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A sample of air consisting of N(2) and ...

A sample of air consisting of `N_(2) and O_(2)` was heated to 2500 K until the equilibrium.
`N _(2) (g) + O _(2) (g) hArr 2NO (g)`
was established with an equilibrium constant `K _(c) = 2.1 xx 10 ^(-3).` At equilibrium the mole % of NO was 1.8. Estimate the initial composition of air in mole fraction of `N_(2) and O_(2).`

A

`0.68, 0.32 `

B

`0.46, 0.54`

C

`0.74, 0.26`

D

`0.83, 0.17`

Text Solution

Verified by Experts

The correct Answer is:
C
Let the total number of moles of `N _(2) and O _(2)` be 100 containing .a. moles of `N _(2)` initially
`{:(N _(2), + , O_(2), hArr , 2 NO ,),(a ,, 100-a,, 0, "initial moles"), ((a-x),, 100- (a-x),, 2x, "moles at equm"):}`
Mole% of NO at equilbrium `= ( 2x )/((a - x) + [10 - (a - x) ] + 2x) xx 100 = 1.8 therefore x = 0.9`
Thus at equilibrium, moles of `N _(2) = (a - 0.9) ` moles of `O_(2) = (100 -a -0.9) = (99.1 -a)`
moles of `NO= 2x = 2 xx 0. 9 = 1. 8`
`K_(c) = (( 2x //V) ^(2))/( ((a - 0. 9)/( V )) ((99.1-a)/( V ))) = ((2x ) ^(2))/( ( a - 0. 9 ) (99.1 -a )) , 2.1 xx 10 ^(-3) = ((1. 8) ^(2))/( (a - 0. 9 )(99.1 -a )) therefore = 79. 46%`
(We kow that percentage of `N _(2)` in the air is more than that of `O _(2))`
Mole fraction of `N _(2) = 0. 7946, ` Mole fraction of `O _(2) = 0. 2054`
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