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Chapter 17: Problem 106

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

Short Answer

Expert verified
\(\Delta H^{\circ} = 82.93 \mathrm{~kJ}/\mathrm{mol}, \Delta S^{\circ} = 269.2 \mathrm{~J}/(\mathrm{K} \cdot \mathrm{mol}), \Delta G^{\circ} = -16.39 \mathrm{~kJ/mol}\). The gaseous form of benzene is more thermodynamically stable at 25°C.

Step by step solution

01

Setting up the problem

We are given the enthalpy of formation (\(\Delta H^{\circ}_{f}\)) of 82.93 kJ/mol and entropy (\(S^{\circ}\)) of 269.2 J/(K.mol) of gaseous benzene. We are asked to calculate the change in enthalpy (\(\Delta H^{\circ}\)), entropy (\(\Delta S^{\circ}\)), and Gibbs free energy (\(\Delta G^{\circ}\)) for the change of state from liquid to gas at 25 °C (which is equivalent to 298.15 K).
02

Calculating the enthalpy change

In this phase transition, the enthalpy change (\(\Delta H^{\circ}\)) is simply equal to the enthalpy of formation (\(\Delta H^{\circ}_{f}\)) of the gaseous benzene because the liquid benzene has an enthalpy of formation of zero (since it's the reference state). Hence, \(\Delta H^{\circ} = \Delta H^{\circ}_{f} = 82.93 \) kJ/mol.
03

Calculating the entropy change

For the phase transition from a liquid to a gas, the entropy change (\(\Delta S^{\circ}\)) is simply equal to the entropy of gaseous benzene (\(S^{\circ}\)) which is already given as 269.2 J/(K.mol), i.e., \(\Delta S^{\circ} = S^{\circ} = 269.2 \) J/(K.mol).
04

Calculating the Gibbs free energy change

The standard Gibbs free energy change (\(\Delta G^{\circ}\)) can be calculated using the formula \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), where T is the absolute temperature (in Kelvin). Plugging the given values into this formula, \(\Delta G^{\circ} = (82.93 \mathrm{~kJ} / \mathrm{mol}) - (298.15 \mathrm{~K}) \times (269.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol})\) converting the enthalpy change in Joules (\(1 \mathrm{~kJ} = 1000 \mathrm{~J}\)), this gets \(\Delta G^{\circ} = (82930 \mathrm{~J}/\mathrm{mol}) - (298.15 \mathrm{~K}) \times (269.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol})\) which simplifies to \(\Delta G^{\circ} = -16.39 \mathrm{~kJ/mol}\)
05

Interpretation of the results

The negative value for \(\Delta G^{\circ}\) implies that the transition of benzene from liquid to gaseous phase is spontaneous at 25°C. Also, the positive values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) suggests an endothermic reaction (absorbs heat) that leads to an increase in disorder as the benzene molecules gain enough energy to transition from the liquid phase to the gas phase.

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Most popular questions from this chapter

Which of the following are not state functions: \(S, H,\) \(q, w, T ?\)

Entropy has sometimes been described as "time's arrow" because it is the property that determines the forward direction of time. Explain.

From the values of \(\Delta H\) and \(\Delta S,\) predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}\) : reaction \(\mathrm{A}: \Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} ;\) reaction \(\mathrm{B}: \Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C}\), at what temperature might it become spontaneous?

A student placed \(1 \mathrm{~g}\) of each of three compounds A, B, and C in a container and found that after 1 week no change had occurred. Offer some possible explanations for the fact that no reactions took place. Assume that \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are totally miscible liquids.

The internal engine of a 1200 -kg car is designed to run on octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right),\) whose enthalpy of combustion is \(5510 \mathrm{~kJ} / \mathrm{mol}\). If the car is moving up a slope, calculate the maximum height (in meters) to which the car can be driven on 1.0 gallon of the fuel. Assume that the engine cylinder temperature is \(2200^{\circ} \mathrm{C}\) and the exit temperature is \(760^{\circ} \mathrm{C},\) and neglect all forms of friction. The mass of 1 gallon of fuel is \(3.1 \mathrm{~kg} .\) [Hint: See the Chemistry in Action essay "The Efficiency of Heat Engines" in Section \(17.5 .\) The work done in moving the car over a vertical distance is \(m g h,\) where \(m\) is the mass of the car in \(\mathrm{kg}\), \(g\) the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right),\) and \(h\) the height in meters.]
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