At 300 K,
`A hArr" Product"DeltaG_(T)^(@)=-"200 kJ mol"^(-1)`
`BhArr" Product "DeltaG_(T)^(@)=-"50 kJ mol"^(-1)`
Thus, the ratio of equilibrium constant at 300 K

Options

A. 100
B. 1000
C. 10000
D. none of these

Answer

To solve the problem, we need to find the ratio of the equilibrium constants \( K_1 \) and \( K_2 \) for the two reactions given their Gibbs free energy changes (\( \Delta G \)) at 300 K.

### Step-by-Step Solution:

1. **Identify the Gibbs Free Energy Changes**:
- For reaction A: \( \Delta G_1 = -200 \, \text{kJ/mol} \)
- For reaction B: \( \Delta G_2 = -50 \, \text{kJ/mol} \)

2. **Convert Gibbs Free Energy to Joules**:
- Since \( R \) (the gas constant) is typically used in J/(mol·K), we convert \( \Delta G \) from kJ to J:
- \( \Delta G_1 = -200 \, \text{kJ/mol} = -200 \times 10^3 \, \text{J/mol} = -200000 \, \text{J/mol} \)
- \( \Delta G_2 = -50 \, \text{kJ/mol} = -50 \times 10^3 \, \text{J/mol} = -50000 \, \text{J/mol} \)

3. **Use the Gibbs Free Energy Equation**:
- The relationship between Gibbs free energy and the equilibrium constant is given by:
\[ \Delta G = -RT \ln K \]
- For reaction A:
\[ \Delta G_1 = -RT \ln K_1 \implies -200000 = -RT \ln K_1 \]
- For reaction B:
\[ \Delta G_2 = -RT \ln K_2 \implies -50000 = -RT \ln K_2 \]

4. **Set Up the Equation for the Ratio of Equilibrium Constants**:
- Rearranging the equations gives:
\[ \ln K_1 = \frac{200000}{RT} \]
\[ \ln K_2 = \frac{50000}{RT} \]
- To find the ratio \( \frac{K_1}{K_2} \), we can subtract the two equations:
\[ \ln K_1 - \ln K_2 = \frac{200000}{RT} - \frac{50000}{RT} \]
\[ \ln \left(\frac{K_1}{K_2}\right) = \frac{200000 - 50000}{RT} = \frac{150000}{RT} \]

5. **Substituting Values**:
- The gas constant \( R = 8.314 \, \text{J/(mol·K)} \) and \( T = 300 \, \text{K} \):
\[ \ln \left(\frac{K_1}{K_2}\right) = \frac{150000}{8.314 \times 300} \]
- Calculate \( RT \):
\[ RT = 8.314 \times 300 = 2494.2 \, \text{J/mol} \]
- Now substitute:
\[ \ln \left(\frac{K_1}{K_2}\right) = \frac{150000}{2494.2} \approx 60.1 \]

6. **Exponentiate to Find the Ratio**:
- To find the ratio \( \frac{K_1}{K_2} \):
\[ \frac{K_1}{K_2} = e^{60.1} \]

### Final Answer:
The ratio of the equilibrium constants at 300 K is:
\[ \frac{K_1}{K_2} \approx e^{60.1} \]