Problem 99 - Chemistry: The Central Science in SI Units, Global Edition

Question

At \(800 \mathrm{~K},\) the equilibrium constant for the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g)\) is \(K_{c}=3.1 \times 10^{-4}\). (a) Assuming both forward and reverse reactions are elementary reactions, which rate constant do you expect to be larger, \(k_{f}\) or \(k_{r} ?\) (b) If the value of \(k_{f}=0.27 \mathrm{~s}^{-1}\), what is the value of \(k_{r}\) at \(800 \mathrm{~K} ?\) (c) Based on the nature of the reaction, do you expect the forward reaction to be endothermic or exothermic? (d) If the temperature is raised to \(1000 \mathrm{~K}\), will the reverse rate constant \(k_{r}\) increase or decrease? Will the change in \(k_{r}\) be larger or smaller than the change in \(k_{f}\) ?

Step-by-Step Solution

Answer

(a) At 800 K, the rate constant \(k_r\) is larger than \(k_f\), as the equilibrium constant Kc is less than 1. (b) The value of \(k_r\) at 800 K is approximately 870.97 s⁻¹. (c) The forward reaction is endothermic, as the reaction favors the reactant side (A2) more than the product side (A). (d) If the temperature is raised to 1000 K, the reverse rate constant \(k_r\) will decrease, and the change in \(k_r\) will be smaller than the change in \(k_f\).

Step 1: Understand the relationship between equilibrium constant and rate constants

The equilibrium constant (Kc) is related to the forward reaction rate constant (kf) and reverse reaction rate constant (kr) by the following equation: \[ K_c = \frac{k_f}{k_r} \] (a)

Step 2: Determine which rate constant will be larger

We know that the equilibrium constant Kc is equal to 3.1 x 10⁻⁴ for the reaction at 800 K. Since Kc is less than 1, this means that kf must be smaller than kr. Therefore, we expect kr to be larger than kf. (b)

Step 3: Calculate the value of kr

Given that kf = 0.27 s⁻¹, we can use the relationship between Kc, kf, and kr to solve for kr: \[ k_r = \frac{k_f}{K_c} \] Substitute the given values, \[ k_r = \frac{0.27 \mathrm{~s}^{-1}}{3.1 \times 10^{-4}}\] Now, calculate kr, \[ k_r \approx 870.97 \mathrm{~s}^{-1} \] (c)

Step 4: Determine the nature of the reaction (endothermic or exothermic)

Since the equilibrium constant Kc is less than 1 and kr is larger than kf, this indicates that the reaction favors the reactant side (A2) more than the product side (A). This suggests that the forward reaction requires energy input, making it endothermic. (d)

Step 5: Effect of increasing temperature on the reverse rate constant kr

Increasing the temperature tends to favor endothermic reactions, shifting the equilibrium towards the products. As a result, the reverse rate constant kr would decrease. The change in kf would be larger than the change in kr, as the forward reaction is endothermic and will be more sensitive to the increase in temperature.

Key Concepts

Rate Constants

In chemical reactions, rate constants play a critical role in determining how fast a reaction proceeds. For any reaction, we define a forward rate constant, denoted as \(k_f\), which represents the rate of the reaction as it converts reactants into products. Conversely, the reverse rate constant, \(k_r\), indicates the rate at which products revert back to reactants. These constants are fundamental in determining the position of equilibrium in a chemical reaction.

The equilibrium constant \(K_c\) for a reaction is given by the ratio of these two constants:

• \( K_c = \frac{k_f}{k_r} \)

This relationship explains why, when \(K_c\) is less than one, the reverse reaction is favored, leading to \(k_r\) being larger than \(k_f\). Understanding these relationships can help predict how changes in conditions, like temperature, impact the reaction's direction and speed.

Endothermic Reactions

Endothermic reactions are chemical processes that absorb energy from their surroundings, usually in the form of heat. This energy absorption is necessary for breaking the bonds of the reactants, which subsequently reforms into products. In our particular example, the reaction \( \mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{~A}(g) \), given the equilibrium constant \(K_c\) is less than one, suggests that more energy is required to form the products from the reactants than is released, classifying the reaction as endothermic.

Some common characteristics of endothermic reactions include:

• They often result in a cooling effect on their immediate environment.
• The product-favored at higher temperatures.
• The standard enthalpy change \( \Delta H \) is positive, meaning it requires heat input to proceed.

Understanding these traits is essential for interpreting and predicting how reaction conditions affect equilibrium and rate phenomena.

Temperature Effect on Reaction Rates

Temperature is a key factor influencing reaction rates and the balance between forward and reverse reactions. As temperature increases, molecules possess higher kinetic energy, causing more frequent and effective collisions between them. This often results in an increase in both forward and reverse reaction rates, which can significantly impact equilibrium dynamics.

For endothermic reactions, raising the temperature typically increases the forward reaction rate more substantially than the reverse. This occurs because endothermic reactions absorb heat, aligning better with higher thermal conditions:

• Forward rate constant \(k_f\) increases since the reaction benefits from supplied energy.
• Reverse rate constant \(k_r\) might also increase but not as significantly as \(k_f\), as there's less reliance on external energy.

Overall, understanding the influence of temperature helps chemists predict equilibrium shifts and optimize conditions for desired chemical processes.