Question

State whether the following reaction is spontaneous or non spontaneous if ΔH = 50 kJ and ΔS = −130 JK⁻¹ at 250 K.

Answer 1

Given: Enthalpy change (ΔH) = 50 kJ = 50000 J

Entropy change (ΔS) = −130 JK⁻¹

Temperature (T) = 250 K

By using the Gibbs free energy change formula:

ΔG = ΔH − TΔS

= 50000 J − (250 K × −130 JK⁻¹)

= 50000 J − (−32500 J)

= 50000 J + 32500 J

= 82500 J

= +82.5 kJ

A reaction is spontaneous only if ΔG < 0. Since the calculated value is positive (+82.5 kJ), the reaction requires energy to proceed and is therefore non-spontaneous.