Upon dissolving \(\operatorname{InCl}(s)\) 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 \(667 \mathrm{~s}\). What is the concentration of \(\operatorname{In}^{+}(a q)\) 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 dilute \(\mathrm{HCl}\) to make \(5.00 \times 10^{2} \mathrm{~mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{~h}\) ?

Step by step solution

01

Calculate the initial concentration of \(In^+\)

First, we need to find the molar mass of InCl, using the atomic masses of In (114.82) and Cl (35.45): Molar mass of InCl = 114.82 + 35.45 ≈ 150.27 g/mol Now, we can calculate the initial moles of InCl: Initial moles of InCl = (2.38 g) / (150.27 g/mol) ≈ 0.0158 mol Since the InCl is fully dissolved in the solution, the initial moles of \(In^+\) will be equal to the moles of InCl: Initial moles of \(In^+\) ≈ 0.0158 mol Now, we can find the initial concentration of \(In^+\) by dividing the moles by the volume: Initial concentration of \(In^+\) = (0.0158 mol) / (5.00 × 10⁻¹ L) ≈ 0.0317 M

02

Calculate the rate constant, k

As the reaction follows first-order kinetics, we can use the formula relating half-life (t) and rate constant (k): t = ln(2) / k Now we can solve for k, using the given half-life of 667 seconds: k = ln(2) / 667 ≈ 1.030 × 10⁻³ s⁻¹

03

Calculate the concentration of \(In^+\) after 1.25 hours

Using the first-order kinetics formula, we can find the concentration of \(In^+\) at any time t: \[C_t = C_0 \times e^{-kt}\] Where: \(C_t\) is the concentration at time t, \(C_0\) is the initial concentration, k is the rate constant, and t is the time in seconds. Now, convert the given time of 1.25 hours to seconds: 1.25 hours = 1.25 × 60 × 60 = 4500 seconds Substitute the values to calculate \(C_t\): \(C_t\) = (0.0317 M) × \(e^{-(1.030 \times 10^{-3} s^{-1})(4500 s)}\) ≈ 0.0242 M

04

Calculate the concentration of \(In(s)\) and \(In^{3+}\) after 1.25 hours

Based on the balanced equation, for each mole of \(In^+\) that reacts, one mole of \(In(s)\) and one mole of \(In^{3+}\) are formed. Thus, the moles of \(In^+\) that reacted are equal to the moles of \(In(s)\) and \(In^{3+}\) formed: Moles of \(In^+\) reacted = (initial moles of \(In^+\)) - (moles of \(In^+\) at time t) = 0.0158 - (0.0242 × 5.00 × 10⁻¹) ≈ 0.0079 mol Thus, the moles of \( In(s)\) and \( In^{3+}\) formed are also 0.0079 mol.

05

Calculate the mass of \(In(s)\) after 1.25 hours

With the moles of \(In(s)\) formed, we can calculate the mass using the molar mass of In: Mass of \(In(s)\) = (moles of \(In(s)\)) × (molar mass of In) = (0.0079 mol) × (114.82 g/mol) ≈ 0.908 g The concentration of \(In^+\) after 1.25 hours is 0.0242 M, and the mass of \(In(s)\) formed after 1.25 hours is 0.908 g.

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