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    HELPP.
    As the volume increases the pressure decreases so the equilibrium position will shift to the side where more moles are present that is forward.
    So obviously the concentrations of products will increase and reactants decrease.
    Arent A and B both right?

    THE CORRECT ANSWER IS B
    Why???
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    (Original post by Insia786)
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    HELPP.
    As the volume increases the pressure decreases so the equilibrium position will shift to the side where more moles are present that is forward.
    So obviously the concentrations of products will increase and reactants decrease.
    Arent A and B both right?

    THE CORRECT ANSWER IS B
    Why???
    You are correct in saying that the equilibrium will shift to the RHS (side of fewer moles of gas) and this will give the RELATIVE amounts of gas with more of the RHS.

    BUT the new equilibrium is in 2 litres so the actual concentration of gases in moles per litre will be lower.

    If you can't see this try putting in some example numbers.

    Imagine that the concentrations in experiment 1 are all 2 mol dm-3

    This gives an equilibrium constant of (2 x 2)/2 = 2

    Now put those moles into the new flask. The new quotient in terms of concentrations = (1 x 1)/1 , which clearly does not equal the equilibrium constant (which must remain the same as the temperature is the same), so the reaction moves to the RHS making more of the products and less of the reactants until the quotient = the equilibrium constant.

    To get the quotient equal 2 you just need to add a little to each term on the top (the products) and take a little of the term at the bottom (the reactant).

    (1 + n)(1 + n)/(1 - n) = 2

    By smple inspection you can see that the concentration cannot ever get as high as 2 (as in the first experiment)

    n = approx 0.25
 
 
 
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