Presentation is loading. Please wait.

Presentation is loading. Please wait.

Equilibrium Chapter 15. At room temperature colorless N 2 O 4 decomposes to brown NO 2. N 2 O 4 (g)  2NO 2 (g) (colorless) (brown)

Similar presentations


Presentation on theme: "Equilibrium Chapter 15. At room temperature colorless N 2 O 4 decomposes to brown NO 2. N 2 O 4 (g)  2NO 2 (g) (colorless) (brown)"— Presentation transcript:

1 Equilibrium Chapter 15

2 At room temperature colorless N 2 O 4 decomposes to brown NO 2. N 2 O 4 (g)  2NO 2 (g) (colorless) (brown)

3 We expect the forward reaction rate to slow over time and the rate of the reverse reaction rate to increase over time. At some time, the color stops changing which indicates and we have an equilibrium mixture of N 2 O 4 and NO 2. Chemical equilibrium is the point at which the concentrations (partial pressures of gases) of all species are constant.

4 A dynamic equilibrium exists when the rates of the forward and reverse reactions are equal. –The reaction does NOT stop, but there is no NET change in the reactant or product concentrations. Use a double arrow (  ) to imply that the process is an equilibrium.

5 Haber process: Industrial preparation of ammonia from nitrogen and hydrogen: N 2 (g) + 3H 2 (g)  2NH 3 (g) Process is carried out at high temperature (500 o C) and pressure (200 atm). Much of the NH 3 produced industrially is used as a fertilizer.

6 N 2 (g) + 3H 2 (g)  2NH 3 (g) If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach equilibrium with constant concentrations of nitrogen, hydrogen and ammonia. If we start with just ammonia, the reaction will reach equilibrium the same equilibrium with constant concentrations of nitrogen, hydrogen and ammonia. No matter what the starting composition of reactants and products is, the equilibrium mixture contains the same relative concentrations of reactants and products. –Equilibrium can be reached from either direction.

7 Consider a simple reaction. A(g)  B(g) Assume that it is an elementary process. We can write rate expressions for each reaction. Forward reaction: A  B Rate = k f [A] k f = rate constant (forward reaction) Reverse reaction:B  A Rate = k r [B]k r = rate constant (reverse reaction) In equilibrium the forward and reverse rates are equal Rearranging:k f [A]= k r [B] This is called the Equilibrium constant! “c” because it’s based on concentrations

8 For gaseous substances we can use the ideal gas equation to convert between concentration and pressure: PV = nRT so M = (n/V) = (P/RT) For substances A and B: [A] = (P A /RT) and [B] = (P B /RT) Rate fwd = k f P A /RT and Rate rev = k r P B /RT At equilibrium:k f P A /RT = k r P B /RT Rearranging, we get: This is called the Equilibrium constant! “p” because it’s based on pressure

9 15.2 The Equilibrium Constant We can write an expression for the relationship between the concentration of the reactants and products at equilibrium. For a general reaction: aA + bB  cC + dD The equilibrium expression (law of mass action) is given by: *equilibrium concentration/pressure of (g) and (aq) only! Where Kc and Kp are equilibrium constants (K eq ). * P in atm

10 Converting between Kp and Kc K p =K c (RT) Δn Δn = (sum of coefficients of gaseous product) – (sum of coefficients of gaseous reactant) R is the gas constant (0.0821) T must be Kelvin When Δn is zero  K p =K c

11 K eq depends on stoichiometry. K eq varies with temperature. K eq does NOT depend on initial concentrations (pressures) K eq does NOT depend on the reaction mechanism. We generally omit the units of K eq

12 The Magnitude of Equilibrium Constants The equilibrium constant, K eq, is the ratio of products to reactants. –the larger K eq the more products are present at equilibrium. –the smaller K eq the more reactants are present at equilibrium. If K eq >> 1, then products dominate at equilibrium and “equilibrium lies to the right” If K eq << 1, then reactants dominate at equilibrium and “equilibrium lies to the left”

13

14 The Direction of the Chemical Equation and K An equilibrium can be approached from any direction. Consider the reaction: N 2 O 4 (g)  2NO 2 (g) The equilibrium constant (at 100 o C) is: However, when we write the equilibrium expression for the reverse reaction: 2NO 2 (g)  N 2 O 4 (g) The equilibrium constant (at 100 o C) is: The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction. A + B  2 C K 2 C  A + B 1/K

15 The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number. A + B  2 C K 2 A + 2 B  4 C K 2 The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps. A + B  2 C K 1 + C + A  D K 2 2 A + B  C + D K 1  K 2

16 Homogeneous equilibria - Equilibria in which all reactants and products are present in the same phase Heterogeneous equilibria - Equilibria in which one or more reactants or products are present in a different phase

17 Heterogeneous equilibria

18 Since the concentration of a pure solid or pure liquid is not included in the equilibrium constant expression. We anticipate that the amount of CO 2 formed will not depend on the amounts of CaO and CaCO 3 present. *As long as CaO and CaCO 3 are both present At equilibrium!

19 Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established!

20 #1 Suppose that the gas phase reactions A → B and B → A are both elementary processes with rate constants of 3.8  s -1 and 3.1  s -1 respectively. (a) What is the value of the equilibrium constant for the equilibrium A  B? (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B? (a) (b)[A] Kc is less than 1 so reactants are favored

21 #2Write the expression for K c for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) 3 NO(g)  N 2 O(g) + NO 2 (g) (b) Ni(CO) 4 (g)  Ni(s) + 4 CO(g) (c) HF(aq)  H +1 (aq) + F -1 (aq) (d) 2 Ag(s) + Zn +2 (aq)  2 Ag +1 (aq) + Zn(s) (a) and (c) are homogeneous

22 #3When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? (a) N 2 + O 2  2 NO K c = 1.5  (b) 2 SO 2 + O 2  2 SO 3 K p = 2.5  10 9 (a)mostly reactants (K c is much less than 1) (b)mostly products (K p is much greater than 1)

23 #4If K c = for PCl 3 (g) + Cl 2 (g)  PCl 5 (g) at 500 K, what is the value of K p for this reaction at this temperature?

24 #5 At 1000 K, K p = 1.85 for the reaction: SO 2 (g) + ½ O 2 (g)  SO 3 (g) (a) What is the value of K p for the reaction SO 3 (g)  SO 2 (g) + ½ O 2 (g)? (b) What is the value of K p for the reaction 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)? (a)1 / 1.85 = (b)(1.85) 2 = 3.42

25 #6Mercury (I) oxide decomposes into mercury and oxygen: 2 Hg 2 O(s)  4 Hg(l) + O 2 (g) (a) Write the equilibrium constant expression for this reaction in terms of partial pressures (b) Explain why we normally exclude pure solids and liquids from equilibrium constant expressions. (a) (b) the concentration/pressure is constant for pure solids and liquids

26 #7Gaseous hydrogen iodide is placed in a closed container at 425 o C, where it partially decomposes to hydrogen and iodine: 2 HI(g)  H 2 (g) + I 2 (g) At equilibrium, it is found that [HI] is 3.53  M, [H 2 ] is 4.79  M, and [I 2 ] is 4.79  M. What is the value of K c at this temperature?

27 #8The equilibrium: 2 NO(g) + Cl 2 (g)  2 NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures of atm, atm, and 0.28 atm for NO, Cl 2, and NOCl respectively. Calculate K p for this reaction at 500 K.

28 Calculating Equilibrium Constants Tabulate initial and equilibrium concentrations (or partial pressures) for all species in the equilibrium. If an initial and an equilibrium concentration is given for a species, calculate the change in concentration. Use the coefficients in the balanced chemical equation to calculate the changes in concentration of all species. Deduce the equilibrium concentrations of all species. Use these equilibrium concentrations to calculate the value of the equilibrium constant.

29 #9A mixture of 0.10 mol of NO and mol of H 2, and 0.10 mol of H 2 O is placed in a 1.0 L vessel at 300 K. The following equilibrium is established: 2 NO(g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O(g) At equilibrium, [NO] = M. (a) Calculate the equilibrium concentrations of H 2, N 2, and H 2 O (b)Calculate K c (a)calculate the change in [NO], 0.10 – = NO + 2 H 2  N 2 +2 H 2 O initial change equil (b)

30 #10A mixture of mol CO 2, mol H 2, and mol H 2 O is placed in a L vessel. The following equilibrium is established at 500 K: CO 2 (g) + H 2 (g)  CO(g) + H 2 O(g) (a) Calculate the initial partial pressures of CO 2, H 2, and H 2 O (b) At equilibrium, P H2O = 3.51 atm. Calculate the equilibrium partial pressures of CO 2, H 2, and CO (c) Calculate K p for the reaction (a) (b)the change in pressure of H 2 O = 3.51 – 3.28 = 0.23 atm CO 2 + H 2  CO+ H 2 O initial change equil (c)


Download ppt "Equilibrium Chapter 15. At room temperature colorless N 2 O 4 decomposes to brown NO 2. N 2 O 4 (g)  2NO 2 (g) (colorless) (brown)"

Similar presentations


Ads by Google