`K_(c)` for the reaction, `A(l)+2B(s)+3C(g)hArr 2D(g)+2E(g)" at "27^(@)C` is `10^(-3)M`.
`K_(P)` at this temperature is

Options

A. `24.63xx10^(-3)`
B. `12.63xx10^(-3)`
C. `1/263xx10^(-3)`
D. `2.46xx10^(-3)`

Answer

To find \( K_p \) for the reaction \( A(l) + 2B(s) + 3C(g) \rightleftharpoons 2D(g) + 2E(g) \) at \( 27^\circ C \), given that \( K_c = 10^{-3} \, M \), we will use the relationship between \( K_p \) and \( K_c \).

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
- The reaction involves:
- \( A \) (liquid)
- \( B \) (solid)
- \( C \) (gas)
- \( D \) (gas)
- \( E \) (gas)

2. **Write the Expression for \( K_c \)**:
- The equilibrium constant \( K_c \) is expressed in terms of the concentrations of the gaseous products and reactants:
\[ K_c = \frac{[D]^2[E]^2}{[C]^3} \]
Since \( A \) and \( B \) are in the liquid and solid states, they do not appear in the expression for \( K_c \).

3. **Determine \( \Delta n_g \)**:
- \( \Delta n_g \) is the change in the number of moles of gas:
- Moles of gaseous products = \( 2 + 2 = 4 \) (from \( 2D \) and \( 2E \))
- Moles of gaseous reactants = \( 3 \) (from \( 3C \))
- Therefore:
\[ \Delta n_g = \text{moles of products} - \text{moles of reactants} = 4 - 3 = 1 \]

4. **Use the Relationship Between \( K_p \) and \( K_c \)**:
- The relationship is given by:
\[ K_p = K_c \cdot R^T \cdot (RT)^{\Delta n_g} \]
where:
- \( R \) (universal gas constant) = \( 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1} \)
- \( T \) (temperature in Kelvin) = \( 27^\circ C + 273 = 300 \, K \)

5. **Calculate \( K_p \)**:
- Plugging in the values:
\[ K_p = K_c \cdot R^T \cdot (RT)^{\Delta n_g} \]
\[ K_p = (10^{-3}) \cdot (0.0821) \cdot (300)^{1} \]
\[ K_p = (10^{-3}) \cdot (0.0821 \cdot 300) \]
\[ K_p = (10^{-3}) \cdot (24.63) \]
\[ K_p = 24.63 \times 10^{-3} \, atm \]

### Final Answer:
\[ K_p = 24.63 \times 10^{-3} \, atm \]