Catalyst Effect on Activation Energy
Test your understanding of the Arrhenius equation and thermodynamic principles.
Question:
For a reversible reaction $A \rightleftharpoons B$, the enthalpy of the forward reaction $\Delta H_{\text{forward}} = 20 \text{ kJ mol}^{-1}$.
The activation energy of the uncatalysed forward reaction is $300 \text{ kJ mol}^{-1}$.
When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at $27^\circ\text{C}$ is found to be same as that of the uncatalysed reaction at $327^\circ\text{C}$.
The activation energy of the catalysed backward reaction is ________ $\text{kJ mol}^{-1}$.
Detailed Solution
Step 1: Understand the rate condition given
We are given that the rate of the catalysed reaction at $27^\circ\text{C}$ equals the rate of the uncatalysed reaction at $327^\circ\text{C}$.
Since the concentrations are the same, the rate constants must be equal ($k_{\text{cat}} = k_{\text{uncat}}$). Using the Arrhenius equation ($k = Ae^{-E_a/RT}$), and assuming the pre-exponential factor ($A$) remains the same:
Taking the natural logarithm on both sides and cancelling terms, we get a very useful relation:
Step 2: Calculate Activation Energy of Catalysed Forward Reaction
First, convert the given temperatures to Kelvin:
- $T_1$ (catalysed) = $27^\circ\text{C} + 273 = 300\text{ K}$
- $T_2$ (uncatalysed) = $327^\circ\text{C} + 273 = 600\text{ K}$
We know $E_{a(f, \text{ uncat})} = 300 \text{ kJ mol}^{-1}$. Substitute these into our derived relation:
Key Concept for Reversible Reactions
The enthalpy of reaction ($\Delta H$) depends only on the initial and final states of the reactants and products. A catalyst DOES NOT change the enthalpy of the reaction ($\Delta H$). It lowers the activation energy of both the forward and backward reactions by the exact same amount.
Step 3: Calculate Activation Energy of Catalysed Backward Reaction
The relationship between enthalpy and activation energies is:
This holds true whether the reaction is catalysed or uncatalysed. Using the values for the catalysed path:
- $20 = E_{a(f, \text{ cat})} - E_{a(b, \text{ cat})}$
- $20 = 150 - E_{a(b, \text{ cat})}$
- $E_{a(b, \text{ cat})} = 150 - 20$
- Final Answer: 130 kJ mol⁻¹
