. The tracer nuclide that is used is oxygen-15. Its short half-life of two minutes makes it so that PET scans of the brain must be processed immediately after the oxygen-15 is made. The brain’s rapid use of glucose produces very vividly colored images that indicate where the most activities in the brain are. This makes it relatively easy to narrow down where abnormal activity is occurring in a patient with a brain disease because the color on the image will be different in the areas where a disease has settled in.
3. Cardiology
Though relatively new, detecting patients at risk of stroke may have been suggested to be the next focus of the process of PET scanning. The tracer nuclides that are used include krypton-79 and nitrogen-13.
4. Psychiatry
Carbon-11 and fluorine-18 are two radionuclide tracers that selectively bind to receptors in the brain, thus are a target for further research in the biological aspects of psychiatry. The states of dopamine, serotonin, and opioid receptors have been studied in humans because of their affect in patients with psychiatric conditions such as schizophrenia and mood disorders.
Problems
1. Given the symbol for nitrogen-13 is 13N, using knowledge of positron emission, find which element the decay of nitrogen produces in the following nuclear reaction that takes place in PET scanning.
$\ce{^{13}N \rightarrow \underline{\;\;\;} + _{+1}\beta}$
1. Write the nuclear reaction for positron emission by oxygen-15, which takes place in a PET scan used to assess the efficiency of the lungs.
$\ce{^{15}O \rightarrow \underline{\;\;\;} + _{+1}\beta}$
1. 3. Write the nuclear reaction for positron emission of a chemical used in bone scanning and studying the sugar metabolism of the cerebral (positron emission by fluorine-18).
2. 4. Write the nuclear reaction for the reaction necessary to assess cardiovascular function in the human body, given that the nuclide krypton-79 emits a positron in the reaction.
Solutions
1. $\ce{^{13}N \rightarrow \underline{^{13}O} + _{+1}\beta}$
2. $\ce{^{15}O \rightarrow \underline{^{15}F} + _{+1}\beta}$
3. $\ce{^{18}F \rightarrow \underline{^{18}Ne} + _{+1}\beta}$
4. $\ce{^{79}Kr \rightarrow \underline{^{79}Rb} + _{+1}\beta}$
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What comes to mind when you think of radiation? If you ask the average American, radiation would conjure images of deformed humans and malignant diseases, namely cancer. Despite radiation's notorious history, radiation has revolutionized the medical field, enhancing the ability of medical professionals to treat and diagnose diseases.
Introduction
Imagine a patient enters the hospital complaining of severe chest pain. After consulting the patient, the doctors decide that the source of the patient's problems is a constricted artery. However, the human body contains several long arteries. The doctors could use surgery to find the source of the clot, but this could take many hours and with no definite guarantee of success.This is exactly where radiation helps medical professionals bridge the gap that once existed. Now, CT scans, short for computed tomography, use X-ray particles along with advanced computer technology in order to produce highly detailed cross sectional images of the body. Using these scans, medical professionals can accurately find clots and other other harmful medical conditions.
In PET scans, short for positron imagery tomography, tracers are injected into the brain. These tracers emit gamma rays, which are picked up by a computer which use the gamma radiation in order to construct a two dimensional image of the brain. With the radiation coupled with an advanced computer, a two dimensional image is displayed, providing medical professionals with a view to the metabolic activity of the brain. On top of assisting medical professionals with diagnosing diseases, radiation also has therapeutic powers.
Radiation Therapy is used as a treatment to control malignant cells within cancer patients. Oncologists (specialists that deals with cancer) used radiation therapy frequently to help slow or cure the spread of cancer within inidivduals. Radiation is specifically applied to malignant tumors in order to shrink them in size, while hopefully not causing more damage to surrounding healthy cells. Medical professionals, mainly radiation oncologists, administer a variety of dosages to patient, contingent to the patients current health, as well as other treatments such as chemotherapy, success of surgery, etc.
Radiation in Imaging
X-Ray Imaging
German physicist, Wilhem Röntgen, discovered eletromagnetic radiation on 8 November 1895. In addition to Röntgen receiving the first Nobel Prize in Physics (1901), the use of the x-ray has become instrumental in modern medicine. Lastly, because of his discovery Röntgen is considered the father of diagnositic radiology.
X-Ray imaging is one of the most basic and routine forms of imaging within modern medicine. Using electromagnetic radiation, scientists and doctors have the ability to visualize internal situations. X-Ray imaging takes advantage of the fact that dense structures (i.e. bone) absorb more x-rays than the softer tissues that surround it. By projecting radiation onto special film paper, the denser substance casts a shadow where less x-rays pass through the body, and a contrasting image is produced. The radiation emitted by an X-Ray is of the wavelength of 10 to 0.01 nanometers(nm). Because some parts of the body may be thicker than others, X-Rays are used in "soft" (10 to 0.10 nm) and "hard" (0.10 to 0.01 nm) dosages or radiation.
X-rays are produced by passing an electron beam through a tungsten anode in a vacuum. The electromagnetic waves that are released can be focused and adjusted according to what is imaged. X-Rays can image in very high definition at a low cost, but they provide poor contrast in soft tissues. Contrast agents can also be added to the body prior to imaging, which allows different tissues to absorb more radiation, creating a better contrasting image. This allows doctors and scientists to work with much more specific images within their diagnosis or research (i.e. the ability to determine different aspects of soft-tissue).
CT Scanning
CT/ CAT (Computerized Axial Tomography) scans have the same fundamentals of x-rays, but they produce 3-D images using tomography. Using digitial geometric processing, the CT scan creates a 3-D image compiling a large number of 2-D x-rays. These 2-D images are thin cross sections of the body taken from a single axis of rotation. A CT scanner consists of a table where the patient lies on, an x-ray device which rotates at high speed around the patient taking hundreds of cross sections of the body, and a computer with appropriate software to compile and render the 3-D image. The table in Figure 1 moves through the tube, allowing the machine to take hundreds of pictures from head to toe.
Just like with x-ray, contrast agents are typically added to the body before imaging occurs so that different tissues are more defined. Due to the fact that many more images are being taken in order to form a complete three dimensional image, the radiation dose is much higher. While a chest x-ray is around .1 mSv or radiation, a chest CAT scan subjects the patient to about 8 mSv of radiation; over twice the amount that the body would absorb from its environment in a year. For this reason, it is uncommon and unsafe to have multiple CAT scans taken unless completely necessary.
Ironically, while X-Rays are used for medicine, they do pose health risks upon patients, especially CT Scans. Stated by the International Agenct for Research on Cancer, X-Ray radiation is considered a carcinogen (cancer causing). Measured in Sievert(s) (Sv) or Gray(s) (Gy), are the units of how much radiation energy is desposited within the body. For example a chest X-Ray has a mSv of 0.1. In contrast, the average American will absorb 3 mSv of background radiation in a year. Despite the X-Ray of a chest posing a small amount or radiation in contrast with our annual intake of background radiation, there exist risks involved with radiation, especially of reproductive organs and the brain.
PET/SPECT Scanning
PET/ SPECT (Positron Emission Tomograpy/ Single Photon Emission Computerized Tomography) uses radioactively tagged molecules to image the body. These radioactive "radionuclides" give off gamma radiation, which eventually annihilates an electron. These radionuclides are isotopes with short half-lives, usually less than 30 minutes. This annihilation sends two gamma rays in exactly opposite directions with energy of 511keVs, which is picked up by a 5,000 to 20,000 gamma detectors that surround the patient. When two detectors on opposite sides of the patients body absorb gamma rays at the same time, the computer marks where the substance is and using this data can render a 3-D image (Figure 4).
Figure 3: upload.wikimedia.org/Wikipedia/commons/c/c1/PET-schema.png
PET Scanning, is used to image the physiological aspects of the body rather than the anatomy. It images the function of the body rather than the form, such as where tagged molecules go and how they are used. For instance, if you were to image the brain of a deceased person, nothing would show up on a PET scan opposed to a CAT scan, as the brain is no longer functional. Pet Scanning is very useful in imaging tumors, which can be done when patients are injected with certain tracers. Often times PET scanners are used in collaboration with CAT scanners to create a composite image that shows both the function and form of the body.
Figure 4:upload.wikimedia.org/Wikipedia/commons/3/3d/PET-MIPS-anim.gif
Contributors and Attributions
• Mike Reed (UC Davis), Eugene Kwon (UC Davis)
Radiocarbon Dating: The Shroud of Turin
The Shroud of Turin is a linen wrapping cloth that appears to possess the image of Jesus Christ. Some people believe this to be the cloth that he was wrapped in following his crucifixion. In 1988, several groups of scientists were allowed samples of the shroud to subject these samples to 14C dating. The carbon-14 to carbon-12 ratio was found to be 92% of that in living organisms.
On the above graph, which depicts the decay curve for carbon-14, you can draw a line from 1988 up to the curve and then from this intersection over to the percent value on the Y axis. The value obtained from the curve is about 80%. This means that the Shroud of Turin may be younger than was previously thought.
The age of the shroud can be obtained from the graph by starting at 92% and drawing a line to the curve. Draw a line from this intersection down to the years and the value obtained is about 1000 AD, which means that the Shroud of Turin was probably created in the Middle Ages.
Contributors
• Prof. Richard Banks, Boise State University, Emeritus
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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.
Radiology
Radiology
Radiology is the science of high energy radiation and of the sources, including the study of the chemical, physical, and biological effects of radiation. Its a term usually refers to the diagnosis and treatment of disease using radiation. It is also a scientific discipline of medical imaging using ionizing radiation, radionuclides, nuclear magnetic resonance, and ultrasound.
One of the most useful technique in radiology is the computed tomography (CT) or computerized axial tomography (CAT). It gives anatomical information from a cross-sectional plane. Each image is generated by a computer synthesis of transmitted X-ray obtained in many different directions in a given plane.
Because it provides detailed, cross-sectional views of all types of tissue, CT is one of the best tools for studying the chest and abdomen. It is often the preferred method for diagnosing many different cancers, including lung, liver, and pancreatic cancer, since the image allows a physician to confirm the presence of a tumor and to measure its size, precise location, and the extent of the tumor's involvement with other nearby tissue. CT examinations are often used to plan and properly administer radiation treatments for tumors, and to guide biopsies and other minimally invasive procedures.
Currently, diagnoses have been carried out routinely with the following systems.
• Central Nervous System: Brain,Spine
• Cardiovascular System: heart, blood vessels
• Musculoskeletal System: bone, muscles, and joints
• Digestive, Urinary, and Respiratory System: intestines, kidneys, liver, stomach, lungs
• Reproductive System and mammography: male and female reproductive organs and breasts
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The energy harnessed in nuclei is released in nuclear reactions. Fission is the splitting of a heavy nucleus into lighter nuclei and fusion is the combining of nuclei to form a bigger and heavier nucleus. The consequence of fission or fusion is the absorption or release of energy.
• Contrasting Nuclear Fission and Nuclear Fusion
Nuclear fusion and nuclear fission are two different types of energy-releasing reactions in which energy is released from high-powered atomic bonds between the particles within the nucleus. The main difference between these two processes is that fission is the splitting of an atom into two or more smaller ones while fusion is the fusing of two or more smaller atoms into a larger one.
• Fission and Fusion
The energy harnessed in nuclei is released in nuclear reactions. Fission is the splitting of a heavy nucleus into lighter nuclei and fusion is the combining of nuclei to form a bigger and heavier nucleus. The consequence of fission or fusion is the absorption or release of energy.
• Fission Chain Reaction
A chain reaction is a series of reactions that are triggered by an initial reaction. An unstable product from the first reaction is used as a reactant in a second reaction, and so on until the system reaches a stable state. It usually follows a pattern of initiation (initial spark), propagation (continuation of the process) and termination (reaching a stable state).
Fission and Fusion
Nuclear fusion and nuclear fission are two different types of energy-releasing reactions in which energy is released from high-powered atomic bonds between the particles within the nucleus. The main difference between these two processes is that fission is the splitting of an atom into two or more smaller ones while fusion is the fusing of two or more smaller atoms into a larger one.
Nuclear Fission Nuclear Fusion
Definition: Fission is the splitting of a large atom into two or more smaller ones. Fusion is the fusing of two or more lighter atoms into a larger one.
Natural occurrence of the process: Fission reaction does not normally occur in nature. Fusion occurs in stars, such as the sun.
Byproducts of the reaction: Fission produces many highly radioactive particles. Few radioactive particles are produced by fusion reaction, but if a fission "trigger" is used, radioactive particles will result from that.
Conditions: Critical mass of the substance and high-speed neutrons are required. High density, high temperature environment is required.
Energy Requirement: Takes little energy to split two atoms in a fission reaction. Extremely high energy is required to bring two or more protons close enough that nuclear forces overcome their electrostatic repulsion.
Energy Released: The energy released by fission is a million times greater than that released in chemical reactions; but lower than the energy released by nuclear fusion. The energy released by fusion is three to four times greater than the energy released by fission.
Nuclear weapon: One class of nuclear weapon is a fission bomb, also known as an atomic bomb or atom bomb. One class of nuclear weapon is the hydrogen bomb, which uses a fission reaction to "trigger" a fusion reaction.
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Nuclear Chain Reactions are a simple, yet powerful method which to produce both constructive and destructive forces. Only understood to a significant degree within the last century, nuclear chain reactions have many practical uses in the modern era. Chain reactions can be addressed into two categories: first, controlled (like a nuclear power plant) and uncontrolled (an atomic bomb). Both are motivated by fission reactions, which are elaborated in this section.
Key words
• Fission: The event wherein a larger nucleus breaks apart into smaller nuclei, presumably, but not always, more stable.
• Transuranium / Transuranic: Any element with atomic number (z) greater than 92.
• Mega Electron Volt (MeV): The unit of measurement used to express the loss in mass of a given nuclear sample, equal to $1.6022 \times 10^{-13} \;J$.
• Excitation Energy: Energy required to change the quantum state of a nucleus.
• Fissile: Element that can fission through neutron capture.
• Fertile: Element that can through neutron capture become a fissile material.
• Heavy Water: D2O water made with heavier isotope of hydrogen.
• Chance Fission: Process of energetics which atoms produce fission (marked by 1st, 2nd, 3rd... etc.).
Introduction
Nuclear chain reactions are one of the modern applications of the fission process. Capitalizing upon the large amounts of energy, nuclear chain reactions can be used for constructive and destructive means. To understand how a chain reaction operates, it is best to search in depth of the nature of fission reactions. By understanding fission reactions, a chain reaction is then just an extrapolation of the excess particles produced by fission for a large quantity of an element.
The neutron was discovered by Sir James Chadwick in 1932. In 1939 atomic fission had been discovered by Lise Meitner (1878 to 1968) with her nephew Otto Frisch and was documented in her paper Disintegration of Uranium by Neutrons. A New Type of Nuclear Reaction where the term fission was coined. On December 2, 1942 Dr. Enrico Fermi, a Nobel Prize winner, produced the first sustained chain reaction at the University of Chicago. His assumption was that said reactions produce transuranium products. Otto Hahn and F. Strassmann reproduced Fermi's experiments and found contrary. By 1943 the first controlled chain reaction had taken place and by 1945, the first atomic weapon was produced. By 1951 nuclear chain reactions took a more practical route and became a means to produce electric power.
Chain Reaction
Chain Reactions are basically fission reactions which through the products produce more chain reactions. One of the most well-known and useful examples of a chain reaction is of U235 which is used to harness nuclear energy. For U235 on average 2.5 neutrons are emitted, starting on average two more fission reactions. Below is a simple fission process:
Note
The elements which the daughter nuclei are composed of is based upon probability (compare the fragments of Figures 4 and 5).
Understanding Fission
Understanding Fission is the key for chain reactions. Hence, below is a brief overview of the different aspects of fission. Below are scientific approaches to fission reactions, and the process which fission is produced. Fission Products can be produced by alpha, gamma, beta, charged particles, and through spontaneous reactions. A fission product is affected by the initial mass of nuclide, and excitation energy (expressed in MeV).
Fission Product Yields
Yield is the result is for any fission event, and is highly motivated by probability. This section covers experimental yields and charge yields. The yield can be expressed in two different experimental categories:
Absolute Yield
which can be expressed by the equation:
$Y{(\%)} = \left(\dfrac{N}{P}\right)*{100}$
Where Y is the absolute yield, N is the number of atoms a given nucleus formed, and P is the number of fission events. Absolute Yield can be broken down into three different categories:
1. Independent
2. Cumulative
3. Total
• Independent Yield: The probability that a given isotope will form directly.
$A + n \rightarrow B + C$
• Cumulative Yield: The probility that a given isotope will form directly (via independent yield) and indirectly by decay of a parent nuclide.
$A + n \rightarrow B + C$
$A \rightarrow D + n$
$D \rightarrow E + \beta$
$E \rightarrow C + \alpha$
(this process is through natural decay, where Element A decomposes to Element C)
• Total Yield: Probability that final product of the radioactive decay chain for a specific mass will form.
Note: Sum of total yields for binary fission will be 200%
Relative Yield
Relative Yield $\gamma_i$ is the ratio of absolute yields formed in fission, which can be expressed through the equation:
$\gamma_i = \dfrac{Y_i}{Y}$
where $\gamma_i$ is the relative yield, Yi is the absolute yield of specific nuclide, which is daughter to a specific nuclei Y also expressed as a yield. Relative yield also has respective independent, cumulative, total yields.
Charge Yield
An approximation of charge ratio (Z/A) is the same for both daughter nuclides as it was for the parent atom. The distribution of charge for daughter nuclides for a fixed mass A can be approximated with the equation:
$fragment\;charge = e^{-\dfrac{(Z-Z_p)^2}{c}}$
where Z is proton number, Zp is approximately the charge ratio of the parent nuclide , and c is a number between 0.8 and 1.0 (depending on fission reaction).
Trends of Fission Products
Fission products behave in specific trends, for example: fission is almost always a binary process, as ternary fission is on average 1000 less times probable (In most cases Ternary fission will appear for 1 to 3 events per 1000). Being able to predict the outcome of a fission event is vital for harvesting nuclear energy from power plants. Other trends that can be found is the nature of symmetric and asymmetric fission products. Fission will be symmetric or asymmetric depending on how many nucleons are required to form closed shells. Symmetric peaks can coexist in asymmetric peaks, such as Ra226. Also, a symmetric mass distribution leads to an asymmetric fission process. Note that asymmetric and Symmetric fission are both present in heavy nuclides.
Asymmetric Fission Products: The preferred form of fission leading to two daughter nuclides of unequal masses. Asymmetric fission has an average yield 600 times greater than symmetric fission. Number of outer nucleons drastically affects shape of nucleus, which in turn may change the nature of fission symmetry. The probability of asymmetric fission is high for fission motivated by thermal neutrons (such as U233, U235, Pu239).
Symmetric Fission Products:Preferred for spontaneous reactions (such as Th232, U238, Cm242, and Cf252), symmetric fission is caused by the preference in atomic nuclei to follow the 'magic numbers' most commonly of 50 proton 82 neutron nucleus. Symmetric Produces two daughter nuclides of equal or near equal masses. The range for atomic mass numbers that will be the product of symmetric fission are in the range of 127 to 139 amu. Symmetric fission also is preferred for high excitation energies (at or over 40 MeV). As a function of initial excitation energy, the probability of symmetric fission increases more rapidly than asymmetric fission. Note that multiple combinations are possible for fission chain reactions.
Probability of Fission
An unstable nucleus by the liquid drop model can be expressed in the inequality:
$\dfrac{Z^2}{A} > {18}$
Which for any proton number Z and Atomic Mass A, holding this inequality to be true will release energy through symmetric fission. A notable exception is Zr50 which upon the border of this inequality is stable. Although bear in mind that the above inequality is limited because of binding energy and therefore is extremely slow if not impossible. A far more probable inequality would is:
$\dfrac{Z^2}{A} > {47}$
Which the second inequality holds for half-lives for super heavy atoms. Note that spontaneous Fission is extremely rare for even radioactive samples.
Favored Reactions
Neutron binding energy is approximately 6 to 7 MeV. Neutron emission of atoms is favored over the fission process when energetically favorable. For the sake of uranium, the neutron emission and fission thresholds are only a few MeV apart. Thus odd-neutron atoms are 'slow neutron fission' and even neutron atoms are 'fast neutron fission'. With this implication it can be deduced that with enough initial excitement energy, multi neutron emission instead of fission is possible.
Fission According to 'Chance'
One of the intriguing portions of operating fission according to chance is the objective of producing more than one neutron before the fission reaction takes place. Hence, is a competition between binding energies of the molecule and neutrons. For example of a second neutron binding energy is greater than the first neutron binding energy, then second chance fission thresholds occur before second neutron emission.
Kinetics of Fission
The major energy of fission lies in the kinetics of the daughter nuclides. The magnitude of the kinetic energy is equal to the coulomb potential energy of fragments. The rest of the energy is released in excitation energy, which is usually neutron emission, and finally gamma radiation for the residual energy. Yield of neutrons that accompany fission is usually a linear relationship expressed as: 1 neutron to 8.5 MeV. A neutron on average carries 5.5 MeV in binding energy and 1.5 MeV in Kinetic energy. Hence, 7 MeV. Neutron yield of lighter fragments is three times greater than heavier fragments.
Note: Excitation energy is placed more into fragments and thus neutron emission than kinetics.
Controlled Chain Reactions
Nuclear Reactors
One of the best examples of Chain reactions is within nuclear power plants. As a fact, nuclear power provides 17 percent of the world's electricity. There are many types of reactors, which all use realatively the same materials, and components to complete a nuclear fission chain reaction.
Terms of Neutron Chain Reaction
• Scattering: When the neutron bounces off the desired fissionable nucleus.
• Leakage: When the neutron does not impact a nucleus, thus not producing a reaction (escapes).
• Capture: When the neutron is absorbed into the nucleus, but said nucleus does not react.
• Thermal Neutron: Neutron with low kinetic energy, less than 0.1 electron volts (eV). More efficient for fission process.
• Fast Neutron: Neutron as a product of fission, with high kinetic energy of 10 million electron volts (eV).
• Critical Mass: The quantity of a radioactive element to sustain a chain reaction.
• Subcritical: Any quantity of a radioactive element below critical mass.
Components of Reactors
• Fuel Element (or Fuel Rod): Contains the fissile material, typically uranium or plutonium, which is used as the fuel to undergo fission and provide the nuclear energy. The fissile material is encased in a solid cladding, made of Zircalloy, to contain both the fuel and the resulting fission products and keep then from escaping into the moderator, coolant, or anywhere outside the cladding. The moderator and coolant flows between the fuel elements (or rods) moderating the neutrons and carrying away the heat. The region inside the nuclear reactor where the fuel elements undergo fission to generate heat is called the nuclear reactor core.
• Moderator: Slows down fast neutrons to thermal energy range. Moderators must be light materials to slow down neutrons without causing capture. Some moderators are: water, carbon, heavy water.
• Coolant: Absorbs and removes the heat produced by nuclear fission. In most current commercial-scale nuclear reactors, purified regular water, called light water, is used as the coolant. Some other coolants are: heavy water, carbon dioxide or helium gas, or molten metals such as sodium, lead, or bismuth.
• Control Rods: Absorbs neutrons, designed to reduce the amount of neutrons available to continue the chain reaction. The control rods, interspersed between the fuel elements in the reactor core, can be inserted into or out of the core as needed to control conditions or shut down the reactor. Some materials used for control rods are: boron, silver, indium, cadmium, or hafnium.
Types of Reactors
• PWR: Pressurized Water Reactor, as of 2003, 212 were active worldwide. Water is both the coolant and moderator and is kept at a high pressure 70 to 150 atm.
• BWR: Boiling Water Reactor, like PWR, water is both the coolant and moderator. Although the water is kept at a lower pressure 70 atm and thus produces steam. The steam directly powers the turbine, thus simplifying the design. The only downside is that over time the turbine accumulates radioactivity (Nitrogen 17 with a half-life of seven seconds).
• Breeder Reactor: Uses both fissile and fertile elements to produce reaction. Because of this, breeder reactors can use materials that are more widely available. Operates by using fast neutrons to convert fertile elements into fissile elements.
Uncontrolled Chain Reactions
The other spectrum of chain reactions, which produce enough energy to cause an explosion. Below are destructive chain reactions by accident.
• Chernobyl: A level 7 on the Nuclear Event Scale, Chernobyl represents a chain reaction gone awry. During an experiment which through numerous errors, the coolant was turned off eventually burning the carbon control rods. A complete meltdown, radioactive smoke spread across Europe causing extensive damage.
• Three Mile Island: In Middleton Pennsylvania 1979, the Three Mile Island reactor experienced a level 5 Nuclear Event. The chain reaction stopped because of too few slow neutrons. Although the chain reaction ceased, the fission process continued sending the fuel rods to extreme temperatures. A small fissure developed in one of the reactors, thus sending radioactive steam into the atmosphere.
• Fukushima BWR reactors: Due to a severe earthquake and tsunami in Japan in March 2011, several BWR nuclear reactors at the Fukushima power plant lost electrical power for cooling, underwent explosions, and suffered reactor core damage as workers eventually pumped seawater into the reactors to cool them down and limit any further damage.
International Nuclear Event Scale (INES): The scale is written by degree of danger (lower in the list is greater danger)
0 No safety significance Deviation
1 Anomaly " "
2 Incident Incident
3 Serious incident " "
4 Accident without significant off-site risk " "
5 Accident with off-site risk Accident
6 Serious Accident " "
7 Major Accident " "
References
1. Nordborg, Claes, Hans Riotte, Miroslav Hrehor, Ted Lazo, Stefan Mundigl, Peter Wilmer, Julia Schwartz, Carol Kessler, Jaques De La Ferte, and Robert Price. "Overview of Nuclear Energy Today, Basic Principles of Nuclear Energy." Nuclear Energy Today. Ed. Cynthia Picot. Paris, France: Nuclear Energy Agency, Organisation for Economic Co-operation and Development, 2003.
2. Print. Petrucci, Ralph H., Jeffrey D. Madura, Carey Bissonnette, and F. Geoffrey Herring. "25-8 Nuclear Fission." General Chemistry: Principles and Modern Applications. 10th ed. Upper Saddle River: Pearson Canada, 2011. 1137-140. Print.
3. Wilets, Lawrence. Theories of Nuclear Fission. Oxford: Clarendon P, 1964. Print.
4. Zysin, Ju A., A. A. Lbov, and L. I. Sel'chenkov. Fission Product Yields and Their Mass Distribution: Transl. from Russian. New York, 1964. Print.
Problems
Easy Problems
1. (True/False) Fission cannot be induced by gamma radiation because it is too small to affect nucleus.
2. After five successful binary fission events, how many daughter fragments are produced?
Moderate Problems
3. A sample of Element Lol produced 15 fragment nuclides in 45 fission events, what is the absolute Yield Y(%)?
4. A new isotope of Xenon has been discovered with an atomic mass of 62 by the practical equation for spontaneous fission, will Xe62 undergo spontaneous fission? (Ignoring that this is absolutely impossible)
Hard Problems
5. For a yield of a given mass, would the end link in a radioactive chain equal the total yield?
Answers
1. False.
2. $2^5 = 32$
3. $\dfrac{15}{45}*{100} = {.333}*{100} = {33.3}\%$
4. $\dfrac{54^2}{62} = 47.03 > 47$
5. By using the definition of Cumulative Yield and Total Yield the answer would be YES.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Fission_and_Fusion/Fission_Chain_Reaction.txt
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The energy harnessed in nuclei is released in nuclear reactions. Fission is the splitting of a heavy nucleus into lighter nuclei and fusion is the combining of nuclei to form a bigger and heavier nucleus. The consequence of fission or fusion is the absorption or release of energy.
Introduction
Protons and neutrons make up a nucleus, which is the foundation of nuclear science. Fission and fusion involves the dispersal and combination of elemental nucleus and isotopes, and part of nuclear science is to understand the process behind this phenomenon. Adding up the individual masses of each of these subatomic particles of any given element will always give you a greater mass than the mass of the nucleus as a whole. The missing idea in this observation is the concept called nuclear binding energy. Nuclear binding energy is the energy required to keep the protons and neutrons of a nucleus intact, and the energy that is released during a nuclear fission or fusion is nuclear power. There are some things to consider however. The mass of an element's nucleus as a whole is less than the total mass of its individual protons and neutrons. The difference in mass can be attributed to the nuclear binding energy. Basically, nuclear binding energy is considered as mass, and that mass becomes "missing". This missing mass is called mass defect, which is the nuclear energy, also known as the mass released from the reaction as neutrons, photons, or any other trajectories. In short, mass defect and nuclear binding energy are interchangeable terms.
Nuclear Fission and Fusion
Nuclear fission is the splitting of a heavy nucleus into two lighter ones. Fission was discovered in 1938 by the German scientists Otto Hahn, Lise Meitner, and Fritz Strassmann, who bombarded a sample of uranium with neutrons in an attempt to produce new elements with Z > 92. They observed that lighter elements such as barium (Z = 56) were formed during the reaction, and they realized that such products had to originate from the neutron-induced fission of uranium-235:
$_{92}^{235}\textrm U+\,_0^1\textrm n \rightarrow \,_{56}^{141}\textrm{Ba}+\,_{36}^{92}\textrm{Kr}+3_0^1\textrm n \label{21.6.11}$
This hypothesis was confirmed by detecting the krypton-92 fission product. As discussed in Section 20.2, the nucleus usually divides asymmetrically rather than into two equal parts, and the fission of a given nuclide does not give the same products every time.
In a typical nuclear fission reaction, more than one neutron is released by each dividing nucleus. When these neutrons collide with and induce fission in other neighboring nuclei, a self-sustaining series of nuclear fission reactions known as a nuclear chain reaction can result (Figure 21.6.2). For example, the fission of 235U releases two to three neutrons per fission event. If absorbed by other 235U nuclei, those neutrons induce additional fission events, and the rate of the fission reaction increases geometrically. Each series of events is called a generation. Experimentally, it is found that some minimum mass of a fissile isotope is required to sustain a nuclear chain reaction; if the mass is too low, too many neutrons are able to escape without being captured and inducing a fission reaction. The minimum mass capable of supporting sustained fission is called the critical mass. This amount depends on the purity of the material and the shape of the mass, which corresponds to the amount of surface area available from which neutrons can escape, and on the identity of the isotope. If the mass of the fissile isotope is greater than the critical mass, then under the right conditions, the resulting supercritical mass can release energy explosively. The enormous energy released from nuclear chain reactions is responsible for the massive destruction caused by the detonation of nuclear weapons such as fission bombs, but it also forms the basis of the nuclear power industry.
Nuclear fusion, in which two light nuclei combine to produce a heavier, more stable nucleus, is the opposite of nuclear fission. As in the nuclear transmutation reactions discussed in Section 20.2, the positive charge on both nuclei results in a large electrostatic energy barrier to fusion. This barrier can be overcome if one or both particles have sufficient kinetic energy to overcome the electrostatic repulsions, allowing the two nuclei to approach close enough for a fusion reaction to occur. The principle is similar to adding heat to increase the rate of a chemical reaction. As shown in the plot of nuclear binding energy per nucleon versus atomic number in Figure 21.6.3, fusion reactions are most exothermic for the lightest element. For example, in a typical fusion reaction, two deuterium atoms combine to produce helium-3, a process known as deuterium–deuterium fusion (D–D fusion):
$2_1^2\textrm H\rightarrow \,_2^3\textrm{He}+\,_0^1\textrm n \label{21.6.12}$
In another reaction, a deuterium atom and a tritium atom fuse to produce helium-4 (Figure $1$), a process known as deuterium–tritium fusion (D–T fusion):
$_1^2\textrm H+\,_1^3\textrm H\rightarrow \,_2^4\textrm{He}+\,_0^1\textrm n \label{21.6.13}$
Initiating these reactions, however, requires a temperature comparable to that in the interior of the sun (approximately 1.5 × 107 K). Currently, the only method available on Earth to achieve such a temperature is the detonation of a fission bomb. For example, the so-called hydrogen bomb (or H bomb) is actually a deuterium–tritium bomb (a D–T bomb), which uses a nuclear fission reaction to create the very high temperatures needed to initiate fusion of solid lithium deuteride (6LiD), which releases neutrons that then react with 6Li, producing tritium. The deuterium-tritium reaction releases energy explosively. Example 21.6.3 and its corresponding exercise demonstrate the enormous amounts of energy produced by nuclear fission and fusion reactions. In fact, fusion reactions are the power sources for all stars, including our sun.
To calculate the energy released during mass destruction in both nuclear fission and fusion, we use Einstein’s equation that equates energy and mass:
$E=mc^2 \label{1}$
with
• $m$ is mass (kilograms),
• $c$ is speed of light (meters/sec) and
• $E$ is energy (Joules).
Example $1$: Neutron Induced Fission
Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when the neutron-induced fission of 235U produces 144Cs, 90Rb, and two neutrons:
$_{92}^{235}\textrm U+\,_0^1\textrm n\rightarrow \,_{55}^{144}\textrm{Cs}+\,_{37}^{90}\textrm{Rb}+2_0^1\textrm n$
Given: balanced nuclear reaction
Asked for: energy released in electronvolts per atom and kilojoules per mole
Strategy:
A Following the method used in Example 21.6.1, calculate the change in mass that accompanies the reaction. Convert this value to the change in energy in electronvolts per atom.
B Calculate the change in mass per mole of 235U. Then use Equation 21.6.3 to calculate the change in energy in kilojoules per mole.
Solution
A The change in mass that accompanies the reaction is as follows:
\begin{align}\Delta m&=\mathrm{mass_{products}}-\mathrm{mass_{reactants}}=\textrm{mass}(_{55}^{144}\textrm{Cs}+\,_{37}^{90}\textrm{Rb}+\,_0^1\textrm n)-\textrm{mass }\,_{92}^{235}\textrm U \&=(143.932077\textrm{ amu}+89.914802\textrm{ amu}+1.008665\textrm{ amu})-\textrm{235.043930 amu} \&=-0.188386\textrm{ amu}\end{align}
The change in energy in electronvolts per atom is as follows:
$\Delta E=(-0.188386\textrm{ amu})(931\textrm{ MeV/amu})=-175\textrm{ MeV}$
B The change in mass per mole of $_{92}^{235}\textrm{U}$ is −0.188386 g = −1.88386 × 10−4 kg, so the change in energy in kilojoules per mole is as follows:
\begin{align}\Delta E&=(\Delta m)c^2=(-1.88386\times10^{-4}\textrm{ kg})(2.998\times10^8\textrm{ m/s})^2 \&=-1.693\times10^{13}\textrm{ J/mol}=-1.693\times10^{10}\textrm{ kJ/mol}\end{align}
Exercise $1$
Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when deuterium and tritium fuse to give helium-4 and a neutron:
$_1^2\textrm H+\,_1^3\textrm H\rightarrow \,_2^4\textrm{He}+\,_0^1\textrm n$
Solution
ΔE = −17.6 MeV/atom = −1.697 × 109 kJ/mol
Figure $1$: Binding energy per nucleon of common isotopes.
Fission
Fission is the splitting of a nucleus that releases free neutrons and lighter nuclei. The fission of heavy elements is highly exothermic which releases about 200 million eV compared to burning coal which only gives a few eV. The amount of energy released during nuclear fission is millions of times more efficient per mass than that of coal considering only 0.1 percent of the original nuclei is converted to energy. Daughter nucleus, energy, and particles such as neutrons are released as a result of the reaction. The particles released can then react with other radioactive materials which in turn will release daughter nucleus and more particles as a result, and so on. The unique feature of nuclear fission reactions is that they can be harnessed and used in chain reactions. This chain reaction is the basis of nuclear weapons. One of the well known elements used in nuclear fission is $\ce{^{235}U}$, which when is bombarded with a neutron, the atom turns into $\ce{^{236}U}$ which is even more unstable and splits into daughter nuclei such as Krypton-92 and Barium-141 and free neutrons. The resulting fission products are highly radioactive, commonly undergoing $\beta^-$ decay.
Nuclear fission is the splitting of the nucleus of an atom into nuclei of lighter atoms, accompanied by the release of energy, brought on by a neutron bombardment. The original concept of this nuclei splitting was discovered by Enrico Femi in 1934—who believed transuranium elements might be produced by bombarding uranium with neutrons, because the loss of Beta particles would increase the atomic number. However, the products that formed did not correlate with the properties of elements with higher atomic numbers than uranium (Ra, Ac, Th, and Pa). Instead, they were radioisotopes of much lighter elements such as Sr and Ba. The amount of mass lost in the fission process is equivalent to an energy of $3.20 \times 10^{-11}\; J$.
Example $1$
Consider the neutron bonbardment
$\ce{_{92}^{235}U + _{1}^{0}n \rightarrow _{92}^{236}U} \rightarrow \; \text{fission products}$
which releases $3.20 \times 10^{-11}\; J$ per $\ce{^{235}U}$ atom.
How much energy would be released if $1.00\;g$ of $\ce{^{235}U}$ were to undergo fission?
Solution
$(1.00\;\rm{g}\; \ce{^{235}U} ) \times \left(\dfrac{1\; mol\; \ce{^{235}U}}{235\; g\; \ce{^{235}U}}\right) \times \left(\dfrac{ 6.022 \times 10^{23}\; \text{atoms}\; \ce{^{235}U}}{1\; mol\; \ce{^{235}U}} \right) \times \left(\dfrac{3.20 \times 10^{-11}\; J}{1\; atom \; \ce{^{235}U}}\right) = 8.20 \times 10^{10}\; J$
Clearly, the fission of a small amount of atoms can produce an enormous amount of energy, in the form of warmth and radiation (gamma waves). When an atom splits, each of the two new particles contains roughly half the neutrons and protons of the original nucleus, and in some cases a 2:3 ratio.
Critical Mass
The explosion of a bomb only occurs if the chain reaction exceeds its critical mass. The critical mass is the point at which a chain reaction becomes self-sustaining. If the neutrons are lost at a faster rate than they are formed by fission, the reaction will not be self-sustaining. The spontaneous nuclear fission rate is the probability per second that a given atom will fission spontaneously--that is, without any external intervention. In nuclear power plants, nuclear fission is controlled by a medium such as water in the nuclear reactor. The water acts as a heat transfer medium to cool down the reactor and to slow down neutron particles. This way, the neutron emission and usage is a controlled. If nuclear reaction is not controlled because of lack of cooling water for example, then a meltdown will occur.
Fusion
Nuclear fusion is the joining of two nuclei to form a heavier nuclei. The reaction is followed either by a release or absorption of energy. Fusion of nuclei with lower mass than iron releases energy while fusion of nuclei heavier than iron generally absorbs energy. This phenomenon is known as iron peak. The opposite occurs with nuclear fission.
The power of the energy in a fusion reaction is what drives the energy that is released from the sun and a lot of stars in the universe. Nuclear fusion is also applied in nuclear weapons, specifically, a hydrogen bomb. Nuclear fusion is the energy supplying process that occurs at extremely high temperatures like in stars such as the sun, where smaller nuclei are joined to make a larger nucleus, a process that gives off great amounts of heat and radiation. When uncontrolled, this process can provide almost unlimited sources of energy and an uncontrolled chain provides the basis for a hydrogen bond, since most commonly hydrogen is fused. Also, the combination of deuterium atoms to form helium atoms fuel this thermonuclear process. For example:
$\ce{^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n} + \text{energy}$
However, a controlled fusion reaction has yet to be fully demonstrated due to many problems that present themselves including the difficulty of forcing deuterium and tritium nuclei within a close proximity, achieving high enough thermal energies, and completely ionizing gases into plasma. A necessary part in nuclear fusion is plasma, which is a mixture of atomic nuclei and electrons that are required to initiate a self-sustaining reaction which requires a temperature of more than 40,000,000 K. Why does it take so much heat to achieve nuclear fusion even for light elements such as hydrogen? The reason is because the nucleus contain protons, and in order to overcome electrostatic repulsion by the protons of both the hydrogen atoms, both of the hydrogen nucleus needs to accelerate at a super high speed and get close enough in order for the nuclear force to start fusion. The result of nuclear fusion releases more energy than it takes to start the fusion so ΔG of the system is negative which means that the reaction is exothermic. And because it is exothermic, the fusion of light elements is self-sustaining given that there is enough energy to start fusion in the first place.
Figure $1$: Scientists have yet to find a method for controlling fusion reactions. Fission reactions on the other hand is the type used in nuclear power plants and can be controlled. Atomic bombs and hydrogen bombs are examples of uncontrolled nuclear reactions.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Fission_and_Fusion/Fission_and_Fusion.txt
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Nuclear energy is the energy that holds together the nuclei of atoms. Atoms are the most simple blocks that make up matter. Every atom has in its center a very small nucleus. Normally, nuclear energy is hidden inside the atoms. However, some atoms are radioactive and send off part of their nuclear energy as radiation. Radiation is given off from the nucleus of unstable isotopes of radioactive substances.
• Energetics of Nuclear Reactions
Nuclear reactions are associated with changes in both mass and energy. In this module, the relationship between these two concepts are examined on a nuclear level.
• Nuclear Magic Numbers
Nuclear Stability is a concept that helps to identify the stability of an isotope. The two main factors that determine nuclear stability are the neutron/proton ratio and the total number of nucleons in the nucleus.
Nuclear Energetics and Stability
Nuclear reactions are associated with changes in both mass and energy. In this module, the relationship between these two concepts are examined on a nuclear level.
Mass Defect and Binding Energy
Albert Einstein’s mass-energy equivalence relates energy and mass in nuclear reactions:
$E=mc^2 \label{Einstein}$
Each time an energy change occurs, there is also a mass change that is related by the constant c2 (the speed of light squared). Compared to the amount of energy due to the nuclear reaction, energy changes in chemical reactions are small, making the mass change insignificant for chemical reactions. However, on a nuclear level, there is a significant amount of energy change in comparison and therefore a discernible mass change. In Albert Einstein’s mass-energy equivalence (Equation \ref{Einstein}), “m” is the net change in mass in kilograms and “c” is a constant (the speed of light) in meters per second. Two common units to express nuclear energy are joules ($J$) and megaelectronvolts ($MeV$).
$1.6022 \times 10^{-13}\, J = 1 \,MeV$
The energy equivalent of 1 atomic mass unit (u), via Equation \ref{Einstein}, is:
$\text{1 atomic mass unit (u)} = 1.4924 \times 10^{-10} J = 931.5\, MeV$
By knowing the mass change in amu, the energy released can be directly calculated using these conversion factors, which have already taken into account mass conversions and the value of $c^2$. Keep in mind that the sum of mass-energy equivalents of reactants and products must equal each other. Figure $1$ provides an example of a mass defect. That is, the sum of the mass of the individual protons and neutrons differs from the overall mass of the nucleus. The nucleus weighs less than the masses of the individual subatomic particles.
The sum of the mass and energy of the reactants are equivalent to the sum of the mass and energy of the products.
Where does this mass go when a nucleus is formed? Recall Einstein's mass-energy equivalence and how matter and energy are essentially different configurations of one another. Mass is lost and as a result, energy is released as the nucleons come together to form the nucleus. This energy is known as the nuclear binding energy. Einstein's mass-energy equivalence can be rewritten in the following terms:
$\text{Nuclear Binding Energy} = \text{Mass Defect} \times c^2$
or
$E = Δm \times c^2 \label{Einstein2}$
The mass is converted into the energy required to bind the protons and neutrons together to make a nuclei. Nuclear binding energy is also the defined as the energy required to break apart a nucleus.
Note that Δm can be either positive or negative (depends on where you set your zero point), but for our purposes, it will not really matter, and we should simply always use a positive mass deficit and say that is the energy released.
One additional thing to keep in mind is what a nucleus is and what a nucleon is. A nucleus (plural nuclei) is the center of the atom, and is composed of nucleon (which are both the protons and neutrons: a nucleon is what makes up the nuclei). This means the nucleus of an atom of Carbon-14 (14C)would contain 14 nucleons a nucleus of Uranium-238 (238U) would contain 238 nucleons etc.
Releasing Binding Energy
The mass number 60 is the maximum binding energy for each nucleon. (In other words, nuclei of mass number of approximately 60 require the most energy to dismantle). This means that the binding energy increases when small nuclei join together to form larger nuclei in a process known as nuclear fusion. For nuclei with mass numbers greater than 60, the heavier nuclei will break down into smaller nuclei in a process known as nuclear fission.
For fusion processes, the binding energy per nucleon will increase and some of the mass will be converted and released as energy (Figure $1$). Fission processes also release energy when heavy nuclei decompose into lighter nuclei. The driving force behind fission and fusion is for an atomic nuclei to become more stable. So nuclei with a mass number of approximately 60 will be the most stable, which explains why iron is the most stable element in the universe. Elements with mass numbers around 60 will also be stable elements, while elements with extremely large atomic masses will be unstable.
• Fusion of elements lighter than iron may release energy to generate nuclei with greater binding energy (per nucleon).
• Fission of elements heavier than iron may release energy to generate nuclei with greater binding energy (per nucleon).
Nuclear fusion can release more energy than nuclear fission, especially when fusing small nuclei like hydrogen and helium into bigger nuclei.
Exercise $1$
A certain nuclear reaction gives off 22.1 MeV. Calculate the energy released in Joules.
Answer
This is a simple conversion problem. Use 1.6022 x 10-13J = 1MeV.
22.1 Mev x 1.6022 x 10-13J/1 MeV = 3.54 x 10-12 J
Exercise $2$
Suppose that the nuclear mass of 14N is reported as 13.998947 amu. Calculate the binding energy per nucleon.
Answer
Nitrogen-14 has 7 protons and 7 neutrons. The combined mass of the subatomic particles is
7 * (proton mass) + 7 * (neutron mass)
7 * (1.00728 amu) + 7 * (1.00866 amu ) = 14.11158 amu
The reported mass of Nitrogen is 13.998947 amu, so the mass defect (difference) is
14.11158 amu - 13.998947 amu = 0.112633 amu
Using Equation \ref{Einstein2},
E = ((0.112633g/1mol)(1kg/1000g)(1mol/6.022*1023nuclei)(1nuclei/14nucleon)) x c2 = 1.24 x 10-12 J/nucleon
Exercise $3$
You have a pebble with a mass of 1.0 gram. If all of the mass was suddenly turned into energy, how much energy would be released?
Answer
The mass-energy equivalent of 1.0 gram is
E = .001 kg x ( 2.9979x108 m/s )2 = 9.0 x 1013 J.
This is roughly equal to the energy released by the Fat Man atomic bomb! Note that this problem tells us that it doesn't matter what's turned into the energy, if it happens, it will release the same amount of energy.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Nuclear_Energetics_and_Stability/Energetics_of_Nuclear_Reactions.txt
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