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Chapter 14: Problem 68

The mechanism proposed for the reaction of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g})\) to form \(\mathrm{HI}(\mathrm{g})\) consists of a fast reversible first step involving \(\mathrm{I}_{2}(\mathrm{g})\) and \(\mathrm{I}(\mathrm{g}),\) followed by a slow step. Propose a two-step mechanism for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g}),\) which is known to be first order in \(\mathrm{H}_{2}\) and first order in \(\mathrm{I}_{2}.\)

Short Answer

Expert verified
A proposed mechanism that fulfills the given conditions is: Step 1 (Fast): \( I_{2}(g) \longleftrightarrow 2I(g) \), Step 2 (Slow): \( H_{2}(g) + I(g) \rightarrow 2HI(g) \). Adding these reactions gives the original reaction ( \( H_{2}(g) + I_{2}(g) \rightarrow 2HI(g) \) ) and the first order dependencies on \( H_{2} \) and \( I_{2} \) are observed as expected.

Step by step solution

01

Propose possible fast and reversible reaction

We start with the first step which is the fast reversible reaction involving \( I_{2}(g) \) and \( I(g) \), we write this as: \( I_{2}(g) \longleftrightarrow 2I(g) \)
02

Propose slow reaction

Then we need to propose a reaction that is slow. Given that the overall reaction involves \( H_{2} \) and \( I_{2} \), and that \( I(g) \) is generated in our fast step, a logical second slow step could be the reaction between \( H_{2} \) and \( I(g) \): \( H_{2}(g) + I(g) \rightarrow 2HI(g) \)
03

Validate the proposed mechanism

The overall reaction from adding the two steps yields: \( H_{2}(g) + I_{2}(g) \rightarrow 2HI(g) \) This matches the overall reaction given in the problem. Also the rate law of the rate-determining step is first order in \( H_{2} \) and first order in \( I_{2} \), which is as required.

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Most popular questions from this chapter

Certain gas-phase reactions on a heterogeneous catalyst are first order at low gas pressures and zero order at high pressures. Can you suggest a reason for this?

Three different sets of data of \([\mathrm{A}]\) versus time are giv the following table for the reaction \(A \longrightarrow\) prod [Hint: There are several ways of arriving at answer each of the following six questions. $$\begin{array}{cccccc} \hline \text { I } & & \text { II } & & \text { III } & \\ \hline \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } & \begin{array}{c} \text { Time, } \\ \text { s } \end{array} & \text { [A], M } \\ \hline 0 & 1.00 & 0 & 1.00 & 0 & 1.00 \\ 25 & 0.78 & 25 & 0.75 & 25 & 0.80 \\ 50 & 0.61 & 50 & 0.50 & 50 & 0.67 \\ 75 & 0.47 & 75 & 0.25 & 75 & 0.57 \\ 100 & 0.37 & 100 & 0.00 & 100 & 0.50 \\ 150 & 0.22 & & & 150 & 0.40 \\ 200 & 0.14 & & & 200 & 0.33 \\ 250 & 0.08 & & & 250 & 0.29 \\ \hline \end{array}$$ Which of these sets of data corresponds to a (a) zero-order, (b) first-order, (c) second-order reaction?

In the first-order reaction \(A \longrightarrow\) products, it is found that \(99 \%\) of the original amount of reactant \(A\) decomposes in 137 min. What is the half-life, \(t_{1 / 2}\), of this decomposition reaction?

The rate of a chemical reaction generally increases rapidly, even for small increases in temperature, because of a rapid increase in (a) collision frequency; (b) fraction of reactant molecules with very high kinetic energies; (c) activation energy; (d) average kinetic energy of the reactant molecules.

One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.
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