The equilibrium constant, Kc, for the following
reaction is 1.80×10-2 at698 K.
2HI(g) → H2(g)
+ I2(g)
Calculate the equilibrium concentrations of reactant and products
when 0.223 moles of HI are
introduced into a 1.00 L vessel at 698 K.
[HI]= ?
[H2]= ?
[I2]= ?
ICE Table:
Equilibrium constant expression is
Kc = [H2]*[I2]/[HI]^2
0.018 = (1*x)^2/(0.223-2*x)^2
sqrt(0.018) = (1*x)/(0.223-2*x)
0.13416 = (1*x)/(0.223-2*x)
2.992*10^-2-0.2683*x = 1*x
2.992*10^-2-1.268*x = 0
x = 0.02359
At equilibrium:
[HI] = 0.223-2x = 0.223-2*0.02359 = 0.17582 M
[H2] = +1x = +1*0.02359 = 0.02359 M
[I2] = +1x = +1*0.02359 = 0.02359 M
Answer:
[HI] = 0.176 M
[H2] = 0.0236 M
[I2] = 0.0236 M
The equilibrium constant, Kc, for the following reaction is 1.80×10-2 at698 K.
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