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

We can solve this by combining the thermodynamic equations into a single expression and then performing the calculations with the provided logs.

Starting equations:
1. $\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}$
2. $\Delta G^{\theta} = -RT \ln K$

Combining them:
$$\Delta H^{\theta} - T\Delta S^{\theta} = -RT \ln K$$
Rearranging to solve for $\Delta S^{\theta}$:
$$T\Delta S^{\theta} = \Delta H^{\theta} + RT \ln K$$
$$\Delta S^{\theta} = \frac{\Delta H^{\theta} + RT \ln K}{T} = \frac{\Delta H^{\theta}}{T} + R \ln K$$

Substitute $\ln K = 2.3 \log K$:
$$\Delta S^{\theta} = \frac{28400}{300} + 8.3 \times 2.3 \times \log(1.8 \times 10^{-7})$$

Calculate $\log(1.8 \times 10^{-7})$:
$$\log(1.8) + \log(10^{-7}) = \log(18/10) - 7 = \log 18 - 1 - 7 = \log(2 \times 3^2) - 8$$
$$= \log 2 + 2\log 3 - 8 = 0.30 + 0.96 - 8 = -6.74$$

Now substitute into the $\Delta S^{\theta}$ equation:
$$\Delta S^{\theta} = 94.67 + 19.09 \times (-6.74)$$
$$\Delta S^{\theta} = 94.67 - 128.6666 = -33.9966$$
Rounding to the nearest integer, $\Delta S^{\theta} = -34$.

The magnitude of $\Delta S^{\theta}$ is 34.