For the reaction 2 HI(g) <--> H2(g) + I2(g), Kc = 0.016 at 520oC. Calculate the equilibrium concentration of I2 if the initial concentration of HI is 0.50 M and the other two gases are not initially present.
Give your answer in number form only with the correct sig figs. Do not use scientific notation.
Calculate the equilibrium concentration present.?
2016-02-26 6:30 pm
回答 (2)
2016-02-26 7:43 pm
✔ 最佳答案
2HI(g) ⇌ H₂(g) + I₂(g)Initial :
[HI]₀ = 0.50 M
[H₂]₀ = [I₂]₀ = 0 M
At eqm :
[HI] = (0.50 - 2y) M
[H₂] = [I₂] = y M
Kc = [H₂] [I₂] / [HI]
y²/(0.50 - 2y)² = 0.016
y/(0.50 - 2y) = √0.016
0.50 - 2y = y/√0.016
0.50 - 2y = 7.91y
9.91y = 0.50
y = 0.050
At eqm, [I₂] = 0.050 M
2016-02-26 7:13 pm
Use an ICE table for the equilibrium. This gives
Kc = x^2/(0.50 - 2x)^2 = 0.016
solve for x, which is [I2] at equilibrium.
Kc = x^2/(0.50 - 2x)^2 = 0.016
solve for x, which is [I2] at equilibrium.
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