The value of `K_(c)` for the following equilibrium is
`CaCO_(3(s))hArrCaO_((s))+CO_(2(g))`.
Given `K_(p)=167` bar at 1073 K.

Options

A. `1.896" mol L"^(-1)`
B. `4.38xx10^(-4)" mol L"^(-1)`
C. `6.3xx10^(-4)" mol L"^(-1)`
D. `6.626" mol L"^(-1)`

Explanation

To find the value of \( K_c \) for the equilibrium reaction:

\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

given that \( K_p = 167 \) bar at \( 1073 \, K \), we can use the relationship between \( K_p \) and \( K_c \):

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

### Step 1: Identify \( \Delta N \)

First, we need to calculate \( \Delta N \), which is defined as the difference in the number of moles of gaseous products and gaseous reactants.

- **Products:**
- \( \text{CaO} \) (solid) - does not count
- \( \text{CO}_2 \) (gas) - counts as 1 mole

- **Reactants:**
- \( \text{CaCO}_3 \) (solid) - does not count

Thus, the number of moles of gaseous products is 1 (from \( \text{CO}_2 \)), and the number of moles of gaseous reactants is 0.

\[ \Delta N = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 1 - 0 = 1 \]

### Step 2: Use the \( K_p \) and \( K_c \) Relationship

Now we can rearrange the equation to solve for \( K_c \):

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

### Step 3: Substitute Values

We know:
- \( K_p = 167 \) bar
- \( R = 0.0821 \, \text{L} \cdot \text{bar} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
- \( T = 1073 \, K \)
- \( \Delta N = 1 \)

Substituting these values into the equation:

\[ K_c = \frac{167}{0.0821 \times 1073^1} \]

### Step 4: Calculate \( K_c \)

Calculating the denominator:

\[ 0.0821 \times 1073 = 88.0733 \]

Now substitute back into the equation:

\[ K_c = \frac{167}{88.0733} \approx 1.896 \, \text{mol/L} \]

### Final Answer

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

\[ K_c \approx 1.896 \, \text{mol/L} \]

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