Problem 99 - Chemistry The Central Science In Si Units

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) \(k_r\) is larger. (b) \(k_r \approx 870.97 \text{ s}^{-1}\). (c) Forward reaction is endothermic. (d) \(k_r\) increases but less than \(k_f\).

Step 1: Analyze Equilibrium Constant Relation

For any reversible reaction, the relationship between the equilibrium constant \(K_c\), and the forward \(k_{f}\) and reverse \(k_{r}\) rate constants is given by \(K_c = \frac{k_f}{k_r}\). Here, \(K_c = 3.1 \times 10^{-4}\), so we can solve this relation to analyze which rate constant is larger.

Step 2: Determine Larger Rate Constant

Since \(K_c\) is much less than 1, this implies that the denominator \(k_r\) must be larger than the numerator \(k_f\). Therefore, \(k_r > k_f\) and the reverse reaction rate constant is expected to be larger.

Step 3: Calculate Reverse Rate Constant

Given \(k_{f} = 0.27 \text{ s}^{-1}\), use the equation \(K_c = \frac{k_f}{k_r}\) to solve for \(k_r\). Rearranging yields \(k_r = \frac{k_f}{K_c} = \frac{0.27}{3.1 \times 10^{-4}} \text{ s}^{-1}\). Thus, \(k_r \approx 870.97 \text{ s}^{-1}\).

Step 4: Assess Reaction Enthalpy

The reaction \( \mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{A}(g) \) involves breaking a bond in \( \mathrm{A}_2 \) to produce two \( \mathrm{A} \) atoms, which requires energy input. Therefore, this forward reaction is endothermic.

Step 5: Effect of Temperature Increase

According to the Arrhenius equation, both \(k_f\) and \(k_r\) generally increase with temperature. However, endothermic reactions typically see a larger relative increase in their rate constants compared to exothermic reactions. Since the forward reaction is endothermic, \(k_f\) is expected to increase more than \(k_r\) as temperature rises from 800 K to 1000 K.

Key Concepts

Reaction Kinetics

In chemical reactions, how fast the reactants are converted to products is governed by reaction kinetics. This involves understanding the rates at which these transformations occur, which is determined by rate constants. The rate of a forward reaction is characterized by the forward rate constant, denoted as \(k_f\), whereas the reverse reaction has a reverse rate constant, \(k_r\). When a reaction reaches dynamic equilibrium, the forward and reverse reaction rates equalize. This equilibrium is quantified by the equilibrium constant \(K_c\), which is the ratio of \(k_f\) to \(k_r\). The magnitude of the equilibrium constant informs us about the relative speeds of the forward and reverse reactions. In our given problem, with \(K_c = 3.1 \times 10^{-4}\), it's clear that the reverse reaction is favored, indicating \(k_r\) must be larger than \(k_f\). This is crucial in predicting how a reaction will behave under different conditions.

Endothermic vs Exothermic Reactions

Reactions can either absorb or release energy, classifying them as endothermic or exothermic.

• Endothermic reactions absorb energy from the surroundings, typically in the form of heat, resulting in a positive enthalpy change (\(\Delta H > 0\)).
• Exothermic reactions release energy, usually as heat, and exhibit a negative enthalpy change (\(\Delta H < 0\)).

Consider the reaction \(\mathrm{A}_{2}(g) \rightleftharpoons 2 \mathrm{A}(g)\). To convert one molecule of \(\mathrm{A}_{2}\) into \(\mathrm{A}\) atoms, energy is required to overcome the bond holding \(\mathrm{A}_2\) together. This makes the forward reaction endothermic, as it necessitates an energy input to break this bond. Understanding this difference is key in anticipating how the reaction might respond to changes, such as temperature variations.

Temperature Effect on Reaction Rates

Temperature plays a pivotal role in influencing reaction rates. According to the Arrhenius equation, an increase in temperature generally results in an increase in reaction rate constants \(k\), as more molecules have the requisite energy to overcome the activation energy barrier.

• For endothermic reactions, this temperature increase typically results in a greater relative increase in \(k_f\), the forward rate constant, compared to \(k_r\), the reverse rate constant.
• The additional heat supplies the necessary energy to the endothermic reaction, enhancing its speed more significantly than the exothermic counterpart.

In the problem scenario, raising the temperature from 800 K to 1000 K is expected to increase both \(k_f\) and \(k_r\). However, because the forward reaction is endothermic, \(k_f\) will likely experience a more noticeable increase, reflecting the inherent nature of energy absorption in endothermic processes. These insights help predict how temperature shifts might affect the balance between forward and reverse reactions.