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

Expert verified
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

01

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.
02

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: \((3.9 \times 10^{-5} \mathrm{~s}^{-1})[\mathrm{PtCl}_{4}^{2-}] = (2.1 \times 10^{-3} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1})[\mathrm{Pt}(\mathrm{H}_{2} \mathrm{O}) \mathrm{Cl}_{3}][\mathrm{Cl}^{-}]\).
03

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 \(K_{\text{eq}}\).
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Amounts of \(0.8 \mathrm{~mol}\) of \(\mathrm{PCl}_{5}\) and \(0.2\) mole of \(\mathrm{PCl}_{3}\) are mixed in a \(1 \mathrm{i}\) flask. At equilibrium, \(0.4\) mole of \(\mathrm{PCl}_{3}\) is present. The equilibrium constant for the reaction, \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) will be (a) \(0.05 \mathrm{~mol}^{-1}\) (b) \(0.13 \mathrm{~mol} 1^{-1}\) (c) \(0.013 \mathrm{~mol} 1^{-1}\) (d) \(0.60 \mathrm{~mol} 1^{-1}\)

At \(0^{\circ} \mathrm{C}\) and 1 atm pressure, \(1 \mathrm{~L}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) decomposes to \(\mathrm{NO}_{2}\) according to the equation \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\). To what extent has the decomposition proceeded when the original volume is \(25 \%\) less than that of existing volume? (a) \(0.67\) (b) \(0.33\) (c) \(0.25\) (d) \(0.75\)

What is the approximate value of \(\log K_{\mathrm{p}}\) for the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\) The standard enthalpy of formation of \(\mathrm{NH}_{3}(\mathrm{~g})\) is \(-40.0 \mathrm{~kJ} / \mathrm{mol}\) and standard entropies of \(\mathrm{N}_{2}(\mathrm{~g}), \mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{NH}_{3}(\mathrm{~g})\) are 191,130 and \(192 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. (a) \(0.04\) (b) \(7.05\) (c) \(8.6\) (d) \(3.73\)

At \(1000^{\circ} \mathrm{C}\) and a pressure of \(16 \mathrm{~atm}\), the equilibrium constant of the reaction: \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\) is such that for every nine moles of \(\mathrm{CO}\), there is one mole of \(\mathrm{CO}_{2}\). For what pressure of the mixture, is the ratio \(\mathrm{CO}: \mathrm{CO}_{2}=4: 1 ?\) The temperature remains \(1000^{\circ} \mathrm{C}\). (a) \(40.5 \mathrm{~atm}\) (b) 81 atm (c) \(33.75 \mathrm{~atm}\) (b) \(6.7 \mathrm{~atm}\)

At constant temperature, the equilibrium constant \(\left(K_{\mathrm{p}}\right)\) for the decomposition reaction: \(\mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}\) is expressed by \(K_{\mathrm{p}}=\frac{4 x^{2} P}{1-x^{2}}\), where \(P=\) total pressure at equilibrium, \(x=\) extent of decomposition. Which one of the following statements is true? (a) \(K_{\mathrm{p}}\) increases with increase of \(P\). (b) \(K_{\mathrm{v}}\) increases with increase of \(x\). (c) \(K\), increases with decrease of \(x\). (d) \(K\), remains constant with change in \(P\) and \(x\).

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