Chp5_PExp5pt6

Chp5_PExp5pt6 - Practice Example 5.6 (A twist) Let us...

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Let us examine practice example 5.6, but from a different “angle.” The reaction is: 2 2 3 2 ) ( 2 O SO g SO (I) Let us suppose we were told that we start off with 12.5 atm of (pure) sulfur trioxide and, at equilibrium, we find (measure) that we have 2.5 atm of molecular oxygen . Now calculate K. We set up the rxn table as before: 2SO 3 = 2SO 2 + O 2 I 12.5 0 0 a +b +c F/Eq d e 2.5 This simply “puts into place” all of the given data; a-e denote entries that we do not have as given. Note that I have made the signs explicit in the delta line. (1) Now we take advantage of the fact that for each column, I + = F. Thus for the O 2 column: 0 + c = 2.5. Thus, c = 2.5 atm. (2) Once we have any single entry in the delta row, we have all the others by relative stoichiometry. For example: b/c = 2/1. Thus, b = 2.5*(2/1) = 5.0 atm . And: a/c = 2/1 so a = 5.0 atm . (3) We now obtain ‘d’ and ‘e’ by:
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Chp5_PExp5pt6 - Practice Example 5.6 (A twist) Let us...

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