One mole of `N_(2)O_(4)` gas at 300 K is kept in a closed container at 1 atm. It is heated to 600 K when `20%` of `N_(2)O_(4)` decomposes to `NO_(2)(g).` The resultant pressure in the container would be

Options

A. 1.2 atm
B. 2.4 atm
C. 2.0 atm
D. 1.0 atm

Answer

B

Explanation

`{:(,N_2O_4 hArr, 2NO_2),("Initial conc.", 1.0,0),("At equi.", 1-0.20 , "0.40 mol"):}`
=0.80 mol
Total moles =0.80 + 0.40 = 1.20 mol
1 mol of vapour has pressure =1 atm at 300 K
Applying pV=nRT
1 x V = 1 x R x 300 …(i)
When n=1.2 mol, T=600 K
p x V =1.2 x R x 600 ...(ii) Dividing (ii) by (i)
p=2.4 atm