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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?

pratiksha agarwal , 13 Years ago
Grade 12
anser 3 Answers
Vikas TU

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

ApprovedApproved
Last Activity: 13 Years ago
Yash Chourasiya
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
Last Activity: 5 Years ago
Devender Chopra
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
Last Activity: 4 Years ago
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