Chemical Equilibrium

  • Thread starter recon
  • Start date
  • #1
399
1
Chemistry is not one of my strong subjects, so please bear with me. Here's one of the questions we were given as homework, of which I'm having some trouble with. I'm sure it's pretty elementary, but the problem is that I'm unable to express the answer clearly enough.

Question:
The reaction of ethanoic acid with ethanol to form ethyl ethanoate and water is an example of dynamic equilibrium. It is catalysed by the presence of [tex]H^+[/tex] ions.

[tex]CH_3CO_2H(l) + C_2H_5OH(l) \rightleftharpoons CH_3CO_2C_2H_5(l) + H_2O(l)[/tex]

State why, in determining the value of [tex]K_c[/tex] for this reaction, it is only necessary to know the number of moles rather than the concentrations of each substance.

My answer (We are given only two lines of answering space):
EDIT: Deleted answer; it's severely wrong. I'll try and come up with a new one.
 
Last edited:

Answers and Replies

  • #2
399
1
Is there another way of expressing the following idea: "The units of volume in the concentration of the substances cancel each other out in the calculation for [tex]K_c[/tex]"?
 
  • #3
GCT
Science Advisor
Homework Helper
1,728
0
Is there another way of expressing the following idea: "The units of volume in the concentration of the substances cancel each other out in the calculation for "?
that's one way to state it
 
  • #4
45
0
For the reaction [tex]aA+bB\rightleftharpoons cC+dD[/tex]
K is calculated by
[tex]K=\frac{[C]^c[D]^d}{[A]^a^b}[/tex]

The capital letters within the brackets denote the molarities of the reagents in the reaction.

The reaction takes place within one system, Does the system change druing the reaction? And in case it doesn't what does that mean for the way you denote the concentrations of the reagents both on the left and on the right of the arrows?
 
Last edited:

Related Threads on Chemical Equilibrium

Replies
1
Views
7K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
6
Views
741
  • Last Post
Replies
9
Views
920
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
242
  • Last Post
Replies
3
Views
538
  • Last Post
Replies
23
Views
10K
  • Last Post
Replies
2
Views
5K
Top