Phosphorus pentachloride decomposes according to the chemical equation.
PCl5(g) <--> PCL3(g)+ CL2(g)
Kc=1.80 at 250*C
A 0.235 mol sample of PCl5(g) is injected into an empty 2.75 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
PCl5= __M
PCl3= __M
NEED CHEM PLEASE HELP CALCULATING CONCENTRATIONS AT EQUILIBRIUM?
2018-03-09 10:49 pm
回答 (2)
2018-03-10 12:07 am
✔ 最佳答案
[PCl₅]ₒ = (0.235 mol) / (2.75 L) = 0.08545 M____________ PCl₅(g) __ ⇌ __ PCl₃(g) __ + __ Cl₂(g) ___ Kc = 1.80
Initial: ____ 0.08545 M ______ 0 M ________ 0 M
Change: ______ -y M _______ +y M _______ +y M
At eqm: _ (0.08545 - y) M ____ y M ________ y M
At eqm:
Kc = [PCl₃] [Cl₂] / [PCl₅]
1.80 = y² / (0.08545 - y)
1.80 (0.08545 - y) = y²
y² + 1.80y - 0.1538 = 0
y = [-1.80 √(1.80² + 4×0.1538)] / 2
y = 0.08173 or y = -1.882 (rejected)
[PCl₅] = 0.08173 M ≈ 0.0817 M (to 3 sig. fig.)
[PCl₃] = (0.08545 - 0.08173) M = 0.00372 M
2018-03-09 11:17 pm
The standard approach to this type of problem is to use an ICE table (initial, change (x), and equilbrium) Doing so gives
Kc = x^2/([PCl5] - x)
where x is eqm concentration of both PCl3 and Cl2, and [PCl5] is the initial concentration of PCl5
so calculate initial [PCl5] (Molarity M = n/V), you have Kc, so solve for x
Kc = x^2/([PCl5] - x)
where x is eqm concentration of both PCl3 and Cl2, and [PCl5] is the initial concentration of PCl5
so calculate initial [PCl5] (Molarity M = n/V), you have Kc, so solve for x
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