Question

# The dissolution of ammonium chloride in water is an endothermic reaction, yet it is a spontaneous process. This is due to the fact that:A.$\Delta H$ is +ve, $\Delta S$ is –ve.B.$\Delta H$ is -ve, $\Delta S$ is +ve.C.$\Delta H$ is +ve, $\Delta S$ is +ve and $\Delta H < T\Delta S$D. $\Delta H$ is +ve and$\;\Delta H\; > T{\text{ }}\Delta S$.

Hint: An endothermic reaction is a reaction in which heat is absorbed and the system feels cold. A spontaneous process is a process which goes in the forward direction and a non-spontaneous process is a process which goes in a backward direction.$\Delta H$ and $\Delta S$ are thermodynamic parameters.

According to Gibbs free energy equation:
$\Delta G{\text{ }} = {\text{ }}\Delta H - T\Delta S$
$\Delta G$ = Gibbs free energy change
$\Delta H$ = enthalpy change
$T$ = temperature
$\Delta S$ = entropy change
The conditions for a process to be spontaneous are:
$\Delta G < 0$ or $\Delta G$ should always be negative.
For an endothermic reaction $\Delta H{\text{ }} > {\text{ }}0$or $\Delta H$ is positive.
So for $\Delta G$ to be negative, $\Delta S$must be positive and the $T{\text{ }}\Delta S$ must be greater than $\Delta H$ as the latter is positive.
Now considering the dissolution of $N{H_4}Cl$, it is an endothermic reaction. Thus $\Delta H{\text{ }} > {\text{ }}0$or $\Delta H$ is positive.
So $\Delta S$ must be positive and the value of $T{\text{ }}\Delta S$must be greater than$\Delta H$. The dissolution of $N{H_4}Cl$ results in a favorable increase of entropy which overcomes the enthalpy change during the dissolution. As a result $T{\text{ }}\Delta S$ become negative and the total of $\Delta H - T\Delta S$ is negative. Hence $\Delta G$is negative. This accounts for the favorable dissolution of ammonium chloride in water.

Note:
Unlike endothermic reaction, an exothermic reaction $\Delta H$is always negative and it makes the reaction go spontaneous. The change in entropy does not affect the spontaneity of the exothermic reaction.