8 mol of AB3(g) are introduced into a 1.0 dm3 vessel. If it dissociates as 2AB3(g) leftharpoons A2(g) + 3B2(g). At equilibrium, 2 mol of A2 are found to be present. The equilibrium constant of this reaction is

Question

8 mol of AB3(g) are introduced into a 1.0 dm3 vessel. If it dissociates as 2AB3(g) leftharpoons A2(g) + 3B2(g). At equilibrium, 2 mol of A2 are found to be present. The equilibrium constant of this reaction is (A) 2 (B) 3 (C) 27 (D) 36

Tardigrade Admin · Accepted Answer

2AB3(g) leftharpoons A2(g) + 3B2(g). beginmatrixat t = 0&8&0&0 at eq.&(8-2×2)&2&3×2 &= 4&2&6 endmatrix now KC = ([A2][B2]3/[AB3]2) = (2/1×[6/1]3/[4/1]2) = 27

Q. 8 mol of $A B_{3}$ (g) are introduced into a 1.0 $d m^{3}$ vessel. If it dissociates as
$2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g) .$ At equilibrium, 2 mol of $A_{2}$ are found to be present. The equilibrium constant of this reaction is

2

3

27

36

$2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g) .$
$a t t = 0 a t e q . 8 (8 - 2 \times 2) = 4 022 0 3 \times 2 6$
now $K_{C} = \frac{[ A _{2} ] [ B _{2} ] ^{3}}{[ A B _{3} ] ^{2}} = \frac{2/1 \times [ 6/1 ] ^{3}}{[ 4/1 ] ^{2}} = 27$

Q. 8 mol of AB3​(g) are introduced into a 1.0 dm3 vessel. If it dissociates as 2AB3​(g)⇌A2​(g)+3B2​(g). At equilibrium, 2 mol of A2​ are found to be present. The equilibrium constant of this reaction is

2

3

27

36

2AB3​(g)⇌A2​(g)+3B2​(g). att=0ateq.​8(8−2×2)=4​022​03×26​ now KC​=[AB3​]2[A2​][B2​]3​=[4/1]22/1×[6/1]3​=27

Q. 8 mol of $A B_{3}$ (g) are introduced into a 1.0 $d m^{3}$ vessel. If it dissociates as
$2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g) .$ At equilibrium, 2 mol of $A_{2}$ are found to be present. The equilibrium constant of this reaction is

$2 A B_{3} (g) ⇌ A_{2} (g) + 3 B_{2} (g) .$
$a t t = 0 a t e q . 8 (8 - 2 \times 2) = 4 022 0 3 \times 2 6$
now $K_{C} = \frac{[ A _{2} ] [ B _{2} ] ^{3}}{[ A B _{3} ] ^{2}} = \frac{2/1 \times [ 6/1 ] ^{3}}{[ 4/1 ] ^{2}} = 27$