Question

for the reaction N2+O2=2NO equilibrium constant kc=2 ,degree of association is ?

Answer 1

N2(g) + O2(g) _ 2NO(g)

1 1 0

1-x 1-x 2x

Kc = (2x)^2 / (1-x)(1-x) =2

i.e (2x)^2 / (1-x)^2=2

2x / (1-x)=√2

2x=√2-√2x

2x+√2x=√2

x(2+√2)=√2

x=√2/(2+√2)

x=1/(√2+1) (we take √2 common and it

gets cancelled)

Hence our answer is X=1/(1+√2)

Answer 2

Given:

The equilibrium constant, Kc = 2

To Find:

The degree of association, i.e., α.

Calculation:

- Let the initial concentration of N2 and O2 be c. Then we have:

N2  +  O2 ⇄ 2 NO

c          c            0    (initial)

c(1-α)  c(1-α)      2αc  (equilibrium)

- From the above data, we have:

Kc = (2αc)² / c²(1-α)²

⇒ 2 = (2α / 1-α)²

⇒ √2 = (2α / 1-α)

⇒ √2 (1-α) = 2α

⇒ √2 - √2α = 2α

⇒ 2α + √2α = √2

⇒ α = √2 / (2 + √2)

⇒ α = 1 / (√2 + 1)

⇒ α = 0.414

- So, the degree of association is 0.414.