**Calculate the Gibbs Free Energy****Objective:** Calculate the Gibbs free energy (in kJ/mol) at 25 degrees C of the following reaction using the data provided in the table.**Reaction:**\[ 4\text{A} + 5\text{B} \rightarrow 5\text{C} + 1\text{D} + 5\text{E} \]**Data Table:**\[\begin{array}{|c|c|c|}\hline\text{Compound} & \Delta H_f^0 (\text{kJ/mol}) & S^0 (\text{J/mol/K}) \\\hline\text{A} & -35.72 & 183.16 \\\text{B} & -59.40 & 141.41 \\\text{C} & -1.94 & 32.74 \\\text{D} & -13.58 & 88.31 \\\text{E} & 51.57 & 82.41 \\\hline\end{array}\]**Solution:****Answer:**\[ 231,812.19 \times (905.59) \, \text{J/mol} \times (\text{kJ/mol}) \](Note: The computation seems to indicate errors, as the multiplication is crossed out, implying a mistake in calculation.)To calculate the Gibbs free energy (ΔG), use the equation:\[ \Delta G = \Delta H - T \Delta S \]Where:- \( \Delta H \) is the change in enthalpy.- \( T \) is the temperature in Kelvin.- \( \Delta S \) is the change in entropy.Ensure calculations adhere to proper unit conversions between J and kJ, and adjust coefficients for stoichiometry of the reaction.

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Calculate the Gibbs free energy (in kJ/mol) at 25 degrees C of the following reaction using the data in the table. 4A +5B → 5C + 1D + 5E AHfo(kJ/mol) A -35.72 B-59.40 C -1.94 D -13.58 E 51.57 Answer: so(J/mol/K) 183.16 141.41 32.74 88.31 82.41 231,812.19 x (905.59) J/molx (kJ/mol)

Calculate the Gibbs free energy (in kJ/mol) at 25 degrees C of the following reaction using the data in the table. 4A +5B → 5C + 1D + 5E AHfo(kJ/mol) A -35.72 B-59.40 C -1.94 D -13.58 E 51.57 Answer: so(J/mol/K) 183.16 141.41 32.74 88.31 82.41 231,812.19 x (905.59) J/molx (kJ/mol)

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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