Substance X has a heat of vaporization of 55.4 kJ/mol at its normal boiling point (423°C). For the process X(l) $\rightarrow$ X(g) at 1 atm and 423°C, calculate the value of:
-$\Delta$S

Question

Quizplus · Accepted Answer

The entropy change (ΔS) for vaporization can be calculated using the formula ΔS = ΔH/T, where ΔH is the heat of vaporization and T is the temperature in Kelvin. Converting 423°C to Kelvin gives 696 K. Thus, ΔS = 55,400 J/mol / 696 K = 79.6 J/K•mol.

Substance X Has a Heat of Vaporization of 55 →\rightarrow→ X(g) at 1 Atm and 423°C, Calculate the Value Of

Substance X Has a Heat of Vaporization of 55 $\rightarrow$ X(g) at 1 Atm and 423°C, Calculate the Value Of