Chapter 14: Chemical Equilibrium. Mrs. Brayfield

Transcription

1 Chapter 14: Chemical Equilibrium Mrs. Brayfield

2 14.2: Dynamic Equilibrium Remember from chapter 13 that reaction rates generally increase with increasing concentration of the reactions and decreases with decreasing reaction concentrations Consider: H 2 g + I 2 (g) 2HI(g) In this reaction, the reaction can proceed in both directions, this is said to be reversible So the concentrations of all species are constantly changing in either direction They are all increasing and decreasing

3 Dynamic Equilibrium Dynamic equilibrium is the condition when the rate of the forward reaction is the same at as the rate of the reverse reaction This means that the concentrations of each species (reactants and products) are no longer changing At equilibrium, the concentrations of all species are NOT equal, they just do not change (are constant) How many products a reaction produces depends on the particular reaction (more on this later)

4 Dynamic Equilibrium

5 14.3: The Equilibrium Constant (K) Look at the general chemical equation: aa + bb cc + dd The equilibrium constant (K) is defined as the ratio at equilibrium of the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients What does this mean? K = [C]c [D] d [A] a [B] b This relationship is known as the law of mass action

6 Equilibrium Constant For example, write the equilibrium constant for the following reactions: 2H 2 g + O 2 (g) 2H 2 O(g) K = C 3 H 8 g + 5O 2 g 3CO 2 g + 4H 2 O g K =

7 Equilibrium Constant Now, what does the equilibrium constant tell us? Well, it tells us how far a reaction will proceed So a large K tells us that the equilibrium point for a certain reaction is far to the right Or there will be high amounts of products and low amounts of reactants And a small K tells us that the equilibrium point for a certain reaction is far to the left Or there will be low amounts of products and high amounts of reactants

8 Equilibrium Constant Which side does equilibrium lie for the following reactions: N 2 g + O 2 g 2NO g K = (at 25 ) H 2 g + Br 2 g 2HBr g K = (at 25 )

9 Equilibrium Constant So for summary: K 1, the reverse reaction is favored (lots of reactants) K 1, neither direction is favored K 1, the forward reaction is favored (lots of products) Homework Problems: #1, 2, 3, 5

10 14.4: The Equilibrium Constant in Terms of Pressure We can relate K c (the equilibrium constant) in terms of pressure with the equation: K p = K c (RT) n Where: K p is the equilibrium constant in terms of pressure K c is the equilibrium constant in terms of concentration R is the ideal gas constant T is the temperature (in K) n is the change in the number moles (product reactant) Note that when the coefficients for the reactants equal the coefficients for the products, K p = K c

11 Equilibrium Constant in Terms of Pressure Using the following reaction: 2SO 3 g 2SO 2 g + O 2 (g) We can also write K p : K p = (P SO 2 ) 2 (P O2 ) (P SO3 ) 2 Where we use the partial pressures of the gases in the chemical equation

12 Pressure Equilibrium Example Consider the following reaction and corresponding value of K c : H 2 g + I 2 g 2HI g K c = (at 25 ) What is the value of K p at this temperature? K p = K c (RT) n Instead of just plugging and chugging, we know that n is 0, so K p = K c =

13 14.5: Heterogeneous Equilibria If we had the following chemical equation: 2CO g CO 2 g + C(s) And were asked for the equilibrium constant, we would write: K c = CO 2 [C] [CO] 2 But this is WRONG! This is because the concentration of a solid does not change no matter what amount is present

14 Heterogeneous Equilibria The reason is this: Solids do not take the shape of their container, so their concentration (no matter how much is there) does not change So any pure solid in the reactants OR products is not included in the equilibrium expression So we would write the equilibrium expression as: K c = CO 2 [CO] 2 Note: we also omit pure liquids (like water)

15 Heterogeneous Equilibria Example Write the equilibrium expression for the following reaction: 4HCl g + O 2 g 2H 2 O l + 2Cl 2 g K c = Homework Problems: #11, 12, 13, 14

16 14.6: Calculating Equilibrium Constant from Measured Equilibrium Concentrations If given any set of equilibrium concentration data, you can just plug and chug into the equilibrium constant equation For example if given: H 2 = 0.11M, I 2 = 0.11M, and HI = 0.78M Calculate the value of K c : K c = Notice that K is unit-less even though the expression has units of M for each concentration

17 Calculating the Equilibrium Constant Regardless of how much material you are given to start with, you should ALWAYS get the right equilibrium constant with the equilibrium concentrations If you don t believe me see table 14.1 on page 526 Now, most math problems here do not give you concentration values you must calculate those yourself We do this using BCA (or ICE) tables BCA: Before, Change, After ICE: Initial, Change, Equilibrium Your pick

18 Calculating Example #1 Consider the reaction: CO g + 2H 2 (g) CH 3 OH(g) If the initial concentrations of CO = 0.27M and H 2 = 0.49M. At equilibrium the concentration of CH 3 OH was 0.11M. Find K c.

19 Calculating Example #2 Consider the reaction: 2CH 4 g 3H 2 g + C 2 H 2 (g) If the initial concentrations of CH 4 = 0.087M. At equilibrium the concentration of H 2 was 0.012M. Find K c.

20 Homework Problems: #15, 16, 18, 20, 22, 24

21 14.7: The Reaction Quotient If we have a reaction that contains both reactants and products and it is not at equilibrium, how do we predict the direction it will go? We do this by calculating the reaction quotient The reaction quotient (Q c ) looks the same as K The only difference between Q and K is that K is only for a particular temperature while Q depends on the current state of the reaction

22 The Reaction Quotient

23 The Reaction Quotient For summary: Q < K, the reaction goes to the right (towards products) Q = K, the reaction is at equilibrium Q > K, the reaction goes to the left (towards reactants)

24 The Reaction Quotient Example Consider the following reaction and its equilibrium constant: N 2 O 4 g 2NO 2 g K c = A reaction mixture contains NO 2 = M and N 2 O 4 = M. Calculate Q and determine the direction in which the reaction will proceed. Q

25 Homework Problems: #26, 28

26 14.8: Finding Equilibrium Concentrations Consider the following reaction: I 2 g 2I g K = at 1200 In an equilibrium mixture, the concentration of I 2 is 0.10M. What is the equilibrium concentration of I? K

27 Finding Equilibrium Concentrations Sometimes we are not given the equilibrium concentrations and are only given the initial concentrations and the equilibrium constant We can still find the equilibrium concentrations, it is just a little harder Prepare your calculators with the quadratic formula solver If not already done so

28 Finding Equilibrium Concentrations If we do not know the equilibrium concentrations, we have to set up our BCA (or ICE) table like so: If we were given the reaction A(g) 2B(g) And say the initial concentration of A was 1.0M and there was none of B. The equilibrium constant is [A] [B] I 1.0M 0.0M C -x +2x E 1.0-x 2x K = [B]2 [A] = (2x)2 1.0 x = = 4x2 1.0 x so we have to solve a quadratic equation to solve for x

29 Finding Equilibrium Concentrations Example 1 For the reaction: N 2 O 4 g 2NO 2 g K = 0.36 at 100 The reaction mixture only contains N 2 O 4 = M, and no products. Find the equilibrium concentrations of all species.

30 Finding Equilibrium Concentrations Example 2 For the reaction: N 2 g + O 2 g 2NO g K = The reaction mixture started only with NO = M, and no reactants. Find the equilibrium concentrations of all species.

31 Finding Equilibrium Pressures We have another equilibrium constant expression, K p Unlike K c, K p only deals with partial pressures These pressures are in atm K p is unitless (just like K c ) and we can do the same math as well

32 Finding Equilibrium Pressures Example For the reaction: I 2 g + Cl 2 (g) 2ICl(g) We know the initial partial pressures and want to know the equilibrium partial pressure of all species. K p = 81.9.

33 Approximations in Calculations Sometimes our equilibrium constant is really small This means that our reaction will not go very far We can approximate our calculations in this scenario: A 2B K = [A] [B] I 1.0 M 0.0 M C - x + x E 1.0 x x K = = [B]2 [A] = (2x)2 1.0 x

34 Approximations in Calculations Since we know that there will not be many products made (small K), we can approximate the bottom to be 1.0. Solve for x (=0.0029) To see if our approximation is valid, we look at the ratio of x to the number it is subtracted from 1.0: % = 0.29% 1.0 If this number is less than 5.0%, we can safely say our assumption is valid If it is larger than 5.0%, then we cannot make the assumption

35 Approximations in Calculations Example 1 For the following reaction: 2H 2 S g 2H 2 g + S 2 (g) We have initial concentrations H 2 S = M, H 2 = 0M, and S 2 = 0M. Find the equilibrium concentration of S 2 if K =

36 Approximations in Calculations Example 2 For the following reaction: 2H 2 S g 2H 2 g + S 2 (g) We have initial concentrations H 2 S = 0.100M, H 2 = 0.100M, and S 2 = 0.00M. Find the equilibrium concentration of S 2 if K =

37 Homework Problems: #30, 32, 34, 36, 38, 40

38 14.9: Le Châtelier's Principle In a chemical system that isn t in equilibrium tends to shift towards equilibrium Le Châtelier's Principle states that this shift will always go in the direction with minimal disturbance Think about a ball going up and down a hill

39 Le Châtelier s Principle Now with chemical systems: If we had the reaction of: 2BrNO g 2NO g + Br 2 (g) What direction would equilibrium shift if we add: Br 2? BrNO? What if we removed some NO?

40 Le Châtelier's Principle What if we changed the pressure? Well, the shift would then be towards the side with the least number of moles For example: N 2 g + 3H 2 (g) 2NH 3 (g) If we increased the pressure, which direction would equilibrium shift?

41 Le Châtelier's Principle Now, what about temperature? Well, as long as we know the change in enthalpy for a reaction (H), we can place its term in the chemical equation and make the same predictions For example, the following reaction is exothermic: 2SO 2 g + O 2 (g) 2SO 3 (g) What would happen if we increased the temperature? Homework Problems: #42, 44, 46, 48, 50

42 Review Problems: #52, 54, 56, 58 L8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr&index=28 8oGr&index=29 Calculations