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Chapter 11: Problem 77

For the reaction between hydrogen and iodine, $$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HI}(\mathrm{g})$$ the experimental rate expression is rate \(=k\left[\mathrm{H}_{2}\right] \times\left[\mathrm{I}_{2}\right] .\) Show that this expression is consistent with the mechanism $$\begin{array}{cc}\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) & \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) & \text { (slow) }\end{array}$$

Short Answer

Expert verified
In this exercise, we were given a reaction between hydrogen and iodine, and an experimental rate expression. We were asked to verify whether the rate expression is consistent with a given two-step reaction mechanism. To do this, we determined the rate law for each step in the mechanism, applied the steady-state approximation method to find the overall rate law, and compared it with the given rate expression. We found that the rate expression is consistent with the mechanism, assuming that the rate constant has the observed relationship.

Step by step solution

01

Write the rate law for each step of the mechanism

For each step in the given mechanism, we can write the rate law based on the reactants involved in each step. For the first step (fast equilibrium): $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ The forward rate: \(rate_1 = k_1[\mathrm{I}_{2}]\) The reverse rate: \(rate_{-1} = k_{-1}[\mathrm{I}]^2\) For the second step (slow): $$\mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g)$$ The rate: \(rate_2 = k_2[\mathrm{H}_{2}][\mathrm{I}]^2\)
02

Apply the steady-state approximation to find the overall rate law

The steady-state approximation states that the rate of the change in concentration of any intermediate species (in this case, I atoms) is zero. Therefore, for I atoms, we have: $$\frac{d[\mathrm{I}]}{dt} = rate_1 - rate_{-1} - rate_2 = 0$$ Substitute the rate laws from Step 1: $$k_1[\mathrm{I}_{2}] - k_{-1}[\mathrm{I}]^2 - k_2[\mathrm{H}_{2}][\mathrm{I}]^2 = 0$$ Now solve for the concentration of I atoms: $$[\mathrm{I}]^2 = \frac{k_1[\mathrm{I}_{2}]}{k_{-1} + k_2[\mathrm{H}_{2}]}$$
03

Express the rate of the overall reaction in terms of the reactants

The overall rate of the reaction is the rate of the slow step, which is given by \(rate_2 = k_2[\mathrm{H}_{2}][\mathrm{I}]^2\). Substitute the expression for \([\mathrm{I}]^2\) obtained in Step 2: $$rate = k_2[\mathrm{H}_{2}]\frac{k_1[\mathrm{I}_{2}]}{k_{-1} + k_2[\mathrm{H}_{2}]}$$
04

Compare the derived rate law with the given rate expression

Now we compare this derived rate law with the given experimental rate expression, which is: $$rate = k[\mathrm{H}_{2}][\mathrm{I}_{2}]$$ To make these rate expressions consistent, the rate constant \(k\) must be equal to: $$k = k_2\frac{k_1}{k_{-1} + k_2[\mathrm{H}_{2}]}$$ Thus, we have shown that the rate expression given in the exercise is consistent with the provided mechanism, assuming that the rate constant \(k\) has the relationship shown above.

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Most popular questions from this chapter

Hypofluorous acid, HOF, is extremely unstable at room temperature. The following data apply to the decomposition of HOF to \(\mathrm{HF}\) and \(\mathrm{O}_{2}\) gases at a certain temperature. $$\begin{array}{cl}\hline \text { Time (min) } & \text { [HOF] } \\\\\hline 1.00 & 0.607 \\\2.00 & 0.223 \\\3.00 & 0.0821 \\\4.00 & 0.0302 \\\5.00 & 0.0111 \\\\\hline \end{array}$$ (a) By plotting the data, show that the reaction is first-order. (b) From the graph, determine \(k\). (c) Using \(k\), find the time it takes to decrease the concentration to \(0.100 \mathrm{M}\) (d) Calculate the rate of the reaction when [HOF] \(=0.0500 M\).

Derive the integrated rate law, \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), for a zero-order reaction. (Hint: Start with the relation \(-\Delta[\mathrm{A}]=k \Delta \mathrm{t}\).)

Consider the reaction $$\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \longrightarrow \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{SCN}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}$$ The following data were obtained: $$\begin{array}{ccc}\hline\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\right] & {\left[\mathrm{SCN}^{-}\right]} & \text {Initial Rate }(\mathrm{mol} /\mathrm{L} \cdot \mathrm{min}) \\ \hline 0.025 & 0.060 & 6.5 \times 10^{-4} \\\0.025 & 0.077 & 8.4 \times 10^{-4} \\\0.042 & 0.077 & 1.4 \times 10^{-3} \\ 0.042 & 0.100 & 1.8 \times 10^{-3} \\\\\hline\end{array}$$ (a) Write the rate expression for the reaction. (b) Calculate \(k\). (c) What is the rate of the reaction when \(15 \mathrm{mg}\) of \(\mathrm{KSCN}\) is added to \(1.50 \mathrm{~L}\) of a solution \(0.0500 \mathrm{M}\) in \(\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}\) ?

In the first-order decomposition of acetone at \(500^{\circ} \mathrm{C}\), $$\mathrm{CH}_{3}-\mathrm{CO}-\mathrm{CH}_{3}(g) \longrightarrow \text { products }$$ it is found that the concentration is \(0.0300 \mathrm{M}\) after \(200 \mathrm{~min}\) and \(0.0200 \mathrm{M}\) after 400 min. Calculate the following. (a) the rate constant (b) the half-life (c) the initial concentration

A catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ}\) to \(206 \mathrm{~kJ} .\) By what factor would you expect the reaction-rate constant to increase at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factors \((\mathrm{A})\) are the same for both reactions. (Hint: Use the formula \(\ln k=\ln \mathrm{A}-E_{2} / R T .\) )
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