4. 5.6 x 10-6 mol of A and 5 x 10-5 mol of B are mixed in a 200 mL flask. The system is represented by the equation:

Question

4.	5.6 x 10-6 mol of A and 5 x 10-5 mol of B are mixed in a 200 mL flask. The system is represented by the equation:
  2A               +       B         -->            3C       
I   2.8x10-5 M      2.5x10-4                 0
C  -2x                   -x                        +3x
E  2.8x10-5-2x   2.5x10-4-x            3x10-5

B = 2.5x10-4-x = 2.4x10-4 
x = 1x10-5
A = 2.8x10-5-2(1x10-5) = 8x10-6
C = 3x10-5 + 3(1x10-5) = 6x10-5
(6x10-5)3/ (8x10-6)2(2.4x10-4) = 14.1

At equilibrium, there is 4.8 x 10-5 mol of B. Calculate the value of the equilibrium constant.

Where have I gone wrong?

DrBob222 · Answer

The problem tells you that A = 5.6E-5 mols and B = 5E-5 mol. You've changed both. Why do that? You didn't do it on the other problem.

Katie · Answer

I did a rough answer to this question with my tutor earlier today and I see what you're saying with the switching, I'm having issues understanding how to do this one properly.