Consider the following equilibrium reaction that is taking place in a 5.00 L container at 25 oC: 2SO2 (g) + O2 (g) --> 2SO3 (g)?

[匿名]

2017-02-12 9:55 pm

It was found that there are 3.00 moles of SO3, 2.00 moles of SO2, and 1.50 moles of O2 when the reaction reached equilibrium. The equilibrium has been disturbed by reducing the volume of the container down to 2.00 L, and by adding some unknown amount of SO3 into the container. When the reaction has established equilibrium again, O2 was found to be 2.00 moles. Determine how many moles of SO3 have been added.

回答 (2)

不用客氣

2017-02-13 12:20 am

✔ 最佳答案

Let y mol/L be [SO₃] at the new equilibrium.

At original eqm:
[SO₂] = 2.00/5.00 mol/L = 0.400 mol/L
[O₂] = 1.50/5.00 mol/L = 0.300 mol/L
[SO₃] = 3.00/5.00 mol/L = 0.600 mol/L

When volume is reduced to 2 L :
[SO₂] = 2.00/2.00 mol/L = 1.00 mol/L
[O₂] = 1.50/2.00 mol/L = 0.750 mol/L
[SO₃] = 3.00/2.00 mol/L = 1.50 mol/L

At new eqm :
[O₂] = 2.00/2.00 mol/L = 1.00 mol/L

_________________ 2SO₂ (g) + O₂(g) → 2SO₃(g) …… Kc = [SO₃]² / {[SO₂]² [O₂]}
Original eqm:__ 0.400 ____ 0.300 ___ 0.600 …… (mol/L)
Volume 2 L: ____ 1.00 _____ 0.750 ___ 1.50 _ …… (mol/L)
SO₃ added: _____ 1.00 _____ 0.750 _____ ? ___ …… (mol/L)
Change: _______ +0.500 __ +0.250 __ -0.50 …… (mol/L)
New eqm: _____ 1.50 ______ 1.00 _______ y _ …… (mol/L)

Temperature is kept constant.
Kc for the new equilibrium = Kc for the original equilibrium
y² / (1.50² × 1.00) = 0.600² / (0.400² × 0.300)
y² = 0.600² × 1.50² × 1.00 / (0.400² × 0.300)
y = 4.11

No. of moles of SO₃ added = (4.11 mol/L) × (2 L) - (3.00 mol) = 5.22 mol

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[匿名]

2017-02-13 9:17 pm

5 moles more

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