Problem 119 - Chemistry The Molecular Science

Question

Use the fact that the equilibrium constant \(K_{\mathrm{c}}\) equals the ratio of the forward rate constant divided by the reverse rate constant, together with the Arrhenius equation \(k=A e^{-E_{\mathrm{a}} / R T}\), to show that a catalyst does not affect the value of an equilibrium constant even though the catalyst increases the rates of forward and reverse reactions. Assume that the frequency factors \(A\) for forward and reverse reactions do not change, and that the catalyst lowers the activation barrier for the catalyzed reaction.

Step-by-Step Solution

Answer

A catalyst speeds up reactions but does not change the equilibrium constant.

Step 1: Understanding the Question

We need to show that a catalyst has no effect on the equilibrium constant \(K_c\), even though it changes the reaction rates. We will use the Arrhenius equation and the fact that \(K_c\) is derived from the ratio of forward and reverse rate constants.

Step 2: Express the Rate Constants Using Arrhenius Equation

For a reaction, the forward rate constant \(k_f\) is given by \(k_f = A_f e^{-E_{a,f} / RT}\). The reverse rate constant \(k_r\) is \(k_r = A_r e^{-E_{a,r} / RT}\), where \(E_{a,f}\) and \(E_{a,r}\) are the activation energies for the forward and reverse reactions, respectively, and \(R\) is the gas constant.

Step 3: Define the Equilibrium Constant

The equilibrium constant \(K_c\) is defined as \(K_c = \frac{k_f}{k_r} = \frac{A_f e^{-E_{a,f} / RT}}{A_r e^{-E_{a,r} / RT}}\). By simplifying, \(K_c = \frac{A_f}{A_r} e^{-(E_{a,f} - E_{a,r}) / RT}\).

Step 4: Effect of Catalyst on Activation Energies

A catalyst lowers both forward and reverse activation energies equally. Let the catalyzed activation energies be \(E_{a,f}'\) and \(E_{a,r}'\), such that \(E_{a,f}' = E_{a,f} - \Delta E_a\) and \(E_{a,r}' = E_{a,r} - \Delta E_a\), with \(\Delta E_a\) being the same for forward and reverse reactions.

Step 5: Equilibrium Constant of Catalyzed Reaction

For the catalyzed reaction, \(K'_c = \frac{k_f'}{k_r'} = \frac{A_f e^{-E_{a,f}' / RT}}{A_r e^{-E_{a,r}' / RT}} = \frac{A_f}{A_r} e^{-(E_{a,f}' - E_{a,r}') / RT}\). Substituting \(E_{a,f}'\) and \(E_{a,r}'\), we get \(K'_c = \frac{A_f}{A_r} e^{-(E_{a,f} - E_{a,r}) / RT}\).

Step 6: Conclusion

Comparing \(K'_c\) with \(K_c\), because the effect of \(\Delta E_a\) cancels out, we find \(K'_c = K_c\). Therefore, the catalyst changes the rates but not the equilibrium constant.

Key Concepts

Catalysis Effect

A catalyst is a chemical substance that speeds up the rate of a chemical reaction without being consumed in the process. It achieves this by lowering the activation energy required for the reaction to proceed. However, it's crucial to understand that while catalysts increase the speed of both the forward and reverse reactions, they do not change the position of equilibrium. Thus, the equilibrium constant, which is a measure of the ratio of concentrations of products to reactants at equilibrium, remains unchanged.
This is because catalysts simply provide an alternative pathway for the reaction with a lower activation energy. Since the equilibrium constant \( K_c \) is determined by the ratio of the rate constants for the forward and reverse reactions, and both these constants are affected equally by the presence of a catalyst, their ratio, and hence the equilibrium constant, remains the same.

• Catalyst speeds up the reaction.
• No change to equilibrium constant \( K_c \).
• Lowers activation energy for both directions.

While catalysts have numerous applications in industry and biochemistry, their role in maintaining the equilibrium constant underscores their importance in processes that require steady-state conditions.

Arrhenius Equation

The Arrhenius Equation is a fundamental formula used to express the temperature dependence of reaction rates. It relates the rate constant \( k \) of a reaction to the temperature \( T \), activation energy \( E_a \), and a pre-exponential factor \( A \). The equation is expressed as: \[ k = A e^{-E_a / RT} \] Where:

• \( k \) is the rate constant.
• \( A \) is the frequency factor, representing the number of times particles collide with the correct orientation per unit time.
• \( E_a \) is the activation energy, the minimum energy required for a successful reaction.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.

Understanding the Arrhenius Equation is crucial for analyzing how reaction rates alter with temperature. According to the equation, as \( T \) increases, the exponential term decreases, thereby increasing \( k \), the rate constant. This reflects the increased kinetic energy of molecules at higher temperatures, making collisions more frequent and effective.
In terms of catalysis, the Arrhenius Equation illustrates how a catalyst lowers \( E_a \), which increases the value of \( k \) without altering the equilibrium constant \( K_c \). This meaningfully demonstrates the power of catalysts in increasing reaction rates.

Activation Energy

Activation energy, often denoted as \( E_a \), is the minimum energy barrier that must be overcome for a chemical reaction to occur. It represents the difference between the energy of the reactants and the highest energy transition state.A lower activation energy implies that more reactant molecules have the necessary energy to reach the transition state, leading to a faster reaction rate.
Catalysts are remarkable because they lower the activation energy for a reaction, thereby allowing more molecules to successfully react, thereby speeding up the reaction. This effect is uniform for both the forward and reverse reactions in a chemical equilibrium. Upon introduction of a catalyst, new pathways involving transition states of lower energy become available, allowing the reaction to proceed more efficiently.

• Low activation energy equals faster reaction.
• Catalysts make reactions more efficient.
• Transition states are made more accessible.

Although activation energy is pivotal to understanding reaction kinetics, the fact that a catalyst lowers the activation energy for both directions without favouring one over the other explains why the equilibrium constant remains unaltered. The changes induced by the catalyst in concert enhance reaction rates, depicting the profound impact of activation energy on chemical processes.