Problem 47 - Chemistry: An Atoms-Focused Approach

Question

What are the signs of \(\Delta S, \Delta H,\) and \(\Delta G\) for the sublimation of dry ice (solid \(\mathrm{CO}_{2}\) ) at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: For the sublimation of dry ice at \(25^{\circ} \mathrm{C}\), the signs are as follows: - \(\Delta S > 0\): The entropy increases due to an increase in disorder. - \(\Delta H > 0\): The enthalpy increases due to the endothermic nature of the sublimation process. - \(\Delta G < 0\): The Gibbs free energy decreases, indicating a spontaneous sublimation process at this temperature.

Step 1: Analyze the state change (sublimation) of dry ice

The sublimation of dry ice involves a change in state from solid \(\mathrm{CO}_{2}\) to gaseous \(\mathrm{CO}_{2}\). This is an endothermic process, as energy is absorbed to transition from the lower energy solid state to the higher energy gaseous state.

Step 2: Determine the sign of \(\Delta S\) (entropy change)

Entropy is a measure of the degree of randomness or disorder in a system. In the sublimation process, the solid-state \(\mathrm{CO}_{2}\) molecules transition to the gaseous state. Gas molecules have greater freedom of movement and more possible configurations than solid molecules. Therefore, the overall entropy of the system increases during sublimation. As a result, \(\Delta S\) for this process is positive (\(\Delta S > 0\)).

Step 3: Determine the sign of \(\Delta H\) (enthalpy change)

Enthalpy is a measure of the total energy in a system. As mentioned before, the sublimation of dry ice is an endothermic process, meaning that energy is absorbed from the surroundings. This increases the overall energy of the system. Therefore, the enthalpy change, \(\Delta H,\) is positive (\(\Delta H > 0\)).

Step 4: Determine the sign of \(\Delta G\) (Gibbs free energy change)

Given that we now have the signs for \(\Delta S\) and \(\Delta H\), we can determine the sign for \(\Delta G\). The relationship between Gibbs free energy, enthalpy, and entropy is given by: \(\Delta G = \Delta H - T\Delta S\) Since \(\Delta H > 0\) and \(\Delta S > 0\): At \(25^{\circ} \mathrm{C}\) (or \(298\,\mathrm{K}\)), \(T\Delta S\) would most likely be greater than \(\Delta H\). In such a condition, the overall \(\Delta G\) would be negative (\(\Delta G < 0\)). A negative \(\Delta G\) means that the sublimation process is spontaneous at \(25^{\circ} \mathrm{C}\). In conclusion, the signs of \(\Delta S\), \(\Delta H\), and \(\Delta G\) for the sublimation of dry ice at \(25^{\circ} \mathrm{C}\) are: - \(\Delta S > 0\): The entropy increases due to an increase in disorder. - \(\Delta H > 0\): The enthalpy increases due to the endothermic nature of the sublimation process. - \(\Delta G < 0\): The Gibbs free energy decreases, indicating a spontaneous sublimation process at this temperature.