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posted by  ss83 on 7/8/2008 8:51:00 PM  |  status: Live  

Equilibrium constant-will give lifesaver if correct!

Course Textbook Chapter Problem
General Chemistry N/A 14 N/A
Question Details:
The equilibrium constant for the reaction {\rm{SO}}_2 \left( g \right)\; + \;{\rm{NO}}_2 \left( g \right)\; \rightleftharpoons \;{\rm{SO}}_3 \left( g \right)\; + \;{\rm{NO}}\left( g \right) is 2.9.

Find the amount of {\rm{NO}}_2 that must be added to 2.2 mol of {\rm{SO}}_2 in order to form 1.3 mol of {\rm{SO}}_3 at equilibrium.

? \rm mol

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posted by Cdutch88 on 7/8/2008 10:00:13 PM  |  status: Live
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ss83's comment:
"Thank you very much! "
Response Details:
Since there are 1.3 mol of SO3 at equilibrium, this means that there must also be 1.3 mol of NO at equilibrium.

This also means that the amount of SO2 at equilibrium will have been reduced to 2.2 - 1.3 = 0.9 mol.
Similarly, the initial amount of NO2 will have been reduced by 1.3 mol as well.
Thus, assuming a volume of 1 L and letting the initial concentration of NO2 be x, we have

K = [SO3][NO] / [SO2][NO2] = (1.3)(1.3) / (0.9)(x - 1.3) = 2.9

Solving for x, we find that x = 1.9 mol (2 significant figures)

Hope that helps!
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