Question

In the equilibrium, $N _{2} O _{4(g)} \rightleftharpoons 2 NO _{2(g)}$ 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)

$N _{2} O _{4}(g) \,\,\,2 NO _{2}(g)$
At. equil. $1\, mol\, 2 \,mol$
Total moles $=1+2=3$
$\therefore n_{N_{2} O_{4}}=\frac{1}{3}$
and $n_{ NO _{2}}=\frac{2}{3}$
$\because p_{N_{2} O_{4}}=n_{N_{2} O_{4}} \times p$
$\therefore p_{N_{2} O_{4}}=\frac{1}{3} \times 1=\frac{1}{3}$
and $p_{ NO _{2}}=\frac{2}{3} \times 1=\frac{2}{3}$
$K_{p}=\frac{\left(p_{N O_{2}}\right)^{2}}{p_{N_{2} O_{4}}}$
$=\frac{\left(\frac{2}{3}\right)^{2}}{\left(\frac{1}{3}\right)}$
$=\frac{4}{3}=1.33 \,atm$