1.1 moles of A and 2.2 moles of B are mixed in a container of one litre volume to obtain the equilibrium A+2 B leftharpoons 2 C+D. At equilibrium 0.2 moles of C are formed. The equilibrium constant for the above reaction is

Question

1.1 moles of A and 2.2 moles of B are mixed in a container of one litre volume to obtain the equilibrium A+2 B leftharpoons 2 C+D. At equilibrium 0.2 moles of C are formed. The equilibrium constant for the above reaction is (A) 0.001 (B) 0.002 (C) 0.003 (D) 0.004

Tardigrade Admin · Accepted Answer

underseta a-x A + underset bb -2 x 2 B leftharpoons underset02 x 2 C + underset0 x D Given, 2 x =0.2, Also, a =1.1 x =0.1 b =2.2 ∴ K c =(4 x 3/( a - x )( b -2 x )2) =(4 ×(0.1)3/(1.1-0.1)(2.2-0.2)2) =0.001

Q. $1.1$ moles of $A$ and $2.2$ moles of $B$ are mixed in a container of one litre volume to obtain the equilibrium $A + 2 B ⇌ 2 C + D$ . At equilibrium $0.2$ moles of $C$ are formed. The equilibrium constant for the above reaction is

0.001

0.002

0.003

0.004

$aa - x A + bb - 2 x 2 B ⇌ 02 x 2 C + 0 x D$
Given, $2 x = 0.2$ ,
Also, $a = 1.1$
$x = 0.1 b = 2.2$
$∴ K_{c} = \frac{4 x ^{3}}{( a - x ) ( b - 2 x ) ^{2}}$
$= \frac{4 \times ( 0.1 ) ^{3}}{( 1.1 - 0.1 ) ( 2.2 - 0.2 ) ^{2}}$
$= 0.001$

Q. 1.1 moles of A and 2.2 moles of B are mixed in a container of one litre volume to obtain the equilibrium A+2B⇌2C+D. At equilibrium 0.2 moles of C are formed. The equilibrium constant for the above reaction is