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April 2, 2015

April 2, 2015

Posted by **alyx** on Sunday, September 14, 2008 at 4:08am.

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

- chemistry - equilibria -
**DrBob222**, Sunday, September 14, 2008 at 12:51pmStarting 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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