Question

At ${60}^{o}C$ and 1 atm of $N_2{O}_4$ is $50\%$ dissociated $N{O}_2$ then $K_p$ is:

• A. $1.33$ atm
• B. $2$ atm
• C. $2.67$ atm
• D. $3$ atm

Answer 1

Answer: (A)

$1.33$ atm

$N_2{O}_4\rightleftharpoons 2{NO}_2$

$1$ $-$ $x=50$% $=0.5$

$1-x$ $2x$ ${\alpha }_n=1-x+2x=1+x$

$K_P=\frac{{\left(n_{{NO}_2}\right)}^2}{n_{N_2{O}_4}}\times \frac{P}{{\alpha }_n}=\frac{{\left(2\left(0.5\right)\right)}^2}{0.5}\times \frac{1}{1.5}=1.33atm$