Question

The air pollutant $NO$ is produced in automobile engines from the high temperature reaction $N_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2NO\left(g\right);K_c=1.7\times {10}^3$ at 2300 K. If the initial concentrations of $N_2$ and $O_2$ at 2300 K are both 1.40 M, the concentration of $NO$, $N_2$ and $O_2$ when the reaction mixture reaches equilibrium is:

• A. $\left[NO\right]=0.056M,\left[N_2\right]=\left[O_2\right]=1.73M$
• B. $\left[NO\right]=0.027M,\left[N_2\right]=\left[O_2\right]=1.37M$
• C. $\left[NO\right]=2.68M,\left[N_2\right]=\left[O_2\right]=0.06M$
• D. none of these

Answer 1

The correct option is C $\left[NO\right]=2.68M,\left[N_2\right]=\left[O_2\right]=0.06M$

	$N_2$	$O_2$	$2NO$
Initial concentration (M)	1.40	1.40	0
Equilibrium concentration (M)	$1.40-x$	$1.40-x$	$2x$

Let $x$ be the change in the concentration of nitrogen and oxygen.

The equilibrium constant expression is:

$K_c=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$

substitute values in the above expression:

$1.7\times {10}^3=\frac{(2x{)}^2}{(1.40-x)(1.40-x)}$

Taking square root on both sides-

$41.23=\frac{2x}{1.40-x}$

$1.40-x=0.0485x$

$x=1.34$

Hence, $\left[NO\right]=2x=2\times 1.34=2.68M$

$\left[N_2\right]=\left[O_2\right]=1.40-x=1.40-1.34=0.06M$