Question

For the reaction ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\left( {\text{s}} \right) \to {\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right) + {\text{HCl}}\left( {\text{g}} \right)$ at ${25^ \circ }{\text{C}}$, enthalpy change $\Delta {\text{H}} = + 177{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ and entropy change is $\Delta {\text{S}} = + 285{\text{ kJ mo}}{{\text{l}}^{ - 1}}$. Calculate free energy change $\Delta {\text{G}}$ at ${25^ \circ }{\text{C}}$ and predict whether the reaction is spontaneous or not.

Answer 1

Hint:To solve this we must know the expression that gives the relation between free energy, entropy and enthalpy. From the expression, calculate the free energy. From the positive or negative value of free energy, we can predict whether the reaction is spontaneous or not.

Formula Used:
$\Delta G = \Delta H - T\Delta S$

 Complete step by step answer:We are given the reaction,
${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\left( {\text{s}} \right) \to {\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right) + {\text{HCl}}\left( {\text{g}} \right)$
In the reaction a solid reactant is converted to two gaseous products. Thus, the entropy increases as the reaction proceeds.
We know the expression that gives the relation between free energy, entropy and enthalpy is as follows:
$\Delta G = \Delta H - T\Delta S$
Where, $\Delta G$ is the change in Gibb’s free energy,
              $\Delta H$ is the change in enthalpy,
                 $T$ is the temperature,
               $\Delta S$ is the change in entropy.
Substitute $ + 177{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ for the change in enthalpy, $ + 285{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ for the change in entropy, ${25^ \circ }{\text{C}} + {\text{273}} = 298{\text{ K}}$ for the temperature and solve for the change in free energy. Thus,
$\Delta G = \left( { + 177{\text{ kJ mo}}{{\text{l}}^{ - 1}}} \right) - \left( {298{\text{ K}}\left( { + 285{\text{ kJ mo}}{{\text{l}}^{ - 1}}} \right)} \right)$
$\Delta G = \left( { + 177{\text{ kJ mo}}{{\text{l}}^{ - 1}}} \right) - \left( { + 84930{\text{ kJ mo}}{{\text{l}}^{ - 1}}} \right)$
$\Delta G = - 84753{\text{ kJ mo}}{{\text{l}}^{ - 1}}$
Thus, the free energy change is $ - 84753{\text{ kJ mo}}{{\text{l}}^{ - 1}}$.
We know that if the free energy change is negative then the reaction is spontaneous. If the free energy change is positive then the reaction is non-spontaneous.
A spontaneous reaction means that the reaction occurs on its own and no external conditions are needed. A non-spontaneous reaction does not occur on its own and some external conditions are needed for the reaction to proceed.
Here, the free energy is $ - 84753{\text{ kJ mo}}{{\text{l}}^{ - 1}}$ which is a negative value.
Thus, the reaction, ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\left( {\text{s}} \right) \to {\text{N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right) + {\text{HCl}}\left( {\text{g}} \right)$ is spontaneous.

 Note: A positive value of $\Delta H$ indicates that the dissolution reaction of ammonium chloride is an endothermic process. In an endothermic process, heat is absorbed and thus, $\Delta H$ is positive.