Given: H₂O(l) $\rightarrow$ H₂O(s), $\Delta$H$\degree$ = -6.02 kJ at 273K, calculate the entropy change of the surroundings ($\Delta$S_surr) when one mole of water freezes at 0$\degree$C and a pressure of one atmosphere.

Question

Quizplus · Accepted Answer

The entropy change of the surroundings ($\Delta S_{surr}$) is calculated using the formula $\Delta S_{surr} = -\frac{\Delta H}{T}$. Here, $\Delta H = -6.02 \text{ kJ} = -6020 \text{ J}$ and $T = 273 \text{ K}$. Thus, $\Delta S_{surr} = -\frac{-6020}{273} = 22.1 \text{ J/K}$.

Given: H2O(l) →\rightarrow→ H2O(s) Δ\DeltaΔ H °\degree° = -602 KJ at 273K, Calculate the Entropy Change of the at 273K

Given: H₂O(l) $\rightarrow$ H₂O(s) $\Delta$ H $\degree$ = -602 KJ at 273K, Calculate the Entropy Change of the at 273K