Question

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 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 dissolving \(2.38 \mathrm{g}\) InCl \((s)\) in dilute HCl to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{h} ?\)

Short answer

The concentration of \(In^+\,(aq)\) after 1.25 h is 0.0143 M, and the mass of \(In\,(s)\) formed after 1.25 h is 0.993 g.

Step-by-step solution

Calculate the initial concentration of In+ (aq)

First, we need to determine the initial concentration of In+ (aq). To do that, we will find the moles of InCl(s) that were dissolved in the solution as follows: 1. Find the molar mass of InCl: \(MM(InCl) = MM(In) + MM(Cl) = 114.82 + 35.45 = 150.27\, g/mol\) 2. Find moles of InCl: \(\frac{2.38\, g}{150.27\, g/mol} = 0.0158\, mol\) Since the proportion between InCl(s) and In+ (aq) is 1:1, we have 0.0158 mol of In+ (aq) at the beginning. Now, calculate the initial concentration of In+ (aq): \(\frac{0.0158\, mol}{0.500\, L} = 0.0316\, M\)

Calculate the concentration of In+ (aq) after 1.25 h

Using the first-order kinetics formula and the provided half-life, we can find the concentration of In+ (aq) after 1.25 h: 1. Convert half-life to rate constant, \(k\): \(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{667\, s} = 1.039 × 10^{-3}\, s^{-1}\) 2. Convert 1.25 h to seconds: \(1.25\, h × \frac{3600\, s}{1\, h} = 4500\, s\) 3. Use the first-order kinetics equation to find the remaining concentration of In+ (aq): \([In^{+}](t) = [In^{+}]_0 e^{-kt} = (0.0316\,M) \times e^{(-1.039 × 10^{-3}\, s^{-1})(4500\,s)} = 0.0143\, M\)

Calculate the mass of In(s) formed

The change in concentration of In+ (aq) represents the amount that converted to In(s) and In3+ (aq). Therefore, we need to find the moles of In+ (aq) that reacted: Change in concentration = Initial concentration - Final concentration = \(0.0316\, M - 0.0143\, M = 0.0173\, M\) Multiplying the change in concentration by the volume of the solution gives us the moles of In(s) formed: \(0.0173\, M \times 0.500\, L = 0.00865\, mol\) Now, convert moles of In(s) to grams: \(0.00865\, mol \times 114.82\, g/mol = 0.993\, g\) So, 0.993 g of In(s) is formed after 1.25 h.

Key concepts

First-Order Kinetics

Understanding first-order kinetics is essential when studying chemical reactions that have a rate dependent on the concentration of a single reactant. In these reactions, the rate of the process is directly proportional to the concentration of the reactant. In mathematical terms, the rate equation for a first-order reaction can be written as:
\( rate = k[Reactant] \),
where \( k \) is the first-order rate constant and \([Reactant]\) symbolizes the concentration of the reactant. A classic feature of first-order reactions is their constant half-life, meaning that the time it takes for half of the reactant to be consumed is fixed, regardless of the initial concentration. In the case of the disproportionation of \(\text{In}^{+}(aq)\), the half-life is used to determine the reaction's kinetics and understand how the concentration of \(\text{In}^{+}(aq)\) changes over time.

Chemical Kinetics

Chemical kinetics deals with the speed or rate at which a chemical reaction proceeds and the factors affecting this rate. It's a crucial aspect of understanding reaction mechanisms and how different conditions influence the reaction speed. In kinetics, we frequently refer to the 'rate law,' which shows the relationship between the reaction rate and the concentration of reactants. For instance, in the disproportionation reaction provided, knowing the kinetics helps determine how rapidly \(\text{In}^{+}(aq)\) reacts to form \(\text{In}(s)\) and \(\text{In}^{3+}(aq)\). This can enable chemists to predict how much product will form over a given time period and under specific conditions.

Half-Life Calculation

The half-life of a reaction, symbolized as \(t_{1/2}\), is the time required for the concentration of a reactant to decrease to half of its initial value. In first-order reactions, the half-life is independent of the initial concentration and can be calculated using the formula:
\(t_{1/2} = \frac{0.693}{k}\),
where \(k\) is the rate constant. This relationship underscores the significance of half-life in first-order reactions, as it remains constant regardless of the concentration, unlike in reactions of other orders. By calculating the half-life or using the given half-life, we can infer the rate constant and later use it to determine the concentration of reactants or products at any time.

Concentration Calculation

Calculating the concentration of reactants or products in a chemical reaction is a fundamental skill in chemistry. Concentration refers to the amount of a substance within a certain volume of solution and is often expressed in molarity (M), which is moles per liter. In the given disproportionation reaction, we use the initial mass of \(\text{InCl}(s)\) to determine the moles and then divide by the volume of the solution in liters to find the initial concentration of \(\text{In}^{+}(aq)\). To find the concentration after a certain time, we apply the first-order kinetic model and use the rate constant \(k\) to calculate how the concentration changes with time.

Molar Mass

Molar mass, represented by \( MM \), is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is a critical value for converting between the mass of a substance and the number of moles. In our exercise, to find the moles of \(\text{InCl}(s)\) dissolved, we need its molar mass, which we obtain by summing the atomic masses of indium (In) and chlorine (Cl) from the periodic table. Knowing the molar mass allows us to perform stoichiometric calculations and understand the relationship between the mass of a substance and the amount in moles.

Moles and Molarity

Moles and molarity are two significant concepts in chemistry that serve as a bridge between the microscopic world of atoms and molecules and the macroscopic world we measure in the lab. A mole represents Avogadro's number of particles, which is \(6.022 \times 10^{23}\) entities, and reflects the quantity of substance. Molarity is the concentration of a solution expressed as the number of moles of solute per liter of solution (mol/L). These concepts are deeply interconnected: to know the molarity, we need the number of moles of solute, which we can calculate using the mass and molar mass of the substance. Once we have the molarity, we can readily engage in various calculations, such as determining reactant and product concentrations at different stages of a reaction.