The Equilibrium State

1. N2O4(g) 2NO2(g) The Equilibrium State Chemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time. Colorless Brown

2. Chemical Equilibrium • No longer “How fast”? But How FAR? • Today—introduce Kc, Equilibrium in general. • Derive Kc Expressions • Simple Calculations • Reverse reactions • Kp expressions (first of MANY K expressions) • Kc values reported without units!!

3. The Equilibrium Constant Kc Why?

4. aA + bB N2O4(g) cC + dD 2NO2(g) [C]c[D]d [NO2]2 [A]a[B]b [N2O4] The Equilibrium Constant Kc For a general reversible reaction: Products Equilibrium equation: Kc = Reactants Equilibrium constant Equilibrium constant expression For the following reaction: Kc = = 4.64 x 10-3 (at 25 °C)

5. (0.0125)2 (0.0141)2 [NO2]2 0.0337 0.0429 [N2O4] The Equilibrium Constant Kc Experiment 1 Experiment 5 Kc = = 4.64 x 10-3 = 4.63 x 10-3

6. 2 P NO2 N2O4(g) 2NO2(g) P N2O4 The Equilibrium Constant Kp P is the partial pressure of that component Kp =

7. If the equilibrium concentrations of COCl2 and Cl2 are the same at 395 °C, find the equilibrium concentration of CO in the reaction: CO(g) + Cl2(g) COCl2(g) Kc = 1.2 x 103 at 395 °C

8. 2 P NO2 N2O4(g) 2NO2(g) P N2O4 The Equilibrium Constant Kp P is the partial pressure of that component Kp =

9. L atm 0.082058 K mol The Equilibrium Constant Kp Kp = Kc(RT)Dn R is the gas constant, T is the absolute temperature (Kelvin). n is the number of moles of gaseous products minus the number of moles of gaseous reactants.

10. Consider the reaction: 2 NO(g) + O2(g) 2 NO2(g) Now consider the reaction: 2 NO2(g) 2 NO(g) + O2(g) Modifying the Chemical Equation [NO2]2 Kc = ––––––––– = 4.67 x 1013 (at 298 K) [NO]2 [O2] What will be the equilibrium constant K'cfor the new reaction? [NO]2 [O2] 1 K'c = ––––––––– = ––––––––––– = [NO2]2 [NO2]2 ––––––––– [NO]2 [O2] 1 1 –– = ––––––––– = 2.14 x 10–14 Kc 4.67 x 1013

11. Modifying the Chemical Equation (cont’d) • For the reverse reaction, K is the reciprocal of K for the forward reaction. • When an equation is divided by two, K for the new reaction is the square root of K for the original reaction. • It should be clear that we must write a balanced chemical equation when citing a value for Kc.

12. The equilibrium constant for the reaction: ½ H2(g) + ½ I2(g) HI(g) at 718 K is 7.07. (a) What is the value of Kc at 718 K for the reaction HI(g) ½ H2(g) + ½ I2(g) (b) What is the value of Kc at 718 K for the reaction H2(g) + I2(g) 2 HI(g)

13. Example: CaCO3(s) CaO(s) + CO2(g) Equilibria Involving PureSolids and Liquids • The equilibrium constant expression does not include terms for pure solid and liquid phases because their concentrations do not change in a reaction. • Although the amounts of pure solid and liquid phases change during a reaction, these phases remain pure and their concentrations do not change. [CaO] [CO2] Kc = –––––––––– [CaCO3] Kc = [CO2]

14. The reaction of steam and coke (a form of carbon) produces a mixture of carbon monoxide and hydrogen, called water-gas. This reaction has long been used to make combustible gases from coal: C(s) + H2O(g)CO(g) + H2(g) Write the equilibrium constant expression for Kc for this reaction.

15. Equilibrium Constants: When Do We Need Them and When Do We Not? • A very large numerical value of Kc or Kp signifies that a reaction goes (essentially) to completion. • A very small numerical value of Kc or Kp signifies that the forward reaction, as written, occurs only to a slight extent. • An equilibrium constant expression applies only to a reversible reaction at equilibrium. • Although a reaction may be thermodynamically favored, it may be kinetically controlled … • Thermodynamics tells us “it’s possible (or not)” • Kinetics tells us “it’s practical (or not)”

16. Is the reaction CaO(s) + CO2(g)CaCO3(s) likely to occur to any appreciable extent at 298 K?

17. The Reaction Quotient, Q • For nonequilibrium conditions, the expression having the same form as Kc or Kp is called the reaction quotient, Qc or Qp. • The reaction quotient is not constant for a reaction, but is useful for predicting the direction in which a net change must occur to establish equilibrium. • To determine the direction of net change, we compare the magnitude of Qc to that of Kc.

18. When Q is smaller than K, the denominator of Q is too big; we have “too much reactants.” When Q =K, equilibrium has been reached. When Q is larger than K, the numerator of Q is too big; we have “too much products.” The Reaction Quotient, Q

19. Le Châtelier’s Principle • When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. • Analogy: Begin with 100 men and 100 women at a dance. • Assume that there are 70 couples dancing, though not always the same couples (dynamic equilibrium). • If 30 more men arrive, what happens? • The equilibrium will shift, and shortly, more couples will be dancing … but probably not 30 more couples.

20. Changing the Amounts of Reacting Species • At equilibrium, Q = Kc. • If the concentration of one of the reactants is increased, the denominator of the reaction quotient increases. • Q is now less thanKc. • This condition is only temporary, however, because the concentrations of all species must change in such a way so as to make Q = Kc again. • In order to do this, the concentrations of the productsincrease; the equilibrium is shifted to the right.