Eight mole of a gas AB 3 attain equilibrium in a closed container of , volume 1 dm 3 as, 2 AB 3 leftharpoons A 2( g )+3 B 2( g ) If at equilibrium 2 mole of A2 are present then, equilibrium constant is:

Question

Eight mole of a gas AB 3 attain equilibrium in a closed container of , volume 1 dm 3 as, 2 AB 3 leftharpoons A 2( g )+3 B 2( g ) If at equilibrium 2 mole of A2 are present then, equilibrium constant is: (A) 72 mol 2 L -2 (B) 36 mol 2 L -2 (C) 3 mol 2 L -2 (D) 27 mol 2 L -2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/8da20633fd6bdb86dd07e36bc2f031b7-.png" /> Thus, Kc=([A2][B2]3/[A B3]2); Also, (a/2)=2 ∴ a=4 ∴ [A B3]=(4/1) ;[A2]=(2/1) ;[B2]=(6/1) Thus, Kc=(2 × 63/42) =27 mol 2 L-2

Q. Eight mole of a gas $A B_{3}$ attain equilibrium in a closed container of , volume $1 d m^{3}$ as, $2 A B_{3} ⇌ A_{2} (g) + 3 B_{2} (g)$ If at equilibrium $2 m o l e$ of $A_{2}$ are present then, equilibrium constant is:

$72 m o l^{2} L^{- 2}$

$36 m o l^{2} L^{- 2}$

$3 m o l^{2} L^{- 2}$

$27 m o l^{2} L^{- 2}$

Thus, $K_{c} = \frac{[ A _{2} ] [ B _{2} ] ^{3}}{[ A B _{3} ] ^{2}}$ ;

Also, $\frac{a}{2} = 2$

$∴ a = 4$

$∴ [A B_{3}] = \frac{4}{1}; [A_{2}] = \frac{2}{1}; [B_{2}] = \frac{6}{1}$

Thus, $K_{c} = \frac{2 \times 6 ^{3}}{4 ^{2}}$
$= 27 m o l^{2} L^{- 2}$

Q. Eight mole of a gas AB3​ attain equilibrium in a closed container of , volume 1dm3 as, 2AB3​⇌A2​(g)+3B2​(g) If at equilibrium 2mole of A2​ are present then, equilibrium constant is:

72mol2L−2

36mol2L−2

3mol2L−2

27mol2L−2