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Chapter 18: Problem 75

Describe two ways that you could measure \(\Delta G^{\circ}\) of a reaction.

Short Answer

Expert verified
\(\Delta G^{\circ}\) of a reaction can be measured by two primary means: 1) Direct measurement of equilibrium constants. The standard Gibbs free energy change is related to the equilibrium constant of the reaction by the equation \(\Delta G^{\circ} = -RT \ln K\). 2) Indirect determination through measurement of enthalpy and entropy changes. This is described by the equation \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\).

Step by step solution

01

Method 1: Using Equilibrium Constants

Given a chemical reaction in equilibrium, the equilibrium constant (K) can be directly measured under standard conditions. According to the Gibbs free energy equation, \(\Delta G^{\circ} = -RT \ln K\), where R is the universal gas constant, T is the absolute temperature, and K is the equilibrium constant. Therefore, measure the equilibrium constant of the reaction, substitute it into the equation along with the known values of R and T, and solve for \(\Delta G^{\circ}\).
02

Method 2: Determination of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\)

For chemical reactions where direct measurement of the equilibrium constant is not feasible, \(\Delta G^{\circ}\) can be determined by indirectly measuring the enthalpy change (\(\Delta H^{\circ}\)) and the entropy change (\(\Delta S^{\circ}\)). The relationship between these quantities is given by \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\). You measure \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction, substitute them into the equation along with the absolute temperature T, and solve for \(\Delta G^{\circ}\).

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Most popular questions from this chapter

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?

Define free energy. What are its units?

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. $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)

Under what conditions does a substance have a standard entropy of zero? Can a substance ever have a negative standard entropy?
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