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?
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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