For the reaction, `A_((g))+2B_((g))hArr 3C_((g))+3_((g)),K_(p)=0.05" atm at "1000K`. The value of `K_(c)` is represented by

Options

A. `5xx10^(-4)R`
B. `(5xx10^(-4))/(R)`
C. `5xx10^(-5)R`
D. `(5xx10^(-5))/(R)`

Answer

To find the value of \( K_c \) for the reaction \( A_{(g)} + 2B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)} \) given that \( K_p = 0.05 \, \text{atm} \) at \( 1000 \, \text{K} \), we can use the relationship between \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Identify the Reaction and Given Values:**
The reaction is:
\[ A_{(g)} + 2B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)} \]
Given:
\[ K_p = 0.05 \, \text{atm} \quad \text{at} \quad T = 1000 \, \text{K} \]

2. **Determine the Change in Moles (\( \Delta n \)):**
\[ \Delta n = \text{(moles of products)} - \text{(moles of reactants)} \]
- Moles of products: \( 3C + 1D = 3 + 1 = 4 \)
- Moles of reactants: \( 1A + 2B = 1 + 2 = 3 \)
\[ \Delta n = 4 - 3 = 1 \]

3. **Use the Relationship Between \( K_p \) and \( K_c \):**
The relationship is given by:
\[ K_p = K_c (R T)^{\Delta n} \]
Where:
- \( R \) (gas constant) = \( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 1000 \, \text{K} \)
- \( \Delta n = 1 \)

4. **Substituting Values into the Equation:**
\[ 0.05 = K_c \times (0.0821 \times 1000)^1 \]
\[ 0.05 = K_c \times 82.1 \]

5. **Solving for \( K_c \):**
\[ K_c = \frac{0.05}{82.1} \]
\[ K_c = 0.000609 \]
\[ K_c = 6.09 \times 10^{-4} \, \text{mol/L} \]

### Final Answer:
The value of \( K_c \) is approximately \( 6.09 \times 10^{-4} \, \text{mol/L} \).