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. (A) `20`
B. (B) `5.0`
C. (C) `0.02`
D. (D) `0.02`

Explanation

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

\[ A_2(g) + 2B_2(g) \rightleftharpoons 2C_2(g) \]

we will follow these steps:

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

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

\[ K_p = \frac{(P_{C_2})^2}{(P_{A_2})(P_{B_2})^2} \]

where \( P_{C_2} \), \( P_{A_2} \), and \( P_{B_2} \) are the partial pressures of the gases at equilibrium.

### Step 2: Identify the given values

From the problem, we know:

- \( P_{A_2} = 0.80 \) atm
- \( P_{B_2} = 0.40 \) atm
- Total pressure \( P_{total} = 2.80 \) atm

### Step 3: Calculate the partial pressure of \( C_2 \)

Using the total pressure, we can find the partial pressure of \( C_2 \):

\[ P_{total} = P_{A_2} + P_{B_2} + P_{C_2} \]

Substituting the known values:

\[ 2.80 = 0.80 + 0.40 + P_{C_2} \]

Now, solving for \( P_{C_2} \):

\[ P_{C_2} = 2.80 - (0.80 + 0.40) = 2.80 - 1.20 = 1.60 \text{ atm} \]

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

Now that we have all the partial pressures, we can substitute them into the \( K_p \) expression:

\[ K_p = \frac{(P_{C_2})^2}{(P_{A_2})(P_{B_2})^2} = \frac{(1.60)^2}{(0.80)(0.40)^2} \]

Calculating the numerator:

\[ (1.60)^2 = 2.56 \]

Calculating the denominator:

\[ (0.40)^2 = 0.16 \quad \text{and} \quad (0.80)(0.16) = 0.128 \]

### Step 5: Calculate \( K_p \)

Now we can calculate \( K_p \):

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

### Final Answer

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

\[ \boxed{20} \]

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