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Chapter 18: Problem 97

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

Expert verified
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

01

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).
02

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.
03

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), \(\Delta G^{\circ}=82.93 \mathrm{~kJ} / \mathrm{mol} - (298.15 \mathrm{~K} * 269.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}) * (1\mathrm{kJ} / 1000 \mathrm{~J})\).
04

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 \(25^{\circ}\mathrm{C}\).

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Most popular questions from this chapter

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s)\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g)\). Give the reasons for your arrangement.

Hydrogenation reactions (for example, the process of converting \(\mathrm{C}=\mathrm{C}\) bonds to \(\mathrm{C}-\mathrm{C}\) bonds in food industry) are facilitated by the use of a transition metal catalyst, such as Ni or \(\mathrm{Pt}\). The initial step is the adsorption, or binding, of hydrogen gas onto the metal surface. Predict the signs of \(\Delta H, \Delta S,\) and \(\Delta G\) when hydrogen gas is adsorbed onto the surface of Ni metal.

Crystallization of sodium acetate from a supersaturated solution occurs spontaneously (see p. 426). What can you deduce about the signs of \(\Delta S\) and \(\Delta H ?\)

Predict whether the entropy change is positive or negative for each of these reactions: (a) \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{O}(g)+\mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

What is the role of ATP in biological reactions?
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