In the reaction C(s) + CO2 ⇌ 2CO(g), the equilibrium pressure is 12 atm. If 50% of CO2 reacts then Kp will be: A. 12 atm B. 16 atm C. 20 atm D. 24 atm

To solve this problem, let's follow these steps:

1. **Write the expression for the equilibrium constant \( K_p \):**
For the reaction:
\[
C(s) + CO_2(g) \rightleftharpoons 2CO(g)
\]
The equilibrium constant \( K_p \) is given by:
\[
K_p = \frac{P_{CO}^2}{P_{CO_2}}
\]
where \( P_{CO} \) is the partial pressure of CO and \( P_{CO_2} \) is the partial pressure of \( CO_2 \).

2. **Determine the initial pressures:**
Let the initial pressure of \( CO_2 \) be \( P_{CO_2,0} \). The initial pressure of \( CO \) is zero.

3. **Determine the changes in pressures at equilibrium:**
Given that 50% of \( CO_2 \) reacts, if the initial pressure of \( CO_2 \) is \( P_{CO_2,0} \), then the change in \( CO_2 \) is:
\[
\Delta P_{CO_2} = -0.5 \times P_{CO_2,0}
\]
Therefore, the pressure of \( CO_2 \) at equilibrium is:
\[
P_{CO_2} = P_{CO_2,0} - 0.5 \times P_{CO_2,0} = 0.5 \times P_{CO_2,0}
\]

For \( CO \), since 2 moles of \( CO \) are produced for every mole of \( CO_2 \) reacted:
\[
\Delta P_{CO} = 2 \times 0.5 \times P_{CO_2,0} = P_{CO_2,0}
\]
Therefore, the equilibrium pressure of \( CO \) is:
\[
P_{CO} = P_{CO_2,0}
\]

4. **Use the equilibrium pressure to find \( K_p \):**
At equilibrium, the total pressure is given as 12 atm. This includes both \( CO \) and \( CO_2 \):
\[
P_{CO} + P_{CO_2} = 12 \text{ atm}
\]
Substituting the pressures:
\[
P_{CO_2,0} + 0.5 \times P_{CO_2,0} = 12 \text{ atm}
\]
\[
1.5 \times P_{CO_2,0} = 12 \text{ atm}
\]
\[
P_{CO_2,0} = \frac{12}{1.5} = 8 \text{ atm}
\]

Thus:
\[
P_{CO_2} = 0.5 \times 8 = 4 \text{ atm}
\]
\[
P_{CO} = 8 \text{ atm}
\]

Now, substitute these values into the expression for \( K_p \):
\[
K_p = \frac{P_{CO}^2}{P_{CO_2}} = \frac{8^2}{4} = \frac{64}{4} = 16
\]

Therefore, the value of \( K_p \) is \( 16 \text{ atm} \), which corresponds to option B.