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Equilibrium constant questions

  1. Feb 21, 2008 #1
    1. The problem statement, all variables and given/known data

    The following reaction is at equilibrium at one atmosphere, in a closed container
    NaOH(s) + CO2(g)-> NaHCO3 (s)
    which, if any, of the following actions will decrease the total amount of CO2 gas present at equilibrium?

    1)decreasing the volume of the container
    2) none of the above
    3) removing half of the solid NaHCO3
    4)adding mlore solid NaOH
    5)adding N2 gas to double the pressure

    3. The attempt at a solution

    I chose (1) and got it wrong, so I chose (5) next but it was wrong again,
    I thought that solid is not included in equilibrium constant which means that
    it has little effect on equilibrium change, that's why I eliminated (3) and (4).
    I guess for, (1), decreasing the volume wouldn't do much
    to shift the equilibrium constant to the right.. then

    I don't understand why (5) is wrong because if you add more gas,
    it will double the pressure thus makeing the reactants to shift to the
    right to re-establish the equilibrium constant,,

    Is the answer none of the above?

    Please help!
    Last edited: Feb 21, 2008
  2. jcsd
  3. Feb 21, 2008 #2


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    5 is wrong because the equilibrium concentration of CO2 in the gas phase is constant regardless of how much pressure there is from N2. The concentration of CO2 is actually the partial pressure of CO2 in that case. Your thinking on #3 is logical but not #4. Remember that the reaction occurs on the surface of the sodium hydroxide solid. What would adding more NaOH surface do?
  4. Feb 21, 2008 #3
    When you add more solid NaOH,, more CO2 molecules will react with NaOH; thus, decreasing the moles of CO2.. Am I right,,?
    But still, does solid really have anything to do with this?

    I don't think I understand the reason why we don't include solid in equilibrium constant
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