Question

The equilirbium constant $K_c$ for the reaction ${SO}_2\left(g\right)+{NO}_2\left(g\right)\rightleftharpoons {SO}_3\left(g\right)+NO\left(g\right)$ is $16$. If $1$ mole of each of all the four gases is taken in $1{dm}^3$ vessel, the equilibrium concentration of $NO$ would be:

• A. $0.4M$
• B. $0.6M$
• C. $1.4M$
• D. $1.6M$

Answer 1

The correct option is D $1.6M$

$S{O}_2+N{O}_3\rightleftharpoons S{O}_3+NO$

Initially:

$1$ $1$ $1$ $1$

Equilibrium:

$1-x$ $1-x$ $1+x$ $1+x$

$K_c=\frac{(1+x{)}^2}{(1-x{)}^2}=16$

$\therefore \frac{1+x}{1-x}=4\Rightarrow x=\frac{3}{5}=0.6M$

$\therefore \left[NO\right]=1+x=1.6M$

Option D is correct.