A vessel contains equal mass of CO2 and N2O . The mole fraction of CO2 is..
Answers
Answer:33.3
Explanation:Initial partial pressures of CO and CO
2
are 3 atm and 2 atm respectively.
2CO(g)⇋CO
2
(g)+C(s)
To reach equilibrium, 2x atm of CO reacts to form x atm of CO
2
The equilibrium partial pressures of CO and CO
2
are 3−2x atm and 2+x atm respectively.
Total equilibrium pressure is (3−2x)+(2+x)=5−x atm. But it is equal to 4.5 atm.
4.5=5−x
x=5−4.5
x=0.5
p
CO
:p
CO
2
=(3−2x):(2+x)=(3−2(0.5)):(2+1(0.5))=2:2.5=4:5
K
p
=
P
CO
2
P
CO
2
K
p
=
2
2
2.5
K
p
=0.625atm
−1
The dissociation of CO =
3
2x
=
3
2(0.5)
=0.333
The % dissociation of CO =100×0.333=33.3
Answer:
Simply apply formula to calculate mole fraction.
You can check the answer by calculating the same for N20 .If on adding the mole fraction of the two components comes out to be 1 then cheers!!! solution is correct...
