Determine AG° for a reaction when AG = -168.4 kJ/mol and Q = 0.035 at 223 K. (R = 8.314 J/mol · K)

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Problem Statement:**

Determine ΔG° for a reaction when ΔG = -168.4 kJ/mol and Q = 0.035 at 223 K.  
(R = 8.314 J/mol · K)

**Solution Explanation:**

To find ΔG°, we use the formula:

\[
ΔG = ΔG° + RT \ln Q
\]

Where:
- ΔG is the Gibbs free energy change for the reaction.
- ΔG° is the standard Gibbs free energy change.
- R is the universal gas constant (8.314 J/mol·K).
- T is the temperature in Kelvin.
- Q is the reaction quotient.

**Steps to Solve:**

1. Convert units if necessary:
   - Since ΔG is given in kJ/mol, convert it to J/mol:
     \(-168.4 \, \text{kJ/mol} = -168400 \, \text{J/mol}\).

2. Plug in the known values and solve for ΔG°:

\[
-168400 = ΔG° + (8.314)(223) \ln 0.035
\]

3. Calculate \(\ln 0.035\) and the entire RT ln Q term.

4. Rearrange the equation to solve for ΔG°.

This process will yield the standard Gibbs free energy change, ΔG°, for this particular reaction at the specified conditions.
Transcribed Image Text:**Problem Statement:** Determine ΔG° for a reaction when ΔG = -168.4 kJ/mol and Q = 0.035 at 223 K. (R = 8.314 J/mol · K) **Solution Explanation:** To find ΔG°, we use the formula: \[ ΔG = ΔG° + RT \ln Q \] Where: - ΔG is the Gibbs free energy change for the reaction. - ΔG° is the standard Gibbs free energy change. - R is the universal gas constant (8.314 J/mol·K). - T is the temperature in Kelvin. - Q is the reaction quotient. **Steps to Solve:** 1. Convert units if necessary: - Since ΔG is given in kJ/mol, convert it to J/mol: \(-168.4 \, \text{kJ/mol} = -168400 \, \text{J/mol}\). 2. Plug in the known values and solve for ΔG°: \[ -168400 = ΔG° + (8.314)(223) \ln 0.035 \] 3. Calculate \(\ln 0.035\) and the entire RT ln Q term. 4. Rearrange the equation to solve for ΔG°. This process will yield the standard Gibbs free energy change, ΔG°, for this particular reaction at the specified conditions.
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