MHTCET 14 May 2023 Shift 1 Solved Paper

To find the entropy change of the surroundings for the given process, we can use the following thermodynamic relation which connects the enthalpy change $(\Delta H)$ of a system to the entropy change of the surroundings ( $\Delta S_{\text{surroundings}}$ ) at constant temperature:$\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}$ In this formula, $\Delta H$ is the enthalpy change for the system, and $T$ is the temperature at which the process takes place. The negative sign indicates that if the process is endothermic $(\Delta H > 0)$ for the system, the entropy of the surroundings decreases ( $\Delta S_{\text{surroundings}} < 0$ ), and vice versa for an exothermic process.For the reaction given:$\mathrm{H}_2\mathrm{O}_{(s)} \rightarrow \mathrm{H}_2\mathrm{O}_{(\ell)}$ the enthalpy change is given as $+7\text{ kJ mol}^{-1}$ or $+7000\text{ J mol}^{-1}$ when converted to joules, since $1\text{ kJ} = 1000\text{ J}$, and the temperature is $300\text{ K}$. Plugging these values into our equation gives:$\Delta S_{\text{surroundings}} = -\frac{+7000\text{ J mol}^{-1}}{300\text{ K}}$To simplify this: $\Delta S_{\text{surroundings}} = -\frac{7000}{300}\text{ J K}^{-1}\text{ mol}^{-1}$$\Delta S_{\text{surroundings}} = -23.333\ldots\text{ J K}^{-1}\text{ mol}^{-1}$Therefore, the entropy change of the surroundings for the melting of ice at $300\text{ K}$ with an enthalpy change of $+7\text{ kJ mol}^{-1}$ is approximately:$\Delta S_{\text{surroundings}} \approx \approx -23.3\text{ J K}^{-1}\text{ mol}^{-1}$This corresponds to Option $B: -23.3\text{ J K}^{-1}$.