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)
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Chapter1: Chemical Foundations
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![**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ae95cc5-a0cd-43b0-8331-9678c9dacca0%2F013cc5b2-4b77-48da-86e9-40066cbcafcf%2Fx0dawq_processed.png&w=3840&q=75)
Transcribed Image Text:**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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