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Chapter 17: Problem 55

Mow by calculation, using Appendix 1, whether dissolving lead(II) chloride $$ \mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ is spontaneous at \(25^{\circ} \mathrm{C}\) (a) when \(\left[\mathrm{Pb}^{2+}\right]=1.0 \mathrm{M} ;\left[\mathrm{Cl}^{-}\right]=2.0 \mathrm{M}\). (b) when \(\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-5} ;\left[\mathrm{Cl}^{-}\right]=2.0 \times 10^{-5} \mathrm{M}\).

Short Answer

Expert verified
Answer: The dissolution of lead(II) chloride is spontaneous in scenario (a) but not spontaneous in scenario (b).

Step by step solution

01

(Step 1: Calculate the reaction quotient Q)

(For each given scenario, the reaction quotient is defined as \(Q=\frac{[\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2}{[\mathrm{PbCl}_{2}]_{(s)}}\). Since the denominator is a solid, it is not included in the expression. Therefore, \(Q=[\mathrm{Pb}^{2+}][\mathrm{Cl}^-]^2\).)
02

(Step 2: Calculate the standard Gibbs free energy change)

(From Appendix 1, the standard Gibbs free energy change (\(ΔG°\)) for the given reaction is obtained. In this case, \(ΔG°=-27.2 \mathrm{kJ/mol}\).) Now we will calculate the actual Gibbs free energy change for each scenario: #a)#
03

(Step 3a: Substitute the given concentrations into the reaction quotient Q)

(Using the given concentrations for scenario (a): \(\left[\mathrm{Pb}^{2+}\right]=1.0 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=2.0 \mathrm{M}\), the reaction quotient Q becomes: \(Q_a=(1.0)(2.0)^2=4.0\).)
04

(Step 4a: Calculate the actual Gibbs free energy change)

(Now we can use the formula for the actual Gibbs free energy change, \(\Delta G=RTln(Q) - \Delta G^\circ\). For this case, we have: \(\Delta G_a=(8.314\times10^{-3}\mathrm{kJ/mol K})(298.15\mathrm{K})(\ln{4})-(-27.2\mathrm{kJ/mol}) \approx{-25.3}\,\mathrm{kJ/mol}\).) Since the actual Gibbs free energy change is negative (\(ΔG_a<0\)), the dissolution of lead(II) chloride is spontaneous in scenario (a). #b)#
05

(Step 3b: Substitute the given concentrations into the reaction quotient Q)

(Using the given concentrations for scenario (b): \(\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=2.0 \times 10^{-5} \mathrm{M}\), the reaction quotient Q becomes: \(Q_b=(1.0 \times 10^{-5})(2.0 \times 10^{-5})^2=8.0 \times 10^{-15}\).)
06

(Step 4b: Calculate the actual Gibbs free energy change)

(Now we can use the formula for the actual Gibbs free energy change, \(\Delta G=RTln(Q) - \Delta G^\circ\). For this case, we have: \(\Delta G_b=(8.314\times10^{-3}\mathrm{kJ/mol K})(298.15\mathrm{K})(\ln(8.0 \times 10^{-15}))-(-27.2\mathrm{kJ/mol}) \approx{19.2}\,\mathrm{kJ/mol}\).) Since the actual Gibbs free energy change is positive (\(ΔG_b>0\)), the dissolution of lead(II) chloride is not spontaneous in scenario (b).

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Most popular questions from this chapter

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\) (b) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g)\)

Consider the reaction $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ At \(500^{\circ} \mathrm{C}\), a flask initially has all three gases, each at a partial pressure of \(0.200\) atm. When equilibrium is established, the partial pressure of HI is determined to be \(0.48\) atm. What is \(\Delta G^{\circ}\) for the reaction at \(500^{\circ} \mathrm{C}\) ?

In each of the following pairs, choose the substance with a lower entropy. (a) \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(10^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}(l)\) at \(30^{\circ} \mathrm{C}\) (b) C (graphite), \(\mathrm{C}\) (diamond) (c) \(\mathrm{Cl}_{2}(l), \mathrm{Cl}_{2}(g)\), both at room temperature

Consider the following reactions at \(25^{\circ} \mathrm{C}\) : $$ \begin{aligned} &\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O} \quad \Delta G^{\circ}=-2870 \mathrm{~kJ} \\ &\mathrm{ADP}(a q)+\mathrm{HPO}_{4}{ }^{2-}(a q)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O} \\ &\Delta G^{\circ}=31 \mathrm{~kJ} \end{aligned} $$ Write an equation for a coupled reaction between glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), and ADP in which \(\Delta G^{\circ}=-390 \mathrm{~kJ}\).

At \(1200 \mathrm{~K}\), an equilibrium mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) gases contains 98.31 mol percent CO and some solid carbon. The total pressure of the mixture is \(1.00 \mathrm{~atm}\). For the system $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) $$ calculate (a) \(P_{\mathrm{Co}}\) and \(P_{\mathrm{CO}_{2}}\) (b) \(K\) (c) \(\Delta G^{\circ}\) at \(1200 \mathrm{~K}\)
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