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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![**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d4e176e-0465-4f48-87a4-78648eedaff5%2Fc5b90950-d5de-4820-8b78-d3291bd930ea%2Fq934zgn_processed.png&w=3840&q=75)
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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