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One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is (a) 1.2 atm (b) 2.4 atm (c) 2.0 atm (d) 1.0 atm choose and explain?

One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is


 (a) 1.2 atm


(b) 2.4 atm


 (c) 2.0 atm


(d) 1.0 atm


choose and explain?

Grade:12

3 Answers

Vikas TU
14149 Points
11 years ago

N2O4 → 2NO2

moles of unreacted N2O4 = 1 (1 - 0.2) = 0.8

moles of NO2 = 2 * 0.2 = 0.4
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (+)
total moles n2 = 0.8 + 0.4 = 1.2

P1/(T1 n1) = P2/(T2 * n2)

1/(300 * 1) = P2/(600 * 1.2)

P2 = 2.4 atm

Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
Dear Student

moles of unreacted N2O4 = 1 (1 - 0.2) = 0.8

moles of NO2 = 2 * 0.2 = 0.4
total moles n2 = 0.8 + 0.4 = 1.2

P1/(T1 n1) = P2/(T2 * n2)
1/(300 * 1) = P2/(600 * 1.2)
P2 = 2.4 atm

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya
Devender Chopra
15 Points
2 years ago
By gay lussac's law P1/T1=P2/T2 
                                  1/300= P2/600
 P2 = 2 atm 
At equilibrium pressure of N2O4 = 2-20% of 2 
= 2-0.4 =1.6
Pressure of NO2 = 2×0.4 = 0.8 
Resultant pressure= 1.6+0.8 = 2.4

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