The combustion of butane in `O_(2)` at 1 bar and `298K` shows a decrease in free energy equal to `2.95xx10^(3)kJ mol ^(-1)` in a fuel cell. `K` and `E^(c-)` of the fuel cell are

Options

A. `9.55xx10^(482),1.096V`
B. `9.55,1.096V`
C. `1.023xx10^(966),2.85V`
D. `5.5xx10^(484),0.55V`

Explanation

To solve the question regarding the combustion of butane in a fuel cell, we need to find the equilibrium constant (K) and the cell potential (E°) for the reaction. The given information states that the change in free energy (ΔG) is -2.95 x 10^3 kJ/mol.

### Step 1: Relate ΔG to K
The relationship between the change in free energy (ΔG) and the equilibrium constant (K) is given by the equation:
\[ \Delta G = -RT \ln K \]
Where:
- \( R \) is the universal gas constant (8.314 J/(mol·K))
- \( T \) is the temperature in Kelvin (298 K)

### Step 2: Convert ΔG to Joules
Since R is in J/(mol·K), we need to convert ΔG from kJ to J:
\[ \Delta G = -2.95 \times 10^3 \text{ kJ/mol} = -2.95 \times 10^6 \text{ J/mol} \]

### Step 3: Solve for K
Now we can rearrange the equation to solve for K:
\[ K = e^{-\Delta G / (RT)} \]
Substituting the values:
- \( R = 8.314 \text{ J/(mol·K)} \)
- \( T = 298 \text{ K} \)
- \( \Delta G = -2.95 \times 10^6 \text{ J/mol} \)

Calculating:
\[ K = e^{-\left(-2.95 \times 10^6\right) / (8.314 \times 298)} \]
\[ K = e^{\left(2.95 \times 10^6\right) / (2477.572)} \]
\[ K = e^{1190.16} \]

### Step 4: Calculate K
Using a calculator:
\[ K \approx e^{1190.16} \]
This value is extremely large, indicating that the reaction strongly favors products at equilibrium.

### Step 5: Relate ΔG to E°
The relationship between ΔG and the standard cell potential (E°) is given by:
\[ \Delta G = -nFE° \]
Where:
- \( n \) is the number of moles of electrons transferred in the reaction
- \( F \) is the Faraday constant (approximately 96485 C/mol)

### Step 6: Solve for E°
Rearranging the equation gives:
\[ E° = -\frac{\Delta G}{nF} \]
Assuming that the combustion of butane involves 2 moles of electrons (this can vary based on the specific reaction):
\[ E° = -\frac{-2.95 \times 10^6 \text{ J/mol}}{2 \times 96485 \text{ C/mol}} \]

Calculating:
\[ E° = \frac{2.95 \times 10^6}{192970} \]
\[ E° \approx 15.29 \text{ V} \]

### Final Results
- The equilibrium constant \( K \) is extremely large, indicating a strong favorability for the products.
- The standard cell potential \( E° \) is approximately 15.29 V.