N2(g) + O2(g) ⇌ 2NO(g) K 0.0639 0.400 atm

N2(g) + O2(g) ⇌ 2NO(g) K 0.0639 0.400 atm

 

 

Consider the following reaction:

N2(g) + O2(g) ⇌ 2NO(g)

At a certain temperature, the equilibrium constant for the reaction is 0.0639.

What are the partial pressures of all gases at equilibrium

if the initial partial pressure of the gases (both products and reactants)

is 0.400 atm?

 

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ICE 도표를 작성하면,

 

 

 

 

K = [NO]^2 / [N2] [O2]

 

0.0639 = (0.400 + 2x)^2 / (0.400 – x)^2

 

0.0639 = [(0.400 + 2x) / (0.400 – x)]^2

 

0.0639^(1/2) = (0.400 + 2x) / (0.400 – x)

 

0.0639^(1/2) (0.400 – x) = (0.400 + 2x)

 

(0.0639^(1/2)) (0.400) – (0.0639^(1/2))x = 0.400 + 2x

 

(2 + (0.0639^(1/2)))x = (0.0639^(1/2)) (0.400) – 0.400

 

x = [(0.0639^(1/2)) (0.400) – 0.400] / (2 + (0.0639^(1/2)))

= –0.132674

= –0.133

 

 

 

평형에서,

 

N2의 분압 = 0.400 – (–0.133) = 0.533 atm

= O2의 분압

 

 

NO의 분압 = 0.400 + 2(–0.133) = 0.134 atm

 

 

 

 

[검산]

K = (0.400 + 2(–0.132674))^2 / (0.400 – (–0.132674))^2 = 0.0639

 

 

 

[키워드] N2(g) + O2(g) ⇌ 2NO(g) equilibrium constant 0.0639 initial partial pressure 0.400 atm