Question

In the equilibrium, $N_2{O}_{4(g)}\rightleftharpoons 2N{O}_{2(g)}$ the $N_2{O}_4$ is fifty per cent dissociated at $60^{\circ }C$. What will be the value of $K_p$ at this temperature and one atmosphere?

• A. 0.33 atm
• B. 1.33 atm
• C. 2.33 atm
• D. 3.33 atm

Answer 1

Answer: (B)

$N_2{O}_4(g)2N{O}_2(g)$
At. equil. $1mol2mol$
Total moles $=1+2=3$
$\therefore n_{N_2{O}_4}=\frac{1}{3}$
and $n_{N{O}_2}=\frac{2}{3}$
$\because p_{N_2{O}_4}=n_{N_2{O}_4}\times p$
$\therefore p_{N_2{O}_4}=\frac{1}{3}\times 1=\frac{1}{3}$
and $p_{N{O}_2}=\frac{2}{3}\times 1=\frac{2}{3}$
$K_p=\frac{{\left(p_{N{O}_2}\right)}^2}{p_{N_2{O}_4}}$
$=\frac{{\left(\frac{2}{3}\right)}^2}{\left(\frac{1}{3}\right)}$
$=\frac{4}{3}=1.33atm$