In the reaction, AB(g) leftharpoons A(g) +B(g) at 30° C,Kp for the dissociation equilibrium is 2.56 × 10-2 atm. if the total pressure at equilibrium is 1 atm, then the percentage dissociation of AB is

Question

In the reaction, AB(g) leftharpoons A(g) +B(g) at 30° C,Kp for the dissociation equilibrium is 2.56 × 10-2 atm. if the total pressure at equilibrium is 1 atm, then the percentage dissociation of AB is (A) 87% (B) 13% (C) 43.5% (D) 6%

Tardigrade Admin · Accepted Answer

AB(g) arrow A(g)+B(g) Kp = (α2 p/1-α2)=2.56 × 10-2 atm = (α2 × 1 atm/1-α2)=2.56 × 10-2 atm α = 0.16 = 16 %

Q. In the reaction, $A B (g) ⇌ A (g) + B (g)$ at $3 0^{\circ} C, K_{p}$ for the dissociation equilibrium is $2.56 \times 1 0^{- 2} a t m .$ if the total pressure at equilibrium is 1 atm, then the percentage dissociation of AB is

87%

13%

43.5%

6%

16%

$A B (g) \to A (g) + B (g)$
$K_{p} = \frac{α ^{2} p}{1 - α ^{2}} = 2.56 \times 1 0^{- 2} a t m$
$= \frac{α ^{2} \times 1 a t m}{1 - α ^{2}} = 2.56 \times 1 0^{- 2} a t m$
$α = 0.16$
$= 16%$

Q. In the reaction, AB(g)⇌A(g)+B(g) at 30∘C,Kp​ for the dissociation equilibrium is 2.56×10−2atm. if the total pressure at equilibrium is 1 atm, then the percentage dissociation of AB is

87%

13%

43.5%

6%

16%

AB(g)→A(g)+B(g) Kp​=1−α2α2p​=2.56×10−2atm =1−α2α2×1atm​=2.56×10−2atm α=0.16 =16%

Q. In the reaction, $A B (g) ⇌ A (g) + B (g)$ at $3 0^{\circ} C, K_{p}$ for the dissociation equilibrium is $2.56 \times 1 0^{- 2} a t m .$ if the total pressure at equilibrium is 1 atm, then the percentage dissociation of AB is

$A B (g) \to A (g) + B (g)$
$K_{p} = \frac{α ^{2} p}{1 - α ^{2}} = 2.56 \times 1 0^{- 2} a t m$
$= \frac{α ^{2} \times 1 a t m}{1 - α ^{2}} = 2.56 \times 1 0^{- 2} a t m$
$α = 0.16$
$= 16%$