Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: spKelpDiver on September 22, 2012, 08:13:59 PM
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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.
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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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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.