1. Part 2: Very Small K Values 1

2.  Students will: 1) Determine the equilibrium concentrations of a chemical equilibrium reaction given the initial concentrations 2

3.  Students will: 1) Apply a problem solving methodology 2) Know if they need to determine the reaction quotient (Q) to solve the question 3) Apply appropriate algebraic skills to solve the problem 3

4. 1)Write the equation and state the K value 2)Determine reaction quotient, Q (if required) 3)Set up an ICE table a) enter initial concentrations b) determine changes in concentration 4)Write K equation 5)Solve for K by entering initial concentrations 6)Use the “hundred rule” to determine if the equation can be simplified, then solve for x 7)Find equilibrium concentrations 8)Check answer by plugging calculated equilibrium concentrations into K equation (values should match) 4

5.  Carbon monoxide is the starting material in the synthesis of many organic compounds, including methanol, CH 3 OH (l). At 2000℃, K is 6.40 x 10 -7 for the decomposition of carbon dioxide into carbon monoxide and oxygen gas. Calculate the concentration of all entities at equilibrium if 0.250 mol of CO 2(g) is placed into a 1.0 L closed container and heated to 2000℃. 5

6.  2 CO 2(g) ⇔ 2 CO (g) + O 2(g) K = 6.40 x 10 -7 6

7.  Since this reaction starts with only reactants, CO 2(g), the reaction must push forward. There is no need to determine Q, since we already know the direction the reaction will move to reach equilibrium. 7

8. a) enter initial concentrations b) determine changes in concentration 8 mol/L2 CO 2(g) ⇔2 CO (g) +O 2(g) I0.25000 C-2x+2x+x E0.250 – 2x2xx

9. K = [CO (g) ] 2 [O 2(g) ] = 6.40 x 10 -7 [CO 2(g) ] 2 9

10. K = [CO (g) ] 2 [O 2(g) ] = 6.40 x 10 -7 [CO 2(g) ] 2 [+2x] 2 [+x] = 6.40 x 10 -7 [0.250-2x] 2 4x 3 = 6.40 x 10 -7 [0.250-2x] 2 10 At this point, we can see that this is a cubic equation and solving for x will be very complicated. However, an assumption can be made → x must be very small if this reaction only proceeds slightly forward. In fact, subtracting 2x from 0.250 would be so small as to be insignificant!!! So, we can assume that x = 0 where subtraction is involved.

11. We have: 4x 3 = 6.40 x 10 -7 [0.250-2x] 2  Compare the concentration value to the K value  Divide:concentration ÷ K value  If the answer is greater than 100, you may assume that -2x is 0 because it is insignificant (i.e. 0.250 – 2x will still equal 0.250 with significant digits)  In the question: [CO (g) ]or 0.250 K 6.40 x 10 -7 = 3.90 x 10 5 We can use the Hundred Rule 11

12. 4x 3 = 6.40 x 10 -7 [0.250-2x] 2 4x 3 = 6.40 x 10 -7 [0.250] 2 4x 3 = 6.40 x 10 -7 (0.250) 2 x 3 = 6.40 x 10 -7 (0.0625) 4 x 3 = 1.00 x 10 -8 x = 2.15 x 10 -3 12

13.  @ equilibrium: [CO 2(g) ] = 0.250 – 2x = 0.250 – 2(2.15 x 10 -3 ) = 0.247 mol/L [CO (g) ] = 2x = 2(2.15 x 10 -3 ) = 4.30 x 10 -3 mol/L [O 2(g) ] = x = 2.15 x 10 -3 mol/L 13

14.  plug calculated equilibrium concentrations into the K equation (values should match) 14

15.  Remember, for the general chemical reaction aA + bB cC + dD 15