For the reaction at 298 K,
2A + B → C
ΔH = 400 kJ mol–1 and ΔS = 0.2 kJ K–1 mol–1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
From the expression,
ΔG = ΔH – TΔS
Assuming the reaction at equilibrium, ΔT for the reaction would be:
`T = (triangle h - triangle G)1/(triangle S)`
= `(triangle H)/ (triangle S) (triangle G = 0 " at equilibrium")`
` = (400 kJmol^(-1))/(0.2 kJ K^(-1) mol^(-1))`
T = 2000 K
For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.
As per the Gibbs Helmholtz equation:
ΔG = Δ H- TΔ S For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS
T = (400 KJ mol-1)/(0.2 KJ K-1 mol-1) = 2000 k
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.