Question

If the pressure of $N_2$ and $H_2$ mixture in a closed apparatus is $100$ atm, and $20\%$ of the mixture then reacts to form ammonia, the pressure at the same temperature would be:

• A. $100$
• B. $90$
• C. $85$
• D. $80$

Answer 1

Answer: (B)

$90$

$N_2+3H_2\to 2N{H}_3$

Moles at $t=0$ $1$ $3$ $0$

Moles at $t=t_{eq}$ $(1-\alpha )$ $(3-3\alpha )$ $2\alpha$

By stoichiometry of the reaction,

Initially, total number of moles, $n_t=1+3=4$

At equilibrium,

Extent of reaction, $\alpha =20$

$\text{Moles of}{N}_2\text{at equilibrium}=1-0.2=0.8$

$\text{Moles of}{H}_2\text{at equilibrium}=3-3\left(0.2\right)=2.4$

$\text{Moles of}N{H}_3\text{at equilibrium}=2\left(0.2\right)=0.4$

Total moles at equilibrium, $n_2=0.8+2.4+0.4=3.6$

Using Ideal gas equation,

$PV=nRT$

As this reaction is carried out in a closed apparatus, the volume of the system remains constant and it is given that the temperature is also the same, hence temperature and volume remain constant.

Initial Conditions, $P=100atm$

$n=4moles$

Equation becomes, $.....\left(1\right)$

Final Conditions, $P=?atm$

$n=3.6moles$

Equation becomes, $.....\left(2\right)$

Dividing equation 1 by equation 2, we get

$\frac{P_1}{P_2}=\frac{n_1}{n_2}$

Putting values in the above equation

$\frac{100}{P_2}=\frac{4}{3.6}$

$P_2=90atm$

$Hencethecorrectanswerisoption\left(B\right).$