Question

Show 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 \(\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M} ;\left[\mathrm{Cl}^{-}\right]=2.0 \times 10^{-5} \mathrm{M}\)

Short answer

Answer: Yes, the dissolution of lead(II) chloride is spontaneous under both given conditions. In the first situation, with concentrations of \([\mathrm{Pb}^{2+}]=1.0 \, M\) and \([\mathrm{Cl}^{-}]=2.0 \, M\), the Gibbs Free Energy Change is \(\Delta G = -24.08 \, kJ/mol\), indicating spontaneity. In the second situation, with concentrations of \([\mathrm{Pb}^{2+}] = 1.0 \times 10^{-5} \, M\) and \([\mathrm{Cl}^{-}] = 2.0 \times 10^{-5} \, M\), the Gibbs Free Energy Change is \(\Delta G = -18.7 \, kJ/mol\), also indicating spontaneity.

Step-by-step solution

Determine the standard Gibbs Free Energy Change, \(\Delta G^{\circ}\)

According to Appendix 1, the standard Gibbs Free Energy Change for the dissolution of lead(II) chloride is \(\Delta G^{\circ} = -27.92 \, kJ/mol\).

Calculate the Reaction Quotient, Q

The reaction quotient, Q, is given by the expression: \(Q = \frac{[\mathrm{Pb}^{2+}] [\mathrm{Cl}^{-}]^2}{1}\), where the denominator is 1 because lead(II) chloride is a solid and the concentration of solids is not taken into account in the reaction quotient. (a) For the given concentrations: \([\mathrm{Pb}^{2+}]=1.0 \, M\) and \([\mathrm{Cl}^{-}]=2.0 \, M\), Q can be calculated as: \(Q = \frac{1.0 \times (2.0)^2}{1} = 4\). (b) For the given concentrations: \([\mathrm{Pb}^{2+}] = 1.0 \times 10^{-5} \, M\) and \([\mathrm{Cl}^{-}] = 2.0 \times 10^{-5} \, M\), Q can be calculated as: \(Q = \frac{1.0 \times 10^{-5} \times (2.0 \times 10^{-5})^2}{1} = 8.0 \times 10^{-15}\).

Calculate the Gibbs Free Energy Change, \(\Delta G\)

Using the equation \(\Delta G = \Delta G^{\circ} + RT \ln Q\), we can calculate the Gibbs Free Energy Change for both situations. R (the gas constant) = 8.314 J/(mol K) T= 25 + 273.15 = 298.15 K (a) For the first situation: \(\Delta G = -27.92 \times 10^3 \, J/mol + (8.314 \, J/(\text{mol K})) (298.15 \, K) \ln(4) \approx -24080 \, J/mol = -24.08 \, kJ/mol\) (b) For the second situation: \(\Delta G = -27.92 \times 10^3 \, J/mol + (8.314 \, J/(\text{mol K})) (298.15 \, K) \ln(8.0 \times 10^{-15}) \approx -18700 \, J/mol = -18.7 \, kJ/mol\)

Determine the spontaneity of the dissolution

Now we can check if the dissolution of lead(II) chloride is spontaneous in both cases. (a) As \(\Delta G = -24.08 \, kJ/mol\), which is a negative value, the dissolution is spontaneous under the given conditions in the first situation. (b) As \(\Delta G = -18.7 \, kJ/mol\), which is a negative value, the dissolution is spontaneous under the given conditions in the second situation.

Key concepts

Gibbs Free Energy

Understanding Gibbs Free Energy (\text{G}) is crucial when determining if a chemical reaction is spontaneous. It combines enthalpy, entropy, and temperature to predict the direction of a chemical process. The equation for Gibbs Free Energy is defined as \( \text{G} = \text{H} - \text{T} \times \text{S} \), where \( \text{H} \) is enthalpy, \( \text{S} \) is entropy, and \( \text{T} \) is temperature in Kelvin.

When \( \text{G} \) is negative, the process is spontaneous as it indicates that the energy required to break bonds is less than the energy released when new bonds form. A positive \( \text{G} \) implies that extra energy is needed to drive the reaction, making it non-spontaneous. At equilibrium, \( \text{G} \) equals zero, indicating that both the forward and reverse reactions occur at the same rate. Thus, calculating the \( \text{G} \) of a reaction provides insightful information regarding the feasibility of the reaction under specified conditions.

To improve comprehension, consider the dissolution of lead(II) chloride: the negative \( \text{G} \) values obtained in the given exercise indicate that, at room temperature, the dissolution process is energetically favorable and would occur without external input of energy.

Reaction Quotient Q

The reaction quotient (\text{Q}) helps in understanding how far a system is from reaching equilibrium. It's calculated using the same formula as the equilibrium constant (\text{K}). Specifically, for a generic reaction \( a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \), \( \text{Q} \) is expressed as \( \text{Q} = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \), where the brackets indicate the molar concentrations of the reactants and products.

The value of \( \text{Q} \) is crucial for predicting the direction of the reaction. If \( \text{Q} < \text{K} \), the reaction moves forward towards equilibrium, indicating that the products are being formed. When \( \text{Q} > \text{K} \), the reaction shifts towards the reactants to reach equilibrium. If \( \text{Q} = \text{K} \), the system is already at equilibrium. The exercise provided illustrates the computation of \( \text{Q} \) for the dissolution of lead(II) chloride in two scenarios, reflecting its transient state before reaching equilibrium.

Solubility Equilibrium

Solubility equilibrium pertains to the balance point where a solute's rate of dissolution equals its rate of precipitation. It's a type of dynamic equilibrium specific to the dissolution process in a saturated solution.

The expression for the solubility product (\text{K}_{sp}) closely relates to the concept of the equilibrium constant, representing the maximum possible concentration of ions in a solution within a dissolution equilibrium. For instance, the dissolution of lead(II) chloride in the exercise has the equilibrium expression \( \text{K}_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \).

Changes in the concentration of solute or solvent, temperature, or addition of other substances can shift this equilibrium. In the context of the given exercise, the solubility equilibrium concept is key in understanding that even when the reaction is spontaneous, as indicated by the negative \( \text{G} \), there is still a limitation to how much lead(II) chloride can dissolve in water, as defined by its \( \text{K}_{sp} \) value.