Question

2 NO (g) equilibrium arrow N2 (g) + O2 (g) Initially, 0.75 M NO is placed...

2 NO (g) equilibrium arrow N2 (g) + O2 (g)

Initially, 0.75 M NO is placed in a reaction vessel and the reaction is allowed to reach equilibrium. If the equilibrium constant for this reaction is 0.463, what is the equilibrium concentration of NO?

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Answer #1

The reaction is 2NO(g)<-----------> N2 (g) + O2 (g)

Initially NO= 0.75

let x= concenttration of N2 at equilibrium

hence at equilibrium

[N2] =[O2] =x and [NO] =0.75-2x

Equilibrium constant = [N2] [O2]/[NO]2 = 0.463

x2/(0.75-2x)2= 0.463

taking square toor on both sides give

x/(0.75-2x)= 0.68

x= 0.68*(0.75-2x)

x*(1+2*0.68)= 0.68*0.75

2.36x= 0.68*0.75

x= 0.216

At equilibrium [NO] =0.75-2*0.216= 0.75-0.432= 0.318M

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