The reaction H2(g) + I2(g) <=> 2HI(g) has Kc = 50.2 at 718K. If a flask...
The reaction H2(g) + I2(g) <=> 2HI(g) has Kc = 50.2 at 718K. If a flask is charged with 4.25M HI, what are the equilibrium concentrations of H2 and I2? at equilibrium, [H^2}+[I^2]= ?????
Homework Answers
ICE Table:
Equilibrium constant expression is
Kc = [HI]^2/[H2]*[I2]
50.2 = (4.25-2*x)^2/(1*x)^2
sqrt(50.2) = (4.25-2*x)/(1*x)
7.085 = (4.25-2*x)/(1*x)
7.085*x = 4.25-2*x
-4.25 + 9.085*x = 0
x = 0.468
At equilibrium:
[H2] = x = 0.468 M
[I2] = x = 0.468 M
Answer:
[H2] = 0.468 M
[I2] = 0.468 M
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