Question

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 \(1.25 \mathrm{~h}\) ?

Short answer

After 1.25 h, the concentration of $\mathrm{In}^{+}(a q)$ is approximately \(0.0256 \textrm{ M}\) and the mass of $\operatorname{In}(s)$ formed is approximately \(0.168 \textrm{ g}\).

Step-by-step solution

Write the balanced chemical equation

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: $$ 2 \mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\mathrm{In}^{3+}(a q) $$

Calculate the initial concentration of In+

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: $$ \text{Initial concentration of In}^{+}(a q) = \frac{0.0157 \textrm{ mol}}{0.500 \textrm{ L}} = 0.0314 \textrm{ M} $$

Calculate the remaining concentration of In+ after 1.25 h

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+: $$ [\textrm{In}^{+}]_{4500s} = 0.0314 \textrm{ M} \times e^{(-1.04 \times 10^{-3} \textrm{ s}^{-1} \times 4500 \textrm{ s})} \approx 0.0256 \textrm{ M} $$

Find the mass of In(s) formed

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.

Key concepts

First-Order Kinetics

In a chemical reaction, first-order kinetics refers to a scenario where the rate of reaction depends linearly on the concentration of one reactant. This means that as the concentration of the reactant decreases, the rate of reaction decreases proportionally. In the disproportionation reaction of \[\mathrm{In}^{+}(aq) \rightarrow \operatorname{In}(s)+\mathrm{In}^{3+}(aq),\]the half-life is given as 667 seconds. The half-life in a first-order reaction remains constant, regardless of the initial concentration of the reactant. A key feature of first-order kinetics is that we can apply an exponential decay model to calculate the concentration of the reactant at any given time using the formula:\[[\textrm{In}^{+}]_{t} = [\textrm{In}^{+}]_{0} \times e^{(-kt)}.\]Here, - \([\textrm{In}^{+}]_{t}\) is the concentration at time \(t\), - \([\textrm{In}^{+}]_{0}\) is the initial concentration,- \(k\) is the rate constant, and - \(e\) is the base of the natural logarithm.Understanding these kinetics helps predict how fast a reaction will proceed given certain conditions, making it possible to estimate the concentration of ions or molecules remaining after a specified period.

Chemical Concentration

Chemical concentration indicates how much of a given substance there is in a certain volume of solution. It's typically measured in moles per liter, also known as molarity (M). For the reaction involving \(\mathrm{In}^{+}(aq)\), calculating both the initial and final concentrations of \(\mathrm{In}^{+}(aq)\) is crucial.First, we determine the moles of the dissolved \(\mathrm{InCl}(s)\) based on its mass and molar mass:- Molar mass of \(\mathrm{InCl} = 151.24 \textrm{ g/mol}\).- Mass provided is 2.38 g.- Moles of \(\mathrm{InCl}\) thus is:\[\text{Moles} = \frac{2.38 \textrm{ g}}{151.24 \textrm{ g/mol}} \approx 0.0157 \textrm{ mol}\]Next, the concentration (initial) of \(\mathrm{In}^{+}(aq)\) is calculated by dividing the number of moles by the solution's volume (in liters), resulting in:\[\text{Initial concentration} = \frac{0.0157 \textrm{ mol}}{0.500 L} = 0.0314 \textrm{ M}\]Knowing these concentrations is essential for predicting how much reactant remains after a given reaction time.

Mole Calculation

Mole calculation is fundamental in chemistry for quantifying substances involved in reactions. When calculating moles, we use the formula:- \[\text{moles} = \frac{\text{mass}}{\text{molar mass}}\]where mass is typically given, and molar mass is derived from the periodic table.In our scenario with \(\mathrm{InCl}(s)\), we started by determining the moles present in 2.38 g with a molar mass of 151.24 g/mol:\[\text{Moles of InCl} = \frac{2.38 \textrm{ g}}{151.24 \textrm{ g/mol}} \approx 0.0157 \textrm{ mol}\]These moles of \(\mathrm{InCl}\) directly relate to the initial number of moles of \(\mathrm{In}^{+}(aq)\) in the solution. Additionally, mole calculations were used again when determining the mass of \(\operatorname{In}(s)\) formed:- From concentration differences calculated earlier and using the stoichiometry of the reaction, we determined the moles of \(\operatorname{In}(s)\),- Then converted this to mass using \(\operatorname{In}(s)\)'s molar mass, yielding:\[\text{Mass of In} = \text{moles} \times 115.79 \textrm{ g/mol} \approx 0.168 \textrm{ g}\]Mole calculations form the bedrock of chemical computations, enabling quantitative predictions in reactions.

Chemical Stoichiometry

Chemical stoichiometry involves the quantitative relationship between reactants and products in a chemical equation. This is necessary for balancing equations and calculating how much product will form from given reactants. For the disproportionation of \(\mathrm{In}^{+}(aq)\):- The balanced equation: \[2 \mathrm{In}^{+}(aq) \rightarrow \operatorname{In}(s)+\mathrm{In}^{3+}(aq)\]implies a 2:1 mole ratio between \(\mathrm{In}^{+}(aq)\) and \(\operatorname{In}(s)\). Understanding this ratio is essential to solve the exercise's final question regarding the mass of \(\operatorname{In}(s)\).The concentration change of \(\mathrm{In}^{+}(aq)\) is used to find the moles of \(\operatorname{In}(s)\) formed:- This change is halved due to the 2:1 ratio specified.Calculating:- After reaction time.\[\Delta [\textrm{In}^{+}] = 0.0314 \textrm{ M} - 0.0256 \textrm{ M} = 0.0058 \textrm{ M}\]Thus, the moles of \(\operatorname{In}(s)\) formed, considering the one-to-one linear relationship of change in concentration and stoichiometry: - This change leads to 0.0029 M for \(\operatorname{In}(s)\) being formed.Chemical stoichiometry allows us to predict the yields of reactions and accurately determine the amounts needed of each reactant.