posted by Jeena .
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)(2.0+1.1)/(4.0-1.1)(.5+x) = 3, but got it wrong...and also tried (4.1-x)(2.0+x)/(4.0-x)(.5+X)=3...and also got it wrong...
Answered below. I can help find it if you can't.
So I found that you would have to set it up using (4.1)(3.1)/(2.9)(x-1.1). The x includes the .5 of NO2 already added, so to find how much was added to the original, find x and just subtract .5. The "x" value was the new initial value used to raise SO3 to 4.1 and that is why it is x-1.1.
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