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?

First, let's understand what is given in the exercise: 1. The dissolution of ammonium nitrate in water is spontaneous - this means the process occurs naturally without any external intervention. 2. The dissolution is endothermic - this implies that heat is absorbed by the system, and its surroundings experience a decrease in temperature.

According to the Second Law of Thermodynamics, the total entropy change of a spontaneous process should be positive (\(\Delta S_{total} > 0\)). For an endothermic reaction, the change in enthalpy (\(\Delta H\)) is positive.

The Gibbs Free Energy equation can be used to determine the spontaneity of a process: \[\Delta G = \Delta H - T\Delta S\] Where: - \(\Delta G\) is the change in Gibbs Free Energy - \(\Delta H\) is the change in enthalpy (positive for an endothermic reaction) - \(T\) is the temperature in Kelvin - \(\Delta S\) is the change in entropy (what we're trying to deduce) For a spontaneous process, \(\Delta G < 0\).

Since \(\Delta G < 0\) for a spontaneous process, and we know that the dissolution of ammonium nitrate in water is a spontaneous endothermic process, we can deduce the sign of \(\Delta S\). Given that \(\Delta H\) is positive (endothermic process), in order for the Gibbs Free Energy equation (\(\Delta G = \Delta H - T\Delta S\)) to be less than 0 (spontaneous), the term \(-T\Delta S\) must be negative and its magnitude should be greater than that of \(\Delta H\). Thus, for the equation to hold true, \(\Delta S\) must be positive. A positive change in entropy suggests an increase in the disorder of the system during the dissolution process. In conclusion, the sign of \(\Delta S\) for the dissolution of ammonium nitrate in water at room temperature is positive.