Find out the value of `K_(c )` for each of the following equilibrium from the value of `K_(p)`:
a. `2NOCl(g) hArr 2NO(g)+Cl_(2)(g), K_(p)=1.8xx10^(-2)` at `500 K`
b. `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g), K_(p)=167` at `1073 K`

Explanation

To find the value of \( K_c \) from the given \( K_p \) values for the equilibria, we will use the relationship between \( K_p \) and \( K_c \):

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

Where:
- \( R \) is the ideal gas constant (0.0831 L·bar/(K·mol) or 0.0821 L·atm/(K·mol))
- \( T \) is the temperature in Kelvin
- \( \Delta N_g \) is the change in the number of moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

### Step 1: Calculate \( K_c \) for the first equilibrium

**Given:**
- Reaction: \( 2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g) \)
- \( K_p = 1.8 \times 10^{-2} \) at \( T = 500 \, K \)

**Calculate \( \Delta N_g \):**
- Moles of gaseous products = \( 2 \, (NO) + 1 \, (Cl_2) = 3 \)
- Moles of gaseous reactants = \( 2 \, (NOCl) = 2 \)
- \( \Delta N_g = 3 - 2 = 1 \)

**Using the relationship:**
\[ K_p = K_c (RT)^{\Delta N_g} \]
\[ K_c = \frac{K_p}{(RT)^{\Delta N_g}} \]

**Substituting values:**
- \( R = 0.0831 \, L \cdot bar/(K \cdot mol) \)
- \( T = 500 \, K \)

\[ K_c = \frac{1.8 \times 10^{-2}}{(0.0831 \times 500)^1} \]
\[ K_c = \frac{1.8 \times 10^{-2}}{41.55} \]
\[ K_c \approx 4.33 \times 10^{-4} \]

### Step 2: Calculate \( K_c \) for the second equilibrium

**Given:**
- Reaction: \( CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \)
- \( K_p = 167 \) at \( T = 1073 \, K \)

**Calculate \( \Delta N_g \):**
- Moles of gaseous products = \( 1 \, (CO_2) = 1 \)
- Moles of gaseous reactants = \( 0 \) (since \( CaCO_3 \) is a solid)
- \( \Delta N_g = 1 - 0 = 1 \)

**Using the relationship:**
\[ K_c = \frac{K_p}{(RT)^{\Delta N_g}} \]

**Substituting values:**
- \( R = 0.0831 \, L \cdot bar/(K \cdot mol) \)
- \( T = 1073 \, K \)

\[ K_c = \frac{167}{(0.0831 \times 1073)^1} \]
\[ K_c = \frac{167}{89.05} \]
\[ K_c \approx 1.87 \]

### Final Answers:
- For the first equilibrium: \( K_c \approx 4.33 \times 10^{-4} \)
- For the second equilibrium: \( K_c \approx 1.87 \)