Chemical EquilibriumHard
Question
One mole of NaO4(g) at300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of decomposes to NO2 (G). The resultant pressure is : NaO4(g)
Options
A.1.2 atm
B.2.4 atm
C.2.0 atm
D.1.0 atm
Solution
1 mole N2O4(i.e.,92g) is lept at 1 atm P and 300
K in a closed container, thus
P2 = 1 atm V = ? n = 1 mole,
R = 0.0821 L. atm. mol -1K -1, T = 300K
V =
= 24.63L
Now, N2O2 decomposes as
N2O2 ⇋ 2NO2
and 20% N2O4
decomposes to NO2, thus
moles of NO2 formed = 2 × moles of N2O4
decomposed
and moles of N2O4 remaining =
= 0.8 mole
and total moles of N2O4 and NO2 = 0.8 + 0.4
= 1.2 moles
Now P2 =
= 2 . 4 atm.
K in a closed container, thus
P2 = 1 atm V = ? n = 1 mole,
R = 0.0821 L. atm. mol -1K -1, T = 300K
V =
= 24.63LNow, N2O2 decomposes as
N2O2 ⇋ 2NO2
and 20% N2O4
decomposes to NO2, thus
moles of NO2 formed = 2 × moles of N2O4
decomposed
and moles of N2O4 remaining =
= 0.8 mole
and total moles of N2O4 and NO2 = 0.8 + 0.4
= 1.2 moles
Now P2 =
= 2 . 4 atm.
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