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 $A+2 B \rightleftharpoons 2 C+D$. 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

Answer 1

Answer: (A)

$\underset{a a-x}{ A }+\underset{ bb -2 x }{2 B } \rightleftharpoons \underset{02 x }{2 C }+\underset{0 x }{ D }$
Given, $2 x =0.2$,
Also, $a =1.1$
$x =0.1 b =2.2$
$\therefore K _{ c }=\frac{4 x ^{3}}{( a - x )( b -2 x )^{2}}$
$=\frac{4 \times(0.1)^{3}}{(1.1-0.1)(2.2-0.2)^{2}}$
$=0.001$