What is the value of `DeltaG(kJ//mol)` at 298 K at some non - equilibrium condition?
Given the concentrations of `[NH_(3)]` is 0.05 M and `[NH_(4)^(+)]=[OH^(-)]=0.002M` in the presence of excess water. Also
`DeltaG_("Reaction")^(@)=+"26.81 KJ mol."`
`NH_(3)(aq)+H_(2)O(l)hArr NH_(4)^(+)(aq)+OH^(-)(aq)`

Options

A. `+3.437`
B. `-9.433`
C. `+50.18`
D. `-50.18`

Answer

To find the value of ΔG at 298 K under the given non-equilibrium conditions, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta G^{\circ} + RT \ln Q \]

### Step 1: Identify the given values
- \(\Delta G^{\circ} = +26.81 \, \text{kJ/mol} = 26810 \, \text{J/mol}\) (convert to Joules)
- \(R = 8.314 \, \text{J/(mol K)}\) (universal gas constant)
- \(T = 298 \, \text{K}\)
- Concentrations:
- \([NH_3] = 0.05 \, \text{M}\)
- \([NH_4^+] = 0.002 \, \text{M}\)
- \([OH^-] = 0.002 \, \text{M}\)

### Step 2: Calculate the reaction quotient \(Q\)
The reaction is:

\[ NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq) \]

The expression for the reaction quotient \(Q\) is given by:

\[ Q = \frac{[NH_4^+][OH^-]}{[NH_3]} \]

Substituting the values:

\[ Q = \frac{(0.002)(0.002)}{0.05} = \frac{4 \times 10^{-6}}{0.05} = 8 \times 10^{-5} \]

### Step 3: Calculate \(\ln Q\)
Now, we need to calculate \(\ln Q\):

\[ \ln(8 \times 10^{-5}) \approx -9.125 (using a calculator or logarithm table) \]

### Step 4: Substitute values into the Gibbs free energy equation
Now we can substitute the values into the Gibbs free energy equation:

\[ \Delta G = 26810 + (8.314)(298)(-9.125) \]

Calculating the second term:

\[ 8.314 \times 298 \approx 2477.572 \, \text{J/mol} \]
\[ 2477.572 \times -9.125 \approx -22500.5 \, \text{J/mol} \]

Now substituting back:

\[ \Delta G = 26810 - 22500.5 = 4309.5 \, \text{J/mol} \]

### Step 5: Convert to kJ/mol
Convert the result back to kJ/mol:

\[ \Delta G \approx 4.31 \, \text{kJ/mol} \]

### Final Answer
Thus, the value of \(\Delta G\) at 298 K under the given non-equilibrium conditions is approximately:

\[ \Delta G \approx 4.31 \, \text{kJ/mol} \]

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