Question

The approach to the following equilibrium was observed kinetically from both directions: $$ \mathrm{PtCl}_{4}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}+\mathrm{Cl} $$ At \(25^{\circ} \mathrm{C}\), it was found that \(-\frac{\mathrm{d}\left[\mathrm{PtCl}_{4}^{2-}\right]}{\mathrm{d} t}=\left(3.9 \times 10^{-5} \mathrm{~s}^{-1}\right)\left[\mathrm{PtCl}_{4}{ }^{2}\right]\) \(-\left(2.1 \times 10^{-3} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\right)\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}_{3}\right][\mathrm{Cl}]\) The value of \(K_{\text {eq }}\) (equilibrium constant) for the complexation of the fourth \(\mathrm{Cl}^{-}\) by \(\mathrm{Pt}(\mathrm{II})\) is (a) \(53.8 \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(0.018 \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(53.8 \mathrm{~L} \mathrm{~mol}^{-1}\) (d) \(0.018 \mathrm{~L} \mathrm{~mol}^{-1}\)

Short answer

The equilibrium constant \(K_{\text{eq}}\) for the reaction is 0.018 \mathrm{~L} \mathrm{~mol}^{-1}, which corresponds to answer choice (d).

Step-by-step solution

Write down the rate expressions for the forward and reverse reactions

For the forward reaction, the rate is given by \(-\frac{d[\mathrm{PtCl}_{4}^{2-}]}{dt} = (3.9 \times 10^{-5} \mathrm{~s}^{-1})[\mathrm{PtCl}_{4}^{2-}]\). For the reverse reaction, the rate is given by \(-\frac{d[\mathrm{Pt}(\mathrm{H}_{2} \mathrm{O}) \mathrm{Cl}_{3}]}{dt} = (2.1 \times 10^{-3} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1})[\mathrm{Pt}(\mathrm{H}_{2} \mathrm{O}) \mathrm{Cl}_{3}][\mathrm{Cl}^{-}]\). Since the system is at equilibrium, the rates of the forward and reverse reactions are equal.

Set the forward and reverse rate expressions equal

At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, so set the two expressions equal to each other: \((3.9 \times 10^{-5} \mathrm{~s}^{-1})[\mathrm{PtCl}_{4}^{2-}] = (2.1 \times 10^{-3} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1})[\mathrm{Pt}(\mathrm{H}_{2} \mathrm{O}) \mathrm{Cl}_{3}][\mathrm{Cl}^{-}]\).

Solve for the equilibrium constant \(K_{\text{eq}}\)

The equilibrium constant \(K_{\text{eq}}\) is the ratio of the reaction rates' coefficients. Thus, \(K_{\text{eq}} = \frac{(3.9 \times 10^{-5})}{(2.1 \times 10^{-3})} \). Simplify to find the value of \(K_{\text{eq}}\).

Calculate the equilibrium constant

By calculating the ratio, we get \(K_{\text{eq}} = \frac{3.9 \times 10^{-5}}{2.1 \times 10^{-3}} = 0.01857 \dots\). The equilibrium constant then is \(K_{\text{eq}} = 0.01857 \mathrm{~L} \mathrm{~mol}^{-1}\), which can be rounded to two significant figures as \(K_{\text{eq}} = 0.018 \mathrm{~L} \mathrm{~mol}^{-1}\), which is one of the answer choices.

Key concepts

Chemical Kinetics

Understanding chemical kinetics is essential when studying how reactions occur and at what speed. Kinetics deals with the factors that affect the rate of a chemical reaction and establishes the relationship between the reaction rate and the concentration of reactants. In our exercise, we observe the complexation of chloride by platinum(II) and determine the rate of the forward reaction based on the concentration of \(\mathrm{PtCl}_{4}^{2-}\).

In general, the rate of reaction can be influenced by several factors such as temperature, concentration of reactants, surface area, and catalysts. For instance, increasing temperature typically speeds up reactions because particles have more energy to overcome the activation energy barrier. The reaction rate expressions provided use the reactant concentrations to define the speed at which concentrations change over time, which is central to chemical kinetics and allows us to understand reaction mechanisms and predict how a system will behave over time.

Reaction Rate Expressions

Reaction rate expressions, or rate laws, describe the relationship between the rate of a chemical reaction and the concentration of its reactants. These expressions are derived from experimental data and can vary depending on the complexity of the reaction mechanism. The rate expression for a reaction typically takes the form \(-\frac{d[A]}{dt} = k[A]^n[B]^m\), where \(k\) is the rate constant, \(A\) and \(B\) are reactants, and \(n\) and \(m\) are the reaction orders with respect to each reactant.

In our exercise, for the forward reaction, the rate expression is first-order with respect to \(\mathrm{PtCl}_{4}^{2-}\), while the reverse reaction is second-order, dependent on both \(\mathrm{Pt}(\mathrm{H}_{2} \mathrm{O}) \mathrm{Cl}_{3}\) and \(\mathrm{Cl}^{-}\). To determine the equilibrium constant, we use these expressions to set the rates of the forward and reverse reactions as equal when the system is at equilibrium. Such an understanding of reaction rate expressions is crucial for solving kinetics problems and determining the dynamics of a reaction.

Chemical Equilibrium

Chemical equilibrium is the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products. It is dynamic, meaning that the reactions are still occurring, but in such a way that the concentrations remain constant.

The equilibrium constant, \(K_{\text{eq}}\), is a value that quantifies the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients. In our case, we solve for \(K_{\text{eq}}\) by setting the forward and reverse reaction rates equal and calculating the ratio of their rate coefficients. It's important to note the units of \(K_{\text{eq}}\) and to understand that a large value indicates products are favored at equilibrium, while a small value suggests reactants are favored. Learning to calculate and interpret \(K_{\text{eq}}\) is fundamental for predicting the extent of chemical reactions and manipulating conditions to control outcomes.