1. Chapter 15 Chemical Equilibrium -occurs when opposing reactions proceed at equal rates -no reactant or product is escaping -when at equilibrium, conc. of reactants and products no longer change with time -results in a reaction that will be reversible -reaction that can proceed in both directions (forward and reverse)

2. Ex: N2O4(g) ⇌ 2NO2(g) forward: N2O4(g)  2NO2(g) ratef= kf[N2O4] reverse: 2NO2(g)  N2O4(g) rater= kr[NO2]2 **page 613 figure 15.2 kf[N2O4] = kr[NO2]2 kf kr = [NO2]2 [N2O4] kf kr = equilibrium constant

3. Law of Mass Action aA + bB ⇌ cC + dD Kc = [C]c[D]d [A]a[B]b products reactants Kc= equilibrium constant with concentrations [ ] = concentration of reactant or product *Kc has no units

4. ex: N2(g) +3H2(g) ⇌ 2NH3(g) Kc = [NH3]2 [N2] [H2]3
ex: N2(g) +3H2(g) ⇌ 2NH3(g) Kc = [NH3]2 [N2] [H2]3 *Kc depends only on the stoichiometry of the reaction and the temp *does not depend on its mechanism *does not depend on initial amounts of reactants and products Sample Exercise 15.1 page 616

5. -when reactants and products are gases we use partial pressures Kp = (PcCPdD) (PaAPbB) *P in atm Kp = Kc(RT)∆n R = Latm/Kmol ∆n = c + d – (a +b) *coefficients from products and reactants (only of gases) *if ∆n = 0 then Kp = Kc *Sample Exercise 15.2 page 618

6. -if K is large (K >> 1), forward reaction is favored -more products than reactants when equilibrium is reached -if K is small (K <<1), reverse reaction is favored -more reactants than products when equilibrium is reached -if K ≈ 1 neither reaction is favored

7. Sample Exercise page 619
inc Kc
*more product
ii < i < iii
i) Kc = 0.60/0.40 = 1.5
ii) Kc = 0.1/0.90 = 0.11
iii) Kc = 0.80/0.20 = 4.0
*verifies order of a) in above
Practice Exercise
-favored more at lower temp b/c Kp is higher

8. if equation is reversed, invert the Kc
2) if the coefficients are multiplied by a factor, raise the Kc to the same factor
if you add two or more individual equations to obtain an overall equation, multiply the equilibrium constants by each other to obtain the overall Kc
Koverall = K1 x K2
**Koverall is the final equation given

9. Sample Exercise 15.4 and Practice page 620
Sample Exercise 15.5 page 622
HF(aq) ⇌ H+(aq) + F-(aq) Kc= 6.8 x 10-4
H2C2O4(aq) ⇌ 2H+(aq) + C2O42-(aq) Kc= 3.8 x 10-6
Find Kc for:
2HF(aq) + C2O42-(aq) ⇌ 2F-(aq) + H2C2O4(aq)
*multiply first by 2
*take reverse of second
Koverall = K1 x K2
Koverall = (K1)2 x 1 K2
(6.8 x 10-4)2 x x 10−6 = 0.12

10. Practice Exercise H2(g) + I2(g) ⇌ 2HI(g) Kp = 54
Practice Exercise H2(g) + I2(g) ⇌ 2HI(g) Kp = 54.0 N2(g) + 3H2(g) ⇌ 2NH3(g) Kp = 1.04 x 10-4 Find Kp for: 2NH3(g) + 3I2(g) ⇌ 6HI(g) + N2(g) *multiply first by 3 *reverse second Koverall = (54.0)3 x x 10−4 = 1.51 x 109

11. homogeneous equilibria- when substances are all in the same phase heterogeneous equilibria- when substances are in different phases -conc. of solids and liquids do not change b/c they do not expand to fill a container -these values are not included in the equilibrium expression Ex- CaCO3(s) ⇌ CaO(s) + CO2(g) Kc = [CO2] and Kp = PCO2 Sample Exercise page 623

12. [A] [B] Initial Change Equilibrium 1.00 -x =0.25 +2x =0.50 0.75 0.50
ICE (initial, change, equilibrium) tables *used if missing some info ex A(g) ⇌ 2B(g) A reaction at 445°C initially contains [A]= 1.00M. At equilibrium, the [A] is found to be 0.75M. Find Kc. Kc = [B]2 [A] Kc = [0.50M]2 [0.75M] = 0.33
[A]
[B]
Initial
Change
Equilibrium
1.00 -x
=0.25
+2x
=0.50
0.75
0.50

13. Predicting the Direction of Reaction Reaction Quotient (Qc or Qp) -does not have to be at equilibrium -will have different values as the reaction proceeds Qc = [C]c[D]d [A]a[B]b Qp = (PcCPdD) (PaAPbB) Q < K reaction favors products Q > K reaction favors reactants Q = K reaction is at equilibrium *page 628 figure 15.8

14. Le Chatelier’s Principle -when a system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance Effect of Concentration on Equilibrium -if [ ] reactants inc reaction shifts to right and vice versa -if [ ] products inc reaction shifts to left and vice versa *right favors products, left favors reactants

15. CaCO3(s) ⇌ CaO(s) + CO2(g) What is the effect of adding additional CO2 to the reaction mixture? *will shift to the left (favors reactants) What is the effect of adding additional CaCO3? *no effect b/c solid

16. Effect of Volume/Pressure on Equilibrium -if V is dec/P inc, reaction shifts to side with fewer moles of gas particles -if V is inc/P dec, reaction shifts to side with greater # of moles of gas particles -if equal # of moles and volume changes, no effect -adding inert gas to mixed has no effect

17. Effect of Temperature on Equilibrium In exothermic, heat is a product: -inc T causes shift to left dec K value -dec T causes shift to right  inc K value In endothermic, heat is a reactant: -inc T causes shift to right  inc K value -dec T causes shift to left  dec K value

18. Effect of Catalysts -catalysts lower Ea for both forward and reverse reaction -inc rate for both forward and reverse reaction -since K is ratio of forward and reverse rate constants, a catalyst does not change numeric value of K -catalysts only inc rate