The value of `K_p` for the reaction at 500 K:
`2NOCl(g) hArr 2NO(g)+Cl_2(g)` is `1.8xx10^(-2) "bar"^(-1)`.
Calculate `K_c` for reaction at this temperature.

Explanation

To calculate \( K_c \) for the reaction \( 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) \) at 500 K, given that \( K_p = 1.8 \times 10^{-2} \, \text{bar}^{-1} \), we can follow these steps:

### Step 1: Write the relationship between \( K_p \) and \( K_c \)

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

\[ K_p = K_c \cdot R^T \cdot \Delta N_g \]

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( \Delta N_g \) is the change in the number of moles of gas.

### Step 2: Calculate \( \Delta N_g \)

To find \( \Delta N_g \), we need to determine the number of moles of gaseous products and reactants:

- On the product side: \( 2 \, \text{NO} + 1 \, \text{Cl}_2 \) gives a total of \( 2 + 1 = 3 \) moles.
- On the reactant side: \( 2 \, \text{NOCl} \) gives a total of \( 2 \) moles.

Now, we can calculate \( \Delta N_g \):

\[ \Delta N_g = \text{(moles of products)} - \text{(moles of reactants)} = 3 - 2 = 1 \]

### Step 3: Substitute values into the equation

Now we can substitute the known values into the equation:

- \( K_p = 1.8 \times 10^{-2} \, \text{bar}^{-1} \)
- \( R = 0.083 \, \text{bar L K}^{-1} \text{mol}^{-1} \)
- \( T = 500 \, \text{K} \)
- \( \Delta N_g = 1 \)

The equation becomes:

\[ 1.8 \times 10^{-2} = K_c \cdot (0.083)^{500} \cdot 1 \]

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

Rearranging the equation to solve for \( K_c \):

\[ K_c = \frac{K_p}{R^T \cdot \Delta N_g} \]

Substituting in the values:

\[ K_c = \frac{1.8 \times 10^{-2}}{0.083 \times 500} \]

Calculating the denominator:

\[ 0.083 \times 500 = 41.5 \]

Now substituting this back into the equation for \( K_c \):

\[ K_c = \frac{1.8 \times 10^{-2}}{41.5} \]

Calculating \( K_c \):

\[ K_c \approx 4.337 \times 10^{-4} \]

### Final Answer

Thus, the value of \( K_c \) at 500 K is approximately:

\[ K_c \approx 4.34 \times 10^{-4} \, \text{mol L}^{-1} \]