Question

If the Pressure of N2 and H2 mixture in a closed apparatus is 100 atm and 20% of the mixture reacts then the pressure at the same temperature would be ?

Answer 1

Answer: Pressure when 20% of the mixture has been reacted is 90 atm.

Explanation: For the reaction,

                                   $N_2+3H_2\rightarrow 2NH_3$

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, $n_1=1+3=4$

At equilibrium,

Degree of dissociation, $\alpha =20%=\frac{20}{100}=0.2$

$\text{Moles of }N_2\text{ at equilibrium}=1-0.2=0.8$

$\text{Moles of }H_2\text{ at equilibrium}=3-3(0.2)=2.4$

$\text{Moles of }NH_3\text{ at equilibrium}=2(0.2)=0.4$

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 same, hence temperature and volume remains constant.

Initial Conditions, P = 100 atm

n = 4 moles

Equation becomes, $P_1V=n_1RT$    .....(1)

Final Conditions, P = ? atm

n = 3.6 moles

Equation becomes, $P_2V=n_2RT$    .....(2)

Dividing equation 1 by equation 2, we get

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

Putting values in the above equation

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

$P_2=90atm$

Answer 2

Answer:

This is the most simple method with easiest answer.

Hope,it will help you.......