`K_(p)` for the reaction given below is `1.36 "at" 499K`. Which of the
following equations can be used to calculate `K_(c)` for this reaction?
`N_(2)O_(5(g))rarrNO_(2(g))+NO_(3(g))`

Options

A. `K_(C)=([(0.0821)xx(499)])/([1.36])`
B. `K_(C)=([(1.36)xx(0.0821)])/([499])`
C. `K_(C)=([1.36])/([(0.0821)xx(499)])`
D. `K_(C)=([(1.36)xx(499)])/([0.0821])`

Answer

To find the equilibrium constant \( K_c \) for the reaction \( N_2O_5(g) \rightleftharpoons NO_2(g) + NO_3(g) \) given \( K_p = 1.36 \) at \( 499 \, K \), we can use the relationship between \( K_p \) and \( K_c \):

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is:
\[ N_2O_5(g) \rightleftharpoons NO_2(g) + NO_3(g) \]

2. **Write the Expression for \( K_p \)**:
The expression for \( K_p \) is given by:
\[ K_p = \frac{P_{NO_2} \cdot P_{NO_3}}{P_{N_2O_5}} \]

3. **Determine \( \Delta N_g \)**:
Calculate \( \Delta N_g \), which is the change in the number of moles of gas:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = (1 + 1) - 1 = 1 \]

4. **Use the Relationship Between \( K_p \) and \( K_c \)**:
The relationship is given by:
\[ K_p = K_c \cdot R \cdot T^{\Delta N_g} \]
Rearranging this to find \( K_c \):
\[ K_c = \frac{K_p}{R \cdot T^{\Delta N_g}} \]

5. **Substitute the Known Values**:
Here, \( K_p = 1.36 \), \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \), \( T = 499 \, K \), and \( \Delta N_g = 1 \):
\[ K_c = \frac{1.36}{0.0821 \cdot 499^1} \]

6. **Calculate \( K_c \)**:
First, calculate \( 0.0821 \cdot 499 \):
\[ 0.0821 \cdot 499 \approx 40.9759 \]
Now, calculate \( K_c \):
\[ K_c = \frac{1.36}{40.9759} \approx 0.0332 \]

### Final Answer:
Thus, the value of \( K_c \) for the reaction at \( 499 \, K \) is approximately \( 0.0332 \).