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 find the correct relationship between \( K_p \) and \( K_c \) for the given reversible reaction at 10 K temperature, we can follow these steps:

### Step 1: Write the Reaction
The given reaction is:
\[ A_2(g) + 3B_2(g) \rightleftharpoons 2AB_3(g) \]

### Step 2: Determine the Change in Number of Moles (\( \Delta n_g \))
To find \( \Delta n_g \), we need to calculate the difference between the total number of moles of products and the total number of moles of reactants.

- **Products**:
- \( 2AB_3 \) contributes 2 moles.

- **Reactants**:
- \( A_2 \) contributes 1 mole.
- \( 3B_2 \) contributes 3 moles.

Thus, the total number of moles of reactants is:
\[ 1 + 3 = 4 \]

Now, we can calculate \( \Delta n_g \):
\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} = 2 - 4 = -2 \]

### Step 3: Use the Relationship Between \( K_p \) and \( K_c \)
The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c (RT)^{\Delta n_g} \]
Where:
- \( R \) is the universal gas constant (0.0821 L·atm/(K·mol)).
- \( T \) is the temperature in Kelvin (10 K in this case).

### Step 4: Substitute \( \Delta n_g \) into the Equation
Substituting \( \Delta n_g = -2 \) into the equation:
\[ K_p = K_c (RT)^{-2} \]
This can be rearranged to:
\[ K_p = \frac{K_c}{(RT)^2} \]

### Step 5: Analyze the Relationship
From the equation \( K_p = \frac{K_c}{(RT)^2} \), we can see that:
- Since \( (RT)^2 \) is a positive value, \( K_p \) will be less than \( K_c \).
- Therefore, we conclude that:
\[ K_c > K_p \]

### Final Answer
The correct relationship between \( K_p \) and \( K_c \) for the given reaction is:
\[ K_c > K_p \]