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

A zero-order reaction is a reaction in which the rate is independent of the concentration of the reactant. The rate law for a zero-order reaction is given by: Rate = k = \(-\frac{\Delta[\mathrm{HI}]}{\Delta t}\) where k is the rate constant, [\(\mathrm{HI}\)] is the concentration of hydrogen iodide, and Δt is the change in time.

Given the zero-order rate law equation, we can calculate the change in concentration of HI from the initial concentration to the time provided: \(\Delta[\mathrm{HI}] = -k \cdot \Delta t\) As we know the rate constant k (\(1.20 \times 10^{-4} \, \mathrm{mol/L \cdot s}\)) and the time given is 25 minutes, we need to convert the time to seconds: \(25 \, \text{minutes} = 25 \cdot 60 = 1500 \, \text{seconds}\) Now, we can use the zero-order rate law equation to calculate the change in concentration of HI: \(\Delta[\mathrm{HI}] = - 1.20 \times 10^{-4} \, \mathrm{mol/L \cdot s} \cdot 1500 \, \text{s}\) \(\Delta[\mathrm{HI}] = -0.18 \, \mathrm{mol/L}\) Since the change in HI concentration is negative, it means that its concentration has decreased. We can calculate the HI concentration at 25 minutes by subtracting the change in HI concentration from the initial concentration provided: [\(\mathrm{HI}\)]_final = [\(\mathrm{HI}\)]_initial + \(\Delta[\mathrm{HI}]\) [\(\mathrm{HI}\)]_final = 0.250 \,\mathrm{mol/L} - 0.18 \,\mathrm{mol/L} [\(\mathrm{HI}\)]_final = 0.07 \, \mathrm{mol/L} So, the HI concentration at 25 minutes after the start of the reaction is 0.07 mol/L.

To find the time required for complete decomposition, we set the final [\(\mathrm{HI}\)] to 0: 0 = [\(\mathrm{HI}\)]_initial + (-k \cdot t) Solving for t: t = \(\frac{[\mathrm{HI}]_\text{initial}}{k}\) t = \(\frac{0.250 \, \mathrm{mol/L}}{1.20 \times 10^{-4} \, \mathrm{mol/L \cdot s}}\) t = \(2083.3 \, \text{seconds}\) We can convert the time to minutes for convenience: t = \(\frac{2083.3 \, \text{seconds}}{60} = 34.7 \, \text{minutes}\) So it will take approximately 34.7 minutes for the complete decomposition of the 0.250 M HI concentration.