K=0.0900, h2o= 0.49 M. And hocl= 0.042 M, What is the equilibrium concentration of Cl2O?
回答 (1)
2016-05-25 2:58 pm
The equation is : Cl₂O(g) + H₂O(g) ⇌ 2HOCl(g) ...... K
K = [HOCl]²/ ([Cl₂O] [H₂O])
Then, [Cl₂O] = [HOCl]²/ (K [H₂O])
At equilibrium, [Cl₂O]
= (0.042)² / (0.0900 × 0.49)
= 0.040 mol/L
K = [HOCl]²/ ([Cl₂O] [H₂O])
Then, [Cl₂O] = [HOCl]²/ (K [H₂O])
At equilibrium, [Cl₂O]
= (0.042)² / (0.0900 × 0.49)
= 0.040 mol/L
收錄日期: 2021-04-18 14:51:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160525064151AAB6RTj