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Topic: Equilibrium constant K from initial concentrations?  (Read 4836 times)

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Offline spKelpDiver

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Equilibrium constant K from initial concentrations?
« on: September 22, 2012, 08:13:59 PM »

Consider the following reaction:
2H2S(g) + SO2(g)  ::equil:: 3S(s) + 2H20 (g)
 
A reaction mixture initially containing 0.510M H2S(g)  and 0.510M SO2(g)  was found to contain 1.1×10−3M H20 (g)  at a certain temperature. A second reaction mixture at the same temperature initially contains  0.250M H2S(g)and  0.325M SO2(g) . Calculate the equilibrium concentration of H20  in the second mixture at this temperature.



 I solved this problem, but I did so by breaking some big rules (or atleast I thought I did).

1.) I took the value of [H20]/[H2S][SO2] (all raised to their respective stoichiometric coefficients) from the first reaction mixture. I arbitrarily called this value Kc, even though I know (or atleast I thought) that you can only obtain Kc from concentrations at equillibrium! The way this problem is worded it appears that these are initial concentrations. This value turned out to be 9.12x10^(-6)

2.) I set the Kc I obtained from the first reaction mixture equal to the concentrations of the second reaction mixture and solved for [H20]...9.12x10^(-6) = [H20]^2 / [H2S]^2[SO2] 


Mastering Chemistry told me I was right even though it felt so wrong! How was I able to obtain Kc from initial concentrations? Why did my final answer turn out to be right? Maybe I misunderstanding the wording of the question itself.

Offline Mitch

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Re: Equilibrium constant K from initial concentrations?
« Reply #1 on: September 22, 2012, 09:17:24 PM »
I think you were right because the amount of water formed was so low that at equilibrium the reactants concentrations are effectively the same as they were initially. Makes sense?
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Offline Borek

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Re: Equilibrium constant K from initial concentrations?
« Reply #2 on: September 23, 2012, 04:04:30 AM »
0.510-1.1x10-3=0.5089

0.5089/0.510*100%=99.8%

Assuming 0.510 you are introducing error of about 0.2%. Not that much.

But technically - as all information is given - seems to me like you could calculate all concentrations from the reaction stoichiometry.
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