Question

The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and 269.2 J/K*mol, respectively. Calculate ΔH° , ΔS° , and ΔG° for the following process at 25.00°C.
C6H6(l) ------> C6H6(g)

Answer 1

Answer:

ΔG° = 2708.4J/mol , ΔS° = 269.2 J/K*mol, ΔH° = 82.93 kJ/mol

Explanation:

The relationship between all three parameters mentioned is given as;

ΔG° (change in free energy) =ΔH (change in enthalpy) −TΔS (temperature * change in entropy)

ΔH° = 82.93 kJ/mol = 82.93 * 1000 = 82930J/mol

ΔS° = 269.2 J/K*mol

T = 25 + 273 = 298K (Converting to Standard unit of Kelvin)

ΔG° = ΔH° - TΔS°

ΔG° = 82930 - 298 * 269.2

ΔG° = 82930J/mol - 80221.6J/mol

ΔG° = 2708.4 J/mol