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An empty steel container is filled with 2.0 atm of H₂ and 1.0 atm of F₂. The system is allowed to reach equilibrium according to the reaction below. If Kp = 0.45 for this reaction, what is the equilibrium partial pressure of HF?

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The equilibrium partial pressure of HF is 0.55 atm.

The equation of the reaction is;

        H2(g) + F2(g) ⇄ 2HF

I         2           1             0

C       -x           -x            +x

E     2 - x        1 - x           x

We know that;

pH2 = 2.0 atm

PF2 =  1.0 atm

pHF = ??

Kp = 0.45

So;

Kp = (pHF)^2/pH2. pF2

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

0.45 =  x^2/x^2 - 3x + 2

0.45(x^2 - 3x + 2) = x^2

0.45x^2 - 1.35x + 0.9 =  x^2

0.55 x^2 + 1.35x - 0.9 = 0

x = 0.55 atm

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