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 of finding the ratio of equilibrium constants at 300 K for the given reactions, we will follow these steps:

### Step 1: Understand the relationship between Gibbs free energy and equilibrium constant
The relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) is given by the equation:
\[ \Delta G° = -2.303RT \log K \]
where:
- \( R \) is the universal gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin.

### Step 2: Write down the given values
From the problem statement, we have:
- For reaction A: \( \Delta G°_A = -200 \, \text{kJ/mol} = -200 \times 10^3 \, \text{J/mol} \)
- For reaction B: \( \Delta G°_B = -50 \, \text{kJ/mol} = -50 \times 10^3 \, \text{J/mol} \)

### Step 3: Set up the equation for the ratio of equilibrium constants
We can express the relationship for both reactions as follows:
\[ \Delta G°_A = -2.303RT \log K_A \]
\[ \Delta G°_B = -2.303RT \log K_B \]

### Step 4: Subtract the equations
To find the ratio of the equilibrium constants, we can subtract the second equation from the first:
\[ \Delta G°_A - \Delta G°_B = -2.303RT (\log K_A - \log K_B) \]
This simplifies to:
\[ \Delta G°_A - \Delta G°_B = -2.303RT \log \left(\frac{K_A}{K_B}\right) \]

### Step 5: Substitute the values into the equation
Substituting the values we have:
\[ (-200 \times 10^3) - (-50 \times 10^3) = -2.303 \times (8.314) \times (300) \log \left(\frac{K_A}{K_B}\right) \]
Calculating the left side:
\[ -200 \times 10^3 + 50 \times 10^3 = -150 \times 10^3 \, \text{J/mol} \]

### Step 6: Calculate the right side
Now, calculate:
\[ -150 \times 10^3 = -2.303 \times 8.314 \times 300 \log \left(\frac{K_A}{K_B}\right) \]
Calculating \( -2.303 \times 8.314 \times 300 \):
\[ -2.303 \times 8.314 \times 300 \approx -5730.57 \]
Thus, we have:
\[ -150 \times 10^3 = -5730.57 \log \left(\frac{K_A}{K_B}\right) \]

### Step 7: Solve for the logarithm
Now, rearranging gives:
\[ \log \left(\frac{K_A}{K_B}\right) = \frac{-150 \times 10^3}{-5730.57} \]
Calculating this value:
\[ \log \left(\frac{K_A}{K_B}\right) \approx 26.11 \]

### Step 8: Find the ratio of equilibrium constants
To find the ratio \( \frac{K_A}{K_B} \):
\[ \frac{K_A}{K_B} = 10^{26.11} \]

### Conclusion
The ratio of the equilibrium constants \( K_A \) and \( K_B \) at 300 K is approximately \( 10^{26.11} \).