Nitrogen chloride (NOCl) decomposes into NO and `Cl_2` as :
`2NOCl (g) hArr 2NO(g)+Cl_2(g)`
The equilibrium partial pressures of NOCl, NO and `Cl_2` at 500 K have been found to be 0.75, 0.11 and 0.84 atm respectively. Calculate `K_p` for the reaction.

Explanation

To calculate the equilibrium constant \( K_p \) for the reaction:

\[ 2NOCl (g) \rightleftharpoons 2NO(g) + Cl_2(g) \]

we will use the equilibrium partial pressures provided:

- Partial pressure of \( NOCl \) = 0.75 atm
- Partial pressure of \( NO \) = 0.11 atm
- Partial pressure of \( Cl_2 \) = 0.84 atm

### Step 1: Write the expression for \( K_p \)

The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the products and reactants:

\[ K_p = \frac{(P_{NO})^2 \cdot (P_{Cl_2})}{(P_{NOCl})^2} \]

where:
- \( P_{NO} \) is the partial pressure of \( NO \)
- \( P_{Cl_2} \) is the partial pressure of \( Cl_2 \)
- \( P_{NOCl} \) is the partial pressure of \( NOCl \)

### Step 2: Substitute the equilibrium partial pressures into the expression

Now, substituting the given values into the expression for \( K_p \):

\[ K_p = \frac{(0.11)^2 \cdot (0.84)}{(0.75)^2} \]

### Step 3: Calculate the values

First, calculate \( (0.11)^2 \):

\[ (0.11)^2 = 0.0121 \]

Next, calculate \( (0.75)^2 \):

\[ (0.75)^2 = 0.5625 \]

Now substitute these values back into the equation for \( K_p \):

\[ K_p = \frac{0.0121 \cdot 0.84}{0.5625} \]

Calculating the numerator:

\[ 0.0121 \cdot 0.84 = 0.010164 \]

Now, divide by the denominator:

\[ K_p = \frac{0.010164}{0.5625} \approx 0.01806 \]

### Final Answer

Thus, the value of \( K_p \) for the reaction at 500 K is approximately:

\[ K_p \approx 0.01806 \]

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