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, we need to find the equilibrium constant \( K_c \) for the reaction:

\[ 2 \text{NOCl}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Cl}_2(g) \]

Given data:
- Initial partial pressure of \( \text{NO} = 0.373 \, \text{atm} \)
- Initial partial pressure of \( \text{Cl}_2 = 0.310 \, \text{atm} \)
- Total pressure at equilibrium \( P_{total} = 0.544 \, \text{atm} \)

### Step 1: Set up the initial conditions and changes at equilibrium

At the start of the reaction (initial conditions):
- \( P_{\text{NO}} = 0.373 \, \text{atm} \)
- \( P_{\text{Cl}_2} = 0.310 \, \text{atm} \)
- \( P_{\text{NOCl}} = 0 \, \text{atm} \)

Let \( 2p \) be the change in pressure for \( \text{NO} \) and \( p \) for \( \text{Cl}_2 \) at equilibrium.

At equilibrium:
- \( P_{\text{NO}} = 0.373 - 2p \)
- \( P_{\text{Cl}_2} = 0.310 - p \)
- \( P_{\text{NOCl}} = 2p \)

### Step 2: Write the total pressure equation

The total pressure at equilibrium can be expressed as:

\[ P_{total} = P_{\text{NO}} + P_{\text{Cl}_2} + P_{\text{NOCl}} \]

Substituting the equilibrium expressions:

\[ 0.544 = (0.373 - 2p) + (0.310 - p) + 2p \]

### Step 3: Simplify the equation

Combine the terms:

\[ 0.544 = 0.373 + 0.310 - 2p - p + 2p \]

This simplifies to:

\[ 0.544 = 0.683 - p \]

### Step 4: Solve for \( p \)

Rearranging gives:

\[ p = 0.683 - 0.544 = 0.139 \, \text{atm} \]

### Step 5: Calculate equilibrium pressures

Now substitute \( p \) back to find the equilibrium pressures:

- \( P_{\text{NO}} = 0.373 - 2(0.139) = 0.095 \, \text{atm} \)
- \( P_{\text{Cl}_2} = 0.310 - 0.139 = 0.171 \, \text{atm} \)
- \( P_{\text{NOCl}} = 2(0.139) = 0.278 \, \text{atm} \)

### Step 6: Calculate \( K_p \)

The expression for \( K_p \) is given by:

\[ K_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2 (P_{\text{Cl}_2})} \]

Substituting the equilibrium pressures:

\[ K_p = \frac{(0.278)^2}{(0.095)^2 \times (0.171)} \]

Calculating this gives:

\[ K_p = \frac{0.077084}{0.009025 \times 0.171} = \frac{0.077084}{0.001543} \approx 50.07 \]

### Step 7: Relate \( K_p \) to \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by:

\[ K_p = K_c (RT)^{\Delta n} \]

Where:
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 300 \, \text{K} \)
- \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - 3 = -1 \)

Substituting into the equation:

\[ K_p = K_c (0.0821 \times 300)^{-1} \]

Calculating \( RT \):

\[ RT = 0.0821 \times 300 = 24.63 \]

Thus,

\[ K_p = K_c \times \frac{1}{24.63} \]

Rearranging gives:

\[ K_c = K_p \times 24.63 \]

Substituting \( K_p \):

\[ K_c = 50.07 \times 24.63 \approx 8.1 \times 10^{-4} \, \text{mol/L} \]

### Final Answer

The value of \( K_c \) for the reaction at 300 K is approximately:

\[ K_c \approx 8.1 \times 10^{-4} \, \text{mol/L} \]