Question

For the reaction $\mathrm{A}\to \mathrm{B}$at 298 K, the change in enthalpy is$-7{\mathrm{kJmol}}^{-1}$and the change in entropy is $-25{\mathrm{kJmol}}^{-1}$. Is the reaction spontaneous? If not, should the temperature be increased or decreased to make the reaction spontaneous?

Short answer

$\mathrm{\Delta H}\succ \mathrm{T\Delta S}$thus the temperature is decreased to make the reaction spontaneous.

Step-by-step solution

Given data

The Gibbs free energy can be denoted as,

$\begin{array}{l}\mathrm{\Delta H}=-7\mathrm{kJ}/\mathrm{mol}\\ \mathrm{\Delta S}=-25\mathrm{J}/{\mathrm{molK}}^{-1}\\ \mathrm{T}=298\mathrm{K}\end{array}$

Substitution

$\begin{array}{l}\mathrm{\Delta G}=\mathrm{\Delta H}-\mathrm{T\Delta S}\\ =-7000-\left(-298*\left(-25\right)\right)\\ =450\mathrm{J}/\mathrm{mol}\end{array}$

$\mathrm{\Delta G}$ is not positive spontaneous reaction.

Simplification

It can be denoted as,

$\begin{array}{l}\Delta G=\Delta H-T\Delta S=+ve\\ \Delta G=\Delta H-T\Delta S\succ 0\\ \Delta H\succ T\Delta S\end{array}$

Conclusion

Therefore, the temperature is decreased to make the reaction spontaneous.