9.1 REACTION MECHANISM: AN ANALOGY
Chemical kinetics is the study of how reactions take place and how fast they occur. (Kinetics tells us nothing about the spontaneity of a reaction, however! We’ll study that a little later.)
Consider this scenario: A group of people are washing a pile of dirty dishes and stacking them up as clean, dry dishes. Our “reaction” has dirty dishes as starting material, and clean, dry dishes as the product:
dirty dish → clean-and-dry dish
But what about a soapy dish? We know it’s part of the process, but the equation doesn’t include it. When we break down the pathway of a dirty dish to a clean-and-dry dish, we realize that the reaction happens in several steps, a sequence of elementary steps that show us the reaction mechanism:
1) dirty dish → soapy dish
2) soapy dish → rinsed dish
3) rinsed dish → clean-and-dry dish
The soapy and rinsed dishes are reaction intermediates. They are necessary for the conversion of dirty dishes to clean-and-dry dishes, but don’t appear either in the starting material or products. If you add up all the reactants and products, the intermediates cancel out, and you’ll have the overall equation.
In the same way, we write chemical reactions as if they occur in a single step:
2 NO + O2 → 2 NO2
But in reality, things are a little more complicated, and reactions often proceed through intermediates that we don’t show in the chemical equation. The truth for the reaction above is that it occurs in two steps:
1) 2 NO → N2O2
2) N2O2 + O2 → 2 NO2
The N2O2 comes and goes during the reaction, but isn’t part of the starting material or products. N2O2 is a reaction intermediate.
Just as the soapy dishes and rinsed dishes are produced and then consumed, we can identify an intermediate in a series of elementary steps as a substance that is produced in one elementary step and then consumed in a subsequent step. Although the two elementary steps don’t need to be sequential, they often are. As above, note that intermediates will not be part of the overall balanced chemical reaction. Depending on the rate of the elementary step that consumes the intermediate, the concentration of the intermediate will vary in solution. As the consuming elementary step becomes faster, the steady-state concentration of the intermediate becomes smaller, and it becomes harder to detect the intermediate.
What determines the rate of a reaction? Consider our friends doing the dishes.
1) dirty dish → soapy dish
Bingo washes at 5 dishes per minute.
2) soapy dish → rinsed dish
Ringo rinses at 8 dishes per minute.
3) rinsed dish → clean-and-dry dish
Dingo dries at 3 dishes per minute.
What will be the rate of the overall reaction? Thanks to Dingo, the dishes move from dirty to clean-and-dry at only 3 dishes a minute. It doesn’t matter how fast Bingo and Ringo wash and rinse; the dishes will pile up behind Dingo. The rate-determining step is Dingo’s drying step, and true to its name, it determines the overall rate of reaction.
The slowest step in a process determines the overall reaction rate.
This applies to chemical reactions as well. For our chemical reaction given above, we have
2 NO → N2O2 (fast)
N2O2 + O2 → 2 NO2 (slow)
The second step is the slowest, and it will determine the overall rate of reaction. No matter how fast the first step moves along, the intermediates will pile up in front of the second step as it plods along. The slow step dictates the rate of the overall reaction.
Once again, there’s an important difference between our dishes analogy and a chemical reaction: While the dishes pile up behind Dingo, in a chemical reaction the intermediates will not pile up. Rather they will shuttle back and forth between reactants and products until the slow step takes it forward. This would be like taking a rinsed dish and getting it soapy again, until Dingo is ready for it!
Example 9-1: Which of the following is the best example of a rate?
A) rate = ∆[A]/ ∆t
B) rate = ∆[A]/ ∆[B]
C) rate = ∆ [A] ∆[B]
D) rate = ∆[A]2
Solution: Regardless of the topic, rate is always defined as change in something over change in time. Choice A is the answer.
9.2 REACTION RATE
The rate of a reaction indicates how fast reactants are being consumed or how fast products are being formed. The reaction rate depends on several factors. Since the reactant molecules must collide and interact in order for old bonds to be broken and new ones to be formed to generate the product molecules, anything that affects these collisions and interactions will affect the reaction rate. The reaction rate is determined by the following:
1) How frequently the reactant molecules collide
2) The orientation of the colliding molecules
3) Their energy
Every chemical reaction has an activation energy (Ea), or the minimum energy required of reactant molecules during a molecular collision in order for the reaction to proceed to products. If the reactant molecules don’t possess this much energy, their collisions won’t be able to produce the products and the reaction will not occur. If the reactants possess the necessary activation energy, they can reach a high-energy (and short-lived!) transition state, also known as the activated complex. For example, if the reaction is A2+ B2 → 2 AB, say, the activated complex might look something like this:
Now that we have introduced all species that might appear throughout the course of a chemical reaction, we can illustrate the energy changes that occur as a reaction occurs in a reaction coordinate diagram. Consider the following two-step process and its reaction coordinate graph below:
A → X
X → B
A → B
Notice that the transition state is always an energy maximum, and is therefore distinct from an intermediate. Remember that reaction intermediates (shown as X in this case) are produced in an early step of the mechanism, and are later used up so they do not appear as products of the overall reaction. The intermediate is shown here as a local minimum in terms of its energy, but has more energy than either the reactants or products. The high energy intermediate is therefore highly reactive, making it difficult to isolate.
Since the progress of the reaction depends on the reactant molecules colliding with enough energy to generate the activated complex, we can make the following statements concerning the reaction rate:
1) The lower the activation energy, the faster the reaction rate. The reaction coordinate above suggests that the second step of the mechanism will therefore be the slow step, or the rate-determining step, since the second “hill” of the diagram is higher.
2) The greater the concentrations of the reactants, the faster the reaction rate. Favorable collisions are more likely as the concentrations of reactant molecules increase.
3) The higher the temperature of the reaction mixture, the faster the reaction rate. At higher temperatures, more reactant molecules have a sufficient energy to overcome the activation-energy barrier, and molecules collide at a higher frequency, so the reaction can proceed at a faster rate.
Notice in the reaction coordinate diagram above that the ∆G° of the reaction has no bearing on the rate of the reaction, and vice versa. Thermodynamic factors and kinetic factors do not affect each other (a concept the MCAT loves to ask about).
Catalysts provide reactants with a different route, usually a shortcut, to get to products. A catalyst will almost always make a reaction go faster by either speeding up the rate-determining step or providing an optimized route to products. A catalyst that accelerates a reaction does so by lowering the activation energy of the rate-determining step, and therefore the energy of the highest-energy transition state:
The key difference between a reactant and a catalyst is that the reactants are converted to products, but a catalyst remains unchanged at the end of a reaction. A catalyst can undergo a temporary change during a reaction, but it is always converted back to its original state. Like reaction intermediates, catalysts aren’t included in the overall reaction equation.
Our dish crew could use a catalyst. Picture Dingo walking to pick up each wet dish, drying it on both sides, and walking back to place it in the clean dish stack. Now imagine a helper, Daisy, who takes the wet dish from Ringo, then walks it over to Dingo, and while he dries and stacks, she returns with another wet dish. This way, Dingo can dry 5 dishes a minute instead of 3, and the overall dish-cleaning rate increases to 5 dishes a minute. Daisy is the catalyst, but the chain of events in the overall reaction remains the same.
In the same way, chemical reactions can be catalyzed. Consider the decomposition of ozone:
O3(g) + O(g) → 2 O2(g)
This reaction actually takes place in two steps and is catalyzed by nitric oxide (NO):
1) NO(g) + O3(g) → NO2(g) + O2(g)
2) NO2(g) + O(g) → NO(g) + O2(g)
NO(g) is necessary for this reaction to proceed at a noticeable rate, and even undergoes changes itself during the process. But NO(g) remains unchanged at the end of the reaction and makes the reaction occur much faster than it would in its absence. NO(g), a product of automobile exhaust, is a catalyst in ozone destruction.
It is important to note that the addition of a catalyst will affect the rate of a reaction, but not the equilibrium or the thermodynamics of the reaction. A catalyst provides a different pathway for the reactants to get to the products, and lowers the activation energy, Ea. But a catalyst does not change any of the thermodynamic quantities such as ∆G, ∆H, and ∆S of a reaction.
Example 9-2: Which of the following statements is true?
A) Catalysts decrease the activation energy of the forward reaction only.
B) Catalysts decrease the activation energy of the reverse reaction only.
C) Catalysts decrease the activation energy of both the forward and reverse reactions.
D) Catalysts decrease the activation energy of the forward reaction and increase the activation energy of the reverse reaction.
Solution: Catalysts decrease the activation energy of the forward reaction and reverse reaction—that’s why they have no net effect on a system that’s already at equilibrium. Choice C is the correct choice.
Example 9-3: Which of the following is true about the hypothetical two-step reaction shown below?
1) + 2 B → C + 2 D (fast)
2) C → A + E (slow)
A) A is a catalyst, and C is an intermediate.
B) A is a catalyst, and D is an intermediate.
C) B is a catalyst, and A is an intermediate.
D) B is a catalyst, and C is an intermediate.
Solution: C is an intermediate; it’s formed in one step and consumed in the other. This eliminates choices B and C. Now, is A or B the catalyst? Notice that B is consumed but not reformed; therefore, it cannot be the catalyst. The answer is A.
Example 9-4: A reaction is run without a catalyst and is found to have an activation energy of 140 kJ/mol and a heat of reaction, ∆H, of 30 kJ/mol. In the presence of a catalyst, however, the activation energy is reduced to 120 kJ/mol. What will be the heat of reaction in the presence of the catalyst?
A) —10 kJ/mol
B) 10 kJ/mol
C) 30 kJ/mol
D) 50 kJ/mol
Solution: Catalysts affect only the kinetics of a reaction, not the thermodynamics. The heat of the reaction will be the same with or without a catalyst. The answer is C.
9.4 RATE LAWS
On the MCAT, you might be given data about the rate of a particular reaction and be asked to derive the rate law. The data for rate laws are determined by the initial rates of reaction and typically are given as the rate at which the reactant disappears. You’ll rarely see products in a rate law expression, usually only reactants. What does a rate law tell us? Although a reaction needs all the reactants to proceed, only those that are involved in the rate-determining step (the slow step) are part of the rate law expression. Some reactants may not affect the reaction rate at all, and so they won’t be a part of the rate law expression.
Let’s look at a generic reaction, a A + b B → c C + d D, and its rate law:
rate = k [A]x[B]y
x = the order of the reaction with respect to A
y = the order of the reaction with respect to B
(x + y) = the overall order of the reaction
k = the rate constant
The rate law can only be determined experimentally. You can’t get the orders of the reactants, not to mention the rate constant k, just by looking at the balanced equation. The exception to this rule is for an elementary step in a reaction mechanism. The rate law is first order for a unimolecular elementary step and second order for a bimolecular elementary step. The individual order of the reactants in a rate law will follow from their stoichiometry in the rate-determining step (similar to the way they’re included in an equilibrium constant).
Let’s look at a set of reaction rate data and see how to determine the rate law for the reaction
A + B + C → D + E
From the experimental data, we can determine the orders with respect to the reactants—that is, the exponents x, y, and z in the equation
rate = k[A]x[B]y[C]z
and the overall order of the reaction, x + y + z.
Let’s first find the order of the reaction with respect to Reactant A. As we go from Experiment 1 to Experiment 2, only [A] changes, so we can use the data to figure out the order of the reaction with respect to Reactant [A]. We notice that the value of [A] doubled, and the reaction rate doubled. Therefore, the reaction rate is proportional to [A], and x = 1.
Next, let’s look at [B]. As we go from Experiment 1 to Experiment 3, only [B] changes. When [B] is doubled, the rate is quadrupled. Therefore, the rate is proportional to [B]2, and y = 2.
Finally, let’s look at [C]. As we go from Experiment 1 to Experiment 4, only [C] changes. When [C] is doubled, the rate is unaffected. This tells us that the reaction rate does not depend on [C], so z = 0.
Therefore, the rate law has the form
rate = k[A][B]2
The reaction is first order with respect to [A], second order with respect to [B], zero order with respect to [C], and third order overall. In general, if a reaction rate increases by a factor f when the concentration of a reactant increases by a factor c, and f = c x, then we can say that x is the order with respect to that reactant.
The Rate Constant
From the experimental data, you can also calculate the rate constant, k. For the reaction we looked at above, we found that the rate law is given by: rate = k[A][B]2. Solving this for k, we get
Now, just pick any experiment in the table, and using the results of Experiment 1, you’d find that
Any of the experiments will give you the same value for k because it’s a constant for any given reaction at a given temperature. That is, each reaction has its own rate constant, which takes into account such factors as the frequency of collisions, the fraction of the collisions with the proper orientation to initiate the desired bond changes, and the activation energy. This can be expressed mathematically with the Arrhenius equation:
k = Ae−(Ea/RT)
Here, A is the Arrhenius factor (which takes into account the orientation of the colliding molecules), Ea is the activation energy, R is the gas-law constant, and T is the temperature in kelvins. If we rewrite this equation in the form ln k = ln A — (Ea/RT), we can more clearly see that adding a catalyst (thus decreasing Ea) or increasing the temperature will increase k. In either case, the expression Ea/RT decreases, and subtracting something smaller gives a greater result, so ln k (and thus k itself) will increase. (By the way, a rough rule of thumb is that the rate will increase by a factor of about 2 to 4 for every 10-degree (Celsius) increase in temperature.)
The units of the rate constant are not necessarily uniform from one reaction to the next. Reactions of different orders will have rate constants bearing different units. In order to obtain the units of the rate constant one must keep in mind that the rate, on the left side of the equation, must always have units of M/s as it measures the change in concentration of a species in the reaction over time. The units given to the rate constant must, when combined with the units of the concentrations in the rate equation, provide M/sec.
Below is a generic second order rate equation.
Rate = k[A][B]
Assuming that the concentrations of both A and B are in molarity (M), then in order to give the left side of the equation units of M/s, the units of the rate constant must be M−1s−1. If the rate were third order, the units would be M−2s−1, or if first order then simply s−1.
Example 9-5: Based on the data given above, determine the rate law for the reaction A + B → C.
A) Rate = k[B]
B) Rate = k[A][B]
C) Rate = k[A]2[B]
D) Rate = k[A][B]2
Solution: Comparing Experiments 1 and 2, we notice that when [B] doubled (and [A] remained unchanged), the reaction rate doubled. Therefore, the reaction is first order with respect to [B]; this eliminates choice D. Now, comparing Experiments 3 and 4, we notice that when [A] doubled (and [B] remained unchanged), the reaction rate also doubled. This means that the reaction is first order with respect to [A] as well. Therefore, the answer is B.
Example 9-6: Which of the following gives the form of the rate law for the balanced reaction
4 A + 2 B → C + 3 D?
A) Rate = k[A]4[B]2
B) Rate = k[A]2[B]
C) Rate = k[C][D]3/[A]4[B]2
D) Cannot be determined from the information given
Solution: Unless the given reaction is the rate-determining, elementary step, we have no way of knowing what the rate law is. The answer is D.
Example 9-7: Using the data given above, determine the numerical value of the rate constant for the reaction A + B → C.
Solution: First, let’s find the rate law. Comparing Experiments 3 and 4, we notice that when [A] doubled (and [B] remained unchanged), the reaction rate increased by a factor of 4. This means the reaction is second order with respect to [A]. Comparing Experiments 1 and 3, we notice that when [B] doubled (and [A] remained unchanged), the reaction rate increased by a factor of 2. This means the reaction is first order with respect to [B]. Therefore, the rate law is rate = k[A]2[B]. Finally, using any of the experiments, we can solve for k; using the data in Experiment 1, say, we find that
Chapter 9 Summary
• Kinetics is the study of how quickly a reaction occurs, but does not determine whether or not a reaction will occur.
• All rates are experimentally determined by measuring a change in the concentration of a reactant or product compared to a change in time (often given in M/s).
• Molecules must collide in order to react, and the frequency and energy of these collisions determines how fast the reaction occurs.
• Increasing the concentration of reactants often increases the reaction rate due to an increased number of collisions.
• Increasing the temperature of a reaction always increases the reaction rate since molecules move faster and collide more frequently; the energy of collisions also increases.
• Activation energy (Ea) is the minimum energy required to start a reaction and decreases in the presence of a catalyst, thereby increasing the reaction rate.
• Transition states are at energy maxima, while intermediates are at local energy minima along a reaction coordinate.
• A reaction mechanism must agree with experimental data, and suggests a possible pathway by which reactants and intermediates might collide in order for a chemical reaction to occur.
• The sum of all elementary steps of a mechanism will add to give the overall chemical reaction.
• The slow step of the mechanism is the rate limiting step, and determines the rate of the overall reaction.
• A rate law can only be determined from experimental data or if given a mechanism, and has the general form: Rate = k [reactants]x, where x is the order of the reaction with respect to the given reactant, and k is the rate constant.
• The overall order of a reaction is the sum of all exponents in the rate law.
• The value of the rate constant, k, depends on temperature and activation energy, and its units will vary depending on the reaction order.
• Coefficients of the reactants in the rate limiting step of a mechanism can be used to determine the order of a reaction in the rate law; coefficients from the overall reaction alone CANNOT be used to find the order of a reaction.
CHAPTER 9 FREESTANDING PRACTICE QUESTIONS
1. In the reaction A + 2 B → C, the rate law is experimentally determined to be rate = k[B]2. What happens to the initial rate of reaction when the concentration of A is doubled?
A) The rate doubles.
B) The rate quadruples.
C) The rate is halved.
D) The rate is unchanged.
2. Which of the following statements is always true about the kinetics of a chemical reaction?
A) The rate law includes all reactants in the balanced overall equation.
B) The overall order equals the sum of the reactant coefficients in the overall reaction.
C) The overall order equals the sum of the reactant coefficients in the slow step of the reaction.
D) The structure of the catalyst remains unchanged throughout the reaction progress.
3. Which of the following is represented by a localized minimum in a reaction coordinate diagram?
A) Transition state
C) Activated complex
4. Which factor always affects both thermodynamic and kinetic properties?
B) Transition state energy level
C) Reactant coefficients of the overall reaction
D) No single factor always affects both thermodynamics and kinetics.
5. Which of the following best describes the role of pepsin in the process of proteolysis?
A) It stabilizes the structure of the amino acid end products.
B) It lowers the energy requirement needed for the reaction to proceed.
C) It increases the Keq of proteolysis.
D) It lowers the free energy of the peptide reactant.
6. Based on the reaction mechanism shown below, which of the following statements is correct?
2 NO + O2 → 2 NO2
1) 2 NO → N2O2 (fast)
2) N2O2 + O2 → 2 NO2 (slow)
A) Step 1 is the rate-determining step and the rate of the overall reaction is k[N2O2].
B) Step 1 is the rate-determining step and the rate of the overall reaction is k[NO]2.
C) Step 2 is the rate-determining step and the rate of the overall reaction is k[NO2]2.
D) Step 2 is the rate-determining step and the rate of the overall reaction is k[N2O2][O2].
7. When table sugar is exposed to air it undergoes the following reaction:
C12H22O11 + 12 O2 → 12 CO2 + 11 H2O
(∆G = −5693 kJ/mol)
When this reaction is observed at the macroscopic level, it appears as though nothing is happening, yet one can detect trace amounts of CO2 and H2O being formed. These observations are best explained by the fact that the reaction is:
A) thermodynamically favorable but not kinetically favorable.
B) kinetically favorable but not thermodynamically favorable.
C) neither kinetically nor thermodynamically favorable.
D) both kinetically and thermodynamically favorable.
CHAPTER 9 PRACTICE PASSAGE
One way to determine a rate law is to look at the slowest elementary step in a reaction mechanism. The rate law is equal to the rate constant times the initial concentrations of the reactants in the slowest step raised to the power of their coefficients in the balanced equation. If a chemical appears in the rate law raised to the X power, we say the reaction is X order for that chemical.
In cases where the slow step is not the first step, the rate law will likely depend on the concentration of intermediate species. This is experimentally inconvenient, since the concentration of intermediates is not as straightforward to control as starting materials. As such, rate laws are often rewritten substituting terms consisting solely of starting materials, when possible. For example, consider the decomposition of nitramide:
O2 NNH2 (aq) → N2O(g) + H2O(l)
This reaction consists of three elementary steps (shown below), with step 2 as the slow step.
Step 1 (fast equilibrium):
O2NNH2(aq) O2NNH−(aq) + H+(aq)
Step 2 (slow):
O2NNH−(aq) → N2O(g) + OH−(aq)
Step 3 (fast):
H+(aq) + OH−(aq) → H2O(l)
One could write a valid rate law for this reaction of the form:
rate = k[O2NNH−]
However, the inclusion of the intermediate term is not ideal. The fast equilibrium in Step 1 allows the substitution of [O2NNH−] according to the equilibrium condition:
Solving for [O2NNH−], and substituting into the rate law gives an equally valid expression, detailing how the rate may be altered by varying the concentration of starting material and the pH of the reaction mixture:
The mechanism for this reaction consists of three elementary steps. The first step is an equilibrium reaction with a significant back reaction and the last step is an equilibrium reaction lying so far to the right that we consider it to go to completion:
1. What is the order of the decomposition of nitramide in water with respect to H+?
A) Negative first order
B) One half order
C) First order
D) Second order
2. If Step 1 were simply a fast reaction and not a fast equilibrium, what would be the expected rate law for the decomposition of nitramide in water?
A) Rate = k[O2NNH2]
B) Rate = k[O2NNH−]
C) Rate = k[H+][OH−]
D) Rate =
3. If separately synthesized Na+[O2NNH−] were added to a reaction in progress (assuming total solubility of the salt), what effect would this have on the rate?
A) No reaction would be observed.
B) The rate of the reaction would decrease.
C) The rate of the reaction would increase.
D) It is impossible to tell without experimental data.
4. If the [H+] goes up by a factor of four, the reaction rate will
A) increase by a factor of four.
B) increase by a factor of two.
C) decrease by a factor of two.
D) decrease by a factor of four.
5. Considering Step 1 in isolation, if a known amount of O2NNH2 is dissolved in water, which of the following plays a role in determining how fast the reaction reaches equilibrium?
A) The pH of the solution
B) The reaction temperature
C) The magnitude of the equilibrium constant
D) The stability of O2NNH2 compared to O2NNH−−and H+
6. What is true regarding the enthalpy and entropy changes for Step 3 of the mechanism?
A) ∆H > 0, ∆S > 0
B) ∆H > 0, ∆S < 0
C) ∆H < 0, ∆S > 0
D) ∆H < 0, ∆S < 0
7. Which of the following best describes how a catalyst might speed up a reaction?
A) It raises the temperature of the reaction mixture.
B) It raises the activation energy, making it easier to overcome the energy barrier.
C) It increases the rate of the slowest step.
D) It increases the frequency of collisions between molecules.
SOLUTIONS TO CHAPTER 9 FREESTANDING PRACTICE QUESTIONS
1. D Since the rate law is independent of [A], (i.e., rate is only dependent on the concentration of B), changing the amount of A will have no effect on the rate.
2. C Choice A is incorrect because rate laws are dependent on the slowest step. If a reactant does not participate in the slow step, it will not be included in the overall rate law. Choice B is incorrect because rate laws of overall reactions can only be determined experimentally. Choice D is incorrect because while it is true that a catalyst comes out of a reaction unchanged, it can undergo temporary transformations during the reaction and revert back into its original form at the end. Choice C is the best option because rate laws can be determined from elementary steps of a reaction mechanism by simply raising the reactants to their respective coefficients.
3. D It should be noted that choices A and C are the same and should therefore be eliminated. Additionally, transition states are localized maximums, not minimums. Choice B is incorrect because the product for a spontaneous reaction is the absolute minimum and not a localized minimum. Intermediates are formed and then used. They have a certain lifespan represented by a local minimum on the reaction coordinate diagram.
4. A Choice B is purely a kinetic factor and can be eliminated. Choice C is eliminated because it dictates the thermodynamic quantity Keq but not necessarily the kinetics of the overall reaction (only of the rate limiting step). Gibbs free energy, a thermodynamic property, is defined as ∆G = ∆H — T∆S, and the Arrhenius equation defines the rate constant k, a kinetic property, as k = Ae(—Ea/RT). Both equations contain the T variable representing temperature. Therefore, choice A is correct and choice D must be incorrect.
5. B Pepsin is an enzyme, a biological catalyst. Catalysts lower the activation energy by providing the correct orientation of reactants for a reaction to proceed. Enzymes make a reaction go faster and affect the kinetics of the reaction, making choice B the best answer. Stability of the products, Keq, and free energy of the reactants are all thermodynamic properties, so choices A, C, and D are eliminated.
6. D The rate-determining step (RDS) of a reaction mechanism is the slowest step of that mechanism, eliminating choices A and B. The rate law of an elementary step can be determined from the coefficients of the reactants in the elementary step. Because Step 2 is the RDS, the overall rate law will be equivalent to the rate law for the RDS. Therefore, rate = k[N2O2][O2].
7. A Given the very negative ∆G value, this is a very thermodynamically favorable, spontaneous chemical reaction (eliminate choices B and C). It is important to make the distinction in this case between kinetics and thermodynamics. The reason only trace amounts of products are formed is that the reaction proceeds at an incredibly slow rate (therefore NOT kinetically favorable) due to a high activation energy.
SOLUTIONS TO CHAPTER 9 PRACTICE PASSAGE
1. A The passage states that if a chemical is raised to the X power, the reaction is X order with respect to that chemical. The rate law can be written as Rate = k[O2NNH2][H+]−1, so the reaction is negative first order with respect to H+.
2. B If the first step was simply a fast step, then we would be able to make the normal assumptions about elementary steps and rate laws. More specifically, the rate law could be determined by the stoichiometry of the slow Step 2, which would yield the rate law in choice B.
3. C Recall that re-writing the rate law in terms of observable starting conditions does not make the rate including [O2NNH−] invalid. The two rate laws are equivalent to one another. As such, increasing the concentration of [O2NNH−] will increase the reaction rate.
4. D Since [H+] is in the denominator of the rate law expression and raised to the first power, if [H+] goes up by a factor of four, the rate will go down by a factor of four.
5. B The question requires an answer related to the kinetics of the reaction. Choices A, C, and D are not kinetic factors. The temperature of a reaction is factored into the rate constant (k = Ae(−Ea/RT)), so it will play a role in determining the speed of progress to equilibrium.
6. D Neutralization reactions release large amounts of heat, so ∆H must be less than zero, eliminating choices A and B. From the point of view of entropy, two molecules become one, increasing the order of the system. Moreover, two aqueous species turning into a pure liquid increases order. Therefore, disorder decreases and ∆S must be less than zero.
7. C Catalysts don’t work by raising the temperature of the reaction mixture, so choice A is eliminated. They lower the activation energy, not raise it, so choice B is wrong. Catalysts sometimes work by speeding up the slowest step (and sometimes by providing an alternate, faster reaction mechanism), so choice C is a viable answer. Catalysts can’t change the frequency of collisions between molecules, leaving choice C as the best answer.