Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \mathrm{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$ \mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\mathrm{In}^{3+}(a q) $$ This disproportionation follows first-order kinetics with a halflife of \(667 \mathrm{~s}\). What is the concentration of \(\mathrm{In}^{+}(a q)\) after \(1.25 \mathrm{~h}\) if the initial solution of \(\mathrm{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{~g} \operatorname{InCl}(s)\) in \(5.00 \times 10^{2} \mathrm{~mL}\) dilute HCl? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{~h}\) ?

We already have the unbalanced equation, which is: $$ \mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\mathrm{In}^{3+}(a q) $$ As the In+ ion is being disproportionated into In(s) and In3+, the balanced equation will simply be: $$ 2 \mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\mathrm{In}^{3+}(a q) $$

To calculate the initial concentration of In+ (aq), we need to find the moles of InCl and then divide by the total volume in liters. First, find the moles of InCl: $$ \text{Moles of InCl} = \frac{\text{mass of InCl}}{\text{molar mass of InCl}} $$ InCl has a molar mass of \(115.79+35.45 = 151.24 \textrm{ g/mol}\). Therefore: $$ \text{Moles of InCl} = \frac{2.38 \textrm{ g}}{151.24 \textrm{ g/mol}} \approx 0.0157 \textrm{ mol} $$ Now, divide by the total volume in liters: $$ \text{Initial concentration of In}^{+}(a q) = \frac{0.0157 \textrm{ mol}}{0.500 \textrm{ L}} = 0.0314 \textrm{ M} $$

Since the disproportionation follows first-order kinetics, we can use the following equation to relate the concentration at a given time with the initial concentration: $$ [\textrm{In}^{+}]_{t} = [\textrm{In}^{+}]_{0} \times e^{(-kt)} $$ Using the half-life given, we can calculate the rate constant, \(k\): $$ k = \frac{\ln{2}}{t_{1/2}} = \frac{\ln{2}}{667 \textrm{ s}} \approx 1.04 \times 10^{-3} \textrm{ s}^{-1} $$ Now, convert the given time to seconds: $$ 1.25 \textrm{ h} = 1.25 \times 3600 = 4500 \textrm{ s} $$ Calculate the remaining concentration of In+: $$ [\textrm{In}^{+}]_{4500s} = 0.0314 \textrm{ M} \times e^{(-1.04 \times 10^{-3} \textrm{ s}^{-1} \times 4500 \textrm{ s})} \approx 0.0256 \textrm{ M} $$

We can find the mass of solid indium formed by considering the change in concentration of In+ and the stoichiometry of the balanced equation. At the beginning, the concentration of In+ was 0.0314 M, and after 1.25 h, it is 0.0256 M. The concentration of In(s) formed is: $$ \Delta [\textrm{In}^{+}] = 0.0314 \textrm{ M} - 0.0256 \textrm{ M} = 0.0058 \textrm{ M} $$ From the balanced equation, the molar ratio between In+ consumed and In(s) formed is 2:1. Hence, the change in concentration of In(s) is half the change in In+ concentration: $$ [\textrm{In}(s)] = \frac{1}{2} \times 0.0058 \textrm{ M} = 0.0029 \textrm{ M} $$ Find the moles of In(s): $$ \text{Moles of In}(s) = 0.0029 \textrm{ M} \times 0.500 \textrm{ L} \approx 0.00145 \textrm{ mol} $$ Finally, we can calculate the mass of In(s) formed by multiplying the moles by its molar mass (115.79 g/mol): $$ \text{Mass of In}(s) = 0.00145 \textrm{ mol} \times 115.79 \textrm{ g/mol} \approx 0.168 \textrm{ g} $$ So, after 1.25 h, the concentration of In+ is approximately 0.0256 M and the mass of In(s) formed is approximately 0.168 g.