平衡常數計算問題

2013-05-04 5:19 am
1. At 25℃ the value of the equilibrium constant for a reaction is 20.5 and △H is -5.29 kJ/mole. What is the value of K at 100℃?

2. 反應2HI (g)↹H2 (g) + I2 (g),在764K時平衡常數微2.18X10^-2,在667K平衡常數微1.64X10^-2,試估計此反應之△H值?

3.有1莫爾的AB和1莫爾的CD,依照下列方程式反應AB+CD=AD+CB,當平衡建立時,發現AB和CD均有3/4莫爾轉變成AD和CB,在體積上無變化,該反應之平衡常數為?

4.Gaseous phosphorous pentachloride decompose to hosphorous trichloride and chlorine at a temperature where K=1.0X10^-3 mole/L. Suppose 2.00 mole of phosphorous pentachloride in a 2.00 L vessel is allowed to come to equilibrium. The concentration of phosphorous pentachloride will be about ?M

5. 在常溫時,以下反應的平衡常數 Kp=0.2atm,PCl5 (g)=PCl3 (g)+ Cl2 (g),平衡時,總壓為3atm,求PCl5的解離率?

回答 (2)

2013-05-04 8:48 am
✔ 最佳答案
1.
T1 = 273 + 25 = 298 K
T2 = 273 + 100 = 373 K

ln(K1/K2) = -(ΔH/R)[(1/T1) - (1/T2)]
ln(20.5/K2) = -(-5290/8.314) x [(1/298) - (1/373)]
The value of K at 100°C, K2 = 13.3 kJ/mol


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2.
ln(K1/K2) = -(ΔH/R)[(1/T1) - (1/T2)]
ln[(2.18 x 10⁻²)/(1.64 x 10⁻²)] = -(ΔH/8.314)[(1/764) - (1/667)]
ΔH = +12400 J/mol = +12.4 kJ/mol


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3.
AB + CD = AD + CB

設體積為 V L。

平衡時:
[AB] = [CD] = [1- (3/4)]/V = 1/4V mol/L
[AD] = [CB] = (3/4)/V = 3/4V mol/L

平衡常數 K
= [AD][CB] / [AB][CD]
= (3/4V)² / (1/4V)²
= 9


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4.
PCl5 = PCl3 + Cl2

Initial concentrations :
[PCl5]o = 2.00/2.00 = 1 mol/L, [PCl3]o= [Cl2]o = 0 mol/L

At equilibrium :
Let [PCl5] = y mol/L, then [PCl3] = [Cl2] = (1- y) mol/L

Kc = [PCl3][Cl2] / [PCl5]
(1 - y)² / y = 1.00 x 10⁻³
1 - 2y + y² = 0.001y
y² - 2.001y + 1 = 0
y = 0.969 or y = 1.03 (rejected)

At equilibrium, [PCl5] = 0.969 mol/L


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5.
PCl5 = PCl3 + Cl2

設解離率為 α。
則平衡時的莫耳數比 PCl5 : PCl3 : Cl2 =(1 - α) : α : α

平衡時:
PPCl5 = 3 x [(1 - α)/ (1 - α + α + α)] = 3(1 - α) / (1 + α) atm
PPCl3 = PCl2 = 3 x [α/ (1 - α + α + α)] = 3α / (1 + α)atm

Kp = PPCl3 x PCl2 / PPCl5
[3α / (1 + α)]² / [3(1 - α) / (1 + α)] = 0.2
3α² / (1 - α²) = 0.2
3α² = 0.2 - 0.2α²
3.2α² = 0.2
α² = 0.0625
α = 0.25

解離率 = 0.25
參考: 不用客氣
2013-05-13 7:18 pm
沒甚麼好困擾吧!
把 2950 J/mol 代入題中驗算,便知除非題目打錯字,否則 2950 kJ/mol 是錯誤的答案。


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