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 \(1.20 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s}\) $$\begin{array}{c} 2 \mathrm{HI}(g) \stackrel{\mathrm{Au}}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t}=k=1.20 \times 10^{-4} \mathrm{mol} / \mathrm{L} \cdot \mathrm{s} \end{array}$$ a. If the initial HI concentration was 0.250 mol/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 \(0.250 \mathrm{M}\) HI to decompose?

For a zero-order reaction, the integrated rate law is given by the formula: \[ [A]_t = [A]_0 - kt \] where [A]_t is the concentration of A at time t, [A]_0 is the initial concentration of A, k is the rate constant, and t is the time.

We are given the initial concentration of HI, [HI]_0 = 0.250 mol/L, the rate constant k = \(1.20 \times 10^{-4}\) mol/L·s, and the time t = 25 minutes. Convert minutes to seconds: \[ 25 \, \text{min} \times \frac{60 \, \text{s}}{1 \, \text{min}} = 1500 \, \text{s} \] Now, use the integrated rate law to calculate the concentration of HI at 25 minutes: \[ [\mathrm{HI}]_{t} = [\mathrm{HI}]_0 - k \cdot t \] \[ [\mathrm{HI}]_{t} = 0.250 \, \mathrm{M} - (1.20 \times 10^{-4} \, \mathrm{M}\, \mathrm{s}^{-1})(1500 \, \mathrm{s}) \] \[ [\mathrm{HI}]_{t} = 0.250 \, \mathrm{M} - 0.180 \, \mathrm{M} \] \[ [\mathrm{HI}]_{t} = 0.070 \, \mathrm{M} \] So, the concentration of HI at 25 minutes after the start of the reaction is 0.070 M.

To find the time it takes for all the 0.250 M HI to decompose, we need to determine when the concentration of HI reaches zero. Using the integrated rate law formula, we set [HI]_t = 0 and solve for t: \[ 0 = [\mathrm{HI}]_0 - kt \] \[ kt = [\mathrm{HI}]_0 \] \[ t = \frac{[\mathrm{HI}]_0}{k} \] \[ t = \frac{0.250 \, \mathrm{M}}{1.20 \times 10^{-4} \, \mathrm{M} \, \mathrm{s}^{-1}} \] \[ t = 2083.33 \, \mathrm{s} \] Now, convert seconds to minutes: \[ t = 2083.33 \, \text{s} \times \frac{1 \, \text{min}}{60 \, \text{s}} \] \[ t = 34.72 \, \text{min} \] So, it will take approximately 34.72 minutes for all the 0.250 M HI to decompose.