Question

Calculate the change in pressure (in atm) when 2 mole of NO and $16g{O}_2$ in a 6.25 litre originally at ${27}^{0}C$ react to produce the maximum quantity of $N{O}_2$ possible according to the equation.
$2NO\left(g\right)+O_2\left(g\right)\to 2N{O}_2\left(g\right)$

• A. 10 atm
• B. 8 atm
• C. 2 atm
• D. 4 atm

Answer 1

Answer: (C)

2 atm

Solution:- (C) $2atm$

Molecular weight of $O_2=32g$

Given weight of $O_2=16g$

No. of moles of $O_2=\frac{\text{Given weight}}{\text{Mol. wt.}}=\frac{16}{32}=0.5\text{mol}$

Given amount of $NO=2\text{mol}$

$2{NO}_{\left(g\right)}+{O_2}_{\left(g\right)}⟶2{N{O}_2}_{\left(g\right)}$

From the above reaction,

$2$ moles of $NO$ react with $1$ mole of $O_2$ to produce $2$ mole of $N{O}_2$.

But the given amount of $O_2$ is $0.5$ mole.

Thus, $O_2$ is limiting reagent.

Now again from the above reaction,

$1$ mole of $O_2$ produces $2$ moles of $N{O}_2$.

Therefore,

$0.5$ mole of $O_2$ will produce $1$ mole of $N{O}_2$.

From ideal gas equation-

$PV=nRT$

$\Rightarrow P=\frac{nRT}{V}$

Initially:-

$2$ mole of $NO$ and $0.5$ mole of $O_2$ were present.

Total no. of moles initially $=2+0.5=2.5$

Therefore,

Initial pressure $\left(P_i\right)=\frac{2.5RT}{V}$

Afte reaction:-

$1$ mole of $N{O}_2$ formed and $1$ mole of $NO$ left.

Total no. of moles after reaction $=1+1=2$

Therefore,

Final pressure $\left(P_f\right)=\frac{2RT}{V}$

Therefore,

Change in pressure $\left(\Delta P\right)=P_i-P_f$

$\Rightarrow \Delta P$

$=\frac{2.5RT}{V}-\frac{2RT}{V}$

$=\frac{RT}{2V}$

Given that $T=27℃=(27+273)=300K$

$V=6.25L$

$\therefore \Delta P=\frac{0.0821\times 300}{2\times 6.25}=2atm$

Hence the change in pressure is $2atm$.