Question

For a particular reaction, ΔH∘=−28.4 kJΔH∘=−28.4 kJ and ΔS∘=−87.9 J/K.ΔS∘=−87.9 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?

Answer 1

$\Delta G^{0}$  = $\Delta H^{0}$ - T$\Delta S^{0}$

For, spontaneous reaction , $\Delta G^{0}$ < 0 ( negative)

for nonspontaneous reaction ,

$\Delta G^{0}$ > 0 ( positive).

Given, $\Delta H^{0}$

= -28.4 KJ = -28.4 × 1000 J = 28400 J

and $\Delta S^{0}$ = - 87.9 J/K.

Therefore, T$\Delta S^{0}$ is postive .

Hence, the reaction will be spontaneous when

absolute value of enthalpy change will be greater than absolute value of T$\Delta S^{0}$ term.

|$\Delta H^{0}$ | > T |$\Delta S^{0}$|

Or, T < (|$\Delta H^{0}$|/|$\Delta S^{0}$ |)

Or, T < (28400/87.9)

Or, T < 323.1 K.

Reaction will be spontaneous at below 323.1 K.