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Consider the balanced equation 2A(g) + 3G(g) „\ 2X(g) + Z(g). To a 20.0 L container maintained at a temperature of 127 C were added 0.200 mol of A, 0.500 mol of G, 0.400 mol of X, and 0.600 mol of Z. After equilibrium was established, it was found that 0.340 mol of X was present. Calculate K for the reaction at 127 C.

  • chemistry -

    I assume the "funny" symbol between G and 2X is an arrow. So it should look like this.
    2A(g) + 3G(g) ==> 2X(g) + Z(g)

    mols X changed from 0.400 mol before equilibrium was established to 0.34 at equilibrium. Thus X changed by 0.400-0.340 = 0.060 mols. That means that 0.060 mol X was used (and obviously the point of equilibrium is to the left). Now watch carefully, I will do this for one and you follow through with the others. If 0.060 mol was used from X, then 1/2 that number or 0.030 mol Z was used making the final mols Z = 0.600 - 0.030 = 0.570 mol Z at equilibrium. Then you know that A must have increased by 0.030 and G must have increased by 0.030 x (3mols G/2 mols X) = ??
    Change all the mols to molarity (M = mols/20 L)and substitute into K = (X)^2(Z)/(A)^2(G)^3 = ?? to calculate Kc.

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