Question

For the reaction 3C2H2(g)---> C6H6(l) at 25 C the standard enthalpy change is -631 kj and the standard entropy change is -430 J/K.Calculate the standard free energy change at 25 C?

Answer 1

Given :

Temperature , T = 25°C = (25 + 273) K = 298 K .

The standard enthalpy change , $\Delta H^{\circ}=-631\ kJ/mol$ .

The standard entropy change , $\Delta S^{\circ}=-430\ J/K=-0.430\ kJ/K$ .

To Find :

The standard free energy change at 25° C .

Solution :

Standard free energy is given by :

$\Delta G^o=\Delta H^o-T\Delta S^o\\\\\Delta G^o=-631-[298\times (-0.430)]\ kJ\\ \\\Delta G^o=-631-(-128.14)\ kJ\\\\\Delta G^o=-502.86\ kJ$

Therefore , the standard free energy change at 25° C is -502.86 kJ .

Hence , this is the required solution .