According to the following reaction
PCl3. + Cl2 ---->. PCl5
In a 3 liters flask there are 1 mol PCl3 and 2 mol Cl2.
After equilibrium has been reached, only 0,70 mol of PCl3 are left in the flask.
Calculate equilibrium constant
Equilibrium constant help! Thanks?
2016-05-23 11:25 am
回答 (2)
2016-05-23 1:56 pm
✔ 最佳答案
PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) ... Equilibrium constant = KcMole ratio PCl₃ : Cl₂ : PCl₅ = 1 : 1 : 1
Initial concentrations :
[PCl₃]ₒ = 1/3 mol/L = 0.333 mol/L
[Cl₂]ₒ = 2/3 mol/L = 0.667 mol/L
[PCl₅]ₒ = 0 mol/L
At equilibrium :
Concentration of PCl₃ consumed = (1 - 0.70)/3 = 0.1 mol
[PCl₃] = (0.333 - 0.1) mol/L = 0.233 mol/L
[Cl₂] = (0.667 - 0.1) mol/L = 0.567 mol/L
[PCl₅] = (0 + 0.1) mol/L = 0.1 mol/L
Kc = [PCl₅] / ([PCl₃] [Cl₂])
Kc = 0.1 / (0.233 × 0.567)
Equilibrium constant, Kc = 0.757
(To 2 sig. fig., Kc = 0.76)
2016-05-23 12:46 pm
Write down the BALANCED reaction equation.
2PCl3 + Cl2 <=> 2PCl5
moles initially 1M , 2M = 0M
moles equil'm 0.70, (2 - 0.15)M , 0.3M
moles(equil'm ; 0.7M , 1.85M , 0.30 M
Kc = [PCl5/3]^2 / [PCl3/3]^2[Cl2/3]
Kc = [0.30^2/3] / [0.70/3]^2* [1.85/3]
Kc = 0.893546607832322118036403750689460
Kc = 0.89 dm^3 mol^-1
2PCl3 + Cl2 <=> 2PCl5
moles initially 1M , 2M = 0M
moles equil'm 0.70, (2 - 0.15)M , 0.3M
moles(equil'm ; 0.7M , 1.85M , 0.30 M
Kc = [PCl5/3]^2 / [PCl3/3]^2[Cl2/3]
Kc = [0.30^2/3] / [0.70/3]^2* [1.85/3]
Kc = 0.893546607832322118036403750689460
Kc = 0.89 dm^3 mol^-1
收錄日期: 2021-04-18 14:59:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160523032540AAQPAHq