Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions.
Question:
Calculate the entropy change in surroundings when $1.00 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}_{(f)}$ is formed under standard conditions. $\Delta \mathrm{f}^{\theta}=-286 \mathrm{~kJ}$ mol $^{-1}$.
Solution:
It is given that $286 \mathrm{~kJ} \mathrm{~mol}^{-1}$ of heat is evolved on the formation of $1 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}_{(j)}$. Thus, an equal amount of heat will be absorbed by the surroundings.
$q_{\text {surr }}=+286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Entropy change $\left(\Delta S_{\text {surr }}\right)$ for the surroundings $=\frac{q_{\text {surr }}}{7}$
$=\frac{286 \mathrm{~kJ} \mathrm{~mol}^{-1}}{298 \mathrm{k}}$
$\therefore \Delta S_{\text {surr }}=959.73 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$