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Chem II - Equilibrium

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3.9 moles of NO and 0.88 moles of CO2 are allowed to react as
NO + CO2 <--> NO2 + CO
At equilibrium 0.11 moles of CO2 was present. Kc for the reaction is _____
0.27
1.7
3.14
6.03

Don't you need to know how many liters, to figure Kc don't you have to change everything to M, so don't you need liters, or do you just assume 1.0 L?

  • Chem II - Equilibrium - ,

    Assume any value for volume? 1.0 L is the easy way go go.

  • Chem II - Equilibrium - ,

    Still confused, I have two concentrations for CO2 and none for NO2. If I change everything I have to M, how can I sub into the Kc formula without knowing the conc. of NO2 or haveing Kc?

  • Chem II - Equilibrium - ,

    NO + CO2 ==> NO2 + CO

    CO2 was 0.88 to begin.
    It was 0.11 at the end. So how much was used? That must be 0.88 - 0.11 = 0.77
    Since all of the coefficients in the balanced equation are 1, then you must have used 0.77 of NO and must have formed 0.77 NO2 and 0.77 CO.
    Just another ICE CHART.
    Substitute the new values into the Kc expression and solve for Kc.

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