Chemical Equilibrium Chapter 15

Homework problems pages 493- 496 1, 4, 5, 6, 8, 12, 15, 16, 19, 20, 28, 30, 32, 34, 36 – 40, 42 – 44, 46, 48 EH Unit 17 Due Feb ? Study Spring 2001 Test #3 1999 Test #2 (1-17) Test 4(26-46)

Physical equilibrium Chemical equilibrium Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H2O (l) H2O (g) Chemical equilibrium N2O4 (g) 2NO2 (g) 14.1

N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2 Start with NO2 & N2O4 14.1

constant 14.1

Equilibrium Will N2O4 (g) 2NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3 aA + bB cC + dD K = [C]c[D]d [A]a[B]b Equilibrium Expression Equilibrium Will K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants 14.1

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N2O4 (g) 2NO2 (g) Kp = NO2 P2 N2O4 P Kc = [NO2]2 [N2O4] In most cases Kc  Kp aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b) 14.2

Homogeneous Equilibrium (Aqueous solution of acetic acid.) CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) [CH3COO-][H3O+] [CH3COOH][H2O] Kc = ‘ [H2O] = constant [CH3COO-][H3O+] [CH3COOH] = Kc [H2O] ‘ Kc = General practice not to include units for the equilibrium constant. 14.2

Ex 15.1 Write expressions for Kcand Kp for 2 NO(g) + O2(g) 2 NO2(g) CH3COOH(aq) + CH3OH(aq) CH3COOCH3(aq) + H2O(l)

Our experiment: mix acetic acid and propyl alcohol CH3COOH + CH3CH2CH2OH CH3COOCH2CH2CH3 + H2O(l)

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g) COCl2 (g) [COCl2] [CO][Cl2] = 0.14 0.012 x 0.054 Kc = = 220 Kp = Kc(RT)Dn 14.2

The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm? 2NO2 (g) 2NO (g) + O2 (g) 2 Kp = 2 PNO PO PNO PO 2 = Kp PNO PO 2 = 158 x (0.400)2/(0.270)2 = 347 atm 14.2

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO3 (s) CaO (s) + CO2 (g) Kc = ‘ [CaO][CO2] [CaCO3] [CaCO3] = constant [CaO] = constant Kc x ‘ [CaCO3] [CaO] Kc = [CO2] = Kp = PCO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 14.2

does not depend on the amount of CaCO3 or CaO CaCO3 (s) CaO (s) + CO2 (g) PCO 2 = Kp PCO 2 does not depend on the amount of CaCO3 or CaO 14.2

Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction? NH4HS (s) NH3 (g) + H2S (g) Kp = P NH3 H2S P = 0.265 x 0.265 = 0.0702 14.2

‘ ‘ ‘ ‘ ‘ Kc = [C][D] [A][B] Kc = [E][F] [C][D] Kc A + B C + D Kc C + D E + F [E][F] [A][B] Kc = A + B E + F Kc Kc = Kc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

Writing the Reaction Quotient for an Overall Reaction Problem: Oxygen gas combines with nitrogen gas in the internal combustion engine to produce nitric oxide, which when out in the atmosphere combines with additional oxygen to form nitrogen dioxide. (1) N2 (g) + O2 (g) 2 NO(g) Kc1 = 4.3 x 10-25 (2) 2 NO(g) + O2 (g) 2 NO2 (g) Kc2 = 6.4 x 109 (a) Show that the overall Kc for this reaction sequence is the same as the product of the Kc’s for the individual reactions. (b) Calculate Kc for the overall reaction.

‘ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4] = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. 14.2

Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2

Qc > Kc system proceeds from right to left to reach equilibrium The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. IF Qc > Kc system proceeds from right to left to reach equilibrium Qc = Kc the system is at equilibrium Qc < Kc system proceeds from left to right to reach equilibrium 14.4

[N2O4] = 0.12 M and [NO2] = 0.55 M. Is the Sample Problem Problem: For the reaction N2O4(g) 2 NO2 (g) Kc = 0.21 at 100°C. At a point during the reaction [N2O4] = 0.12 M and [NO2] = 0.55 M. Is the reaction at equilibrium? If not, in which direction is it progressing?

Predicting Reaction Direction and Calculating Equilibrium Concentrations Problem: Two components of natural gas can react according to the following chemical equation: CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g) In an experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S, and 2.00 mol H2 are mixed in a 250 mL vessel at 960°C. At this temperature, Kc = 0.036. (a) In which direction will the reaction go? (b) If [CH4] = 5.56 M at equilibrium, what are the concentrations of the other substances? [H2S] = 8.00 M, [CS2] = 4.00 M and [H2 ] = 8.00 M 1.00 mol 0.250 L [CH4] = = 4.00 M

[I2] = 0.0884 M (at equilibrium) Calculating Kc from Concentration Data Problem: Hydrogen iodide decomposes at moderate temperatures by the reaction below: When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc ? Set up a reaction table (also called a change table). 2 HI(g) H2 (g) + I2 (g) [HI] = 0.800 M (initial) [I2] = 0.0884 M (at equilibrium)

Determining Equilibrium Concentrations from Kc Problem: One laboratory method of making methane is from carbon disulfide reacting with hydrogen gas, and Kc this reaction at 900°C is 27.8. CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g) At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol CS2, 1.10 mol of H2, and 0.45 mol of H2S, how much methane was formed? [CS2] = 0.0532 M [H2] = 0.234 M [H2S] = 0.0957 M

Calculating Equilibrium Concentrations Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. Having solved for x, calculate the equilibrium concentrations of all species. 14.4

Determining Equilibrium Concentrations from Initial Concentrations and Kc Problem: Given that the reaction to form HF from molecular hydrogen and fluorine has a Kc of 115 at a certain temperature. If 3.000 mol of each component is added to a 1.500 L flask, calculate the equilibrium concentrations of each species. H2 (g) + F2 (g) 2 HF(g) 3.000 mol [H2] = = 2.000 M 1.500 L Kc = = 115 [HF]2 [H2] [F2] 3.000 mol [F2] = = 2.000 M 1.500 L 3.000 mol [HF] = = 2.000 M 1.500 L

ICE At 12800C the equilibrium constant (Kc) for the reaction Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br2 (g) 2Br (g) Let x be the change in concentration of Br2 Br2 (g) 2Br (g) Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x [Br]2 [Br2] Kc = Kc = (0.012 + 2x)2 0.063 - x = 1.1 x 10-3 Solve for x 14.4

Kc = (0.012 + 2x)2 0.063 - x = 1.1 x 10-3 4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x2 + 0.0491x + 0.0000747 = 0 -b ± b2 – 4ac  2a x = ax2 + bx + c =0 x = -0.0105 x = -0.00178 Br2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.063 0.012 -x +2x 0.063 - x 0.012 + 2x At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M At equilibrium, [Br2] = 0.062 – x = 0.0648 M 14.4

Test on Chapters 14 and 15 Monday June 2

Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Changes in Concentration N2 (g) + 3H2 (g) 2NH3 (g) Add NH3 Equilibrium shifts left to offset stress 14.5

Le Châtelier’s Principle Changes in Concentration continued Remove Add Add Remove aA + bB cC + dD Change Shifts the Equilibrium Increase concentration of products(s) left Decrease concentration of products(s) right Increase concentration of reactants(s) right Decrease concentration of reactants(s) left 14.5

None of these changes affect Kc. Predicting the Effect of a Change in Concentration on the Position of the Equilibrium Problem: Carbon will react with water to yield carbon monoxide and and hydrogen, in a reaction called the water gas reaction that was used to convert coal into a fuel that can be used by industry. C(s) + H2O (g) CO(g) + H2 (g) What happens to: (a) [CO] if C is added? (c) [H2O] if H2 is added? (b) [CO] if H2O is added? (d) [H2O] if CO is removed? None of these changes affect Kc.

FeSCN2+(aq) Fe3+(aq) + SCN-(aq) red pale yellow colorless What happens when some NaSCN is added? What happens when some Na2C2O4 is added? (C2O42- reacts with Fe3+ to form a complex ion.) 1. What was stress? 2. Which direction does rxn shift? 3. How does this relieve stress? 4. How does this explain observations?

The Effect of a Change in Pressure (Volume) Pressure changes are mainly involving gases as liquids and solids are nearly incompressible. For gases, pressure changes can occur in three ways: Changing the concentration of a gaseous component Adding an inert gas (one that does not take part in the reaction) Changing the volume of the reaction vessel An increase in pressure by a decrease in volume favors the net reaction that decreases the total number of moles of gases. WHY?? Kc does not change.

Consider the following equilibrium systems. 2 PbS(s) + 3 O2(g) 2 PbO(s) + 2 SO2(g) PCl5(g) PCl3(g) + Cl2(g) H2(g) + CO2(g) H2O(g) + CO(g) Predict the direction of the net reaction as a result of increasing the pressure by decreasing the volume.

The Effect of a Change in Temperature Only temperature changes will alter the equilibrium constant, and that is why we always specify the temperature when giving the value of Kc. When you raise the temperature you are adding heat. In order to use up some of this heat the reaction will shift in the endothermic direction. If you have an exothermic reaction, it will shift in the reverse direction when the temperature is increased. (What about Kc?) O2 (g) + 2 H2 (g) 2 H2O(g) + Energy = Exothermic A temperature increase favors the endothermic direction and a temperature decrease favors the exothermic direction. WHY?? A temperature rise will increase Kc for a system with a + Δ H0rxn A temperature rise will decrease Kc for a system with a - ΔH0rxn

Sample problem How does an increase in temperature affect the equilibrium concentration of the underlined substance and Kc? ΔH(kJ) CaO(s) + H2O Ca(OH)2(aq) -82 CaCO3(s) CaO(s) + CO2(g) 178 SO2(g) S(s) + O2(g) 297

Predicting the Effect of Temperature and Pressure Problem: How would you change the volume (pressure) and temperature in the following reactions to increase the chemical yield of the products? (a) 2 SO2 (g) + O2 (g) 2 SO3 (g); Ho = 197 kJ (b) CO(g) + 2 H2 (g) CH3OH(g); Ho = -90.7 kJ (c) C(s) + CO2 (g) 2 CO(g); Ho = 172.5 kJ

The Haber process is used to produce 110 million tons of ammonia each year. N2(g) + 3 H2(g) 2 NH3(g) ΔH = -91.8 kJ What conditions of temperature and pressure favor the maximum yield of ammonia?

Effect of Temperature for Ammonia Synthesis T (K) Kc 200. 7.17 x 1015 300. 2.69 x 108 400. 3.94 x 104 500. 1.72 x 102 600. 4.53 x 100 700. 2.96 x 10 -1 800. 3.96 x 10 -2

Percent Yield of Ammonia vs. Temperature (°C) at five different operating pressures.

Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas 14.5

Le Châtelier’s Principle Changes in Temperature Change Exothermic Rx Endothermic Rx Increase temperature K decreases K increases Decrease temperature K increases K decreases Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner 14.5

Catalyst lowers Ea for both forward and reverse reactions. uncatalyzed catalyzed Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. 14.5

Le Châtelier’s Principle Change Equilibrium Constant Change Shift Equilibrium Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no 14.5