wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A sample of air consisting of N2 and O2 was heated to 2500 K until the equilibrium
N2(g)+O2(g) 2NO(g)
was established with an equilibrium constant Kc=2.1 ×103. At equilibrium, the mole% of NO was 1.8. Estimate the initial composition of air in mole fraction of N2 and O2

Open in App
Solution

N2(g)+O2(g)2NO(g)
At t=0- a b 0
At t= equilibrium- (ax) (bx) 2x
KC=(2x)2(ax)(bx)=2.1×103 equation 1
But 2xax+bx+2x=0.018
2xa+b=0.018 equation 2
By solving 1 and 2 we get,
aa+b=0.79= Mole fraction of N2
and ba+b=0.21= Mole fraction of O2
Therefore, mole fraction of N2 and O2 are 0.79 and 0.21 respectively.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Monosaccharides
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon