**Calculate \( K_{eq} \) for the following reaction at 25 °C.**\[ A + B \rightleftharpoons C + D \]\(\Delta G = -1.45 \, \text{kcal/mol}\)\[\Delta G^0 = -RT \ln K_{eq} \]\( R = 0.001987 \, \text{kcal/mol.K} \)---This image presents a chemical equilibrium problem where you are asked to calculate the equilibrium constant (\( K_{eq} \)) for a given reaction at 25 °C. The reaction involves the conversion of reactants A and B into products C and D. The Gibbs free energy change (\( \Delta G \)) for the reaction is provided as -1.45 kcal/mol. The formula to calculate \( K_{eq} \) involves the standard Gibbs free energy change (\( \Delta G^0 \)), the universal gas constant (\( R \)), and the absolute temperature in Kelvin. The relationship is shown by the equation:\[\Delta G^0 = -RT \ln K_{eq}\]Where:- \( R \) is the gas constant (given as 0.001987 kcal/mol.K).- \( T \) is the temperature in Kelvin.- \( K_{eq} \) is the equilibrium constant.The challenge is to apply this equation and solve for \( K_{eq} \).

Given that: dG = -1.45 kcal/mol T = 25°C = 25°C + 273 = 298 K

Answered: Calculate Keq for the following…

Calculate Keq for the following reaction at 25 °C. A + B C + D AG = -1.45 kcal/mol = AG°= - RT LnKea R= 0.001987 kcal/mol.K

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

**Calculate \( K_{eq} \) for the following reaction at 25 °C.**

\[ A + B \rightleftharpoons C + D \]

\(\Delta G = -1.45 \, \text{kcal/mol}\)

\[
\Delta G^0 = -RT \ln K_{eq}
\]

\( R = 0.001987 \, \text{kcal/mol.K} \)

---

This image presents a chemical equilibrium problem where you are asked to calculate the equilibrium constant (\( K_{eq} \)) for a given reaction at 25 °C. The reaction involves the conversion of reactants A and B into products C and D. The Gibbs free energy change (\( \Delta G \)) for the reaction is provided as -1.45 kcal/mol.

The formula to calculate \( K_{eq} \) involves the standard Gibbs free energy change (\( \Delta G^0 \)), the universal gas constant (\( R \)), and the absolute temperature in Kelvin. The relationship is shown by the equation:

\[
\Delta G^0 = -RT \ln K_{eq}
\]

Where:
- \( R \) is the gas constant (given as 0.001987 kcal/mol.K).
- \( T \) is the temperature in Kelvin.
- \( K_{eq} \) is the equilibrium constant.

The challenge is to apply this equation and solve for \( K_{eq} \).

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