1. 7.3 The Equilibrium Constant: Measuring Equilibrium Concentrations Equilibrium Calculations (ICE Charts) Qualitatively Interpreting the Equilibrium Constant Meaning of Small Equilibrium Constant (pp.339-353) SCH4U – Grade 12 Chemistry, University Preparation Ms. Papaiconomou & Ms. Lorenowicz 1

2. QUIZ: Equilibrium Expressions Write the equilibrium expression for the following chemical equations: [__/6] 1.CO (g) + 1/2 O 2(g) CO 2(g) 2.2 HBr (g) H 2(g) + Br 2(g) 3.4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (g) 2

3. We learned in previous lessons that… LAW OF CHEMICAL EQUILIBRIUM: At equilibrium, there is a constant ratio between the concentration of reactants and products in any change. Equilibrium constant (K eq or K c ) is the ratio of the forward rate constant and reverse rate constant. k f = K eq krkr The equilibrium expression is aP + bQ cR+ dS so: K c = [R] c [S] d [P] a [Q] b Equilibrium constant is effected by temperature. 3

4. What if you don’t know the equilibrium constant? 4

5. Measuring Equilibrium Concentrations You can determine some equilibrium concentrations if you know: ▫the initial concentration(s) of the reactant(s) & ▫ the concentration of one product at equilibrium [which will tell you the change in concentration(s)]. Once you have the equilibrium concentrations, you can find the equilibrium constant!! Use an ICE table to perform these calculations: ▫ I = initial concentration ▫ C = change in concentration ▫ E = equilibrium concentration 5

6. Fe 3+ (aq) + SCN - (aq) Fe(SCN) 2+ (aq) Fe 3+ (aq) + SCN - (aq) are both colourless. Fe(SCN) 2+ (aq) is red. Because this reaction involves a colour change, you can determine the equilibrium concentration of Fe(SCN) 2+ (aq) by measuring the change in the intensity of colour (absorbance). Knowing the initial concentrations of Fe 3+ (aq) and SCN - (aq), you can calculate the equilibrium concentration of each ion using the chemical equation. Using the equilibrium concentrations, you can calculate the equilibrium constant!! colourlessred 6

7. Fe 3+ (aq) + SCN - (aq) Fe(SCN) 2+ (aq) Concentration (mol/L) Fe 3+ (aq) SCN - (aq) Fe(SCN) 2+ (aq) Initial0.00640.00100 Change-4.5 x 10 -4 4.5 x 10 -4 Equilibrium0.0065.5 x 10 -4 4.5 x 10 -4 K c = [Fe(SCN) 2+ ] = [4.5x10 -4 ] = 136.36 ≈ 140 [Fe 3+ ][SCN - ] [0.006][5.5x10 -4 ] 7

8. What if you don’t know the equilibrium concentrations? 8

9. Measuring Equilibrium Concentrations You can determine some equilibrium concentrations if you know: ▫the initial concentration(s) of the reactant(s) & ▫ the equilibrium constant. Using this information and an ICE table, you can calculate the equilibrium concentrations of all the chemical compounds. Use an ICE table to perform these calculations: ▫ I = initial concentration ▫ C = change in concentration ▫ E = equilibrium concentration 9

10. Sample Problem At 700 K, the equilibrium constant is 8.3. Suppose you start with 1.0 mol of CO (g) and 1.0 mol of H 2 O (g) in a 5.0 L container. What amount of each substance will be present in the container when the gases are at equilibrium at 700 K? 10

11. ReactantsProducts Concentration (mol/L) CO (g) H 2 O (g) H 2(g) CO 2(g) Initial 1/5 = 0.20 00 Change -x +x Equilibrium 0.20 - x xx 11

12. Doesn’t make sense 12

13. So… ReactantsProducts Concentration (mol/L) CO (g) H 2 O (g) H 2(g) CO 2(g) Initial 1/5 = 0.20 00 Change -0.15 +0.15 Equilibrium 0.05 0.15 So, in a 5.0 L container: Amount (n) at Equilibrium 0.05*5 = 0.25 mol 0.05*5 = 0.25 mol 0.15*5 = 0.75 mol 0.15*5 = 0.75 mol 13

14. The previous problem had perfect squares. It was easy to find the square root both sides and simplify for x. Many problems do not involve perfect squares. You will be required to use the quadratic equation to find the value of x. 14

15. Sample Problem The above reaction has an equilibrium constant of 25.0 at 1100 K. In a 1.00 L reaction vessel, 2.00 mol H 2(g) and 3.00 mol I 2(g) react. What is the equilibrium concentration of each gas? You need to find [H 2 ], [I 2 ], and [HI]. You need to ensure you have a balanced equation. Set up an ICE table, and calculate the initial concentrations of the gases. Let x equal all the change in the concentration of all chemical species. Write the equilibrium expression. Substitute and rearrange to solve for x. 15

16. ReactantsProducts Concentration (mol/L) H 2(g) I 2(g) HI (g) Initial 2.003.000 Change -x +2x Equilibrium 2.00 - x3.00 -x2x 16

17. Doesn’t make sense 17

18. ReactantsProducts Concentration (mol/L) H 2(g) I 2(g) HI (g) Initial 2.003.000 Change -1.7 +2(1.7) Equilibrium 2.00 – 1.7 = 0.3 3.00 -1.7 = 1.3 3.4 So… 18

19. So what does K mean?? 19

20. Why is the equilibrium constant important? K describes the extent of a reaction! A large K >1 means: [products] > [reactants] ▫  position of equilibrium lies to the right ▫  equilibrium favours the products A K c = 1 means: [products] = [reactants] A small K c <1 means: [products] < [reactants] ▫  position of equilibrium lies to the left ▫  equilibrium favours the reactants 20

21. 21 Video: F:\Courses\SCH4U1 - Chemistry Gr12 Univ\~ Resources ~\Nelson Chemistry 12 Textbook\Chem_12\Attac hments\d)_Animations\16 M05VD1.mov F:\Courses\SCH4U1 - Chemistry Gr12 Univ\~ Resources ~\Nelson Chemistry 12 Textbook\Chem_12\Attac hments\d)_Animations\16 M05VD2.mov

22. Sample Problem Consider the reaction of carbon monoxide and chlorine gas to make phosgene. At 870K, the value of K c is 0.20 and at 370 K, the value of K c is 4.6 x 10 7. Based only on the values of K c, is the production of COCl 2(g) more favourable at the higher or lower temperature? ▫K c = 0.20 at 870 K ▫K c = 4.6 x 10 7 at 370 K Is COCl 2(g) a product or reactant? Which K c favours COCl 2(g ) ? 22

23. So what does a really small K c mean?? When K c is really small compared to the initial concentration, the initial value minus x is approximately equal to the initial value, so you can ignore x! How do you know when you can do this? K c = 4.2 x 1o -8 & [NO] = 0.085 mol/L Divide the initial concentration by the value of K c. ▫If the answer is > 500, the approximation is justified. ▫If the answer is 100, it may be justified. ▫If the answer is < 100, you can’t ignore x and need to solve the equilibrium expression in full. 23

24. Homework Please re-read Section 7.3 (pp.339-353) and answer: ▫pp.347-348 Q.11-15 ▫pp.349-350 Q.16-20 ▫p.352 Q.21-25 ▫p.353 Q.2-5 Look ahead… ▫Quiz on Chapter 7 (7.1, 7.2, 7.3) coming up next period Check the website for homework/ downloadable files/ interesting YouTube chemistry videos: http://papaiconomou.weebly.com 24