Problem 17 - Chemistry: An Atoms-Focused Approach

Question

Suppose the reaction \(A \rightleftharpoons B\) in the forward direction is first order in \(\mathrm{A}\) with \(k_{\mathrm{f}}=1.50 \times 10^{-2} \mathrm{s}^{-1} .\) The reverse reaction is first order in \(\mathrm{B}\) with \(k_{\mathrm{r}}=4.50 \times 10^{-2} \mathrm{s}^{-1}\) at the same temperature. What is the valuc of the cquilibrium constant for the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\) at this temperature?

Step-by-Step Solution

Answer

Question: Given the rate constants for the forward reaction, \(k_{\mathrm{f}} = 1.50 \times 10^{-2} \mathrm{s}^{-1}\), and the reverse reaction, \(k_{\mathrm{r}} = 4.50 \times 10^{-2} \mathrm{s}^{-1}\), for the reaction A \(\rightleftharpoons\) B, calculate the equilibrium constant, \(K_c\). Answer: Based on the calculations, the equilibrium constant for the reaction A \(\rightleftharpoons\) B at given temperature is 0.333.

Step 1: Write the rate laws for the forward and reverse reactions

For a first-order reaction, the rate law gives the rate of the reaction as the product of the rate constant and the concentration of the reactant. For the forward reaction: Rate\(_\mathrm{f} = k_{\mathrm{f}}[\mathrm{A}]\) For the reverse reaction: Rate\(_\mathrm{r} = k_{\mathrm{r}}[\mathrm{B}]\)

Step 2: Define the equilibrium constant in terms of the rate constants

At equilibrium, the rate of the forward reaction (Rate\(_\mathrm{f}\)) is equal to the rate of the reverse reaction (Rate\(_\mathrm{r}\)). Using the expressions for Rate\(_\mathrm{f}\) and Rate\(_\mathrm{r}\) from Step 1, we can write the following equation: \( k_{\mathrm{f}}[\mathrm{A}] = k_{\mathrm{r}}[\mathrm{B}] \) The equilibrium constant, \(K_c\), is defined as the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium: \( K_c = \frac{[\mathrm{B}]}{[\mathrm{A}]} \) Rearranging the equation for the rates at equilibrium, we can express \(K_c\) in terms of the rate constants: \( K_c = \frac{k_{\mathrm{f}}}{k_{\mathrm{r}}} \)

Step 3: Calculate the equilibrium constant

Now we can use the values of \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) given in the problem to calculate the equilibrium constant. \( K_c = \frac{k_{\mathrm{f}}}{k_{\mathrm{r}}} = \frac{1.50 \times 10^{-2} \mathrm{s}^{-1}}{4.50 \times 10^{-2} \mathrm{s}^{-1}} = \frac{1}{3} = 0.333 \) Thus, the equilibrium constant for the reaction A \(\rightleftharpoons\) B at this temperature is 0.333.

Key Concepts

Rate Constant

The rate constant, often denoted as \( k \), plays a crucial role in the study of chemical kinetics as it helps determine how fast a chemical reaction proceeds. It is a proportionality factor that appears in the rate law of a reaction. Each chemical reaction has its own specific rate constant, which can vary with changes in temperature and presence of catalysts.

For a first-order reaction, the rate constant has units of \( \text{s}^{-1} \). This implies that the speed at which the reaction occurs can be directly influenced by the rate constant itself. A higher rate constant means a faster reaction, while a lower rate constant indicates a slower reaction.

It is important to remember that the rate constant is not affected by the concentration of reactants. Instead, it is primarily influenced by

• Temperature: Higher temperatures usually increase the rate constant as reactant molecules gain energy and collide more frequently.
• Presence of catalysts: Catalysts can lower the energy barrier for a reaction, increasing the rate constant.

First-Order Reaction

A first-order reaction is a type of chemical reaction where the rate depends linearly on the concentration of one reactant. In simple terms, if you were to double the concentration of the reactant, the rate of reaction would also double. The general form of a first-order reaction can be represented by the expression
\[ \text{Rate} = k[A] \]where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

In the context of the given exercise, both the forward and reverse reactions are first-order. This means that for each direction, the reaction rate depends solely on the concentration of one substance:

• Forward: Depends on \([A]\), the concentration of reactant \(A\).
• Reverse: Depends on \([B]\), the concentration of reactant \(B\).

Understanding first-order kinetics is essential when predicting the behavior of reactions in dynamic systems, such as those found in chemical equilibria.

Rate Law

The rate law is a mathematical expression that describes the rate of a chemical reaction - how quickly or slowly the reaction proceeds. It typically depends on the concentration of one or more reactants, each raised to a power determined experimentally. The general form of the rate law can be written as:
\[ \text{Rate} = k[A]^m[B]^n \]where \( k \) is the rate constant, \([A]\) and \([B]\) are concentrations of the reactants, and \( m \) and \( n \) are the orders of the reaction with respect to each reactant.

For a first-order reaction, the rate law simplifies to \( \text{Rate} = k[A] \), meaning the reaction rate depends linearly on the concentration of one reactant. This was the case in the original exercise where both forward and reverse reactions had rate laws written as:

• \( \text{Forward: Rate}_f = k_f[A] \)
• \( \text{Reverse: Rate}_r = k_r[B] \)

At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, helping us to define the equilibrium constant \( K_c \) based on the ratio of these rate constants. Thus, rate laws not only provide insights into the speed of a reaction but also crucially connect to the concept of dynamic equilibrium in reversible reactions.