chemistry  equilibria
posted by alyx .
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
xx

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.