Given: H₂O(l) → H₂O(s) ΔH° = −6.02 kJ at 273K Calculate the entropy change of the surroundings (ΔS_surr) when one mole of water freezes at 0°C and a pressure of one atmosphere.

Question

Quizplus · Accepted Answer

The entropy change of the surroundings (ΔS_surr) when a process occurs at constant temperature and pressure can be calculated using the formula ΔS_surr = -ΔH/T, where ΔH is the enthalpy change of the system and T is the temperature in Kelvin. Given ΔH° = −6.02 kJ (or −6020 J for one mole of water freezing) and T = 273K, ΔS_surr = -(-6020 J) / 273 K = 22.05 J/K, which rounds to 22.1 J/K.

Given: H2O(l) → H2O(s) ΔH° = −6

Given: H₂O(l) → H₂O(s) ΔH° = −6