Steps plz :)
1. The equilibrium constant, Kc, for the dissociation of PCl5(g) is 0.04 mol dm-3 at 600K.
PCl5(g) <==> PCl3(g) + Cl2(g)
1 mol of PCl5(g) is allowed to dissociate to reach equilibrium in a 2 dm3 closed container.
Calculate the concentration of PCl5(g) in mol dm-3, in the equilibrium mixture at 600 K.
2. At 3000 K, the equilibrium constant for the following reaction is 1.5X10-3.
N2(g) + O2(g) <==> 2NO(g)
In an experiment, 1 mol of N2(g) and 2 mol of NO(g) are mixed in a 1 dm3 closed flask at 3000 K to reach equilibrium. Calculate the concentration of O2(g) in mol dm-3 in the equilibrium mixture.
F5 chem chemical equilibrium
2014-03-26 5:12 am
回答 (2)
2014-03-26 6:34 am
✔ 最佳答案
The answer is as follows :圖片參考：https://s.yimg.com/lo/api/res/1.2/goug5s87R04oNx5PkXYJdg--/YXBwaWQ9dHdhbnN3ZXJzO3E9ODU-/http://dc599.4shared.com/img/50uIidUSba/s7/144f9a78020/2014-03-25A.png?async&rand=0.14750782286487096
http://dc599.4shared.com/img/50uIidUSba/s7/144f9a78020/2014-03-25A.png?async&rand=0.14750782286487096
參考: pingshek
2014-03-26 5:46 am
1. PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Let x mole of PCl₅ dissociated
[PCl₅] = (1-x)/2
[PCl₃] = x/2
[Cl₂] = x/2
Kc = [PCl₃][Cl₂] / [PCl₅] = 0.04 = (x/2)(x/2) / [(1-x) / 2]
(0.04)(1-x) = x²
x² + 0.04x -0.04 = 0
x = 0.181
[PCl₅] = (1-x)/2 = (1-0.181)/2 = 0.41 mol dm⁻³
2014-03-25 21:56:13 補充：
Correction :
Kc = [PCl₃][Cl₂] / [PCl₅] = 0.04 = (x/2)(x/2) / [(1-x) / 2]
(0.08)(1-x) = x²
x² + 0.08x -0.08 = 0
x = 0.2457
[PCl₅] = (1-x)/2 = (1-0.2457)/2 = 0.3772 mol dm⁻³
2014-03-25 22:05:02 補充：
2. N₂(g) + O₂(g) ⇄ 2NO(g) Kc = 1.5X10⁻³
Let 2x mole of NO dissociated
[N₂] = (1+x) / 1 dm³
[O₂] = x / 1 dm³
[NO] = (2 - 2x) / 1 dm³
Kc = [NO]² / [N₂][O₂] = 1.5X10⁻³
1.5X10⁻³ = (2-2x)² / [(1+x)x]
(1.5X10⁻³)[(1+x)x] = (2-2x)²
(1.5X10⁻³)[x² + x] = 4x² - 8x + 4
2014-03-25 22:25:08 補充：
(1.5X10⁻³)x² + (1.5X10⁻³)x = 4x² - 8x + 4
(4 - 1.5X10⁻³)x² - (8 + 1.5X10⁻³)x + 4 = 0
3.9985x² - 8.0015x + 4 = 0
x = 1.028 (reject)
x = 0.9732
[O₂] = x / 1 dm³ = 0.9732 mol dm⁻³
Let x mole of PCl₅ dissociated
[PCl₅] = (1-x)/2
[PCl₃] = x/2
[Cl₂] = x/2
Kc = [PCl₃][Cl₂] / [PCl₅] = 0.04 = (x/2)(x/2) / [(1-x) / 2]
(0.04)(1-x) = x²
x² + 0.04x -0.04 = 0
x = 0.181
[PCl₅] = (1-x)/2 = (1-0.181)/2 = 0.41 mol dm⁻³
2014-03-25 21:56:13 補充：
Correction :
Kc = [PCl₃][Cl₂] / [PCl₅] = 0.04 = (x/2)(x/2) / [(1-x) / 2]
(0.08)(1-x) = x²
x² + 0.08x -0.08 = 0
x = 0.2457
[PCl₅] = (1-x)/2 = (1-0.2457)/2 = 0.3772 mol dm⁻³
2014-03-25 22:05:02 補充：
2. N₂(g) + O₂(g) ⇄ 2NO(g) Kc = 1.5X10⁻³
Let 2x mole of NO dissociated
[N₂] = (1+x) / 1 dm³
[O₂] = x / 1 dm³
[NO] = (2 - 2x) / 1 dm³
Kc = [NO]² / [N₂][O₂] = 1.5X10⁻³
1.5X10⁻³ = (2-2x)² / [(1+x)x]
(1.5X10⁻³)[(1+x)x] = (2-2x)²
(1.5X10⁻³)[x² + x] = 4x² - 8x + 4
2014-03-25 22:25:08 補充：
(1.5X10⁻³)x² + (1.5X10⁻³)x = 4x² - 8x + 4
(4 - 1.5X10⁻³)x² - (8 + 1.5X10⁻³)x + 4 = 0
3.9985x² - 8.0015x + 4 = 0
x = 1.028 (reject)
x = 0.9732
[O₂] = x / 1 dm³ = 0.9732 mol dm⁻³
收錄日期: 2021-04-27 20:37:02
原文連結 [永久失效]:
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