Question

At room temperature, the following reaction proceeds nearly to completion:

$2NO+O_2⟶2N{O}_2⟶N_2{O}_4$

The dimer, $N_2{O}_4$, solidifies at 262 K.

A 250 mL flask and a 100 mL flask are separated by a stopcock. At 300 K, the nitric oxide in the larger flask exerts a pressure of 1.053 atm and the smaller one contains oxygen at 0.789 atm. The gases are mixed by opening the stopcock and after the end of the reaction, the flasks are cooled to 220 K. Neglecting the vapour pressure of the dimer, find out the pressure of the gas remaining at 220 K.

If the value is $22\times {10}^{-x}$, then what is the value of x?

[Assume the gases to behave ideally].

Answer 1

$PV=nRT$

or $n=\frac{PV}{RT}$

Moles of NO

$a=\frac{1.053\times 250}{0.0821\times 300\times 1000}=0.01069$

Moles of $O_2$

$b=\frac{0.789\times 100}{0.0821\times 300\times 1000}=0.003203$

The number of moles of oxygen is very small, so it must be the limiting reagent.

$2NO+O_2⟶2N{O}_2⟶N_2{O}_4$

$a$ $b$ 0
$(a-2b)$ 0 $b$

All oxygen is consumed.
$\text{Moles of NO left}=(a-2b)$
$=0.01069-2\left(0.003203\right)$
$=0.004284$

At 220 K, the dimer is solidified and it does not exert any pressure (neglecting vapour pressure).

All the pressure is due to the left NO only.

$PV=nRT$

or $P=\frac{n}{V}RT$

$=\frac{0.004284\times 1000}{(250+100)}\times 0.0821\times 220$

$=0.221=22\times {10}^{-2}atm$

Thus, $\text{x}$ is 2.