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

From the given problem, we know that the initial amount of InCl(s) was 2.38 g dissolved in a 500 mL (=0.500L) solution. Let's first find the moles of \(\mathrm{InCl}\) and then the initial concentration of \(\operatorname{In}^{+}(a q)\). Molar mass of \(\mathrm{InCl} = \mathrm{In} +(1\times \mathrm{Cl}) = 114.82+35.45=150.27\) g/mol. Moles of \(\mathrm{InCl}= \frac{\text{mass}}{\text{molar mass}}= \frac{2.38 \,g}{150.27\,g \,mol^{-1}}=0.01583\) mol. Since, 1 mol of \(\mathrm{InCl}\) will give 1 mol of \(\operatorname{In}^{+}(a q)\) upon dissolution, initial \(\operatorname{In}^{+}(a q)\) concentration in the 500 mL solution is: $$[ \operatorname{In}^{+}]=\frac{\text{moles of}\, \mathrm{In}^{+}}{\text{volume of solution}}= \frac{0.01583 \,mol}{0.500 \, L} = 0.03166 \mathrm{M}$$

We are given that the disproportionation of \(\operatorname{In}^{+}(a q)\) follows first-order kinetics with a half-life of 667 s. We can use the first-order formula for the concentration after a given time: $$ \frac{[\operatorname{In}^{+}]_\text{remaining}}{[\operatorname{In}^{+}]_0} = e^{-kt} $$ Where \([\operatorname{In}^{+}]_0 = 0.03166 \mathrm{M}\) is the initial concentration, \([\operatorname{In}^{+}]_\text{remaining}\) is the remaining concentration after a certain time, \(k\) is the rate constant, and \(t\) is the time after which the concentration is measured. We are given the half-life \(t_\frac{1}{2} = 667\,\mathrm{s}\), and \(t = 1.25\,\mathrm{h} = 4500\,\mathrm{s}\). The rate constant \(k\) can be calculated from the half-life using the formula: $$ k = \frac{ln(2)}{t_\frac{1}{2}} = \frac{ln(2)}{667} = 0.00104\,\mathrm{s}^{-1} $$ Now we can plug these values into the first-order formula and find \([\operatorname{In}^{+}]_\text{remaining}\). $$ \frac{[\operatorname{In}^{+}]_\text{remaining}}{0.03166} = e^{-0.00104\times4500} $$ $$ [\operatorname{In}^{+}]_\text{remaining} = 0.03166 \times e^{-0.00104\times4500} = 0.01583 \mathrm{M} $$

Since the reaction is a disproportionation reaction, for every mole of \(\operatorname{In}^{+}(a q)\) consumed, a mole of \(\mathrm{In}(s)\) is formed. We can calculate the moles of \(\mathrm{In^{+}}\) consumed as: $$ \text{moles consumed} = [\operatorname{In}^{+}]_\text{initial} \times \text{initial volume} - [\operatorname{In}^{+}]_\text{remaining} \times \text{final volume} $$ $$ = 0.03166 \times 0.5 \, L - 0.01583 \times 0.5 \, L = 0.007915 \, mol $$ Finally, we can calculate the mass of In\((s)\) formed after 1.25 hours by multiplying the moles consumed with the molar mass of In. $$ \text{Mass of In(s) formed} = \text{moles consumed} \times \text{molar mass of In} = 0.007915 \times 114.82 = 0.908\, \mathrm{g} $$ To summarize, after 1.25 hours, the concentration of \(\operatorname{In}^{+}(a q)\) is 0.01583 M and the mass of In(s) formed is 0.908 g.