Question

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?

Short answer

The dissolution of ammonium nitrate in water is spontaneous and endothermic, so \(\Delta G < 0\) and \(\Delta H > 0\). Using the Gibbs free energy equation (\(\Delta G = \Delta H - T \Delta S\)), we deduce that the change in entropy (\(\Delta S\)) must be positive for the inequality to hold true. Therefore, \(\Delta S > 0\) for this solution process.

Step-by-step solution

Recall the Gibbs free energy equation

The Gibbs free energy equation relates the enthalpy, entropy, and temperature of a process to its spontaneity. The equation is given by: \[ \Delta G = \Delta H - T \Delta S\] where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy. A spontaneous process is characterized by a negative value of \(\Delta G\), i.e., \(\Delta G < 0\).

Analyze the given information

We are given that the dissolution of ammonium nitrate is spontaneous and endothermic. This means that \(\Delta G < 0\) and \(\Delta H > 0\) (endothermic implies an increase in enthalpy or a positive value of \(\Delta H\)). From the Gibbs free energy equation, we have: \[\Delta G = \Delta H - T \Delta S < 0\]

Deduce the sign of \(\Delta S\)

From the inequality \(\Delta G = \Delta H - T \Delta S < 0\), we have: \[T \Delta S > \Delta H\] Since the temperature \(T\) is positive (room temperature) and \(\Delta H > 0\), we can deduce the sign of \(\Delta S\). For the inequality to hold true, the change in entropy (\(\Delta S\)) must be positive. Thus, the dissolution of ammonium nitrate in water has a positive change in entropy (\(\Delta S > 0\)).