Posted by Jeena on Monday, March 29, 2010 at 3:34pm.
SO2(g) + NO2(g) reverse reaction arrow SO3(g) + NO (g)
At a given temperature, analysis of an equilibrium mixture found [SO2] = 4.00 M, [NO2] = 0.500 M, [SO3] = 3.00 M, and [NO] = 2.00 M.
---How many moles/liter of NO2 would have to be added to the original equilibrium mixture to increase the equilibrium concentration of SO3 from 3.00 M to 4.10 M at the same temperature?
How would you set this up? I tried to set it up using (4.1)(3.1)/(2.9)(.5+x) = 3, but got it wrong...and also tried (4.1)(2.0+x)/(4.0-1.1)(.5+X)=3...and also got it wrong...
Chem Equilbirum - DrBob222, Monday, March 29, 2010 at 5:23pm
SO3 = 3.00
NO = 2.00
NO2 = 0.500
If you want SO3 to be 4.10, that means adding 1.10 to it. Therefore, NO will be 2.00 + 1.10 = 3.30 at new equilibrium.
SO2 = 4.00+x at new eq.
NO2 = 0.500+x at new eq.
Solve for x and don't forget to add the x amount to 4.00 and to 0.500.
Chem Equilbirum - Jeena, Monday, March 29, 2010 at 5:31pm
i thought you would have to decrease x from SO2?
Chem Equilbirum - DrBob222, Monday, March 29, 2010 at 6:29pm
I think you are right. Let me think about it and get back. I increased BOTH SO2 and NO2 and that isn't the question. I'll be gone about the next hour or so, then I'll get back.
Chem Equilbirum - DrBob222, Monday, March 29, 2010 at 7:25pm
OK. Ate supper, watched the news, thought about the problem and you are absolutely correct. I appreciate you bringing this to my attention. I just didn't think and I increased BOTH reactants. That isn't what the problem asked us to do.
So we make a minor change.
NO = 3.10
SO3 = 4.10
SO2 = 4.00-x
NO2 = 0.500+x
That way we are increasing NO2, which the problem asked us to do, and the SO3 is adjusting itself to whatever value it needs so that the increased amounts of NO2, SO2, and NO will be off-set and Kc still equals 3.00
I obtained x = 0.841 so the new equilibrium amounts should be
SO2 = 4.00-0.841
NO2 = 0.5 + 0.841
NO = 3.10
SO2 = 4.10
I checked Kc and it is 3.00 with these values.
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