Question

If ΔH = 400 kjoule mole⁻¹ and ΔS = 0.2 kjoule K⁻¹ mole⁻¹ for a reaction 2x + y --> z, at which minimum temperature the reaction will be spontaneous?

Answer 1

Answer:

2000K

Explanation:

ΔH = 400 kjoule mole⁻¹

ΔS = 0.2 kjoule K⁻¹

For the reaction at 298 K, the equation formed will be -

2A + B → C ∆H = 400 kJ mol–1 and

∆S = 0.2 kJ K–1 mol–1

Therefore, from the expression,  ∆G = ∆H – T∆S  

Assuming the reaction at equilibrium, ∆T for the reaction would be

T = ∆H - ∆G × 1/∆S

= ∆H /∆S

= 400/0.2kJ ( ∆G = 0 at equilibrium)

T = 2000 K  

Thus, for the reaction to be spontaneous, ∆G must be negative.

Hence, at a temperature of 2000 K the reaction will be spontaneous.