Question

Calculate the change in pressure when $1.04$ mole of $NO$ and $20g{O}_2$ in $20$ litre vessel originally at ${27}^{o}C$ react to produce the maximum quantity of $N{O}_2$ possible according to the equation. $2NO\left(g\right)+O_2\left(g\right)⟶2N{O}_2\left(g\right)$

• A. 0.2113 atm
• B. 1.2321 atm
• C. 0.6396 atm
• D. 0.5687 atm

Answer 1

The correct option is C 0.6396 atm

According to equation, 2 moles of $NO$ reacts with one mole of oxygen.

Number of moles of $NO$ = 1.04 mole

Number of moles of oxygen = $\frac{20}{32}=0.625mol$

So, according to reaction, $1.04$ mol of $NO$ reacts with $0.52$ mol of oxygen

So, leftover oxygen = $0.625-0.52=0.105$

Now for calculation of pressure

(1). Before reaction, 1.04 moles of $NO$ and 0.625 moles of oxygen are present.

So, total number of moles = $1.04+0.625=1.665$

Since, $PV=nRT$

So, $P_{initial}=\frac{nRT}{V}{P}_{initial}=\frac{1.665RT}{V}$

(2). After reaction, 1.04 mole of ${NO}_2$ is formed and 0.105 moles of $O_2$ is left.

So, total number of moles = $1.04+0.105=1.145$

So, $P_{final}=\frac{1.145RT}{V}$.

Now, change in pressure

$=P_{initial}-P_{final}=\frac{(1.665-1.145)RT}{V}=\frac{0.52\times 0.082\times 300}{20}=0.6396atm$

So, correct answer is option C.