Chemistry

SO2(g) + NO2(g) 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.
(a) What is the new equilibrium concentration of NO when 1.50 moles of NO2 are added to the equilibrium mixture? Assume a 1.00 L container.
[NO] = 2.84M <<answer
(b) 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?
[NO2] =?????????????/

i need help on part b how do u do this? thanks!!

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  1. K = [SO3][NO]/[SO2][NO2]
    K = (3.0)(2.0)/(4.0)(0.5)= 3.0
    (a) Some of the added NO2 will be used up causing a shift to the right. Let that be equal to x.
    At the new equilibrium,
    [SO2]=4.0-x, [NO2]=0.5+1.5-x=4.0-x
    [NO]=2.0+x, [SO3]=3.0+x,
    Set up:
    K = [3.0+x][2.0+x]/[4.0-x][2.0-x]
    Simplify and solve for x
    [NO]=2.0+x

    (b)
    Let the moles of NO2 added = x
    Set up:
    3.0 = [3.0+1.1][2.0+1.1]/[4.0-1.1][0.5+x]
    Solve for x

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  2. I tried to answer the hwk using this tip for part B, but I didn't get it right...is there another way to set it up?

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  3. SO2(g) + NO2(g) SO3(g) + NO (g)

    At a given temperature, analysis of an equilibrium mixture found [SO2] = 0.300M, [NO2] = 0.100 M, [SO3] = 0.600 M, and [NO] = 2.00 M. At the same temp., extra SO2 was added to make SO2 is added to make [SO2}= 0.800M. Calculate the composition of the mixture when equilibrium has been reestablished

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  4. K = [SO3] [NO] / [SO2] [NO2]
    K = (0.6)(2.0) / (0.3)(0.1) = 40
    At the new equilibrium,
    [SO2] = 0.3 + [SO2]add - x = 0.8 M ;
    [NO2] = 0.1-x ; [NO] = 2.0+x ; [SO3] = 0.6+x
    K = [0.6+x] [2.0+x] / ( [0.8] [0.1-x] ) = 40
    x = 0.0577 M
    [SO2] = 0.8 M ; [NO2]=0.1-x = 0.0423 M ;
    [SO3] = 0.6+x = 0.6577 M ; [NO] = 2.0+x = 2.0577 M

    Verif. → (0.6577*2.0577) / (0.8*0.0423) = 39.993

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