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chemistry - equilibria

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1. A(aq) + B(aq) ¡ê C(aq) + D(aq)
Initially, 1 mol of A and 1 mol of B are allowed to react and come to equilibrium. It is found hat there is 0.5 mol of A left at equilibrium. Calculate Kc for the reaction.

how do i work out Kc? when ive done questions b4 ive had volume/conc and the amount of products there are

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  • chemistry - equilibria - ,

    Starting mols:
    A = 1
    B = 1
    C = 0
    D = 0

    equilibrium mols:
    A = 0.5
    B =
    C =
    D =

    change in mols:
    A = -x
    B = -x
    C = +x
    D = +x
    Obviously, if we started with 1 mol A and ended up with 0.5 mols A, then the change (what we have labeled x) is 0.5.
    Thus change in B is -0.5, change in C is +0.5 and change in D is +0.5
    These are mols and not concns and there is no volume listed. Since the mols ratios of the reaction are 1:1:1:1, then we can assume a volume of 1 L which makes the concns and mols the same. Plug into Kc expression and solve for Kc. Check my thinking. I think you can assume ANY value for volume and obtain the same answer for Kc.

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