PCl\(_5\)\((_g\)) → PCl\(_3\)\((_s\)) + Cl\(_2\)\((_g\)) In the equation above, the reaction will be spontaneous if

Answer Details

A reaction is spontaneous when the Gibbs free energy change is negative, that is, \(\Delta G < 0\). The Gibbs equation relates enthalpy, entropy, and temperature:

\[\Delta G = \Delta H - T\Delta S\]

For the decomposition of phosphorus pentachloride:

\[\text{PCl}_5(g) \rightarrow \text{PCl}_3(s) + \text{Cl}_2(g)\]

Consider the entropy change. On the reactant side there is 1 mole of gas, and on the product side there is 1 mole of solid and 1 mole of gas. Since a solid has much lower entropy than a gas, the total entropy of the products is lower than that of the reactant. Therefore \(\Delta S\) is negative.

With \(\Delta S < 0\), the term \(-T\Delta S\) becomes positive, which adds to \(\Delta G\). For \(\Delta G\) to still be negative (spontaneous), \(\Delta H\) must be sufficiently negative to overcome the positive \(-T\Delta S\) contribution:

\[\Delta G = \Delta H - T\Delta S < 0\]

\[\Delta H < T\Delta S \quad (\text{where } \Delta S < 0, \text{ so } T\Delta S < 0)\]

This means \(\Delta H\) must be negative. An exothermic reaction (\(\Delta H\) is negative) releases enough energy to drive the process forward despite the unfavourable entropy change.

Exam tip: When \(\Delta S\) is negative, only a sufficiently negative \(\Delta H\) can make \(\Delta G\) negative, and such reactions tend to be spontaneous only at low temperatures.