Which is correct relationship between `K_(p) and K_(c)` for following reversible reaction at 10 K temperature ? `K_(p) and K_(c)` in untis of atm and M respectively
`A_(2)(g)+3B_(2)(g)hArr 2AB_(3)(g)`

Options

A. `K_(p)gtK_(c)`
B. `K_(p) lt K_(c)`
C. `K_(p)=K_(c)`
D. Relation cannot be predicted

Answer

To determine the correct relationship between \( K_p \) and \( K_c \) for the reaction \( A_2(g) + 3B_2(g) \rightleftharpoons 2AB_3(g) \) at 10 K, we will use the formula that relates these two equilibrium constants.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The given reaction is:
\[ A_2(g) + 3B_2(g) \rightleftharpoons 2AB_3(g) \]

2. **Determine the Change in Moles of Gas (\( \Delta n \))**:
- Count the moles of gaseous reactants:
- For \( A_2 \): 1 mole
- For \( B_2 \): 3 moles
- Total moles of reactants = \( 1 + 3 = 4 \) moles
- Count the moles of gaseous products:
- For \( AB_3 \): 2 moles
- Therefore, the change in moles of gas (\( \Delta n \)) is:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 4 = -2 \]

3. **Use the Relationship Between \( K_p \) and \( K_c \)**:
The relationship is given by:
\[ K_p = K_c (RT)^{\Delta n} \]
where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin (10 K),
- \( \Delta n \) is the change in moles of gas calculated above.

4. **Substitute Values into the Equation**:
Plugging in the values:
\[ K_p = K_c (0.0821 \times 10)^{-2} \]
Simplifying this:
\[ K_p = K_c (0.821)^{-2} \]
\[ K_p = K_c \times \frac{1}{0.67} \quad (\text{since } 0.821^2 \approx 0.67) \]

5. **Analyze the Relationship**:
Since \( K_p = K_c \times \frac{1}{0.67} \), this indicates that:
\[ K_p > K_c \]

### Conclusion:
The correct relationship between \( K_p \) and \( K_c \) for the reaction at 10 K is:
\[ K_p > K_c \]