Question

When 0.1 moles of $N_2{O}_4\left(g\right)$ was placed in a 1.0 litre flask at 400 K, the following equilibrium is reached with a total pressure of 6 bar.

$N_2{O}_4\rightleftharpoons 2N{O}_2$

Assuming ideal behaviour of the gases, the partial pressure of $N_2{O}_4\left(g\right)$ at equilibrium in bar is:

[R= 0.083 bar L mol${}^{-1}{k}^{-1}$]

• A. 3.32
• B. 9.32
• C. 0.66
• D. 2.68

Answer 1

Answer: (C)

0.66

$N_2{O}_4⟶2N{O}_2$
$0.1-x$ $2x$

P1 = partial pressure of $N_2{O}_4$
P2 = partial pressure of $N{O}_2$

Partial pressure of $N_2{O}_4=6\times \frac{(0.1-x)}{(0.1+x)}$

Total moles = $\frac{PV}{RT}$

Total moles = $6\times \frac{1}{(0.083\times 400)}$ $=0.18moles$

Total moles $=0.1+x=0.18$

$x=0.18-0.1=0.08$

Partial pressure of $N_2{O}_4=6\times \frac{(0.1-0.08)}{(0.1+0.08)}$

Partial pressure of $N_2{O}_4=\frac{(0.6-0.48)}{0.18}$

Partial pressure of $N_2{O}_4=0.66bar$

Hence, option C is correct.