Question

Consider the following reversible reaction $A_{(g)} + B_{(g)} \rightleftharpoons AB_{(g)}$. The activation energy of backward reaction exceeds than that of forward reaction by 2RT (in Joule $mole^{-1}$). If the pre-exponential factor of the forward reaction is 4 times than that of reverse reaction, the value of $\Delta G^{\circ}$ for the reaction at 300 K is $-Y \times 10^2$ Joule. The value of 'Y' is ______. [Take R = 8.3J/mol-K and ln2 = 0.693]

Answer 1

84

Answer 2

Here's how to solve the problem:

1. Arrhenius Equation:

   The forward and reverse rate constants are given by the Arrhenius equation:

   $k_f = A_f e^{-E_{a,f}/RT}$

   $k_b = A_b e^{-E_{a,b}/RT}$

2. Equilibrium Constant:

   The equilibrium constant is:

   $K_{eq} = \dfrac{k_f}{k_b} = \dfrac{A_f}{A_b} e^{(E_{a,b} - E_{a,f})/RT}$

3. Given Values:

   We are given:

   $\dfrac{A_f}{A_b} = 4$

   $E_{a,b} - E_{a,f} = 2RT$

4. Substitution:

   Substituting the given values:

   $K_{eq} = 4 \, e^{2RT/RT} = 4e^2$

5. Gibbs Free Energy Change:

   The standard Gibbs free energy change is given by:

   $\Delta G^\circ = -RT \ln K_{eq}$

   Substitute $K_{eq} = 4e^2$:

   $\Delta G^\circ = -RT (\ln 4 + 2)$

6. Calculation:

   Given $R=8.3 \, \text{J/mol·K}$, $T=300\,K$, and $\ln 2=0.693$ (so $\ln 4 = 2\ln2 = 1.386$):

   $RT = 8.3 \times 300 = 2490 \, \text{J/mol}$

   Thus,

   $\Delta G^\circ = -2490 (1.386 + 2) = -2490 \times 3.386 \approx -8430 \, \text{J/mol}$

7. Final Answer:

   Expressed in the form $-Y \times 10^2$ Joule, we have:

   $-8430 \, \text{J/mol} = -84.3 \times 10^2 \, \text{J/mol}$

   Rounded off, $Y \approx 84$.