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Chapter 17: Problem 27

(a) Calculate \(\Delta G^{\circ}\) and \(K_{P}\) for the following equilibrium reaction at \(25^{\circ} \mathrm{C}\). The \(\Delta G_{\mathrm{f}}^{\circ}\) values are 0 for \(\mathrm{Cl}_{2}(g),-286 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{3}(g),\) and \(-325 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{5}(g)\) $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ (b) Calculate \(\Delta G\) for the reaction if the partial pressures of the initial mixture are \(P_{\mathrm{PCl}_{5}}=0.0029 \mathrm{~atm}\) \(P_{\mathrm{PCl}_{3}}=0.27 \mathrm{~atm},\) and \(P_{\mathrm{Cl}_{2}}=0.40 \mathrm{~atm}\)

Short Answer

Expert verified
\(\Delta G^{\circ} = 39 kJ/mol\), \(K_{P} = 1.16 \times 10^{-6} atm^{-1}\) and \(\Delta G = 47.9 kJ/mol\)

Step by step solution

01

Calculate \(\Delta G^{\circ}\)

The standard free energy change of a reaction \(\Delta G^{\circ}\) is calculated using the equation \(\Delta G^{\circ} = \Delta G_{f}^{\circ}(products) - \Delta G_{f}^{\circ}(reactants)\). Using the provided standard Gibbs free energies of formation \(\Delta G_{f}^{\circ}\), we obtain \(\Delta G^{\circ} = [(-286 kJ/mol + 0 kJ/mol) - (-325 kJ/mol)] = 39 kJ/mol\).
02

Calculate \(K_{P}\)

Now using \(\Delta G^{\circ}\), we calculate the equilibrium constant \(K_{P}\) using the relationship \(K_{P} = e^{-\Delta G^{\circ}/RT}\). The gas constant \(R\) is 8.314 J/(mol*K), temperature \(T\) is 25 + 273.15 = 298.15 K and remember to convert \(\Delta G^{\circ}\) to J/mol: \(K_{P} = e^{-39000J/(8.314J/mol*K \* 298.15K)} = 1.16 \times 10^{-6} atm^{-1}\).
03

Calculate \(\Delta G\) for specific partial pressures

To find \(\Delta G\) we first calculate the reaction quotient \(Q = [PCl_{3}][Cl_{2}]/[PCl_{5}]\) using the given partial pressures. So, \(Q = (0.27 atm)(0.40 atm)/(0.0029 atm) = 36.55\). Now, \(\Delta G\) can be calculated by using \(\Delta G = \Delta G^{\circ} + RT ln(Q)\): \(\Delta G = (39 kJ/mol)*1000 + (8.314 J/mol*K)(298.15K)ln(36.55) = (39 kJ/mol)*1000 + (8.314 J/mol*K)(298.15K)(3.598) = (39*10^3) + (8.314*298.15*3.598) = 39000 J/mol + 8927.60 J/mol = 47927.60 J/mol = 47.9 kJ/mol (when converted back to kJ/mol).\nThus, the free energy change for the reaction with the given initial partial pressures is 47.9 kJ/mol.

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Most popular questions from this chapter

Comment on the correctness of the analogy sometimes used to relate a student's dormitory room becoming untidy to an increase in entropy.

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) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)

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

In each of the following reactions, there is one species for which the standard entropy value is not listed in Appendix \(2 .\) Determine the \(S^{\circ}\) for that species. (a) The \(\Delta S_{\mathrm{rxn}}^{\circ}\) for the reaction \(\mathrm{Na}(s) \longrightarrow \mathrm{Na}(l)\) is \(48.64 \mathrm{~J} / \mathrm{K}\) mol. (b) The \(\Delta S_{\mathrm{rxn}}^{\circ}\) for the reaction \(2 \mathrm{~S}\) (monoclinic) \(+\) \(\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{S}_{2} \mathrm{Cl}_{2}(g)\) is \(43.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} .\) (c) The \(\Delta S_{\mathrm{rxn}}^{\circ}\) for the reaction \(\mathrm{FeCl}_{2}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)\) is \(-118.3 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\)

Explain the following nursery rhyme in terms of the second law of thermodynamics. Humpty Dumpty sat on a wall; Humpty Dumpty had a great fall. All the King's horses and all the King's men Couldn't put Humpty together again.
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