For the reaction PCl3(g)+Cl2(g)<-->PCl5(g),kc=96.2 at 400k.What is the concentration of cl2 at equilibrium if the initial concentrations were .20 M for PCl3 and 7.0 M for Cl2?
I keep on getting the wrong answer,the answer is suppose to be 6.8M
Chemistry help-equilibrium?
2016-06-20 8:23 pm
回答 (1)
2016-06-20 9:41 pm
✔ 最佳答案
PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) ...... Kc = 96.2Initial concentrations :
[PCl₃]ₒ = 0.20 mol/L
[Cl₂]ₒ = 7.0 mol/L
[PCl₅]ₒ = 0 mol/L
Equilibrium concentrations :
[PCl₃] = (0.20 - y) mol/L
[Cl₂] = (7.0 - y) mol/L
[PCl₅] = y mol/L
Kc = [PCl₅] / ([PCl₃] [Cl₂])
96.2 = y / (0.20 - y)(7.0 - y)
As Kc is great, the denominator is very small.
0.20 - y ≈ 0
y ≈ 0.2
[Cl₂] at equilibrium = (7.0 - 0.20) mol/L = 6.8 mol/L
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