The ΔH° for a reaction is 80.0 kJ mol⁻¹ and ΔS° is 0.213 kJ mol⁻¹ K⁻¹. Calculate the ΔG° for this reaction at 298 K.**Explanation:**To calculate ΔG° (Gibbs free energy change) at a temperature of 298 K (25°C), we can use the Gibbs free energy equation:\[ \Delta G° = \Delta H° - T\Delta S° \]Where:- \( \Delta G° \) is the change in Gibbs free energy,- \( \Delta H° \) is the change in enthalpy (80.0 kJ/mol),- \( T \) is the temperature in Kelvin (298 K),- \( \Delta S° \) is the change in entropy (0.213 kJ/mol·K).Substituting the given values:\[ \Delta G° = 80.0 \, \text{kJ/mol} - (298 \, \text{K} \times 0.213 \, \text{kJ/mol·K}) \]Calculate to find the value of \( \Delta G° \).

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Answered: The AHⓇ for a reaction is 80.0 kJ mol-¹…

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The AHⓇ for a reaction is 80.0 kJ mol-¹ and AS is 0.213 kJ mol¹ K1¹. Calculate the AG° for this reaction at 298 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

The ΔH° for a reaction is 80.0 kJ mol⁻¹ and ΔS° is 0.213 kJ mol⁻¹ K⁻¹. Calculate the ΔG° for this reaction at 298 K.

**Explanation:**

To calculate ΔG° (Gibbs free energy change) at a temperature of 298 K (25°C), we can use the Gibbs free energy equation:

\[ \Delta G° = \Delta H° - T\Delta S° \]

Where:
- \( \Delta G° \) is the change in Gibbs free energy,
- \( \Delta H° \) is the change in enthalpy (80.0 kJ/mol),
- \( T \) is the temperature in Kelvin (298 K),
- \( \Delta S° \) is the change in entropy (0.213 kJ/mol·K).

Substituting the given values:

\[ \Delta G° = 80.0 \, \text{kJ/mol} - (298 \, \text{K} \times 0.213 \, \text{kJ/mol·K}) \]

Calculate to find the value of \( \Delta G° \).

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