Question

For the reaction $N_2{O}_4\rightleftharpoons 2N{O}_2$, equilibrium mixture contains $N{O}_2$ at $P=1.1atm$ & $N_2{O}_4$ at $P=0.28atm$ at 350 K. The volume of the container is doubled. Calculate the equilibrium pressures of the two gases when the system reaches new equilibrium :

• A. $P_{N{O}_2}=0.32atm,P_{N_2{O}_4}=0.095atm$
• B. $P_{N{O}_2}=0.64atm,P_{N_2{O}_4}=0.049atm$
• C. $P_{N{O}_2}=0.64atm,P_{N_2{O}_4}=0.05atm$
• D. $P_{N{O}_2}=0.32atm,P_{N_2{O}_4}=0.049atm$

Answer 1

The correct option is C $P_{N{O}_2}=0.64atm,P_{N_2{O}_4}=0.05atm$

For the first equilibrium, the total pressure is $1.1+0.28=1.38atm$
The value of the equilibrium constant will be $K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}=\frac{{1.1}^2}{0.28}=4.32$
For the second equilibrium, since the volume of the container is doubled, the total pressure will be one half i.e $\frac{1.38}{2}=0.69atm$
The partial pressures of $N_2{O}_4$ and $N{O}_2$ at equilibrium will be $0.69-x$ atm and $x$ atm respectively.
The value of the equilibrium constant will be $K_p=\frac{P_{N{O}_2}^2}{P_{N_2{O}_4}}=\frac{x^2}{0.69-x}=4.32$
$P_{N{O}_2}=x=0.64atm$ and $P_{N_2{O}_4}=0.69-x=0.05atm$