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Exercises: Pre-Equilibrium Approximation

Practice★☆☆
In a mechanism, the first step is a fast, reversible step: $\text{A} + \text{B} \rightleftharpoons \text{C}$, with equilibrium constant $K$. Write the expression for…

In a mechanism, the first step is a fast, reversible step: $\text{A} + \text{B} \rightleftharpoons \text{C}$, with equilibrium constant $K$. Write the expression for $K$, then solve it for $[\text{C}]$. Explain in one sentence what must be true about the forward and reverse rates of this step for the pre-equilibrium approximation to apply.

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Practice★☆☆
A mechanism for $2\,\text{NO} + \text{O}2 \rightarrow 2\,\text{NO}2$ is: Step 1 (fast equilibrium): $2\,\text{NO} \rightleftharpoons \text{N}2\text{O}2$ Step 2 (slow):…

A mechanism for $2\,\text{NO} + \text{O}_2 \rightarrow 2\,\text{NO}_2$ is:

Step 1 (fast equilibrium): $2\,\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$

Step 2 (slow): $\text{N}_2\text{O}_2 + \text{O}_2 \rightarrow 2\,\text{NO}_2$

Identify the intermediate and write an expression for $[\text{N}_2\text{O}_2]$ in terms of $K$ and $[\text{NO}]$.

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Practice★☆☆
For each mechanism, state whether the rate law can be written directly from the slow step or whether you must use a pre-equilibrium to eliminate an intermediate. (i)…

For each mechanism, state whether the rate law can be written directly from the slow step or whether you must use a pre-equilibrium to eliminate an intermediate.

(i) Step 1 slow: $\text{A} + \text{B} \rightarrow \text{X}$; Step 2 fast: $\text{X} + \text{B} \rightarrow \text{P}$.

(ii) Step 1 fast, reversible: $\text{A} + \text{B} \rightleftharpoons \text{X}$; Step 2 slow: $\text{X} + \text{A} \rightarrow \text{P}$.

(iii) Step 1 slow: $\text{A} \rightarrow \text{X}$; Step 2 fast: $\text{X} \rightarrow \text{P}$.

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Practice★★☆
Use the mechanism from the previous section, Step 1 (fast equilibrium) $2\,\text{NO} \rightleftharpoons \text{N}2\text{O}2$ with constant $K$, and Step 2 (slow)…

Use the mechanism from the previous section, Step 1 (fast equilibrium) $2\,\text{NO} \rightleftharpoons \text{N}_2\text{O}_2$ with constant $K$, and Step 2 (slow) $\text{N}_2\text{O}_2 + \text{O}_2 \rightarrow 2\,\text{NO}_2$ with rate constant $k_2$.

(a) Derive the rate law in terms of $[\text{NO}]$ and $[\text{O}_2]$ only. (b) Write the observed rate constant $k_{\text{obs}}$ in terms of $k_2$ and $K$. (c) If $K = 0.040\ \text{M}^{-1}$ and $k_2 = 5.0 \times 10^{3}\ \text{M}^{-1}\text{s}^{-1}$, calculate $k_{\text{obs}}$ with units.

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Practice★★☆
The decomposition of ozone in the upper atmosphere is proposed to follow this mechanism: Step 1 (fast equilibrium): $\text{O}3 \rightleftharpoons \text{O}2 + \text{O}$…

The decomposition of ozone in the upper atmosphere is proposed to follow this mechanism:

Step 1 (fast equilibrium): $\text{O}_3 \rightleftharpoons \text{O}_2 + \text{O}$

Step 2 (slow): $\text{O} + \text{O}_3 \rightarrow 2\,\text{O}_2$

(a) Derive the rate law. (b) State the order with respect to $\text{O}_3$ and with respect to $\text{O}_2$. (c) By what factor does the rate change if $[\text{O}_3]$ is tripled and $[\text{O}_2]$ is doubled? (d) Explain why adding $\text{O}_2$ slows the reaction, using the equilibrium in Step 1.

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Practice★★☆
A mechanism for $\text{H}2 + \text{Br}2 \rightarrow 2\,\text{HBr}$ is: Step 1 (fast equilibrium): $\text{Br}2 \rightleftharpoons 2\,\text{Br}$ Step 2 (slow):…

A mechanism for $\text{H}_2 + \text{Br}_2 \rightarrow 2\,\text{HBr}$ is:

Step 1 (fast equilibrium): $\text{Br}_2 \rightleftharpoons 2\,\text{Br}$

Step 2 (slow): $\text{Br} + \text{H}_2 \rightarrow \text{HBr} + \text{H}$

Step 3 (fast): $\text{H} + \text{Br}_2 \rightarrow \text{HBr} + \text{Br}$

(a) Derive the rate law from the slow step and the pre-equilibrium. (b) State the order with respect to each reactant and the overall order. (c) Identify all intermediates.

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Practice★★★
The reaction $2\,\text{NO} + \text{Br}2 \rightarrow 2\,\text{NOBr}$ is proposed to proceed by: Step 1 (fast equilibrium):…

The reaction $2\,\text{NO} + \text{Br}_2 \rightarrow 2\,\text{NOBr}$ is proposed to proceed by:

Step 1 (fast equilibrium): $\text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2$, constant $K$

Step 2 (slow): $\text{NOBr}_2 + \text{NO} \rightarrow 2\,\text{NOBr}$, rate constant $k_2$

(a) Derive the rate law. (b) With $K = 0.30\ \text{M}^{-1}$ and $k_2 = 2.0 \times 10^{2}\ \text{M}^{-1}\text{s}^{-1}$, calculate $k_{\text{obs}}$ and the rate when $[\text{NO}] = 0.10\ \text{M}$ and $[\text{Br}_2] = 0.050\ \text{M}$. (c) Experiments show that the rate decreases slightly as temperature increases. If the activation energy of Step 2 is small, explain how this can happen when Step 1 is exothermic.

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Practice★★★
For the reaction $2\,\text{A} + \text{B} \rightarrow \text{C}$, two mechanisms are proposed. Mechanism 1: $2\,\text{A} \rightleftharpoons \text{A}2$ (fast equilibrium,…

For the reaction $2\,\text{A} + \text{B} \rightarrow \text{C}$, two mechanisms are proposed.

Mechanism 1: $2\,\text{A} \rightleftharpoons \text{A}_2$ (fast equilibrium, constant $K$); $\text{A}_2 + \text{B} \rightarrow \text{C}$ (slow, $k_2$).

Mechanism 2: $2\,\text{A} \rightarrow \text{A}_2$ (slow); $\text{A}_2 + \text{B} \rightarrow \text{C}$ (fast).

Initial-rate data at a fixed temperature:

Experiment $[\text{A}]$ (M) $[\text{B}]$ (M) Initial rate (M/s)
1 0.10 0.10 $4.0 \times 10^{-4}$
2 0.20 0.10 $1.6 \times 10^{-3}$
3 0.10 0.20 $8.0 \times 10^{-4}$

(a) Derive the rate law predicted by each mechanism. (b) Determine the experimental rate law and $k_{\text{obs}}$ with units, and decide which mechanism is supported. (c) If $K = 0.50\ \text{M}^{-1}$, calculate $k_2$. (d) Explain what dependence on $[\text{B}]$ the other mechanism would have predicted.

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