The following reaction, in CCl4 solvent, has been studied at 25°C.
2BrCl ↔ Br2 + Cl2
The equilibrium constant Kc is known to be 2.7x10-5. If the initial concentration of chlorine is 0.021 M and of bromine monochloride is 0.022 M, what is the equilibrium concentration of bromine?
ICE Table question??
2018-05-13 2:02 pm
回答 (1)
2018-05-13 10:07 pm
__________ 2BrCl(aq) _____ ⇌ _____ Br₂ _____ + _____ Cl₂ _______ Kc = 2.7 × 10⁻⁵
Initial: ____ 0.022 M _____________ 0 M __________ 0.021 M
Change: ____ -2y M _____________ +y M ___________ +y M
Eqm: ___ (0.022 - 2y) M ___________ y M ________ (0.021 + y) M
As Kc is very small and due to the presence of Cl₂, only a very small amount of Br₂ is formed.
We can assume that 0.021≫ y, and thus
[BrCl] at eqm = (0.022 - 2y) M ≈ 0.022 M
[Cl₂] at eqm = (0.021 + y) M ≈ 0.021 M
At eqm:
Kc = [Br₂] [Cl₂] / [BrCl]²
2.7 × 10⁻⁵ = y × 0.021 / 0.022²
y = (2.7 × 10⁻⁵) × 0.022² / 0.021 = 6.2 × 10⁻⁷
[Br₂] at eqm = 6.2 × 10⁻⁷ M
Initial: ____ 0.022 M _____________ 0 M __________ 0.021 M
Change: ____ -2y M _____________ +y M ___________ +y M
Eqm: ___ (0.022 - 2y) M ___________ y M ________ (0.021 + y) M
As Kc is very small and due to the presence of Cl₂, only a very small amount of Br₂ is formed.
We can assume that 0.021≫ y, and thus
[BrCl] at eqm = (0.022 - 2y) M ≈ 0.022 M
[Cl₂] at eqm = (0.021 + y) M ≈ 0.021 M
At eqm:
Kc = [Br₂] [Cl₂] / [BrCl]²
2.7 × 10⁻⁵ = y × 0.021 / 0.022²
y = (2.7 × 10⁻⁵) × 0.022² / 0.021 = 6.2 × 10⁻⁷
[Br₂] at eqm = 6.2 × 10⁻⁷ M
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