Question

For the reaction at \(298\ K\),
\(2A+B→C\)
\(∆H = 400\ kJ mol^{-1}\) and \(∆S = 0.2\ kJ K^{-1} mol^{-1}\)
At what temperature will the reaction become spontaneous considering \(∆H\) and \(∆S\) to be constant over the temperature range.

Answer 1

From the expression,
\(∆G = ∆H – T∆S\)
Assuming the reaction at equilibrium, \(∆T\) for the reaction would be:
\(T = (△H-△G)\frac {1}{△S}\)
\(T= \frac {△H}{△S}\)         (\(∆G = 0\) at equilibrium)

\(T=\frac { 400\ kJ mol^{-1}}{0.2\ kJK^{-1} mol^{-1}}\)
\(T = 2000\ K\)
For the reaction to be spontaneous, \(∆G\) must be negative. Hence, for the given reaction to be spontaneous, \(T\) should be greater than \(2000\ K\).