1. Equilibrium Notes Mrs. Stoops Chemistry

2. Eqm day 1 Chapter problems p 660 – 665: 14, 16, 20, 28, 32, 38, 42, 46, 50, 52, 59, 61, 70,

3. Eqm Day 1 Equilibrium Opposing reactions (forward and reverse) are going at equal rates A + B  C + D Ex: Haber process – N 2 + 3H 2  2NH 3 There is a constant, K, for the process. K c is for concentration (M) & K p is pressure

4. Eqm Day 1 Finding K c Law of mass action –relationship between reactants and products concentrations Ex:aA + bB  pP + qQ K c = [P] p [Q] q Products [A] a [B] b Reactants K c only depends on the stoichiometry (coefficients) of the reaction. The mechanism does not matter. Does vary with temperature

5. Eqm Day 1 Finding K c Find K c for N 2 O 4  2 NO 2 K c = [NO 2 ] 2 [N 2 O 4 ] If [NO 2 ] = 0.0172M & [N 2 O 4 ] = 0.0014M, find the K c K c = [0.0172] 2 = 0.211 [0.0014] NO UNITS!!!!

6. Eqm Day 1 Rules FOR THE CONSTANT, this time No Units The constant does not change unless –The temperature changes –Reaction is different

7. Pressure For gases: K p does not always equal K c Use PV = nRT(R = 0.0821 for atm) K p = K c (RT)  n  n = change in moles = P – R N 2 O 4  2NO 2  n = 2-1= 1 K p = K c (RT) 1

8. Eqm Day 1 Magnitude of Eqm Constant Usually very large or very small Ex: CO + Cl 2  COCl 2 K c = 4.57 x 10 9 (big!) K c = [COCl 2 ]top bigger than [CO] [Cl 2 ]bottom number K  1; Products are favored; positive exponents K  1; reactants are favored; negative exponents

9. Eqm Day 1 Direction of Eqm and K N 2 O 4  2NO 2 K c = [NO 2 ] 2 = 0.211 [N 2 O 4 ] 2NO 2  N 2 O 4 K c ’ = [N 2 O 4 ] = 0.0014 = 4.72 [NO 2 ] 2 (0.0172) 2 The reverse reaction is 1 K c

10. Eqm Day 1 Heterogeneous Eqm Having all the same phases; they are homogeneous Different phases are called heterogeneous Ex:CaCO 3(s)  CaO (s) + CO 2(g) K c = [CaO] [CO 2 ] = [CO 2 ] = K c [CaCO 3 ] When a substance is pure solid or liquid, its concentration is constant. Exclude (not use) solids and liquids from the K expression. Use only gases and solutions (aq)

11. Eqm Day 1 Example of multiple states Ex: SnO 2(s) + 2CO (g)  Sn (s) + 2CO 2(g) K c = [CO 2 ] 2 [CO] 2 Note when combining steps (like Hess’s law) K total = K 1 x K 2 Using a multiple of the reaction, exponents if you need 2 x the reaction, you take K 2

12. Eqm Day 1 Homework Page 660: 13, 15, 17, 19, 21, 23, 25

13. Eqm Day 2 ICE Chart Use this chart to find values for the reactants and product in eqm expressions Ex: Mixture of 5.00 x 10 -3 mole H 2 and 1.00 x 10 -2 mol I 2 is placed in a 5.00 L container at 448  C. The concentration of HI is 1.87 x 10 -3 M at eqm. Calculate K c at 448  C. H 2 + I 2  2HI M I2 = 5.00x10 -3 /5.00 = 1.00x10 -3 M M H2 = 1.00x10 -2 /5.00 = 2.00x10 - 3 M H 2 + I2 I2  2HI I 1.00x10 - 3 M 2.00x10 - 3 M 0 C E 1.87x10 - 3 M -0.935x10 - 3 M 1.065x10 -3 M 0.065x10 - 3 M K c = [HI] 2 = (1.87x10 -3 ) 2 = 51 [H 2 ][I 2 ] (0.065x10 -3 )(1.065x10 -3 )

14. Eqm Day 2 Applications of K Predict the direction of the reaction (which side is favored) Calculate the concentrations Reaction Quotient = Q Q is found just like K, but use initial concentrations. Q = K; system is a t eqm (balanced) Q  K; form the reactants; right to left Q  K; form the products; left to right

15. Eqm Day 2 Homework Page 660-1: 27, 29, 31, 33

16. Eqm Day 3 Calculate Eqm Concentrations Ex: A 1 L flask has 1.00 mol H 2, 2.00 mol I 2 at 448  C. K c = 50.5 for H 2 + I 2  2HI. What is the concentrations at eqm? H 2 + I2 I2  2HI I 1.00 M2.00 M0 C E 2x + 2x- x 2.00 - x1.00 - x K c = 50.5 = [HI] 2 [H 2 ][I 2 ] 50.5 = [2x] 2 [1.00-x][2.00-x] 50.5 = 4x 2 2.00 - 3.00x + x 2 50.5(2.00 - 3.00x + x 2 ) = 4x 2 101.0 – 151.5x + 50.5x 2 = 4x 2 101.0 – 151.5x + 46.5x 2 = 0 QUADRATIC EQUATION

17. Eqm Day 3 Quadratic Equation 101.0 – 151.5x + 46.5x 2 = 0 H 2 + I2 I2  2HI I 1.00 M2.00 M0 C E + 2x 2x - x 2.00 - x1.00 - x abc x = -b +/- (b 2 – 4ac) 1/2 2a x = -(-151.5) +/- [(-151.5) 2 – 4(46.5)(101.5)] 1/2 2(46.5) x = 2.323 or 0.935 [H 2 ] = 1.00 – 0.935 = 0.065 M [I 2 ] = 2.00 – 0.935 = 1.065 M [HI] = 2.00 (0.935) = 1.87 M

18. Eqm Day 3 A Different Example At 90  C, 2H 2(g) + S 2(s)  2H 2 S (g) K c = 6.8 x 10 -6 If 2.0 mol of H 2 and 1.0 mol S 2 are heated in a 1L flask, what will be the concentration of H 2 S at eqm? 2H 2(g) + S 2(s)  2H 2 S (g) I 2 M0 C E +x- x 2 - x x 6.8 x10 -6 = [H 2 S] 2 = x 2 [H 2 ] 2 [S 2 ] [2 – x] 2 X 2 = (6.8 x10 -6 )(4) X 2 = 0.000272 X = 0.00522 Check assumption: 2-0.00522 = 1.99 (no affect), assumption passes Assume x = is much smaller than 2; ignore that x

19. Eqm Day 3 Homework Page 661-663: 37, 39, 41, 43, 49, 68

20. Eqm Day 4 Le Châtlelier’s Principle If a system at equilibrium is disturbed by a change in temperature, pressure, or concentration of one of its components, the system will shift its equilibrium position to counteract the effect.

21. Eqm Day 4 Le Châtlelier’s Principle Concentration effectEx: N 2(g) + 3H 2(g)  2NH 3(g) The shift will occur in the direction to minimize or reduce the effect. Adding a substance, the eqm will shift to consume Remove a substance, the eqm will shift to form more Ex: –add H 2 –Remove N 2 –Add NH 3 To the right To the left

22. Eqm Day 4 Le Châtlelier’s Principle Volume or PressureEx: N 2(g) + 3H 2(g)  2NH 3(g) Increase the pressure, (compress – volume decreases) reduce the total number of molecules Reduce the volume causes a shift in the direction that reduces the number of molecules of gas. Ex: –Increase P (reduce V) –Increase V To the right (less moles) To the left

23. Eqm Day 4 Le Châtlelier’s Principle Temperature Treat heat as a product or reactant based on  H –  H negative, product –  H positive, reactant When temp increases, eqm will shift to absorb it. Ex: N 2(g) + 3H 2(g) + heat  2NH 3(g) eqm will shift right

24. Eqm Day 4 Le Châtlelier’s Principle Catalyst N 2(g) + 3H 2(g)  2NH 3(g) Increase rate but does NOT affect the composition or equilibrium Pt

25. Eqm Day 4 Homework Page 662: 45, 51, 53 Packet 17.63, 17.68, 17.71, 17.76