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Chapter 12: Problem 44

The decomposition of hydrogen iodide on finely divided gold at \(150^{\circ} \mathrm{C}\) is zero order with respect to HI. The rate defined below is constant at \(1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) $$ \begin{array}{r} 2 \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t}=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{array} $$ a. If the initial HI concentration was \(0.250 \mathrm{~mol} / \mathrm{L}\), calculate the concentration of HI at 25 minutes after the start of the reaction. b. How long will it take for all of the \(0.250 M\) HI to decompose?

Short Answer

Expert verified
a. The concentration of HI after 25 minutes is 0.070 mol/L. b. It takes approximately 34.7 minutes for all of the 0.250 mol/L HI to decompose.

Step by step solution

01

Identify the reaction order and integrated rate law

Since the decomposition of hydrogen iodide (HI) is zero-order with respect to HI, we can write the integrated rate law for this reaction, as follows: $$ [\mathrm{HI}] = -kt + [\mathrm{HI}]_0 $$ where [HI] is the concentration of HI at time t, k is the rate constant, t is time, and [HI]\(_0\) is the initial concentration of HI.
02

Calculate the concentration of HI at 25 minutes

We are given the initial concentration of HI, which is 0.250 mol/L, and the rate constant, which is 1.20 x 10^{-4} mol/L⋅s. We convert 25 minutes to seconds, and then we can plug these values into our integrated rate law and calculate the concentration of HI at 25 minutes. $$ t = 25\,\text{minutes} \times \frac{60\,\text{s}}{1\,\text{minute}} = 1500\,\text{s} $$ $$ [\mathrm{HI}] = -kt + [\mathrm{HI}]_0 \\ [\mathrm{HI}] = -(1.20 \times 10^{-4}\,\text{mol/L⋅s})(1500\,\text{s}) + (0.250\,\text{mol/L}) \\ [\mathrm{HI}] = 0.070\,\text{mol/L} $$ So, the concentration of HI after 25 minutes is 0.070 mol/L.
03

Calculate the time required for complete decomposition of HI

To find the time it takes for all of the 0.250 mol/L HI to decompose, we can set the HI concentration at time t to 0 and solve for t using the integrated rate law. $$ 0 = -kt + [\mathrm{HI}]_0 \\ t = \frac{[\mathrm{HI}]_0}{k} \\ t = \frac{0.250\,\text{mol/L}}{1.20 \times 10^{-4}\,\text{mol/L⋅s}} $$ $$ t = 2083.3\,\text{s} $$ Converting this to minutes: $$ t = \frac{2083.3\,\text{s}}{60\,\text{s/min}} \approx 34.7\,\text{minutes} $$ So, it takes approximately 34.7 minutes for all of the 0.250 mol/L HI to decompose.

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Most popular questions from this chapter

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{~s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\), respectively. What is the value of the activation energy?

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