llbm Calculations Day 1 - a revisit

The lesson was all about calculating llbm concentrations.
(Finding 'x')

The handouts from the day's lesson are up on moodle (llbm calcn's page 1 & page 2)

If you visit moodle, you may also see the homework solutions to the Le Chatelier stuff [pages 457 - 460]

From the day's lesson - homework:
page 466(3,4)  page 472 (5,6)  page480(10) and page 481(3abc)

Most (if not all) of the questions require ICE charts.

Consider the generic reaction A (g)  +  B (g)  <=> 2C (g)  Kc = 37.9
A typical question may be: Determine the llbm concentration of C if 1.80 mol of A and 1.80 mol of B are placed in a 2.00L container and allowed to reach llbm.
 - complete an Initial Change Equilibrium chart and write the llbm law (K= expression)
 - when completing the chart, express values as concentrations
 - since no amount of product is mentioned, it will = 0
 - once you know one change, use the mole ratios of the balanced chemical equation to determine all other changes.
       A      +       B    <=>   2C
I    0.900M    0.900M        0
C    -X              -X           +2X
E  0.900-X   0.900-X        2X

Kc = [C]2 / [A][B] = 37.9 = (2X)2 /(0.900-X)(0.900-X)  
- notice that both numerator and denominator involve 'perfect squares'
- square root both sides

6.156 = 2X / (0.900-X)
5.540 - 6.156X = 2X
5.540 = 8.156X
0.679 = X

Do not leave the answer at 'X',
the question asked for [C]llbm = 2X = 1.36M

Demos were today- :<)

Have you submitted the clock reaction lab yet?

Finally, something (sort of) in honour of the 40th anniversary of Sesame Street. A monster of a cookie joke. Why did the cookie go to the doctor? [Ans: It was feeling crumby.]