For the reaction `A_(2)(g) + 2B_(2)hArr2C_(2)(g)` the partial pressure of `A_(2)` and `B_(2)` at equilibrium are `0.80` atm and `0.40` atm respectively.The pressure of the system is `2.80` atm. The equilibrium constant `K_(p)` will be

Options

A. 20
B. `5.0`
C. `0.02`
D. `0.2`

Explanation

To find the equilibrium constant \( K_p \) for the reaction

\[ A_{2}(g) + 2B_{2}(g) \rightleftharpoons 2C_{2}(g) \]

given the partial pressures of \( A_{2} \) and \( B_{2} \) at equilibrium, we can follow these steps:

### Step 1: Write down the given information.
- Partial pressure of \( A_{2} \) (\( P_{A} \)) = 0.80 atm
- Partial pressure of \( B_{2} \) (\( P_{B} \)) = 0.40 atm
- Total pressure of the system (\( P_{total} \)) = 2.80 atm

### Step 2: Calculate the partial pressure of \( C_{2} \).
The total pressure is the sum of the partial pressures of all gases involved in the reaction:

\[ P_{total} = P_{A} + P_{B} + P_{C} \]

Substituting the known values:

\[ 2.80 = 0.80 + 0.40 + P_{C} \]

### Step 3: Solve for \( P_{C} \).
Rearranging the equation gives:

\[ P_{C} = P_{total} - (P_{A} + P_{B}) = 2.80 - (0.80 + 0.40) \]

Calculating the right side:

\[ P_{C} = 2.80 - 1.20 = 1.60 \text{ atm} \]

### Step 4: Write the expression for \( K_p \).
The equilibrium constant \( K_p \) for the reaction is given by:

\[ K_p = \frac{(P_{C})^{2}}{(P_{A})^{1}(P_{B})^{2}} \]

### Step 5: Substitute the values into the \( K_p \) expression.
Substituting the partial pressures we found:

\[ K_p = \frac{(1.60)^{2}}{(0.80)^{1} \times (0.40)^{2}} \]

Calculating the numerator and denominator:

\[ K_p = \frac{2.56}{0.80 \times 0.16} \]

Calculating the denominator:

\[ 0.80 \times 0.16 = 0.128 \]

### Step 6: Final calculation of \( K_p \).
Now substituting back into the equation:

\[ K_p = \frac{2.56}{0.128} = 20 \]

Thus, the equilibrium constant \( K_p \) is:

\[ \boxed{20} \]

---