H2(g) + I2(g)⇋ 2HI(g)
The system is at equilibrium at a given temperature in a
2 L container. If the system has 0.2 mol of H2, 0.2 mol
of I2, and 0.6 mol of HI at equilibrium; how many moles
of HI should be added to increase the mol number of
H2 to 0.4?
Problem involving chemical equilbrium?
2018-02-28 8:41 pm
回答 (3)
2018-03-01 1:24 am
✔ 最佳答案
Let y mol be the number of moles of HI should be added.As the total coefficient on the left is equal to the total coefficient on the right, numbers of moles should be used instead of concentrations in the equilibrium law.
____________ H₂ (g) __ + __ I₂(g) __ ⇌ __ 2HI(aq) ___ Kc
At eqm: _____ 0.2 mol ___ 0.2 mol _____ 0.6 mol
Disturbed: __ 0.2 mol ____ 0.2 mol ___ (0.6 + y) mol
Change: ____ +0.2 M ____ +0.2 mol _____ -0.4 mol
New eqm:___ 0.4 mol ____ 0.4 mol ___ (0.2 + y) mol
Kc = 0.6² / (0.2 × 0.2) = (0.2 + y)² / (0.4 × 0.4)
(0.2 + y)² = 0.6² × (0.4 × 0.4) / (0.2 × 0.2)
(0.2 + y)² = 1.44
0.2 + y = 1.2
y = 1.0
Hence, number of moles of HI should be added = 1.0 mol
2018-03-01 12:23 am
Ke = [0.6]/[0.2][0.2]
Ke = 15.0
Hence
15.0 = [HI]/ [0.2][0.4]
[HI] = 15.0 x 0.2 x 0.4
[HI] = 1.2 The conc'n of HI a equilibrium
Hence moles to add is 1.2 - 0.6 = 0.6 moles to be added.
This assuming that the moles of H2 remain the same at 0.2
Ke = 15.0
Hence
15.0 = [HI]/ [0.2][0.4]
[HI] = 15.0 x 0.2 x 0.4
[HI] = 1.2 The conc'n of HI a equilibrium
Hence moles to add is 1.2 - 0.6 = 0.6 moles to be added.
This assuming that the moles of H2 remain the same at 0.2
2018-02-28 10:08 pm
Kc = [HI]^2 / [H2][I2]
Initial equilibrium:
[H2] = [I2] = 0.2 mol / 2 L = 0.1 mol/L
[HI] = 0.6 mol/2 L = 0.3 mol/L
Kc = (0.3)^2 / (0.1)(0.1) = 9
In new equilibrium, [H2] = 0.4 mol/2 L = 0.2 mol/L = [I2]. Let [HI] = x. Then
Kc = 9 = x^2 /0.2(0.2)
x = [HI] = 0.6 mol/L X 2 L = 1.2 mol
To reach new equilibrium condition, because moles H2 increased by 0.2, 0.4 moles of HI had to decompose.
Also, moles HI at equilibrium is increased by 0.6. So, 1.0 mol HI had to be added to the container to establish the new equilibrium.
Initial equilibrium:
[H2] = [I2] = 0.2 mol / 2 L = 0.1 mol/L
[HI] = 0.6 mol/2 L = 0.3 mol/L
Kc = (0.3)^2 / (0.1)(0.1) = 9
In new equilibrium, [H2] = 0.4 mol/2 L = 0.2 mol/L = [I2]. Let [HI] = x. Then
Kc = 9 = x^2 /0.2(0.2)
x = [HI] = 0.6 mol/L X 2 L = 1.2 mol
To reach new equilibrium condition, because moles H2 increased by 0.2, 0.4 moles of HI had to decompose.
Also, moles HI at equilibrium is increased by 0.6. So, 1.0 mol HI had to be added to the container to establish the new equilibrium.
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