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PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 14

PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 14. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 14 CHEMICAL EQUILIBRIUM. CHEMICAL EQUILIBRIUM. - Many reactions do not go to completion

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PRINCIPLES OF CHEMISTRY II CHEM 1212 CHAPTER 14

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  1. PRINCIPLES OF CHEMISTRY II CHEM 1212CHAPTER 14 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

  2. CHAPTER 14 CHEMICAL EQUILIBRIUM

  3. CHEMICAL EQUILIBRIUM - Many reactions do not go to completion - Amount of products formed or reactants consumed cannot be predicted from stoichiometry alone - These reactions achieve a condition of equilibrium Equilibrium - A state of balance between opposing processes

  4. CHEMICAL EQUILIBRIUM - Occurs when there is product build-up during a chemical reaction - The product molecules interact with one another to re-produce reactants forward reaction A + B C + D reverse reaction Chemical Equilibrium - When the rate of product formation (forward reaction) is equal to the rate of reactant formation (reverse reaction)

  5. CHEMICAL EQUILIBRIUM - Reactant and product concentrations are usually not equal - Such reactions are known as reversible reactions - Forward reaction rate decreases with time as reactants are used up - Reverse reaction rate increases with time as products are being formed - Concentrations are reached when both forward and reverse rates become equal

  6. CHEMICAL EQUILIBRIUM Homogeneous Equilibrium - Substances are in the same phase 2NO2(g) + O2(g) ↔ 2NO3(g) Heterogeneous Equilibrium - Substances are in more than one phase KOH(s) + SO3(g) ↔ KHSO4(s)

  7. EQUILIBRIUM CONSTANT - Describes the extent of reaction in a given system - For a chemical reaction of the form aA + bB ↔ cC + dD - The equilibrium constant (Keq) is given by

  8. EQUILIBRIUM CONSTANT - [ ] denotes equilibrium concentration in mol/liter or M - Product concentrations in the numerator - Reactant concentrations in the denominator - Concentrations are raised to the powers of the respective coefficients (a, b, c, d)

  9. EQUILIBRIUM CONSTANT - Substances in solution (in mol/L) are written in Keq expressions - Pure solids, pure liquids (e.g. water), and solvents are not included since their concentrations are constant - Keq changes with change in temperature - For exothermic forward reactions (heat released) Keq decreases with increasing temperature - For endothermic forward reactions (heat absorbed) Keq increases with increasing temperature

  10. EQUILIBRIUM CONSTANT Large Keq - Greater product concentrations than reactant concentrations - Equilibrium position lies to the right Small Keq - Smaller product concentrations than reactant concentrations - Equilibrium position lies to the left Intermediate Keq (near unity) - Both products and reactants are in significant amounts - Equilibrium position lies neither to the right nor to the left

  11. EQUILIBRIUM CONSTANT - Longer arrows may be used to indicate the predominant species - Longer forward reaction arrow for large Keq - Longer reverse reaction arrow for small Keq CO2 + H2O H2CO3

  12. EQUILIBRIUM CONSTANT - Any Keq expression refers to a particular form of a chemical equation - Consider the following aA + bB ↔ cC + dD - If the reaction is written in the reverse order cC + dD ↔ aA + bB

  13. EQUILIBRIUM CONSTANT - Any Keq expression refers to a particular form of a chemical equation - This implies that

  14. EQUILIBRIUM CONSTANT - Any Keq expression refers to a particular form of a chemical equation - Consider the following aA + bB ↔ cC + dD - If the reaction is multiplied by a factor n ncC + ndD ↔ naA + nbB

  15. EQUILIBRIUM CONSTANT - Any Keq expression refers to a particular form of a chemical equation - This implies that K'' = Kn

  16. EQUILIBRIUM CONSTANT - If two or more chemical equations are added to yield a new equation 1. 2NO2 ↔ N2O4 K1 2. N2O4 + O2 ↔ 2NO3 K2 3. 2NO2 + O2 ↔ 2NO3 K3 - Reactions 1 and 2 are added to give reaction 3 - This implies that K3 = K1K2

  17. EQUILIBRIUM CONSTANT At a given temperature, K = 1.3 x 10-2 for the reaction N2(g) + 3H2(g) ↔ 2NH3(g) a) Write an equilibrium expression for the reaction b) Calculate the values of K for the following reactions 2NH3(g) ↔ N2(g) + 3H2(g) 1/2N2(g) + 3/2H2(g) ↔ NH3(g) 3N2(g) + 9H2(g) ↔ 6NH3(g) K = 1/(1.3 x 10-2) = 77 K = (1.3 x 10-2)1/2 = 0.11 K = (1.3 x 10-2)3 = 2.2 x 10-6

  18. PRESSURE AND CONCENTRATION - The concentration of gases are generally expressed in terms of their partial pressures - Atmospheres (atm) or bars may be used in equilibrium calculations involving gases For the reaction below 2NO2(g) + O2(g) ↔ 2NO3(g)

  19. PRESSURE AND CONCENTRATION - Kc is used for concentrations (mol/L) - Kp is used for pressures (atm or bar) - Keq is for general use when the differences are not important

  20. PRESSURE AND CONCENTRATION - From the ideal gas law PV = nRT - It is proved that [n/V] = P/RT - Pressure (in atm) is converted to concentration (in mol/L) when divided by RT R= 0.08206 L·atm/mol∙K

  21. PRESSURE AND CONCENTRATION - It is also be proved that Kp = Kc(RT)∆n ∆n = change in number of moles of gas = (number of moles of gaseous products) – (number of moles of gaseous reactants) Example 2NO2(g) + O2(g) ↔ 2NO3(g) ∆n = 2 – (2 + 1) = -1

  22. PRESSURE AND CONCENTRATION For the reaction 2NO(g) + 2H2(g) ↔ N2(g) + 2H2O(g) It is determined that, at equilibrium at a given temperature, the concentrations are as follows: [NO(g)] = 8.1 x 10-3 M, [H2(g)] = 4.1 x 10-5 M, [N2(g)] = 5.3 x 10-2 M and [H2O(g)] = 2.9 x 10-3 M Calculate the equilibrium constant for the reaction at this temperature 4.0 x 106

  23. PRESSURE AND CONCENTRATION At a particular temperature, a 3.0-L flask contains 2.4 mol Cl2, 1.0 mol NOCl, and 4.5 x 10-3 mol NO. Calculate the equilibrium constant at this temperature for the reaction 2NOCl(g) ↔ 2NO(g) + Cl2(g) 1.7 x 10-5

  24. PRESSURE AND CONCENTRATION The following equilibrium pressures at 27 oC were observed for the reaction 2NO2(g) ↔ 2NO(g) + O2(g) PNO2 = 0.55 atm, PNO = 6.5 x 10-5 atm, PO2 = 4.5 x 10-5 atm Calculate the value for the equilibrium constant Kp at this temperature Also calculate Kc 6.3 x 10-13 2.6 x 10-14

  25. PRESSURE AND CONCENTRATION At 54 oC, the equilibrium concentrations are [CH3OH(g)] = 0.15 M, [CO(g)] = 0.24 M, and [H2(g)] = 1.1 M for the reaction CH3OH(g) ↔ CO(g) + 2H2(g) Calculate the value for the equilibrium constant Kp at this temperature 1.4 x 103

  26. REACTION QUOTIENT (Q) - Used when the system is not in equilibrium - Concentrations at the time rather than equilibrium concentrations are used Used to determine whether - more products will form - more reactants will form or - system is at equilibrium

  27. REACTION QUOTIENT (Q) - Consider the following which is not at equilibrium aA + bB ↔ cC + dD

  28. REACTION QUOTIENT (Q) - Q tells which direction the reaction proceeds to reach equilibrium - Concentrations change to bring Q closer to Keq If Q < Keq - Reaction proceeds to the right - Concentration of products will increase - Concentration of reactants will decrease - Q will increase to get closer to Keq

  29. REACTION QUOTIENT (Q) - Q tells which direction the reaction proceeds to reach equilibrium - Concentrations change to bring Q closer to Keq If Q > Keq - Reaction proceeds to the left - Concentration of products will decrease - Concentration of reactants will increase - Q will decrease to get closer to Keq

  30. REACTION QUOTIENT (Q) The Keq is 2.4 x 103 at a certain temperature for the reaction 2NO(g) ↔ N2(g) + O2(g) For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a) A 1.0-L flask contains 0.024 mol NO, 2.0 mol N2, 2.6 mol O2 b) A 2.0-L flask contains 0.032 mol NO, 0.62 mol N2, 4.0 mol O2 c) A 3.0-L flask contains 0.060 mol NO, 2.4 mol N2, 1.7 mol O2 a) Q > Keq , reaction shifts left b) Q = Keq , reaction at equilibrium c) Q < Keq , reaction shifts right

  31. LE CHATELIER’S PRICIPLE - If a stress (change of conditions) is applied to a system in equilibrium the system will readjust (change the equilibrium position) in the direction that best reduces the stress imposed on the system - Change of such conditions as Concentration, Pressure, Temperature - If more products form as a result of the applied stress the equilibrium is said to have shifted to the right - If more reactants form as a result of the applied stress the equilibrium is said to have shifted to the left

  32. LE CHATELIER’S PRICIPLE Concentration Changes For a reaction mixture at equilibrium - Addition of reactant(s) shifts the equilibrium position to the right - Removal of product(s) shifts the equilibrium position to the right - Addition of product(s) shifts the equilibrium position to the left - Removal of reactant(s) shifts the equilibrium position to the left

  33. LE CHATELIER’S PRICIPLE Volume and Pressure Changes - Gases must be involved in the chemical reaction - The total number of moles of the gaseous state must change - Equilibrium is shifted in the direction of fewer moles - Volume and pressure are inversely related - Decrease in volume implies increase in pressure - Increase in volume implies decrease in pressure

  34. LE CHATELIER’S PRICIPLE Volume and Pressure Changes Higher moles of gaseous reactants than products (∆n is negative) - Increase in pressure shifts the equilibrium position to the right - Decrease in pressure shifts the equilibrium position to the left Higher moles of gaseous products than reactants (∆n is positive) - Increase in pressure shifts the equilibrium position to the left - Decrease in pressure shifts the equilibrium position to the right

  35. LE CHATELIER’S PRICIPLE Volume and Pressure Changes No change in equilibrium position occurs if - There is no reactant nor product in the gaseous state - Number of moles of gaseous reactants equals number of moles of gaseous products(∆n is zero) - Pressure is increased by adding a nonreactive (inert) gas

  36. LE CHATELIER’S PRICIPLE Temperature Changes Exothermic Reactions - Heat is a product - Increase in temperature shifts the equilibrium position to the left - Decrease in temperature shifts the equilibrium position to the right N2(g) + 3H2(g) ↔ 2NH3(g) ∆H = -92 kJ

  37. LE CHATELIER’S PRICIPLE Temperature Changes Endothermic Reactions - Heat is a reactant - Increase in temperature shifts the equilibrium position to the right - Decrease in temperature shifts the equilibrium position to the left H2(g) + I2(g) ↔ 2HI(g) ∆H = +52 kJ

  38. LE CHATELIER’S PRICIPLE Addition of Catalysts - Catalysts do not change equilibrium positions - Catalysts speed up both forward and reverse reactions so have no net effect - Catalysts allow equilibrium to be established more quickly by lowering the activation energy

  39. LE CHATELIER’S PRICIPLE In which direction will the position of the equilibrium 2HI(g) ↔ H2(g) + I2(g) be shifted for each of the following changes? HI(g) is added HI(g) is removed H2(g) is added H2(g) removed I2(g) is added I2(g) removed Some Ar(g) is added The volume of the container is tripled The temperature is decreased (the reaction is exothermic)

  40. EQUILIBRIUM CALCULATIONS To determine the equilibrium constant from the concentrations of species - Write the balanced chemical equation - Complete the iCe table - Write the equilibrium constant expression - Substitute values using the iCe table - Solve for Keq

  41. EQUILIBRIUM CALCULATIONS The iCe table aA + bB ↔ cC + dD intital concentration, M Change in concentration, M equilibrium concentration, M Fill in iCe concentrations (mol/L or M) for all speices

  42. EQUILIBRIUM CALCULATIONS A 1.00-L flask was filled with 2.00 mol gaseous SO2 and 2.00 mol gaseous NO2 and heated. After equilibrium is reached, it was found that 1.30 mol gaseous NO was present. Calculate Keq for the reaction if it occurs under these conditions. SO2(g) + NO2(g) ↔ SO3(g) + NO(g) 3.4

  43. EQUILIBRIUM CALCULATIONS To determine the concentrations of species from the equilibrium constant - Intitial concentrations of reactants are usually known - Keq is also known - Concentrations of products are written using a variable - Fill in the iCe table - Write the equilibrium expression - Solve for the unknown variable

  44. EQUILIBRIUM CALCULATIONS At 35 oC, Keq = 1.6 x 10-5 for the reaction 2NOCl(g) ↔ 2NO(g) + Cl2(g) Calculate the concentrations of all species at equilibrium if the initial amounts are a) 1.0 M NOCl b) 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0-L flask a) [NOCl] = 1.0 M, [NO] = 0.032 M, [Cl2] = 0.016 M b) [NOCl] = 2.0 M, [NO] = 0.0080 M, [Cl2] = 1.0 M

  45. HETEROGENEOUS EQUILIBRIA - Substances are in more than one phase - Solids and pure liquids are excluded in equilibrium calculations because their concentrations are constant Example CaCO3(s) ↔ CaO(s) + CO2(g) Kc = [CO2] Kp = PCO2

  46. HETEROGENEOUS EQUILIBRIA Consider the following reaction at a certain temperature 4Fe(s) + 3O2(g) ↔ 2Fe2O3(s) An equilibrium mixture contains 1.0 mol Fe, 1.0 x 10-3 mol O2, and 2.0 mol Fe2O3 all in a 2.0-L container. Calculate Keq for this reaction. 8.0 x 109

  47. SOLUBILITY EQUILIBRIA - Reactions involving dissolution of solids into solution and formation of solids from solution - Precipitation reactions Example Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) Net Ionic Equation Pb2+(aq) + 2I-(aq) → PbI2(s)

  48. SOLUBILITY PRODUCT CONSTANT - Ksp is the equilibrium constant for the reaction in which a solid salt (an ionic compound) dissolves to give its constituent ions in solution - Solid is in its standard state so is omitted from the Ksp expression Example PbI2(s) ↔ Pb2+ + 2I- Ksp = [Pb2+][I-]2 = 7.9 x 10-9

  49. SOLUBILITY PRODUCT CONSTANT - The solution is said to be saturated if all the solid is capable of being dissolved - Electrostatic interaction causes the formation of some other types of ions (ion pairing) but is not discussed here

  50. SOLUBILITY PRODUCT CONSTANT Use is made of the iCe table for Ksp calculations PbI2(s) ↔ Pb2+ + 2I- i solid 0 0 C -s +s +2s e solid s 2s Ksp = (s)(2s)2 = 4s3

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