Given: H₂O(l) $\rightarrow$ H₂O(g), $\Delta$H$\degree$ = 40.7 kJ at 373K, what is the entropy change in the system ($\Delta$S) when one mole of water vaporizes at 100$\degree$C and a pressure of one atmosphere?

Question

Quizplus · Accepted Answer

The entropy change ($\Delta$S) can be calculated using the formula $\Delta$S = $\Delta$H/T. Here, $\Delta$H = 40.7 kJ = 40700 J and T = 373 K. So, $\Delta$S = 40700 J / 373 K = 109 J/K.

Given: H2O(l) →\rightarrow→ H2O(g) Δ\DeltaΔ H °\degree° = 407 KJ at 373K, What Is the Entropy Change in at 373K

Given: H₂O(l) $\rightarrow$ H₂O(g) $\Delta$ H $\degree$ = 407 KJ at 373K, What Is the Entropy Change in at 373K