Fall 2016 #6C  [ENDORSED]

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Michael Lonsway 3O
Posts: 43
Joined: Wed Sep 21, 2016 2:57 pm

Fall 2016 #6C

Postby Michael Lonsway 3O » Fri Nov 11, 2016 6:42 pm

Consider the following reaction at 800k, for which you know kc=2.1x10^-3: I2(g)<-->I(g)
If you measure [I2] to be 3.18x10^-4M, calculate the equilibrium concentration of I for this experiment.

The answer is 8.2x10^-4M but I got a number much larger than that

nikita bhat 2D
Posts: 29
Joined: Wed Sep 21, 2016 2:59 pm

Re: Fall 2016 #6C  [ENDORSED]

Postby nikita bhat 2D » Fri Nov 11, 2016 11:25 pm

I am not sure how you reached the answer larger, but I will explain what I did and maybe you see what differs in our work.

1. Balance the equation
I2(g)<-->2I(g)

2. Find the equilibrium constant expression.

Kc=[product]/[reactant] = [I]^2/[I2]

3. Plug the known values in.
2.1x10^-3 = [I]^2/ (3.18x10-4 M)

4. solve for [I].

[I]^2 = 6.678x10^-7
I = 8.1719x10^-4 M or 8.2x10^-4 M

Hope this helps.

lorelrodriguez4c
Posts: 9
Joined: Fri Sep 25, 2015 3:00 am

Re: Fall 2016 #6C

Postby lorelrodriguez4c » Mon Nov 14, 2016 5:16 pm

^ Great explanation! Another reason you could be getting a larger number is because of the way you input numbers in your calculator. The first time I did this problem I got a larger number as well, but I went back and realized I had inputed a parentheses wrong. Be careful and always double check!


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