Question

The catalyst in a second order reversible reaction increases the rate of the:

• A. Forward reaction and decreases that of backward reaction
• B. Forward and backward reactions equally
• C. Forward reaction only
• D. Forward reaction to a greater extent than that of the backward reaction

Hint:

Core Catalyst Rules: - A catalyst lowers the activation energy by the same amount for both directions: \(\Delta E_{af} = \Delta E_{ab}\). - It increases the forward and backward reaction rates equally. - It does not change the thermodynamic equilibrium constant (\(K_c\)) or the final equilibrium conversion.

Answer 1

The Correct Option is B

Concept: A catalyst is a chemical substance that increases the rate of a chemical reaction without being consumed in the process. It functions by providing an alternative reaction pathway or mechanism that has a lower activation energy ($E_a$) than the uncatalyzed pathway. Thermodynamically, a catalyst does not alter the net energies of the initial reactants or the final products. Consequently, it has no effect on the standard Gibbs free energy change ($\Delta G^\circ$) or the net enthalpy change ($\Delta H$) of the reaction.

Step 1: Examining the effect of a catalyst on Activation Energy.
Let us consider a generic reversible reaction system: \[ \text{Reactants} \rightleftharpoons \text{Products} \] Let $E_{af}$ be the activation energy for the forward reaction, and $E_{ab}$ be the activation energy for the backward reaction. The net enthalpy change of the reaction is given by: \[ \Delta H = E_{af} - E_{ab} \] When a catalyst is introduced, it lowers the activation energy for both the forward and backward paths by the exact same amount ($\Delta E_a$). The new activation energies are: \[ E_{af}' = E_{af} - \Delta E_a \quad \text{and} \quad E_{ab}' = E_{ab} - \Delta E_a \]

Step 2: Evaluating the kinetic rate constants using the Arrhenius equation.
Let us substitute these modified activation energies into the Arrhenius equation to evaluate the new rate constants for the forward ($k_f'$) and backward ($k_b'$) reactions: \[ k_f' = A_f \cdot \exp\left( -\frac{E_{af} - \Delta E_a}{R \cdot T} \right) = k_f \cdot \exp\left( \frac{\Delta E_a}{R \cdot T} \right) \] \[ k_b' = A_b \cdot \exp\left( -\frac{E_{ab} - \Delta E_a}{R \cdot T} \right) = k_b \cdot \exp\left( \frac{\Delta E_a}{R \cdot T} \right) \] The equations show that both rate constants are multiplied by the exact same acceleration factor, $\exp\left( \frac{\Delta E_a}{R \cdot T} \right)$.

Step 3: Evaluating the impact on the Equilibrium Constant.
The chemical equilibrium constant ($K_c$) is defined as the ratio of the forward rate constant to the backward rate constant: \[ K_c' = \frac{k_f'}{k_b'} = \frac{k_f \cdot \exp\left( \frac{\Delta E_a}{R \cdot T} \right)}{k_b \cdot \exp\left( \frac{\Delta E_a}{R \cdot T} \right)} = \frac{k_f}{k_b} = K_c \] Because both the forward and backward reaction rates are increased by the exact same factor, the equilibrium constant $K_c$ remains unchanged. This confirms that a catalyst increases the rates of the forward and backward reactions equally, allowing the system to reach its equilibrium state faster without altering the final equilibrium composition.