1. GOES WITH CHAPTER 17: SILBERBERG; PRINCIPLES OF GENERAL CHEMISTRY AP CHEMISTRY MRS. LAURA PECK Topic 12: Equilibrium 1

2. Objectives/Study Guide Write equilibrium expressions for a given reaction Calculate Q and compare it to K to determine if a reaction is at equilibrium Manipulate K if a reaction is reversed or multiplied by a coefficient Calculate K from given equilibrium concentrations, or if given K and all except one equilibrium concentration, solve for the missing value. Calculate equilibrium concentration (or one of the two missing variables) if given any two of the following values: K, the initial concentrations, one equilibrium concentration Do calculations involving gaseous equilibria, partial pressures, and Kp Use Le Chatelier’s principle to determine in what direction the position of equilibrium will shift when a change is imposed. Calculate the value of K from thermodynamic values such as ΔG° 2 In Section II of the AP exam, the first question will ALWAYS Be an equilibrium problem, worth 20% of the F.R. score. It can Be one of the types of problems found in any of the three Equilibrium sections in this topic.

3. Equilibrium Condition 3 Many reactions are reversible. When equilibrium is reached, the rate of the forward rxn equals the rate of the reverse rxn. The ratio of the product concentrations to the reactant concentrations is constant at equilibrium.

4. The Equilibrium Expression 4 The equilibrium constant, K, measures the ratio of product concentrations to reactant concentrations. For a hypothetical reaction aA + bB  cC + dD K = [A] a [B] b [C] c [D] d K is the equilibrium constant, usually given without units. [] represents the concentrations of the reactants and products at equilibrium in moles/liter The concentrations of products and reactants are raised to the powers of their respective coefficients in the balanced chemical equation.  You will be expected to write the equilibrium expression for a given reaction.  Remember, the concentrations of solids and liquids are not included.

5. Example #1 5 Write the equilibrium expression for the reaction P 4(s) + 5O 2(g)   P 4 O 10(s) *the [] of the solids are not included in the expression because the [] of a solid or liquid is a constant, so it is included in the value of K The value of the equilibrium constant measures the extent to which a rxn occurs. K>1: the [products] > [reactants] at equilibrium, rxn proceeds towards completion. K [products] at equilibrium, so the rxn hardly proceeds toward completion. Solution: The equilibrium expression for the Reaction is – K = 1 [O 2 ] 5

6. Reaction Quotient 6 When reactants and products are mixed together, they may not be at equilibrium.  The reaction quotient, Q, compared to the equilibrium constant, K, will determine which way a system will shift to reach equilibrium.  If a system is not at equilibrium, it will move in a direction to reach equilibrium. If a reactant or product in a reaction is not present and is at zero concentration, the reaction will move in the direction that produces the missing component. If all reactants and products are present and have an initial concentration, you must determine the value of Q, the reaction quotient. To determine if the reaction is at equilibrium or the direction it will shift to attain equilibrium, plug all of the initial concentrations into the reaction quotient, which is the same as the equilibrium expression (Law of Mass Action) and compare the value of Q to K.

7. Comparison of Q to K 7 If Q = K, the reaction is at equilibrium If Q>K, the reaction will shift to the left. A shift toward the reactants will consume products. If Q<K, the reaction will shift to the right. A shift to produce more products will consume reactants.

8. Example #2 8 For the reaction 2NO (g)   N 2(g) + O 2(g) K=2.4x10 3 at a particular temperature.  A) If the initial concentrations are 0.024M NO, 2.0M N 2, and 2.6 M O 2, is the system at equilibrium?  B) If it is not at equilibrium, in which direction will the reaction shift? The reaction is not at equilibrium, Q>K, so it will shift to the left. Q = [N2][O2] = (2.0)(2.6) = 9.0x10 3 [NO] 2 (0.024) 2 9.0x10 3 >2.4x10 3

9. Types of Equilibrium Problems 9 1) manipulation of the equilibrium constant, K  If the reaction is reversed, the equilibrium expression is the reciprocal of the expression for the forward reaction.  A   B: K = [B]/[A] B   A: K’ = 1/K = [A]/[B]  If the coefficients in a balanced equation are multiplied by a number, n, the equilibrium constant is raised to the power n.  A   B: K=[B]/[A] 2A   2B: K” = K 2 = [B] 2 /[A] 2 2) If given all equilibrium concentrations, calculate the value of the constant, K 3) If given the value of the equilibrium constant, K, and all but one of the equilibrium concentrations, solve for the missing concentration. 4) If given the value of the initial concentrations of the reactants and one of the equilibrium concentrations of either the reactants or products, solve for all equilibrium concentrations and the value of K 5) If given the initial concentrations and the value of K, solve by approximation for the equilibrium concentrations.

10. Example #3 10 (Type 1) For the reaction N 2(g) + 3H (g)  2NH 3(g) K = 1.3x10 -2 at a certain temperature. Calculate the value of K, called K’, for the reaction: NH 3(g)  1/2N 2(g) + 3/2H 2(g) The reaction is the reverse and one half of the one which is given. K’ = [N 2 ] 1/2 [H 2 ] 3/2 / [NH 3 ] = (1/K) 1/2 ; (1/(1.3x10 -2 )) 1/2 = 8.8

11. The ICE chart Many times you will use an ‘ICE’ chart to help arrange and solve equilibrium problems. I = initial concentration in mol/L (note that units are omitted from chart) C = change to reach equilibrium represented by +x or –x.  A minus sign indicates a decrease in concentration, a plus sign indicate an increase.  The coefficient in front of the reactant in the balanced equation is placed in front of the x in the change line. E = Equilibrium concentrations which are obtained by adding the I and C line together. 11

12. Example #4 (using ICE) 12 (type 4) At a particular temperature, K = 1.00x10 2 for the reaction:  H 2(g) + I 2(g)   2HI (g) In an experiment, 1.00 mol H 2, 1.00 mol I 2. and 1.00 mol HI are introduced into a 1.00L container. Calculate the equilibrium concentrations of all reactions and products. H 2(g) + I 2(g)   2HI (g) K = [HI] 2 [H 2 ][I 2 ] I: 1.00 1.00 1.00 C: -x -x +2x E: 1.00-x 1.00-x 1.00+2x K=100 = (1.00+2x) 2 take square root  10.0=(1.00+2x) (1.00-x) 2 (1.00-x)  10.0 – 10.0x = 1.00 + 2x  12x = 9.0  x = 0.75M *Use this value of x to solve for the equilibrium concentrations of all Reactants and products. [H 2 ]=[I 2 ] = 1.00 – 0.75 = 0.25M [HI] = 1.00 + 2(0.75) = 2.50M *Last, check your equilibrium concentrations by making sure that they Equal the correct value for K.

13. Example #5 (using ICE) (type 5) For the reaction N 2 O 4(g)   2NO 2(g) K=4.0x10 -7 at a specific temperature In an experiment, 1.0 mol of N2O4 is placed in a 10.0L vessel. Calculate the equilibrium concentrations of NO2 and N2O4. 13 At equilibrium, [N 2 O 4 ] = 0.10M and [NO 2 ] = 2.0x10 -4 M. To solve this problem, Proceed as in the previous example. N 2 O 4(g)   2NO 2(g) K=[NO 2 ] 2 = (2x) 2 = 4.0x10 -7 I: 0.10 0 [N 2 O 4 ] (0.10 – x) C: -x +2x **since the value of K is much smaller than 1, E: 0.10-x 2x you can assume the change from the initial concentration, x in 0.10-x, is so small that it is negligible, that is, 0.10-x is about = 0.10 Simplifying the math to K = (2x) 2 /(0.10) = 4.0x10 -7 4x 2 = 4.0x10 -8  x = 1.0x10 -4 M *It is okay to make this assumption if the change from the initial concentration, In this case, x, is less than 5% of the initial concentration x/0.10 x 100% = (1.0 x 10 -4 )/0.10 x 100% = 0.10% (less than 5%) [N 2 O 4 ] = 0.10-x = 0.10-(1.0x10 -4 ) = 0.10M [NO 2 ] = 2x = 2(1.0x10 -4 ) = 2.0x10 -4 M

14. Gaseous Equilibrium For equilibrium in the gas phase, the equilibrium expression can be written in terms of the partial pressures of the gases  For the reaction AsH 2(g)   2As (s) + 3H 2(g)  K p = (P H2 ) 3  P AsH3 K p is the equilibrium constant in terms of partial pressures of the gases. P represents the partial pressure of the gases raised to their coefficients in the balanced chemical equation. 14

15. Example #6 Given the following reaction:  2NO (g) + Br 2(g)   2NOBr (g) K p =109 @ 25°C The equilibrium partial pressure of Br2=0.0159atm and NOBr=0.0768atm. Calculate the equilibrium partial pressure of NO  K p = (P NOBr ) 2  (P NO ) 2 (P Br2 ) 15 109 atm -1 = (0.768 atm) 2 (P NO ) 2 (0.0159atm)  P NO = 0.0583 atm

16. Modifying the Ideal Gas Law The relationship between K p and K is given by K p =K(RT) Δn K p is the equilibrium gas constant Δn is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. 16 Example #7: 2NO (g) + Cl 2(g)   2NOCl (g) K p = 1.9x10 3 @ 25°C Calculate the value for K at 25°C Δn = 2 – (2+1) = -1 K p = K(RT) -1 = K/RT K = K p RT = (1.9x10 3 )(0.08206)(298K) = 4.6x10 4

17. LeChatelier’s Principle The value of K is a constant at a particular temperature. The only factor that changes the value of K is temperature. Pressure, a catalyst, and changes in concentration will not affect the value of K LeChatelier’s Principle states that is a change is imposed in a system at equilibrium, the position of the equilibrium will shift in a direction that will counteract the change. 17

18. Example #8 (conceptual) Consider the following changes on the system below at equilibrium. 2SO 3(g)   2SO 2(g) + O 2(g) ΔH = 197 kJ 18 a)Addition of O 2 b)Removal of SO 2 The equilibrium will shift to the left to form more reactants. The Addition of oxygen increases the rate of the reverse rxn so More reactants will form, until the rate of the forward reaction Again equals the rate of the reverse reaction. Another way to Explain the shift is the Q>K so that must decrease the products And increase reactants for Q=K again. Removal of an equilibrium component such as SO2 at constant Pressure and temperature will cause the equilibrium to shift Toward the removed component to increase its concentration. The reaction in this example will shift to the right. c) Increase in Temp d) Increase in pressure To consider the effect of changes in temperature on the Equilibrium constant, treat the heat of the reaction as a reactant (endothermic) or as a product (exothermic). The direction of the Shift can be predicted in the same way as the addition/removal Of a reactant or product. Since this rxn is endothermic, you would treat heat as a Reactant. Thus, an increase in temp will cause the reaction to Shift right, producing more products, increasing value of K. Decreasing temp will cause a shift to the left, increasing reactant Concentrations and lowering value of K. If the pressure in increased in a equilibrium system, the reaction will shift Toward the side with fewer moles of gas. Decreasing the volume has the same effect because the only way pressure can be increased w/out Changing temp or number of moles is to decrease volume. In this Example, the rxn will shift to the left when pressure is increased or Volume decreased. If the moles of gas are the same on both sides of the reaction, no shift will occur. e) Addition of a solid or inert gas, such as Ne f) An inert gas, such as Ne, is added at constant pressure. If a solid or an inert gas (with no change in volume) is added to the reaction, there Will be no shift in equilibrium. Neither the solid nor the inert gas are part of the Equilibrium expression. The concentrations of all the components of the Equilibrium expression remain unchanged. There will be a shift in the equilibrium to the side of the equation with more moles Of gas. To add Ne at constant pressure, the volume of the container must Increase so the concentrations or partial pressures of all gases have decreased. If the moles of gas are greater in the reactants, then Q<K, so equilibrium can Be reestablished only be increasing the products.

19. Relationship of K to Free Energy, ΔG° The equilibrium constant at standard temperature can be determined from the thermodynamic values of a reaction using the equation: ΔG° = -RTlnK R = 8.314 J/(mol*K) 19 Example #9: Calculate the value of K at 25.0°C for the reaction: 2NO2(g)   N2O4(g) Values of ΔH° = -58.03 kJ/mol and ΔS° = -176.6 J/K*mol At 25°C, ΔG° = ΔH° - TΔS° = (158.03x10 3 J/mol)-(298K)(-176.6J/(K*mol)) = -5.40x10 3 J/mol ΔG° = -RTlnK  lnK = -ΔG°/RT = 2.18  K = 8.8

20. The end…… 20