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))=0.6 atm `pO2=0.2 atm

Explanation

To solve the problem, we need 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 Data:
- Partial pressure of CO, \( p_{CO} = 0.4 \, \text{atm} \)
- Partial pressure of CO2, \( p_{CO_2} = 0.6 \, \text{atm} \)
- Partial pressure of O2, \( p_{O_2} = 0.2 \, \text{atm} \)
- Temperature, \( T = 3000 \, \text{K} \)

### Step 1: Calculate \( K_p \)

The equilibrium constant \( K_p \) is given by the expression:

\[ K_p = \frac{p_{CO_2}}{p_{CO} \cdot (p_{O_2})^{1/2}} \]

Substituting the values:

\[ K_p = \frac{0.6}{0.4 \cdot (0.2)^{1/2}} \]

Calculating \( (0.2)^{1/2} \):

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

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

\[ K_p = \frac{0.6}{0.4 \cdot 0.4472} = \frac{0.6}{0.17888} \approx 3.34 \]

### Step 2: Calculate \( K_c \)

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

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

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

For the given reaction:
- Moles of products = 1 (from CO2)
- Moles of reactants = 1 (from CO) + 0.5 (from O2) = 1.5

Thus,

\[ \Delta n = 1 - 1.5 = -0.5 \]

Now substituting into the equation:

\[ K_c = \frac{K_p}{R T^{\Delta n}} = \frac{3.34}{0.0821 \cdot 3000^{-0.5}} \]

Calculating \( 3000^{-0.5} \):

\[ 3000^{-0.5} = \frac{1}{\sqrt{3000}} \approx 0.0183 \]

Now substituting this value:

\[ K_c = \frac{3.34}{0.0821 \cdot 0.0183} = \frac{3.34}{0.0015} \approx 2226.67 \]

### Final Results:
- \( K_p \approx 3.34 \)
- \( K_c \approx 2226.67 \)

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