To find the value of ΔG (Gibbs free energy change) at 298 K under non-equilibrium conditions for the given reaction, we will follow these steps:
### Step 1: Write the Reaction and Identify Given Data
The reaction is:
\[ \text{NH}_3(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{NH}_4^+(aq) + \text{OH}^-(aq) \]
Given:
- \([NH_3] = 0.05 \, M\)
- \([NH_4^+] = [OH^-] = 0.002 \, M\)
- \(\Delta G^\circ = +26.81 \, kJ/mol\)
### Step 2: Calculate the Reaction Quotient (Q)
The reaction quotient \(Q\) is calculated using the formula:
\[ Q = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Substituting the values:
\[ Q = \frac{(0.002)(0.002)}{0.05} = \frac{0.000004}{0.05} = 8 \times 10^{-5} \]
### Step 3: Convert ΔG° to Joules
Since ΔG° is given in kJ/mol, we convert it to Joules:
\[ \Delta G^\circ = 26.81 \, kJ/mol = 26810 \, J/mol \]
### Step 4: Use the Gibbs Free Energy Equation
The equation to find ΔG is:
\[ \Delta G = \Delta G^\circ + RT \ln(Q) \]
Where:
- \(R = 8.314 \, J/(mol \cdot K)\)
- \(T = 298 \, K\)
### Step 5: Calculate RT
Calculating \(RT\):
\[ RT = 8.314 \, J/(mol \cdot K) \times 298 \, K = 2477.572 \, J/mol \]
### Step 6: Calculate ln(Q)
Now, calculate \(\ln(Q)\):
\[ \ln(8 \times 10^{-5}) \approx -9.125 \]
### Step 7: Substitute Values into the ΔG Equation
Now substitute these values into the ΔG equation:
\[ \Delta G = 26810 \, J/mol + 2477.572 \, J/mol \times (-9.125) \]
Calculating the second term:
\[ 2477.572 \times (-9.125) \approx -22500.5 \, J/mol \]
Thus:
\[ \Delta G = 26810 - 22500.5 \approx 4309.5 \, J/mol \]
### Step 8: Convert ΔG back to kJ
Convert ΔG back to kJ:
\[ \Delta G \approx 4.3095 \, kJ/mol \]
### Step 9: Round to Appropriate Significant Figures
Rounding to three significant figures:
\[ \Delta G \approx 4.31 \, kJ/mol \]
### Final Answer
The value of ΔG at 298 K under the given non-equilibrium conditions is approximately:
\[ \Delta G \approx 4.31 \, kJ/mol \]
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