Question

For the reaction at 298 K,

\[\ce{2A + B → C}\]

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?

Answer 1

From the expression,

ΔG = ΔH – T ΔS

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" kJ mol"^(-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.