`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 can use the relationship between \( K_C \) and \( K_P \):

\[ K_P = K_C \cdot (RT)^{\Delta n} \]

**Step 1: Determine \( \Delta n \)**

- \( \Delta n \) is the change in the number of moles of gas during the reaction.
- Count the moles of gaseous products and reactants:
- Products: \( 2D(g) + 2E(g) \) → total = \( 2 + 2 = 4 \) moles
- Reactants: \( 3C(g) \) → total = \( 3 \) moles
- Therefore,

\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 4 - 3 = 1 \]

**Step 2: Convert temperature to Kelvin**

- The temperature given is \( 27^\circ C \). To convert this to Kelvin:

\[ T(K) = 27 + 273 = 300 \, K \]

**Step 3: Use the ideal gas constant \( R \)**

- The value of \( R \) (ideal gas constant) is \( 0.08205 \, \text{L atm K}^{-1} \text{mol}^{-1} \).

**Step 4: Substitute values into the equation**

- Now, we can substitute \( K_C \), \( R \), \( T \), and \( \Delta n \) into the equation for \( K_P \):

\[ K_P = K_C \cdot (RT)^{\Delta n} \]

Substituting the values:

\[ K_P = 10^{-3} \cdot (0.08205 \cdot 300)^1 \]

**Step 5: Calculate \( RT \)**

- Calculate \( RT \):

\[ RT = 0.08205 \cdot 300 = 24.615 \]

**Step 6: Calculate \( K_P \)**

- Now substitute \( RT \) back into the equation for \( K_P \):

\[ K_P = 10^{-3} \cdot 24.615 \approx 24.615 \times 10^{-3} \]

Thus,

\[ K_P \approx 24.63 \times 10^{-3} \]

**Final Answer:**

\[ K_P \approx 24.63 \times 10^{-3} \, \text{atm} \]

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