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Thread: An equilibrium problem

  1. #1 An equilibrium problem 
    New Member
    Join Date
    Oct 2006
    Chandigarh, India
    When we write the equation for the equilibrium constant and the activities of the reactants and products, we take the activity of solids equal to unity since their density is constant. But consider this scenario:

    I take an evacuated chamber, which is maintained at a constant temp. In this I place a chunk of dry ice. Now it's in equilirium with gaseous CO<sub>2</sub>. Now,
    K<sub>eq</sub>= [P<sub>CO2</sub>]
    Since K is dependent only on temp, pressure of carbon-di-oxide is fixed at a given temperature. At equilibrium, rate of condensation of the gaseous form is equal to the rate of sublimation of dry ice. Suppose I add another similar chunk of dry ice to the chamber. The forward reaction rate doubles; so to maintain equilibrium, the backward reaction rate too will double, which means that pressure of CO2 will correspondingly increase. But K is not dependent on the amount of the reactant, it cannot be equal to pressure, which means that there will have to be some term to take into account the activity of the solid carbon-dioxide, which will not be equal to unity.

    How right or wrong am I? Please help!

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  3. #2  
    Forum Sophomore
    Join Date
    Sep 2006
    Edmonton, AB, Canada
    The forward reaction rate doubles; so to maintain equilibrium, the backward reaction rate too
    Your error is in the statement there. The forward reaction rate doubles to reach equilibrium, and then once equilibrium is reached, they will be equal. At equilibrium, the rates will be different than before - but the relationship between them will remain the same.[/quote]

    It is not so much that I have confidence in scientists being right, but that I have so much in nonscientists being wrong. --- Isaac Asimov
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