Q] One mole of N_{2}O_{4}(g) at 300K is kept in a closed container under one atmosphere. It is heated to 600K when 20% by mass of N_{2}O_{4}(g) decomposes to NO_{2}(g). The resultant pressure is??

N_{2}O_{4 } ⇔ 2NO_{2}

1mole 0 Initial moles

-20% of 1=0.2 +2(20% of 1)=0.40 At equilibrium

1-0.2= 0.8 0.4 Moles at equilibrium

Total moles at equilibrium = 0.8+0.4 = 1.2 moles

1 mole vapour pressure = 1 atm at 300 K

Applying PV = nRT

1xV = 1x R x 300 .......................... (1)

When n=1.2 moles, T = 600 K

P x V = 1.2 x R x 600 ..................... (2)

Dividing (2) by (1),

PxV/(1xV) = (1.2 x R x 600)/ (1 x R x 300)

Therefore, P= 2.4 atm

Hence, resultant pressure of mixture is 2.4 atm

**
**