At 1000 K, the value of `K_(p)` for the reaction: `A(g) + 2B(g)hArr3C(g) + D(g)` is `0.05` atmosphere. The value of `K_(c)` in terms of R would be:

Options

A. 20000 R
B. `0.02 R`
C. `5xx10^(-5) R`
D. `5xx10^(-5)xxR^(-1)`

Explanation

To find the value of \( K_c \) in terms of \( R \) for the given reaction \( A(g) + 2B(g) \rightleftharpoons 3C(g) + D(g) \), we can use the relationship between \( K_p \) and \( K_c \):

1. **Identify the relationship**: The relationship between \( K_p \) and \( K_c \) is given by the equation:
\[ K_p = K_c \cdot R^T \cdot (RT)^{\Delta n} \]
where:
- \( R \) is the ideal gas constant (0.0821 L·atm/(K·mol)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n \) is the change in the number of moles of gas.

2. **Calculate \( \Delta n \)**:
- On the product side, there are 4 moles (3 from \( C \) and 1 from \( D \)).
- On the reactant side, there are 3 moles (1 from \( A \) and 2 from \( B \)).
- Therefore, \( \Delta n = \text{moles of products} - \text{moles of reactants} = 4 - 3 = 1 \).

3. **Substitute the values into the equation**:
\[ K_p = K_c \cdot R^T \cdot (R^T)^{\Delta n} \]
Since \( \Delta n = 1 \), we can simplify the equation to:
\[ K_p = K_c \cdot R^T \]

4. **Rearranging the equation to solve for \( K_c \)**:
\[ K_c = \frac{K_p}{R^T} \]

5. **Substituting the known values**:
- Given \( K_p = 0.05 \) atm,
- Given \( T = 1000 \) K,
- Therefore, we can write:
\[ K_c = \frac{0.05}{R^{1000}} \]

6. **Express \( K_c \) in terms of \( R \)**:
- Since \( 0.05 \) can be expressed as \( 5 \times 10^{-2} \):
\[ K_c = \frac{5 \times 10^{-2}}{R^{1000}} \]

7. **Final expression**:
- Thus, the value of \( K_c \) in terms of \( R \) is:
\[ K_c = 5 \times 10^{-2} R^{-1000} \]