Question

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right);K_C=4$
This reversible reaction is studied graphically as shown in figure. Select correct statements out of I, II and III
(I) Reaction quotient has maximum value at point A.
(II) Reaction proceeds from left to right at a point when,
$\left[N_2{O}_4\right]=\left[N{O}_2\right]=0.1M$.
(III) $K_c=Q$, when point D or F is reached

• A. I, II
• B. II, III
• C. I, III
• D. I, II, III

Answer 1

The correct option is B II, III

$N_2{O}_4\rightleftharpoons 2N{O}_2;K_C=4$
(I) At point - A,
$Q=\frac{\left[\text{Product}\right]}{\left[\text{Reactant}\right]}=0$
So, Q have minimum value at point A.
(II) At point, $\left[N_2{O}_4\right]=\left[N{O}_2\right]=0.1MQ=\frac{[N{O}_2{]}^2}{\left[N_2{O}_4\right]}=\frac{0.1\times 0.1}{0.1}=0.1Q<K_C$
So, reaction proceeds from left to right
(III) Point D and F represent equilibrium because concentration is not changing with time.
So $K_c=Q$ at point [D and F]