Question

Consider the reaction $X \rightleftharpoons Y$ at $300$ K. If $\Delta H^{\theta}$ and $K$ are $28.40$ kJ mol$^{-1}$ and $1.8 \times 10^{-7}$ at the same temperature, then the magnitude of $\Delta S^{\theta}$ for the reaction in J K$^{-1}$ mol$^{-1}$ is ______. (Nearest integer)
(Given : $R = 8.3$ J K$^{-1}$ mol$^{-1}$, $\ln 10 = 2.3$, $\log 3 = 0.48$, $\log 2 = 0.30$)

Hint:

Use the formula $\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}$ and $\Delta G^{\theta} = -RT \ln K$. Calculate $\Delta G^{\theta}$ first using the logarithmic values provided, then solve for $\Delta S^{\theta}$.

Answer 1

Correct Answer: 34

To solve this problem, we need to relate the equilibrium constant ($K$) to the standard Gibbs free energy change ($\Delta G^{\theta}$) and then relate $\Delta G^{\theta}$ to the standard enthalpy ($\Delta H^{\theta}$) and standard entropy ($\Delta S^{\theta}$) changes.

Step 1: Calculate the standard Gibbs free energy change ($\Delta G^{\theta}$) using the formula:
$$\Delta G^{\theta} = -RT \ln K$$
Substituting the relation $\ln x = 2.3 \log x$ as given by $\ln 10 = 2.3$:
$$\Delta G^{\theta} = -RT (2.3 \log K)$$

Step 2: Evaluate $\log K$.
$$K = 1.8 \times 10^{-7} = \frac{18}{10} \times 10^{-7} = 18 \times 10^{-8}$$
$$\log K = \log(18 \times 10^{-8}) = \log(2 \times 3^2) + \log(10^{-8})$$
$$\log K = \log 2 + 2\log 3 - 8$$
Using given values $\log 2 = 0.30$ and $\log 3 = 0.48$:
$$\log K = 0.30 + 2(0.48) - 8 = 0.30 + 0.96 - 8 = 1.26 - 8 = -6.74$$

Step 3: Substitute values back into the $\Delta G^{\theta}$ equation.
$$\Delta G^{\theta} = - (8.3 \text{ J K}^{-1} \text{ mol}^{-1}) \times (300 \text{ K}) \times 2.3 \times (-6.74)$$
$$\Delta G^{\theta} = 2490 \times 2.3 \times 6.74$$
$$\Delta G^{\theta} = 5727 \times 6.74 = 38600.02 \text{ J mol}^{-1} \approx 38600 \text{ J mol}^{-1}$$

Step 4: Use the Gibbs-Helmholtz equation:
$$\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}$$
Convert $\Delta H^{\theta}$ to Joules: $28.40 \text{ kJ mol}^{-1} = 28400 \text{ J mol}^{-1}$.
$$38600 = 28400 - 300 \times \Delta S^{\theta}$$
$$300 \times \Delta S^{\theta} = 28400 - 38600 = -10200$$
$$\Delta S^{\theta} = -\frac{10200}{300} = -34 \text{ J K}^{-1} \text{ mol}^{-1}$$

Step 5: Find the magnitude.
Magnitude of $\Delta S^{\theta} = |-34| = 34$.