In the reaction C(s) + CO2(g) leftharpoons 2CO(g), the equilibrium pressure is 12 atm. If 50 % of CO2 reacts, then Kp will be:

Question

In the reaction C(s) + CO2(g) leftharpoons 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

Tardigrade Admin · Accepted Answer

C(s)+CO2(g) leftharpoons 2CO(g) P -P/2 P =(3P/2)=12 So, KP = (P2/(P/2)) =2P=2 × 8=16 atm

Q. In the reaction $C (s) + C O_{2} (g) ⇌ 2 CO (g),$ the equilibrium pressure is $12 a t m$ . If $50%$ of $C O_{2}$ reacts, then $K_{p}$ will be:

$12$ atm

$16$ atm

$20$ atm

$24$ atm

$C (s) + C O_{2} (g) ⇌ 2 CO (g)$
$P - P /2 P = \frac{3 P}{2} = 12$
So, $K_{P} = \frac{P ^{2}}{( P /2 )} = 2 P = 2 \times 8 = 16$ atm

Q. In the reaction C(s)+CO2​(g)⇌2CO(g), the equilibrium pressure is 12atm. If 50% of CO2​ reacts, then Kp​ will be:

12 atm

16 atm

20 atm

24 atm