Calculate `Delta_(r)G` for the reaction at `27^(@)C`
`H_(2)(g)+2Ag^(+)(aq)hArr2Ag(s)+2H^(+)(aq)`
Given : `P_(H2)=0.5` bar, `[Ag^(+)]=10^(-5)M,`
`[H^(+)]=10^(-3)M,Delta_(r)G^(@)[Ag^(+)(aq)]=77.1kJ//mol`

Options

A. `-154.2kJ//mol`
B. `-178.9kJ//mol`
C. `-129.5kJ//mol`
D. None of these

Answer

To calculate the Gibbs free energy change (Δ_rG) for the given reaction at 27°C, we will follow these steps:

### Step 1: Write the Reaction and Identify Given Data
The reaction is:
\[ H_2(g) + 2Ag^+(aq) \rightleftharpoons 2Ag(s) + 2H^+(aq) \]

Given data:
- Partial pressure of \( H_2 \) = 0.5 bar
- Concentration of \( [Ag^+] \) = \( 10^{-5} \) M
- Concentration of \( [H^+] \) = \( 10^{-3} \) M
- Standard free energy of formation for \( Ag^+(aq) \) = \( \Delta_rG^\circ[Ag^+] = 77.1 \) kJ/mol

### Step 2: Calculate the Reaction Quotient (Q)
The reaction quotient \( Q \) is calculated using the formula:
\[ Q = \frac{[H^+]^2}{[Ag^+]^2 \cdot P_{H_2}} \]

Substituting the values:
\[ Q = \frac{(10^{-3})^2}{(10^{-5})^2 \cdot 0.5} \]

Calculating \( Q \):
\[ Q = \frac{10^{-6}}{10^{-10} \cdot 0.5} = \frac{10^{-6}}{5 \times 10^{-11}} = 2 \times 10^4 \]

### Step 3: Calculate Standard Gibbs Free Energy Change (Δ_rG°)
For the reaction, the standard Gibbs free energy change can be calculated as:
\[ \Delta_rG^\circ = \text{(products)} - \text{(reactants)} \]

Since the standard Gibbs free energy of pure solids is zero:
\[ \Delta_rG^\circ = 0 - (2 \times 77.1 \text{ kJ/mol}) = -154.2 \text{ kJ/mol} \]

### Step 4: Calculate Δ_rG Using the Formula
The Gibbs free energy change at non-standard conditions is given by:
\[ \Delta_rG = \Delta_rG^\circ + RT \ln Q \]

Where:
- \( R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K} \)
- \( T = 27°C = 300 \, \text{K} \)

Now substituting the values:
\[ \Delta_rG = -154.2 \text{ kJ/mol} + (0.008314 \text{ kJ/mol·K} \cdot 300 \text{ K}) \ln(2 \times 10^4) \]

Calculating \( RT \):
\[ RT = 0.008314 \times 300 = 2.4942 \text{ kJ/mol} \]

Now, calculate \( \ln(2 \times 10^4) \):
\[ \ln(2 \times 10^4) \approx 9.903 \]

Now substituting back:
\[ \Delta_rG = -154.2 + (2.4942 \cdot 9.903) \]
\[ \Delta_rG = -154.2 + 24.701 \]
\[ \Delta_rG \approx -129.5 \text{ kJ/mol} \]

### Final Answer
Thus, the Gibbs free energy change for the reaction is:
\[ \Delta_rG \approx -129.5 \text{ kJ/mol} \]

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