Question

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot\) mol, respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}\) and \(\Delta G^{\circ}\) for the process at \(25^{\circ} \mathrm{C}\). Comment on your answers. $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

Short answer

Simply plug in the provided values to the formulas \(\Delta H^{\circ}\)=82.93 kJ/mol, \(\Delta S^{\circ}\)=269.2 J/K.mol and \(\Delta G^{\circ}\)=\(\Delta H^{\circ}-T\Delta S^{\circ}\). The sign of \(\Delta G^{\circ}\) will indicate if the transition from liquid to gas benzene is spontaneous at \(25^{\circ}\mathrm{C}\).

Step-by-step solution

Calculation of \(\Delta H^{\circ}\)

For calculating \(\Delta H^{\circ}\), we avoid enthalpy changes not directly caused by the phase transition from liquid to gas. Thus \(\Delta H^{\circ}=82.93 \mathrm{~kJ} / \mathrm{mol}\) (given value of standard enthalpy of formation).

Calculation of \(\Delta S^{\circ}\)

The system's entropy change \(\Delta S^{\circ}\) during the transformation from liquid benzene to gaseous is given in the problem, which is 269.2 J/K per mol.

Calculation of \(\Delta G^{\circ}\) at \(25^{\circ}\mathrm{C}\)

The Gibbs free energy change can be calculated using the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\). Rearranging the equation and remembering that the temperature should be in Kelvin (T = 25 + 273.15 = 298.15 K), \(\Delta G^{\circ}=82.93 \mathrm{~kJ} / \mathrm{mol} - (298.15 \mathrm{~K} * 269.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}) * (1\mathrm{kJ} / 1000 \mathrm{~J})\).

Commenting on the results

Once we calculate the values, we can analyze the nature of the process, for instance, if \(\Delta G^{\circ}\) results negative, the process will be spontaneous at \(25^{\circ}\mathrm{C}\).

Key concepts

Enthalpy Change

Enthalpy change, represented as \(\Delta H^{\circ}\), is an essential concept when examining thermodynamic processes. In the exercise, the enthalpy change refers to the energy needed to transform benzene from its liquid to gaseous state under standard conditions. For benzene, this change is given directly as \(82.93 \,\mathrm{kJ/mol}\).
This value is a measure of heat absorbed or released during the phase transition, indicating that the system gains energy during the process in this case. Since the value is positive, it suggests that the transition from liquid to gas for benzene is endothermic, requiring input of energy to overcome the molecular forces in the liquid phase.
In practical terms, this means we need to add heat to liquid benzene to make it evaporate into gas under these specific conditions.
A positive enthalpy change like this is common for phase transitions from liquid to gas, as energy is required to separate molecules more significantly.

Entropy Change

Entropy change, denoted as \(\Delta S^{\circ}\), is a key concept to understand the degree of disorder or randomness in a system. For our exercise, the entropy change from liquid to gaseous benzene is given as \(269.2 \,\mathrm{J/K \, mol}\).
This indicates that the disorder within the system increases as benzene transitions from liquid to gas. Generally, gases have higher entropy compared to liquids because gas molecules have more space to move freely and randomly. This heightened randomness reflects the increased entropy value.
In the context of thermodynamics, entropy change tells us about the feasibility and spontaneity of a process.
For benzene, an increase in entropy is expected and makes sense as the transition involves higher molecular freedom and chaos.

Gibbs Free Energy

Gibbs free energy change, \(\Delta G^{\circ}\), integrates both enthalpy and entropy changes to tell us about the spontaneity of a process. It is calculated using the formula \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\). For benzene, inserting our given values, the calculation becomes:\[\Delta G^{\circ} = 82.93 \,\mathrm{kJ/mol} - (298.15 \,\mathrm{K} \times 269.2 \,\mathrm{J/K \cdot mol}) \times \left(\frac{1 \,\mathrm{kJ}}{1000 \,\mathrm{J}}\right)\]This formula considers both the heat absorbed and the entropy at the given temperature of \(298.15 \,\mathrm{K}\). A negative result for \(\Delta G^{\circ}\) indicates that the process is spontaneous under those conditions.
In essence, Gibbs free energy helps us understand whether a reaction or phase change will proceed without outside energy input.

• If \(\Delta G^{\circ} < 0\), the process is spontaneous.
• If \(\Delta G^{\circ} > 0\), it is non-spontaneous.

Utilizing this knowledge allows chemists and engineers to predict and manipulate conditions to facilitate desired reactions or processes.