Calculate the equilibrium constant `K_(p)` and `K_(c )` for the reaction: `CO(g)+1//2O_(2)(g) hArr CO_(2)(g)`. Given that the partial pressure at equilibrium in a vessel at `3000 K` are `p_(CO)=0.4 atm, p_(CO_(2))=0.2 atm`

Explanation

To calculate the equilibrium constants \( K_p \) and \( K_c \) for the reaction:

\[ \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{CO}_2(g) \]

given the partial pressures at equilibrium:

- \( p_{\text{CO}} = 0.4 \, \text{atm} \)
- \( p_{\text{CO}_2} = 0.2 \, \text{atm} \)
- \( p_{\text{O}_2} = 0.2 \, \text{atm} \)

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

The equilibrium constant \( K_p \) for the reaction is given by the formula:

\[ K_p = \frac{p_{\text{CO}_2}}{p_{\text{CO}} \cdot (p_{\text{O}_2})^{1/2}} \]

### Step 2: Substitute the known values into the \( K_p \) expression

Substituting the given partial pressures into the equation:

\[ K_p = \frac{0.2}{0.4 \cdot (0.2)^{1/2}} \]

### Step 3: Calculate \( (0.2)^{1/2} \)

Calculating the square root of \( 0.2 \):

\[ (0.2)^{1/2} = 0.4472 \, \text{(approximately)} \]

### Step 4: Substitute back into the \( K_p \) expression

Now substitute this value back into the equation for \( K_p \):

\[ K_p = \frac{0.2}{0.4 \cdot 0.4472} \]

### Step 5: Calculate the denominator

Calculating the denominator:

\[ 0.4 \cdot 0.4472 = 0.17888 \, \text{(approximately)} \]

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

Now calculate \( K_p \):

\[ K_p = \frac{0.2}{0.17888} \approx 1.118 \]

### Step 7: Calculate \( K_c \)

To find \( K_c \), we can use the relationship between \( K_p \) and \( K_c \):

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

Where:
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 3000 \, \text{K} \)
- \( \Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 1 - (1 + \frac{1}{2}) = -\frac{1}{2} \)

### Step 8: Calculate \( K_c \)

Substituting the values into the equation:

\[ K_p = K_c \cdot (0.0821 \cdot 3000)^{-\frac{1}{2}} \]

Calculating \( (0.0821 \cdot 3000)^{-\frac{1}{2}} \):

\[ 0.0821 \cdot 3000 = 246.3 \]
\[ (246.3)^{-\frac{1}{2}} \approx 0.0635 \]

Now substituting back into the equation:

\[ 1.118 = K_c \cdot 0.0635 \]

Solving for \( K_c \):

\[ K_c = \frac{1.118}{0.0635} \approx 17.6 \]

### Final Results

Thus, the equilibrium constants are:

\[ K_p \approx 1.118 \quad \text{and} \quad K_c \approx 17.6 \]