For the reaction,
`2NO(g)+Cl_(2)(g)hArr 2NOCl(g)`
a reaction mixture containing `NO(g)` and `Cl_(2)(g)` at partial pressure of 0.373 atm and 0.310 atm at 300 K respectively, is taken. The total pressure of the system at equilibrium was found to be 0.544 atm at 300 K. The value of `K_(C)` for the reaction,
`2NOCl(g)hArr 2NO(g)+Cl_(2)(g)`
at 300 K would be

Options

A. `4.62xx10^(2)M`
B. `4.62xx10^(-3)M`
C. `4.62xx10^(-4)M`
D. `8.10xx10^(-4)M`

Answer

To solve the problem step by step, we will follow the outlined process to calculate the equilibrium constant \( K_c \) for the reaction \( 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) \).

### Step 1: Write the initial conditions
We are given:
- Partial pressure of \( NO(g) = 0.373 \, \text{atm} \)
- Partial pressure of \( Cl_2(g) = 0.310 \, \text{atm} \)
- Total pressure at equilibrium = 0.544 atm

### Step 2: Set up the change in pressures
Let \( p \) be the change in pressure of \( Cl_2(g) \) that reacts at equilibrium. The stoichiometry of the reaction tells us:
- \( 2NO(g) \) will decrease by \( 2p \)
- \( Cl_2(g) \) will decrease by \( p \)
- \( 2NOCl(g) \) will increase by \( 2p \)

Thus, at equilibrium:
- Partial pressure of \( NO(g) = 0.373 - 2p \)
- Partial pressure of \( Cl_2(g) = 0.310 - p \)
- Partial pressure of \( NOCl(g) = 2p \)

### Step 3: Write the expression for total pressure at equilibrium
The total pressure at equilibrium can be expressed as:
\[ P_{total} = (0.373 - 2p) + (0.310 - p) + 2p = 0.544 \, \text{atm} \]

### Step 4: Simplify the equation
Combining the terms:
\[ 0.373 + 0.310 - 2p - p + 2p = 0.544 \]
This simplifies to:
\[ 0.683 - p = 0.544 \]

### Step 5: Solve for \( p \)
Rearranging gives:
\[ p = 0.683 - 0.544 = 0.139 \, \text{atm} \]

### Step 6: Calculate partial pressures at equilibrium
Now substitute \( p \) back to find the equilibrium partial pressures:
- \( P_{NO} = 0.373 - 2(0.139) = 0.095 \, \text{atm} \)
- \( P_{Cl_2} = 0.310 - 0.139 = 0.171 \, \text{atm} \)
- \( P_{NOCl} = 2(0.139) = 0.278 \, \text{atm} \)

### Step 7: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction \( 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) \) is given by:
\[ K_p = \frac{(P_{NO})^2 (P_{Cl_2})}{(P_{NOCl})^2} \]
Substituting the values:
\[ K_p = \frac{(0.095)^2 (0.171)}{(0.278)^2} \]

### Step 8: Calculate \( K_p \)
Calculating the values:
\[ K_p = \frac{0.009025 \times 0.171}{0.077284} \approx 0.105 \]

### Step 9: Convert \( K_p \) to \( K_c \)
Using the relation \( K_p = K_c(RT)^{\Delta n} \):
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 300 \, \text{K} \)
- \( \Delta n = (2 + 1) - 2 = 1 \)

Thus,
\[ K_p = K_c \times (0.0821 \times 300)^1 \]
\[ K_c = \frac{K_p}{(0.0821 \times 300)} = \frac{0.105}{24.63} \approx 0.00427 \]

### Step 10: Find \( K_c \) for the reverse reaction
Since we need \( K_c \) for the reaction \( 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) \), we take the reciprocal of the calculated \( K_c \):
\[ K_c = \frac{1}{0.00427} \approx 234.3 \]

### Final Answer
The value of \( K_c \) for the reaction \( 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) \) at 300 K is approximately **234.3**.