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?
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.
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.
Based on the given information, the equation for the reaction is:
2A + B --> 3C
You correctly determined the equilibrium concentrations of A, B, and C as follows:
A = 2.8 x 10-5 M - 2(1 x 10-5 M) = 8 x 10-6 M
B = 2.5 x 10-4 M - x
C = 3 x 10-5 M + 3(1 x 10-5 M) = 6 x 10-5 M
To find the equilibrium constant, you correctly set up the expression:
Kc = (6 x 10-5 M)^3 / (8 x 10-6 M)^2 (2.4 x 10-4 M)
However, you made a mistake when calculating the values of A, B, and C. Let's re-evaluate them:
Given that the initial moles of A is 5.6 x 10-6 mol, and B is 5 x 10-5 mol, and the total volume is 200 mL:
Convert the volumes to moles:
Volume of A: (5.6 x 10-6 mol) / (200 mL) = 2.8 x 10-8 M
Volume of B: (5 x 10-5 mol) / (200 mL) = 2.5 x 10-4 M
Now, we can apply these new values to calculate A, B, and C:
A = 2.8 x 10-8 M - 2(1 x 10-5 M)
B = 2.5 x 10-4 M - x
C = 3 x 10-5 M + 3(1 x 10-5 M)
With these corrected values, you can then proceed to calculate the equilibrium constant Kc using the expression mentioned above.