Calculate Keq for a AG° value of 6.9 kcal/mol.K at 25 °C.

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
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**Transcription:**

Calculate \( K_{eq} \) for a \( \Delta G^\circ \) value of 6.9 kcal/mol at 25 °C.

**Explanation:**

This text presents a problem that involves calculating the equilibrium constant (\( K_{eq} \)) for a chemical reaction using the standard free energy change (\( \Delta G^\circ \)) value, which is given as 6.9 kcal/mol at a temperature of 25 °C. The calculation typically involves the relationship between \( \Delta G^\circ \) and \( K_{eq} \) given by the equation:

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

Where:
- \( \Delta G^\circ \) is the standard free energy change.
- \( R \) is the universal gas constant (1.987 cal/mol·K or 8.314 J/mol·K, depending on the units used).
- \( T \) is the temperature in Kelvin.
- \( K_{eq} \) is the equilibrium constant.

To solve this, convert 25 °C to Kelvin (298 K) and use the appropriate units for \( R \).
Transcribed Image Text:**Transcription:** Calculate \( K_{eq} \) for a \( \Delta G^\circ \) value of 6.9 kcal/mol at 25 °C. **Explanation:** This text presents a problem that involves calculating the equilibrium constant (\( K_{eq} \)) for a chemical reaction using the standard free energy change (\( \Delta G^\circ \)) value, which is given as 6.9 kcal/mol at a temperature of 25 °C. The calculation typically involves the relationship between \( \Delta G^\circ \) and \( K_{eq} \) given by the equation: \[ \Delta G^\circ = -RT \ln K_{eq} \] Where: - \( \Delta G^\circ \) is the standard free energy change. - \( R \) is the universal gas constant (1.987 cal/mol·K or 8.314 J/mol·K, depending on the units used). - \( T \) is the temperature in Kelvin. - \( K_{eq} \) is the equilibrium constant. To solve this, convert 25 °C to Kelvin (298 K) and use the appropriate units for \( R \).
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∆G°= - RT lnKeq

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