If values of ΔH° & ΔS° for a process/reaction are 77.2 kJ & 48 J/k respectively. Then find value of log 1/k. Given : Temp is 300K

The correct option is 11.

To find the value of $\log (1/K)$, we can use the thermodynamic relationship between the standard Gibbs free energy change ($\Delta G^{°}$), enthalpy change ($\Delta H^{°}$), and entropy change ($\Delta S^{°}$):

$\Delta G^{°}=\Delta H^{°}-T\Delta S^{°}$

Step 1: Convert the units and calculate $\Delta G^{°}$
Given parameters:
• $\Delta H^{°}=77.2\text{kJ}=77200\text{J}$
• $\Delta S^{°}=48\text{J/K}$
• $T=300\text{K}$

Substituting these values into the Gibbs free energy equation:
$\Delta G^{°}=77200\text{J}-(300\text{K}\times 48\text{J/K})$
$\Delta G^{°}=77200\text{J}-14400\text{J}$
$\Delta G^{°}=62800\text{J}$

Step 2: Relate $\Delta G^{°}$ to the equilibrium constant $K$
The standard Gibbs free energy change is related to the equilibrium constant $K$ by the formula:
$\Delta G^{°}=-2.303RT\log K$

Using the property of logarithms where $-\log K=\log (1/K)$, we can rewrite the equation as:
$\Delta G^{°}=2.303RT\log (1/K)$

Rearranging the equation to solve for $\log (1/K)$:
$\log (1/K)=\frac{\Delta G^{°}}{2.303RT}$

Using the universal gas constant $R=8.314\text{J/(mol K)}$:
$\log (1/K)=\frac{62800}{2.303\times 8.314\times 300}$
$\log (1/K)=\frac{62800}{5744.14}\approx 10.93$

Rounding off to the nearest integer, we get:
$\log (1/K)\approx 11$