The decomposition of hydrogen iodide on finely divided gold at $150^{\circ} \mathrm{C}\( is zero order with respect to \)\mathrm{HI}$. The rate defined below is constant at $1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}$. $$ \begin{aligned} & 2 \mathrm{HI}(g) \stackrel{\text { ?u }}{\longrightarrow} \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \\ \text { Rate }=-\frac{\Delta[\mathrm{HI}]}{\Delta t} &=k=1.20 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ 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?

To calculate the concentration of HI at 25 minutes, we need to plug the given values into the zero-order rate formula: \([HI]_{25} = [HI]_0 - k \cdot t\) where \([HI]_{25}\) is the concentration of HI after 25 minutes, \([HI]_0\) is the initial concentration which is \(0.250\, mol/L\), \(k\) is the rate constant which is \(1.20 \times 10^{-4}\, mol/L \cdot s\), and \(t\) is the time which is 25 minutes. First, we need to convert 25 minutes to seconds: \(25\, min \times \frac{60\, s}{1\, min} = 1500\, s\) Now, we can plug the values into the formula: \([HI]_{25} = 0.250\, mol/L - (1.20 \times 10^{-4}\, mol/L \cdot s)(1500\, s)\) Now, calculate the concentration: \([HI]_{25} = 0.250 - (1.20 \times 10^{-4} \times 1500)\) \([HI]_{25} = 0.250 - 0.180\) \([HI]_{25} = 0.070\, mol/L\) The concentration of HI at 25 minutes after the start of the reaction is \(0.070\, mol/L\).

To calculate the time it will take for all the 0.250 M HI to decompose, we can use the zero-order rate formula again: \([HI]_t = [HI]_0 - kt\) This time we want to find \(t\), and the final concentration \([HI]_t\) will be 0. Rearrange the formula to solve for t: \(t = \frac{[HI]_0 - [HI]_t}{k}\) Plug in the values: \(t = \frac{0.250 - 0}{1.20 \times 10^{-4}}\) \(t = \frac{0.250}{1.20 \times 10^{-4}}\) Now, calculate the time: \(t = 2083.33\, s\) For convenience, we can convert the time from seconds to minutes: \(2083.33\, s \times \frac{1\, min}{60\, s} = 34.72\, min\) It will take approximately 34.72 minutes (rounded) for all of the 0.250 M HI to decompose.