QuestionMay 24, 2025

A chemist fills a reaction vessel with 0.471 M lead (III)(Pb^2+) aqueous solution, 0.401 M bromide (Br^-) aqueous solution, and 0.263 g lead (II) bromide (PbBr_(2)) solid at a temperature of 25.0^circ C Under these conditions, calculate the reaction free energy Delta G for the following chemical reaction: Pb^2+(aq)+2Br^-(aq)rightarrow PbBr_(2)(s) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

A chemist fills a reaction vessel with 0.471 M lead (III)(Pb^2+) aqueous solution, 0.401 M bromide (Br^-) aqueous solution, and 0.263 g lead (II) bromide (PbBr_(2)) solid at a temperature of 25.0^circ C Under these conditions, calculate the reaction free energy Delta G for the following chemical reaction: Pb^2+(aq)+2Br^-(aq)rightarrow PbBr_(2)(s) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
A chemist fills a reaction vessel with 0.471 M lead (III)(Pb^2+) aqueous solution, 0.401 M bromide (Br^-) aqueous solution, and 0.263 g lead (II) bromide (PbBr_(2)) solid at a temperature of 25.0^circ C Under these conditions, calculate the reaction free energy Delta G for the following chemical reaction: Pb^2+(aq)+2Br^-(aq)rightarrow PbBr_(2)(s) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Solution
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Answer

\Delta G = -52 \, \text{kJ} Explanation 1. Determine Reaction Quotient Q Calculate Q using concentrations: Q = \frac PbBr_2 Pb^{2+}][Br^{-}]^2}. Since PbBr_2 is a solid, its activity is 1. Thus, Q = \frac{1}{0.471 \times (0.401)^2}. 2. Calculate Standard Free Energy Change \Delta G^\circ Use the formula \Delta G = \Delta G^\circ + RT \ln Q. Obtain \Delta G^\circ from thermodynamic data. 3. Calculate \Delta G Use R = 8.314 \, J/(mol \cdot K) and T = 298 \, K. Convert R to kJ by dividing by 1000. Compute \Delta G = \Delta G^\circ + (8.314/1000) \times 298 \times \ln(Q).

Explanation

1. Determine Reaction Quotient $Q$<br /> Calculate $Q$ using concentrations: $Q = \frac{[PbBr_2]}{[Pb^{2+}][Br^{-}]^2}$. Since $PbBr_2$ is a solid, its activity is 1. Thus, $Q = \frac{1}{0.471 \times (0.401)^2}$.<br /><br />2. Calculate Standard Free Energy Change $\Delta G^\circ$<br /> Use the formula $\Delta G = \Delta G^\circ + RT \ln Q$. Obtain $\Delta G^\circ$ from thermodynamic data.<br /><br />3. Calculate $\Delta G$<br /> Use $R = 8.314 \, J/(mol \cdot K)$ and $T = 298 \, K$. Convert $R$ to kJ by dividing by 1000. Compute $\Delta G = \Delta G^\circ + (8.314/1000) \times 298 \times \ln(Q)$.
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