Question

$1$mole of $NO$ and $1$ mole of $O_3$ are taken in $10L$ vessel and heated. At equilibrium, $50\%$ of $NO$ (by mass) reacts with $O_3$ according to the equation:

$NO\left(g\right)+O_{3\left(g\right)}\rightleftharpoons N{O}_{2\left(g\right)}+O_{2\left(g\right)}$.

What will be the equilibrium constant for this reaction?

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

The given reaction is :-

$NO\left(g\right)+O_3\left(g\right)\rightleftharpoons {NO}_2\left(g\right)+O_2\left(g\right)$

Initial cone : $1$ $1$ $0$ $0$

At eqm : $1-x$ $1-x$ $x$ $x$

Given initial conc. of $NO\&O_3$ each have reacted.

Now, $x=50$% of $NO=\frac{50}{100}\times 1=\frac{1}{2}$

So, $K_C=\frac{\left[{NO}_2\right]\left[O_2\right]}{\left[NO\right]\left[O_3\right]}$

$K_C=\frac{1/2\times 1/2}{1/2\times 1/2}=1$