For the reaction A2(g) + 2B2 leftharpoons 2C2(g) the partial pressure of A2 and B2 at equilibrium are 0.80 atm and 0.40 atm respectively. The pressure of the system is 2.80 atm. The equilibrium constant Kp will be

Question

For the reaction A2(g) + 2B2 leftharpoons 2C2(g) the partial pressure of A2 and B2 at equilibrium are 0.80 atm and 0.40 atm respectively. The pressure of the system is 2.80 atm. The equilibrium constant Kp will be (A) 20 (B) 5.0 (C) 0.02 (D) 0.2

Tardigrade Admin · Accepted Answer

A2(g )+2B2(g)leftharpoons 2C2(g) PA2 = 0.80 atm PB2 = 0.4 atm Total pressure o f the system = 2.8 atm ∴ Pc2 = 2.8 -0.8 -0.4 =1.6 Kp = (P2C2/PA2 × P3B2) = ((1.6)2/0.8 × (0.4)2)=20

Q. For the reaction $A_{2} (g) + 2 B_{2} ⇌ 2 C_{2} (g)$
the partial pressure of $A_{2}$ and $B_{2}$ at equilibrium are $0.80 a t m$ and $0.40 a t m$ respectively. The pressure of the system is $2.80 a t m$ . The equilibrium constant $K_{p}$ will be

$20$

$5.0$

$0.02$

$0.2$

$A_{2} (g) + 2 B_{2} (g) ⇌ 2 C_{2} (g)$
$P_{A_{2}} = 0.80$ atm
$P_{B_{2}} = 0.4$ atm
Total pressure o f the system $= 2.8$ atm
$∴ P_{c_{2}} = 2.8 - 0.8 - 0.4 = 1.6$
$K_{p} = \frac{P _{C_{2}}^{2}}{P _{A_{2}} \times P _{B_{2}}^{3}} = \frac{( 1.6 ) ^{2}}{0.8 \times ( 0.4 ) ^{2}} = 20$

Q. For the reaction A2​(g)+2B2​⇌2C2​(g) the partial pressure of A2​ and B2​ at equilibrium are 0.80atm and 0.40atm respectively. The pressure of the system is 2.80atm. The equilibrium constant Kp​ will be

20

5.0

0.02

0.2