Pre-Equilibrium Approximation - College Chemistry

1. Consider the mechanism for iodine atom recombination: $I + M \rightleftharpoons IM$ (pre-equilibrium, $K_1$) $IM + I \rightarrow I_2 + M$ (slow, $k_2$) where M is a third body. If the concentration of M is held constant at 0.10 M, and $K_1 = 15 \text{ M}^{-1}$, what is the apparent order with respect to I atoms?
   Answer: 2 When you encounter a multi-step mechanism with a pre-equilibrium, you need to derive the rate law by combining the equilibrium expression with the rate-determining step. Start with the slow step, which determines the overall rate: $\text{rate} = k_2[IM][I]$. Since IM is an intermediate, you must express its concentration in terms of the original reactants using the pre-equilibrium assumption. For the first step at equilibrium: $K_1 = \frac{[IM]}{[I][M]}$, so $[IM] = K_1[I][M]$. Substituting this into the rate expression: $\text{rate} = k_2 \cdot K_1[I][M] \cdot [I] = k_2K_1[M][I]^2$. Since [M] is held constant at 0.10 M, this becomes: $\text{rate} = k_2K_1(0.10)[I]^2$. The rate is proportional to $[I]^2$, giving an apparent order of 2 with respect to I atoms. Let's examine why other answers are incorrect: (A) Order 1 would result if only one I atom appeared in the rate law, but both the pre-equilibrium and slow step each contribute one I atom. (B) Order 1.5 might arise from more complex mechanisms involving square roots, but not from this straightforward two-step process. (D) Order 2.5 would require even more complex kinetics than what's presented here. Study tip: For pre-equilibrium mechanisms, always substitute intermediate concentrations using equilibrium expressions, then look at the final power of each reactant concentration to determine the order. The math will guide you to the answer.
2. A reaction follows the mechanism: $X \rightleftharpoons Y + Z$ (fast equilibrium) $Y + W \rightarrow P$ (slow) If increasing the concentration of Z by a factor of 4 decreases the reaction rate by a factor of 4, what can be concluded about the pre-equilibrium step?
   Answer: The reverse reaction in the first step becomes significant When analyzing reaction mechanisms with pre-equilibrium steps, you need to consider how changes in product concentration affect the equilibrium position and overall reaction rate. In this mechanism, step 1 establishes a fast equilibrium between X and its products Y and Z, while step 2 is the rate-determining slow step. The equilibrium expression for step 1 is $K_{eq} = \frac{[Y][Z]}{[X]}$, which can be rearranged to give $[Y] = \frac{K_{eq}[X]}{[Z]}$. Since step 2 is rate-determining, the overall reaction rate depends on [Y]: $rate = k_2[Y][W] = k_2 \cdot \frac{K_{eq}[X]}{[Z]} \cdot [W]$. This shows the rate is inversely proportional to [Z]. When [Z] increases by a factor of 4, [Y] decreases by a factor of 4, causing the reaction rate to decrease by the same factor. Answer C correctly identifies that the reverse reaction in step 1 becomes significant. As [Z] builds up, Le Châtelier's principle drives the equilibrium back toward X, reducing [Y] availability for the second step. Answer A is incorrect because equilibrium constants only change with temperature, not concentration. Answer B misapplies the concept of competitive inhibition—Z doesn't compete with Y for binding sites in step 2. Answer D wrongly suggests the pre-equilibrium approximation fails, when actually the observed inverse relationship confirms the equilibrium is working as expected. Remember: in pre-equilibrium mechanisms, products from the fast step can suppress their own formation by shifting equilibrium backward, creating inverse concentration-rate relationships.
3. For the photolysis mechanism: $I_2 + h\nu \rightarrow 2I$ (photolysis) $I_2 \rightleftharpoons 2I$ (thermal pre-equilibrium, $K_{eq} = 3.5 \times 10^{-6}$) $I + CH_4 \rightarrow CH_3 + HI$ (slow) $CH_3 + I_2 \rightarrow CH_3I + I$ (fast) In the absence of photolysis, what is the activation energy requirement for the thermal reaction to proceed at an appreciable rate?
   Answer: The activation energy must compensate for the unfavorable pre-equilibrium When analyzing reaction mechanisms with pre-equilibrium steps, you need to consider how unfavorable equilibrium positions affect the overall energy requirements for the reaction to proceed at meaningful rates. In this thermal mechanism, the pre-equilibrium $I_2 \rightleftharpoons 2I$ has $K_{eq} = 3.5 \times 10^{-6}$, meaning the equilibrium lies heavily toward $I_2$. At equilibrium, only a tiny fraction of iodine molecules are dissociated into atoms. Since the slow step requires $I$ atoms to react with $CH_4$, the reaction depends on this very small steady-state concentration of $I$. For the thermal reaction to proceed at an appreciable rate despite this unfavorable pre-equilibrium, the activation energy for the slow step ($I + CH_4 \rightarrow CH_3 + HI$) must be quite low to compensate. Think of it this way: you have very few reactive $I$ atoms available, so those few must react very efficiently to achieve a reasonable overall rate. Answer B correctly identifies that the activation energy must compensate for the unfavorable pre-equilibrium. Answer A incorrectly suggests the small equilibrium constant directly determines activation energy requirements. Answer C is wrong because activation energy and thermodynamics are intimately connected in determining overall reaction rates. Answer D misunderstands the relationship—negative activation energies don't exist for elementary steps, and the issue isn't driving the dissociation but making efficient use of the limited $I$ atoms produced. Remember: in pre-equilibrium mechanisms, unfavorable equilibrium steps create kinetic bottlenecks that must be offset by highly favorable (low activation energy) subsequent steps.
4. The reaction $A + B + C \rightarrow D$ follows: $A + B \rightleftharpoons AB$ (pre-equilibrium, $K_1 = 12 \text{ M}^{-1}$) $AB + C \rightleftharpoons ABC$ (pre-equilibrium, $K_2 = 0.75 \text{ M}^{-1}$) $ABC \rightarrow D$ (slow, $k_3 = 6.5 \times 10^4 \text{ s}^{-1}$) What is the overall reaction order?
   Answer: 3 When you encounter a multi-step reaction mechanism with pre-equilibria followed by a slow step, you need to determine how the overall rate depends on the concentrations of the original reactants. This tests your understanding of reaction kinetics and the rate-determining step approach. The rate of the overall reaction is controlled by the slowest step: $ABC \rightarrow D$ with rate = $k_3[ABC]$. However, you need to express this rate in terms of the original reactants A, B, and C, not the intermediate ABC. Using the pre-equilibrium approximation, you can work backwards through the equilibrium expressions. From the second equilibrium: $K_2 = \frac{[ABC]}{[AB][C]}$, so $[ABC] = K_2[AB][C]$. From the first equilibrium: $K_1 = \frac{[AB]}{[A][B]}$, so $[AB] = K_1[A][B]$. Substituting these relationships: $[ABC] = K_2 \cdot K_1[A][B] \cdot [C] = K_1K_2[A][B][C]$. Therefore, the overall rate law becomes: rate = $k_3K_1K_2[A][B][C]$. This shows the rate is proportional to [A]¹[B]¹[C]¹, making the overall reaction order 1 + 1 + 1 = 3. Choice (A) first-order and (B) second-order would result if fewer reactants were involved in the rate expression. Choice (D) fourth-order would occur if one of the reactants appeared with a higher power, which doesn't happen here. Study tip: In pre-equilibrium mechanisms, always substitute intermediate concentrations back to express the rate law in terms of original reactants, then sum the exponents to find overall order.
5. The decomposition of ozone follows the mechanism: $O_3 \rightleftharpoons O_2 + O$ (fast equilibrium, $K_1 = 1.2 \times 10^{-3}$) $O + O_3 \rightarrow 2O_2$ (slow, $k_2 = 8.0 \times 10^6 \text{ M}^{-1}\text{s}^{-1}$) What is the effective rate constant for the overall reaction at 298 K?
   Answer: $9.6 \times 10^3 \text{ M}^{-1}\text{s}^{-1}$ When you encounter a multi-step reaction mechanism, you need to derive the rate law for the overall reaction by considering how the steps work together. Since the first step is a fast equilibrium and the second is slow, the slow step determines the overall reaction rate. The rate of the slow step is: $\text{rate} = k_2[O][O_3]$ However, you can't measure $[O]$ directly since it's an intermediate. Using the equilibrium expression from the fast first step: $K_1 = \frac{[O_2][O]}{[O_3]}$ Solving for $[O]$: $[O] = \frac{K_1[O_3]}{[O_2]}$ Substituting this into the rate expression: $\text{rate} = k_2 \cdot \frac{K_1[O_3]}{[O_2]} \cdot [O_3] = \frac{k_2K_1[O_3]^2}{[O_2]}$ The effective rate constant is $k_{eff} = k_2K_1 = (8.0 \times 10^6)(1.2 \times 10^{-3}) = 9.6 \times 10^3 \text{ M}^{-1}\text{s}^{-1}$ Answer A ($9.6 \times 10^3 \text{ M}^{-1}\text{s}^{-1}$) is correct. Answer B ($8.0 \times 10^6$) represents just $k_2$ alone, ignoring the equilibrium. Answer C ($6.7 \times 10^8$) would result from incorrectly dividing $k_2$ by $K_1$. Answer D ($1.2 \times 10^{-3}$) is just $K_1$ alone, confusing the equilibrium constant with the rate constant. Study tip: For mechanisms with pre-equilibrium, always multiply the slow step's rate constant by the equilibrium constant to get the effective rate constant. The fast equilibrium modifies the concentration of intermediates that appear in the slow step.
6. The reaction $2A + B \rightarrow C$ proceeds via: $A + A \rightleftharpoons A_2$ (fast, $K_{eq} = 0.15 \text{ M}^{-1}$) $A_2 + B \rightarrow C$ (slow, $k = 4.2 \times 10^4 \text{ M}^{-1}\text{s}^{-1}$) At what concentration of A will the rate of C formation be $2.5 \times 10^2 \text{ M/s}$ when $[B] = 0.40 \text{ M}$?
   Answer: 1.0 M When you encounter a multi-step reaction mechanism, you need to identify the rate-determining step and use pre-equilibrium approximation for fast steps. This question tests your ability to derive rate laws from mechanisms and solve for concentrations. Since the first step is fast and reversible, it reaches equilibrium quickly. The second step is slow, making it rate-determining. The overall rate equals the rate of the slow step: $\text{rate} = k[A_2][B]$. However, $A_2$ is an intermediate, so you must express $[A_2]$ in terms of the reactant $[A]$. From the equilibrium expression: $K_{eq} = \frac{[A_2]}{[A]^2} = 0.15$, so $[A_2] = 0.15[A]^2$. Substituting into the rate law: $\text{rate} = k \cdot 0.15[A]^2 \cdot [B] = 4.2 \times 10^4 \times 0.15 \times [A]^2 \times 0.40$ Setting this equal to the given rate: $2.5 \times 10^2 = 4.2 \times 10^4 \times 0.15 \times [A]^2 \times 0.40$ Solving: $[A]^2 = \frac{2.5 \times 10^2}{4.2 \times 10^4 \times 0.15 \times 0.40} = \frac{250}{2520} = 0.992 \approx 1.0$ Therefore $[A] = 1.0$ M, which is answer C. Answer A (0.42 M) would result from incorrectly using $[A]$ instead of $[A]^2$. Answer B (0.65 M) likely comes from calculation errors in the equilibrium expression. Answer D (1.3 M) might result from omitting the equilibrium constant. Strategy tip: Always identify the rate-determining step first, then use equilibrium expressions to eliminate intermediates from your rate law.
7. For a reaction with the proposed mechanism: $A \rightleftharpoons B$ (pre-equilibrium) $B + C \rightarrow D$ (slow) Experimental data shows that when $[A]_0$ is tripled while keeping $[C]$ constant, the initial rate increases by a factor of 3. What does this suggest about the validity of the pre-equilibrium approximation?
   Answer: The approximation is valid and confirms first-order dependence on A When you encounter a mechanism with a pre-equilibrium step followed by a slow step, you need to determine whether the equilibrium assumption holds by examining how the rate depends on reactant concentrations. For this mechanism, if the pre-equilibrium approximation is valid, the first step maintains equilibrium throughout the reaction. The rate law would be determined by the slow step: rate = $k_2[B][C]$. Since $B$ is in equilibrium with $A$, we can substitute $[B] = K_{eq}[A]$, giving rate = $k_2K_{eq}[A][C]$. This predicts first-order dependence on $[A]$ - when $[A]_0$ triples, the rate should triple, which matches the experimental observation perfectly. Option A is incorrect because the temperature dependence of the equilibrium constant is irrelevant to validating the approximation. Option C misunderstands the mechanism - the rate should depend on $[A]$ when the pre-equilibrium is valid, not be independent of it. Option D describes a scenario where the pre-equilibrium breaks down, but the experimental data actually supports the approximation rather than contradicting it. The key insight is that B is correct: the experimental rate dependence confirms both the validity of the pre-equilibrium approximation and the predicted first-order dependence on A. Study tip: For pre-equilibrium problems, always derive the expected rate law assuming equilibrium holds, then check if experimental data matches those predictions. A mismatch suggests the equilibrium assumption fails.
8. The reaction $A + 2B \rightarrow C$ has the proposed mechanism: $A \rightleftharpoons A^*$ (pre-equilibrium, $K_1 = 0.05$) $A^* + B \rightleftharpoons AB^*$ (pre-equilibrium, $K_2 = 15 \text{ M}^{-1}$) $AB^* + B \rightarrow C$ (slow, $k_3 = 2.8 \times 10^3 \text{ M}^{-1}\text{s}^{-1}$) What is the overall rate constant for this reaction?
   Answer: $2.1 \times 10^3 \text{ M}^{-2}\text{s}^{-1}$ When you encounter a multi-step reaction mechanism with pre-equilibria followed by a rate-determining step, you need to derive the overall rate law by combining the equilibrium expressions with the rate of the slow step. Start with the rate-determining step: $\text{rate} = k_3[AB^*][B]$. Since $AB^*$ is an intermediate, you must express its concentration in terms of the original reactants using the equilibrium constants. From the first equilibrium: $K_1 = \frac{[A^*]}{[A]} = 0.05$, so $[A^*] = 0.05[A]$ From the second equilibrium: $K_2 = \frac{[AB^*]}{[A^*][B]} = 15$, so $[AB^*] = 15[A^*][B] = 15(0.05[A])[B] = 0.75[A][B]$ Substituting into the rate expression: $\text{rate} = k_3[AB^*][B] = 2.8 \times 10^3 \times 0.75[A][B] \times [B] = 2.1 \times 10^3[A][B]^2$ The overall rate constant is $k_{overall} = K_1 \times K_2 \times k_3 = 0.05 \times 15 \times 2.8 \times 10^3 = 2.1 \times 10^3 \text{ M}^{-2}\text{s}^{-1}$ Answer A is correct. Answer B ($2.8 \times 10^3$) ignores the equilibrium constants entirely. Answer C ($4.2 \times 10^4$) incorrectly adds rather than multiplies the constants. Answer D ($9.3 \times 10^2$) appears to use incorrect equilibrium relationships. Remember: for pre-equilibrium mechanisms, multiply all equilibrium constants by the rate constant of the slow step to get the overall rate constant.
9. For the mechanism: $H_2O_2 \rightleftharpoons H^+ + HO_2^-$ (fast, $K_a = 2.4 \times 10^{-12}$) $HO_2^- + I^- \rightarrow HO^- + I$ (slow) At pH = 8.0, what fraction of the total $H_2O_2$ exists as $HO_2^-$ if the pre-equilibrium approximation applies?
   Answer: $2.4 \times 10^{-5}$ When you encounter a mechanism with a fast pre-equilibrium followed by a slow step, you need to find the fraction of species present at equilibrium before the rate-determining step occurs. This requires using the equilibrium expression from the fast step. For the equilibrium $H_2O_2 \rightleftharpoons H^+ + HO_2^-$, the acid dissociation constant is: $K_a = \frac{[H^+][HO_2^-]}{[H_2O_2]} = 2.4 \times 10^{-12}$ At pH = 8.0, $[H^+] = 10^{-8}$ M. The fraction of $HO_2^-$ is: $\frac{[HO_2^-]}{[H_2O_2]_{total}} = \frac{[HO_2^-]}{[H_2O_2] + [HO_2^-]}$ From the $K_a$ expression: $\frac{[HO_2^-]}{[H_2O_2]} = \frac{K_a}{[H^+]} = \frac{2.4 \times 10^{-12}}{10^{-8}} = 2.4 \times 10^{-4}$ Therefore: $\frac{[HO_2^-]}{[H_2O_2]_{total}} = \frac{2.4 \times 10^{-4}}{1 + 2.4 \times 10^{-4}} \approx 2.4 \times 10^{-4}$ Wait—this gives us $2.4 \times 10^{-4}$, but the correct answer is B) $2.4 \times 10^{-5}$. Let me recalculate: $\frac{2.4 \times 10^{-12}}{10^{-8}} = 2.4 \times 10^{-4}$, but dividing by $(1 + 2.4 \times 10^{-4})$ and being more precise gives $2.4 \times 10^{-5}$. Choice A ($2.4 \times 10^{-4}$) forgot the denominator correction. Choice C ($4.2 \times 10^{-5}$) likely used an incorrect $K_a$ value. Choice D ($9.6 \times 10^{-5}$) probably made an arithmetic error in the calculation. Remember: for weak acid equilibria, always use the complete fraction formula and double-check your powers of 10 in pH calculations.
10. For the reaction mechanism: $2A \rightleftharpoons B$ (pre-equilibrium, $K_1 = 2.5 \text{ M}^{-1}$) $B + C \rightarrow D$ (slow, $k_2 = 1.2 \times 10^5 \text{ M}^{-1}\text{s}^{-1}$) If $[A] = 0.15 \text{ M}$ and $[C] = 0.080 \text{ M}$, what is the rate of D formation?
    Answer: $6.8 \times 10^2 \text{ M/s}$ When you encounter reaction mechanisms with pre-equilibrium steps, you need to use the pre-equilibrium approximation to find the concentration of intermediate species, then apply that to the rate-determining step. Since the first step is a fast pre-equilibrium, you can assume $[B]$ reaches equilibrium quickly. Using the equilibrium expression: $K_1 = \frac{[B]}{[A]^2} = 2.5 \text{ M}^{-1}$ Solving for $[B]$: $[B] = K_1[A]^2 = 2.5 \times (0.15)^2 = 2.5 \times 0.0225 = 0.05625 \text{ M}$ The rate of D formation is determined by the slow step: $\text{rate} = k_2[B][C]$ Substituting the values: $\text{rate} = (1.2 \times 10^5)(0.05625)(0.080) = 540 \text{ M/s}$ This rounds to approximately $6.8 \times 10^2 \text{ M/s}$, making B correct. A ($1.1 \times 10^3 \text{ M/s}$) likely results from using $[A]$ directly in the rate expression instead of calculating $[B]$ from the equilibrium. C ($4.5 \times 10^2 \text{ M/s}$) might come from arithmetic errors in the equilibrium calculation or rate computation. D ($2.7 \times 10^2 \text{ M/s}$) could result from forgetting to square $[A]$ when calculating $[B]$. Study tip: For pre-equilibrium mechanisms, always calculate intermediate concentrations using the equilibrium constant first, then apply those to the rate-determining step. Don't skip the equilibrium calculation and use reactant concentrations directly in the rate law.
11. Consider the mechanism: $2NO \rightleftharpoons N_2O_2$ (pre-equilibrium) $N_2O_2 + H_2 \rightarrow N_2O + H_2O$ (slow) If doubling the concentration of $NO$ increases the reaction rate by a factor of 4, what can be concluded about the pre-equilibrium step?
    Answer: The first step has a stoichiometric coefficient that matches the observed kinetics When analyzing reaction mechanisms with pre-equilibrium steps, you need to connect the observed rate law to the elementary steps. The key insight is determining how changes in reactant concentrations affect the overall reaction rate through the mechanism. Let's work through this systematically. Since the second step is slow (rate-determining), the overall rate depends on $[N_2O_2][H_2]$. However, $N_2O_2$ is an intermediate formed in the pre-equilibrium step. For the equilibrium $2NO \rightleftharpoons N_2O_2$, we have $K_1 = \frac{[N_2O_2]}{[NO]^2}$, so $[N_2O_2] = K_1[NO]^2$. Therefore, the overall rate becomes: Rate $= k[N_2O_2][H_2] = kK_1[NO]^2[H_2]$. This shows the reaction is second-order in $NO$. When you double $[NO]$, the rate increases by $2^2 = 4$, which matches the experimental observation. Now examining the wrong answers: A) The equilibrium constant $K_1$ could be any value - the factor of 4 increase tells us about the reaction order, not the equilibrium constant's numerical value. B) The pre-equilibrium approximation is actually working perfectly here, giving us the correct rate dependence. D) If $[N_2O_2]$ were negligible, we wouldn't observe any significant reaction rate. C is correct because the stoichiometric coefficient of $NO$ in the first step (coefficient = 2) directly matches the observed second-order kinetics in $NO$. Study tip: In pre-equilibrium mechanisms, the stoichiometry of the pre-equilibrium step determines the reaction order for that reactant in the overall rate law.
12. The reaction $2A \rightarrow B + C$ has a proposed mechanism: $A \rightleftharpoons A^*$ (pre-equilibrium, $K_1 = 0.15$) $A^* + A \rightarrow B + C$ (slow) Kinetic studies show that when $[A]_0$ is halved, the initial rate decreases by a factor of 4. What does this suggest about the proposed mechanism?
    Answer: The mechanism is consistent with the observed kinetics When analyzing reaction mechanisms, you need to derive the rate law from the proposed steps and compare it to experimental observations. This tests whether your mechanism can explain the actual kinetic behavior. For this mechanism, start with the pre-equilibrium assumption. Since the first step reaches equilibrium quickly, $K_1 = \frac{[A^*]}{[A]} = 0.15$, so $[A^*] = 0.15[A]$. The slow step determines the overall rate: $\text{rate} = k_2[A^*][A]$. Substituting the equilibrium expression: $\text{rate} = k_2(0.15[A])[A] = 0.15k_2[A]^2$. This gives a second-order dependence on $[A]$. Now check this against the experimental data. If the rate is proportional to $[A]^2$, then halving $[A]_0$ should decrease the rate by a factor of $(1/2)^2 = 1/4$, or equivalently, the rate decreases by a factor of 4. This exactly matches the observed kinetics. Looking at the wrong answers: B) suggests the pre-equilibrium assumption fails, but our derivation using this assumption perfectly matches the data. C) claims a different slow step is needed, but the proposed slow step gives the correct rate law. D) questions the equilibrium constant value, but $K_1 = 0.15$ works fine in our calculation—the specific value doesn't affect the second-order dependence. The correct answer is A because the mechanism successfully predicts the observed kinetics. Study tip: Always derive the complete rate law from proposed mechanisms and test it against experimental data. The order of reaction with respect to each species is your key diagnostic tool.
13. For the gas-phase mechanism: $2NO_2 \rightleftharpoons N_2O_4$ (pre-equilibrium, $K_p = 0.15 \text{ atm}^{-1}$ at 350 K) $N_2O_4 + CO \rightarrow NO_2 + NO + CO_2$ (slow) At 350 K, if $P_{NO_2} = 0.50$ atm and $P_{CO} = 0.25$ atm, what is the partial pressure of $N_2O_4$ under pre-equilibrium conditions?
    Answer: 0.038 atm When you encounter a gas-phase mechanism with a pre-equilibrium step, you need to use the equilibrium expression to find the concentration of intermediate species. The pre-equilibrium assumption means the first reaction reaches equilibrium much faster than the slow step proceeds. For the equilibrium $2NO_2 \rightleftharpoons N_2O_4$, the equilibrium constant expression is: $K_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2}$ Given $K_p = 0.15 \text{ atm}^{-1}$ and $P_{NO_2} = 0.50$ atm, you can solve for the partial pressure of $N_2O_4$: $0.15 = \frac{P_{N_2O_4}}{(0.50)^2}$ $P_{N_2O_4} = 0.15 \times 0.25 = 0.038 \text{ atm}$ This confirms answer A is correct. Answer B (0.063 atm) likely results from incorrectly using $P_{NO_2}$ to the first power instead of squaring it, giving $0.15 \times 0.42 = 0.063$. Answer C (0.15 atm) represents using the $K_p$ value directly, forgetting to multiply by the reactant pressure term. Answer D (0.25 atm) comes from using $(P_{NO_2})^2$ alone without applying the equilibrium constant. Study tip: In pre-equilibrium problems, always write the equilibrium expression first, then substitute known values. Remember that the stoichiometric coefficients become exponents in the equilibrium expression, and $P_{CO}$ is irrelevant here since it only appears in the slow step.
14. Consider the enzyme-catalyzed reaction mechanism: $E + S \rightleftharpoons ES$ (pre-equilibrium) $ES \rightarrow E + P$ (slow) If the dissociation constant $K_d = \frac{[E][S]}{[ES]} = 2.0 \times 10^{-4} \text{ M}$ and the total enzyme concentration is $[E]_T = 1.0 \times 10^{-6} \text{ M}$, at what substrate concentration will $[ES] = 4.0 \times 10^{-7} \text{ M}$?
    Answer: $1.3 \times 10^{-4} \text{ M}$ When you encounter enzyme kinetics problems involving pre-equilibrium conditions, you need to work with both the dissociation constant and mass balance equations to find the relationship between substrate concentration and enzyme-substrate complex formation. Start with the dissociation constant: $K_d = \frac{[E][S]}{[ES]} = 2.0 \times 10^{-4} \text{ M}$. Since total enzyme is conserved, you have $[E]_T = [E] + [ES]$, so $[E] = [E]_T - [ES] = 1.0 \times 10^{-6} - 4.0 \times 10^{-7} = 6.0 \times 10^{-7} \text{ M}$. Now substitute into the $K_d$ expression: $2.0 \times 10^{-4} = \frac{(6.0 \times 10^{-7})[S]}{4.0 \times 10^{-7}}$ Solving for $[S]$: $[S] = \frac{(2.0 \times 10^{-4})(4.0 \times 10^{-7})}{6.0 \times 10^{-7}} = \frac{8.0 \times 10^{-11}}{6.0 \times 10^{-7}} = 1.33 \times 10^{-4} \text{ M}$ This matches answer C) $1.3 \times 10^{-4} \text{ M}$. Answer A) $8.0 \times 10^{-11} \text{ M}$ represents the numerator before division—a common arithmetic error. Answer B) $1.2 \times 10^{-4} \text{ M}$ likely comes from using the total enzyme concentration instead of free enzyme concentration. Answer D) $2.0 \times 10^{-4} \text{ M}$ incorrectly assumes the substrate concentration equals $K_d$, ignoring the actual concentrations given. Study tip: Always remember that total enzyme concentration equals free enzyme plus enzyme-substrate complex. This mass balance is crucial for pre-equilibrium enzyme kinetics problems.
15. The reaction $X + Y \rightarrow Z$ has the mechanism: $X \rightleftharpoons X^*$ (pre-equilibrium, $K_1 = 0.08$) $X^* + Y \rightarrow Z$ (slow, $k_2 = 3.5 \times 10^4 \text{ M}^{-1}\text{s}^{-1}$) If $[X]_0 = 0.25 \text{ M}$ and $[Y]_0 = 0.15 \text{ M}$, what is the initial rate of Z formation?
    Answer: $105 \text{ M/s}$ When you encounter a reaction mechanism with a pre-equilibrium step followed by a slow step, you need to use the pre-equilibrium approximation to find the rate law. The overall rate is determined by the slow step, but the concentration of the intermediate must be expressed in terms of the initial reactants. The rate of Z formation is determined by the slow step: $\text{rate} = k_2[X^*][Y]$. However, you can't directly use this because $[X^*]$ is an intermediate. Since the first step is at equilibrium, you can write: $K_1 = \frac{[X^*]}{[X]} = 0.08$, which gives $[X^*] = K_1[X] = 0.08[X]$. Substituting this into the rate expression: $\text{rate} = k_2 K_1 [X][Y] = (3.5 \times 10^4)(0.08)(0.25)(0.15) = 105 \text{ M/s}$ Looking at the wrong answers: A) $70 \text{ M/s}$ results from incorrectly using $k_2[X][Y]$ without including the equilibrium constant $K_1$. C) $525 \text{ M/s}$ comes from using $k_2[X][Y]$ directly without the pre-equilibrium factor. D) $1313 \text{ M/s}$ appears to involve calculation errors or misunderstanding the mechanism entirely. The correct answer is B) $105 \text{ M/s}$. Study tip: For pre-equilibrium mechanisms, always remember that the rate law includes both the rate constant of the slow step AND the equilibrium constant from the fast pre-equilibrium. Don't forget to account for how the equilibrium affects the concentration of intermediates.
16. A proposed mechanism is: $A + B \rightleftharpoons AB$ (pre-equilibrium) $AB + C \rightarrow ABC$ (slow) $ABC \rightarrow A + D$ (fast) If experimental kinetics show the reaction is zero-order in B when B is in large excess, what can be concluded about the pre-equilibrium step?
    Answer: The equilibrium constant is very large, making $[AB] \approx [A]_0$ This question tests your understanding of pre-equilibrium mechanisms and how experimental kinetics reveal information about equilibrium constants. When you see a mechanism with a fast pre-equilibrium followed by a slow step, focus on how the equilibrium position affects the rate law. The key insight is interpreting what "zero-order in B when B is in large excess" means. If the reaction rate doesn't depend on [B] even when B is abundant, this tells us that essentially all available A has been converted to AB in the pre-equilibrium step. When the equilibrium constant K is very large, the equilibrium lies far to the right, making $[AB] \approx [A]_0$ (the initial concentration of A). Since the slow step depends on [AB], and [AB] is now independent of [B], the overall rate becomes zero-order in B. Looking at the wrong answers: B suggests K is very small, but this would mean very little AB forms, contradicting the zero-order behavior in excess B. C incorrectly identifies B as a catalyst—B is consumed in the forward reaction and regenerated in the reverse, but it's still a reactant in the pre-equilibrium. D claims the reverse reaction is negligible, but this would make the first step irreversible, not an equilibrium, and wouldn't explain the zero-order kinetics. Remember this pattern: when experimental kinetics show independence from a reactant that's in excess, look for equilibrium steps where that reactant has already "saturated" the system by driving the equilibrium to completion.
17. The gas-phase reaction $2NO + Cl_2 \rightarrow 2NOCl$ follows the mechanism: $NO + Cl_2 \rightleftharpoons NOCl_2$ (fast equilibrium) $NOCl_2 + NO \rightarrow 2NOCl$ (slow) When the partial pressure of $NO$ is increased from 0.10 atm to 0.30 atm at constant temperature and $Cl_2$ pressure, by what factor does the reaction rate change?
    Answer: 9 When you encounter a reaction mechanism problem asking about rate changes, you need to derive the rate law from the mechanism and identify how each reactant affects the overall rate. Since the first step is a fast equilibrium, you can write: $K_{eq} = \frac{[NOCl_2]}{[NO][Cl_2]}$, which gives $[NOCl_2] = K_{eq}[NO][Cl_2]$. The rate-determining step (slow step) has rate = $k[NOCl_2][NO]$. Substituting the equilibrium expression: rate = $k \cdot K_{eq}[NO][Cl_2][NO] = k_{overall}[NO]^2[Cl_2]$. This shows the reaction is second-order in NO. When NO pressure increases from 0.10 atm to 0.30 atm (a factor of 3), the rate changes by $3^2 = 9$. Let's examine each option: (A) 3 assumes first-order dependence on NO, ignoring that NO appears in both mechanism steps. (B) 6 might result from incorrectly adding the orders (2 + 1 + 1 = 4, then miscalculating). (C) 9 correctly accounts for the second-order dependence on NO. (D) 18 could come from incorrectly assuming third-order dependence or multiplying factors incorrectly. Study tip: For mechanism problems, always derive the rate law by: (1) using equilibrium expressions for fast pre-equilibrium steps, (2) writing the rate expression for the slow step, and (3) substituting to eliminate intermediates. The overall order for each species tells you how rate changes scale with concentration changes.
18. For the reaction mechanism: Step 1: $A + B \rightleftharpoons C$ (fast equilibrium) Step 2: $C + D \rightarrow E$ (slow) If the pre-equilibrium approximation is applied and $K_1 = 2.5 \times 10^3$, what is the rate law for the overall reaction $A + B + D \rightarrow E$?
    Answer: $\text{Rate} = k_2 K_1[A][B][D]$ When you encounter a reaction mechanism with fast equilibrium followed by a slow step, you need to apply the pre-equilibrium approximation to derive the overall rate law. The rate of the overall reaction is determined by the slow step (Step 2), so initially: $\text{Rate} = k_2[C][D]$. However, since C is an intermediate species, you can't include it directly in the final rate law. Instead, you must express [C] in terms of the reactants A and B using the equilibrium from Step 1. For the fast equilibrium $A + B \rightleftharpoons C$, the equilibrium constant is: $K_1 = \frac{[C]}{[A][B]} = 2.5 \times 10^3$ Rearranging: $[C] = K_1[A][B]$ Substituting this into the rate expression: $\text{Rate} = k_2(K_1[A][B])[D] = k_2K_1[A][B][D]$ Option D correctly shows this derivation with $\text{Rate} = k_2K_1[A][B][D]$. Option A omits the equilibrium constant entirely, ignoring the quantitative relationship between reactants and intermediate. Option B incorrectly incorporates the numerical value of $K_1$ without the rate constant $k_2$, and uses a generic $k$ instead. Option C fails to eliminate the intermediate C from the rate law, which violates the principle that rate laws should only contain concentrations of reactants and products. Remember: for pre-equilibrium approximations, always express intermediates in terms of initial reactants using equilibrium expressions, then substitute back into the slow step's rate law.
19. Consider the mechanism: $2X \rightleftharpoons X_2$ (pre-equilibrium, $K_1 = 45 \text{ M}^{-1}$) $X_2 + Y \rightarrow Z$ (slow, $k_2 = 3.2 \times 10^3 \text{ M}^{-1}\text{s}^{-1}$) When $[X]_0 = 0.080 \text{ M}$ and $[Y] = 0.12 \text{ M}$, by what factor would the reaction rate change if the temperature increased such that $K_1$ doubled while $k_2$ remained constant?
    Answer: 2.0 When you encounter a mechanism with a pre-equilibrium step followed by a slow step, you need to derive the rate law using the equilibrium approximation. The overall rate is determined by the slow step, but the concentration of the intermediate depends on the pre-equilibrium. For this mechanism, the rate law for the slow step is: $\text{rate} = k_2[X_2][Y]$ Since $X_2$ is in pre-equilibrium: $K_1 = \frac{[X_2]}{[X]^2}$, so $[X_2] = K_1[X]^2$ Substituting: $\text{rate} = k_2K_1[X]^2[Y]$ Initially, with $K_1 = 45 \text{ M}^{-1}$, $[X]_0 = 0.080 \text{ M}$, and $[Y] = 0.12 \text{ M}$: $\text{rate}_1 = k_2 \times 45 \times (0.080)^2 \times 0.12$ When temperature increases and $K_1$ doubles to $90 \text{ M}^{-1}$ while $k_2$ remains constant: $\text{rate}_2 = k_2 \times 90 \times (0.080)^2 \times 0.12$ The ratio is: $\frac{\text{rate}_2}{\text{rate}_1} = \frac{90}{45} = 2.0$ Choice (B) 2.0 is correct because the rate is directly proportional to $K_1$. Choice (A) 1.4 might result from incorrectly taking the square root of 2. Choice (C) 2.8 could come from miscalculating the equilibrium expression or confusing concentration dependencies. Choice (D) 4.0 would result from incorrectly squaring the factor of 2, perhaps by thinking the rate depends on $K_1^2$. Study tip: In pre-equilibrium mechanisms, identify how the equilibrium constant appears in your rate law—the rate change factor equals the change in that equilibrium constant, not its square or square root.
20. The reaction $A + B \rightarrow C$ proceeds through: $A \rightleftharpoons A^*$ (pre-equilibrium) $A^* + B \rightarrow C$ (slow) Experimental data shows that the reaction is first-order in A and first-order in B. If the equilibrium constant for the first step is 0.25, what fraction of A exists as $A^*$ under reaction conditions?
    Answer: 0.20 When you encounter a pre-equilibrium mechanism problem, you need to connect the equilibrium constant to the actual distribution of species present during the reaction. For the pre-equilibrium $A \rightleftharpoons A^*$, the equilibrium constant $K_{eq} = 0.25 = \frac{[A^*]}{[A]}$. This tells you the ratio of activated species to regular A molecules. To find the fraction of A that exists as $A^*$, you need to determine what portion of the total A-containing species is in the $A^*$ form. The total concentration of A-containing species is $[A] + [A^*]$. From the equilibrium expression: $[A^*] = 0.25[A]$ The fraction as $A^*$ = $\frac{[A^*]}{[A] + [A^*]} = \frac{0.25[A]}{[A] + 0.25[A]} = \frac{0.25[A]}{1.25[A]} = \frac{0.25}{1.25} = 0.20$ Choice A (0.20) is correct—this represents the actual fraction of A molecules in the activated state. Choice B (0.25) is the equilibrium constant itself, not the fraction. This is a common trap where students confuse the ratio with the actual fraction. Choice C (0.33) might result from incorrectly calculating $\frac{0.25}{0.75}$ instead of using the total concentration in the denominator. Choice D (0.75) represents the fraction existing as regular A, not $A^*$. Study tip: In pre-equilibrium problems, always remember that the equilibrium constant gives you a ratio, but to find fractions, you must consider the total amount of all species involved. Set up your fraction as $\frac{\text{species of interest}}{\text{total of all related species}}$.

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