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Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson , Brady, & Hyslop. CHAPTER 15 Chemical Equilibrium. Learning Objectives: Reversible Reactions and Equilibrium Writing Equilibrium Expressions and the Equilibrium Constant (K)

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Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition

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  1. Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

  2. CHAPTER 15 Chemical Equilibrium • Learning Objectives: • Reversible Reactions and Equilibrium • Writing Equilibrium Expressions and the Equilibrium Constant (K) • Reaction Quotient (Q) • KcvsKp • ICE Tables • Quadratic Formula vs Simplifying Assumptions • LeChatelier’s Principle • van’t Hoff Equation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  3. CHAPTER 15 Chemical Equilibrium Lecture Road Map: Dynamic Equilibrium Equilibrium Laws Equilibrium Constant Le Chatelier’s Principle Calculating Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  4. CHAPTER 15 Chemical Equilibrium Calculating Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  5. Calculations Overview • For gaseous reactions, use either KP or KC • For solution reactions, must use KC • Either way, two basic categories of calculations • Calculate K from known equilibrium concentrations or partial pressures • Calculate one or more equilibrium concentrations or partial pressures using known KP or KC Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  6. Calculations Kc with Known Equilibrium Concentrations • When all concentrations at equilibrium are known • Use mass action expression to relate concentrations to KC • Two common types of calculations • Given equilibrium concentrations, calculate K • Given initial concentrations and one final concentration • Calculate equilibrium concentration of all other species • Then calculate K Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  7. Calculations Kc with Known Equilibrium Concentrations Ex. 3 N2O4(g) 2NO2(g) • If you place 0.0350 mol N2O4 in 1 L flask at equilibrium, what is KC? • [N2O4]eq = 0.0292 M • [NO2]eq = 0.0116 M KC = 4.61  10–3 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  8. Group Problem For the reaction: 2A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature? • 14 • 0.15 • 1.5 • 6.75 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  9. Calculations Kc with Known Equilibrium Concentrations Ex. 4 2SO2(g) + O2(g) 2SO3(g) At 1000 K, 1.000 mol SO2 and 1.000 mol O2 are placed in a 1.000 L flask. At equilibrium 0.925 mol SO3 has formed. Calculate K C for this reaction. • First calculate concentrations of each • Initial • Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  10. Calculations Example Continued • Set up concentration table • Based on the following: • Changes in concentration must be in same ratio as coefficients of balanced equation • Set up table under balanced chemical equation • Initial concentrations • Controlled by person running experiment • Changes in concentrations • Controlled by stoichiometry of reaction • Equilibrium concentrations EquilibriumConcentration Change inConcentration InitialConcentration = – Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  11. Calculations Example Continued –0.925 –0.462 +0.925 0.075 0.925 0.538 [SO2] consumed = amount of SO3 formed = [SO3] at equilibrium = 0.925 M [O2] consumed = ½ amount SO3 formed = 0.925/2 = 0.462 M [SO2] at equilibrium = 1.000 – 0.975 = 0.075 [O2] at equilibrium = 1.00 – 0.462 = 0.538 M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  12. Calculations Overview • Finally calculate KC at 1000 K Kc = 2.8 × 102 = 280 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  13. Calculations ICE Table Summary ICE tables used for most equilibrium calculations: • Equilibrium concentrations are only values used in mass action expression • Values in last row of table • Initial value in table must be in units of mol/L (M) • [X]initial = those present when reaction prepared • No reaction occurs until everything is mixed Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  14. Calculations ICE Table Summary ICE tables used for most equilibrium calculations: • Equilibrium concentrations are only values used in mass action expression • Values in last row of table • Initial value in table must be in units of mol/L (M) • [X]initial = those present when reaction prepared • No reaction occurs until everything is mixed • Changes in concentrations always occur in same ratio as coefficients in balanced equation • In “change” row be sure all [reactants] change in same directions and all [products] change in opposite direction. • If [reactant]initial = 0, its change must be an increase (+) because [reactant]final cannot be negative • If [reactants] decreases, all entries for reactants in change row should have minus sign and all entries for products should be positive Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  15. Calculations Calculate [X]equilibrium from Kcand [X]initial • When all concentrations but one are known • Use mass action expression to relate Kc and known concentrations to obtain missing concentrations Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g) • At 1500 °C, Kc= 5.67. An equilibrium mixture of gases had the following concentrations: [CH4] = 0.400 M and [H2] = 0.800 M and [CO] = 0.300 M. What is [H2O] at equilibrium ? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  16. Calculations Calculate [X]equilibrium from Kcand [X]initial Ex. 5 CH4(g) + H2O(g) CO(g) + 3H2(g)Kc= 5.67 [CH4] = 0.400 M; [H2] = 0.800 M; [CO] =0.300 M • What is [H2O] at equilibrium? • First, set up equilibrium • Next, plug in equilibrium concentrations and Kc [H2O] = 0.0678 M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  17. Calculations Calculating [X]Equilibrium from Kc When Initial Concentrations Are Given • Write equilibrium law/mass action expression • Set up Concentration table • Allow reaction to proceed as expected, using “x” to represent change in concentration • Substitute equilibrium terms from table into mass action expression and solve Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  18. Calculations Calculate [X]equilibrium from [X]initial and KC Ex. 6H2(g) + I2(g)2HI(g) at 425 ˚C KC = 55.64 If one mole each of H2 and I2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI? Step 1. Write Equilibrium Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  19. Calculations Calculate [X]equilibrium from [X]initial and KC Step 2: Construct an ICE table • Initial [H2] = [I2] = 1.00 mol/0.500 L =2.00 M • Amt of H2 consumed = Amt of I2 consumed = x • Amount of HI formed = 2x – x – x +2x 2.00 – x 2.00 – x +2x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  20. Calculations Calculate [X]equilibrium from [X]initial and KC Step 3. Solve for x • Both sides are squared so we can take square root of both sides to simplify Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  21. Calculations Calculate [X]equilibrium from [X]initial and KC Step 4. Equilibrium Concentrations • [H2]equil = [I2]equil = 2.00 – 1.58 = 0.42 M • [HI]equil = 2x = 2(1.58) = 3.16 – 1.58 – 1.58 +3.16 0.42 +3.16 0.42 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  22. Calculations Calculate [X]equilibrium from [X]initial and KC Ex. 7H2(g) + I2(g) 2HI(g) at 425 ˚C KC = 55.64 • If one mole each of H2, I2 and HI are placed in a 0.500 L flask at 425 ˚C, what are the equilibrium concentrations of H2, I2 and HI? • Now have product as well as reactants initially Step 1. Write Equilibrium Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  23. Calculations Calculate [X]equilibrium from [X]initial and KC Step 2. Concentration Table – x – x +2x 2.00 – x 2.00 – x 2.00 + 2x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  24. Calculations Calculate [X]equilibrium from [X]initial and KC Step 3. Solve for x • [H2]equil = [I2]equil = 2.00 – x • = 2.00 – 1.37 = 0.63 M • [HI]equil = 2.00 + 2x = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  25. Group Problem N2(g) + O2(g) 2NO(g) Kc = 0.0123 at 3900 ˚C If 0.25 moles of N2 and O2 are placed in a 250 mL container, what are the equilibrium concentrations of all species? A. 0.0526 M, 0.947 M, 0.105 M B. 0.947 M, 0.947 M, 0.105 M C. 0.947 M, 0.105 M, 0.0526 M D. 0.105 M, 0.105 M, 0.947 M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  26. Group Problem Conc (M) N2(g) + O2(g) 2NO(g) • Initial 1.00 1.00 0.00 • Change – x – x + 2x • Equil 1.00 – x 1.00 – x+ 2x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  27. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Ex. 8 CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) + H2O(l)acetic acid ethanol ethyl acetate KC = 0.11 An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  28. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 1. Write equilibrium law • Need to find equilibrium values that satisfy this Step 2: Set up concentration table using “x” for unknown • Initial concentrations • Change in concentrations • Equilibrium concentrations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  29. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 2 Concentration Table • Amt of CH3CO2H consumed = Amt of C2H5OH consumed = – x • Amt of CH3CO2C2H5 formed = + x • [CH3CO2H]eq and [C2H5OH ] = 0.810 – x • [CH3CO2C2H5] = x –x – x +x 0.810 – x 0.810 – x +x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  30. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 3. Solve for x • Rearranging gives • Then put in form of quadratic equation ax2 + bx+ c = 0 • Solve for the quadratic equation using Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  31. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 3. Solve for x • This gives two roots: x = 10.6 and x = 0.064 • Only x = 0.064 is possible • x = 10.6 is >> 0.810 initial concentrations • 0.810 – 10.6 = negative concentration, which is impossible Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  32. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Quadratic Equation Step 4. Equilibrium Concentrations [CH3CO2C2H5]equil = x = 0.064 M [CH3CO2H]equil = [C2H5OH]equil = 0.810 M – x = 0.810 M – 0.064 M = 0.746 M –0.064 –0.064 +0.064 0.746 0.746 +0.064 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  33. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Cubic When KC is very small Ex. 92H2O(g) 2H2(g) + O2(g) • At 1000 °C, KC = 7.3  10–18 • If the initial H2O concentration is 0.100 M, what will the H2 concentration be at equilibrium? Step 1. Write Equilibrium Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  34. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Cubic Step 2. Concentration Table • Cubic equation – tough to solve • Make approximation • KCvery small, so x will be very small • Assume we can neglect x • Must prove valid later – 2x +2x +x 0.100 – 2x +2x +x Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  35. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Cubic Step 3. Solve for x • Assume (0.100 – 2x)  0.100 • Now our equilibrium expression simplifies to – 2x +2x +x 0.100 +2x +x = 7.3 × 10–20 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  36. Calculations Calculate [X]equilibrium from [X]initial and KC Example: Cubic Step 3. Solve for x • Now take cube root • x is very small • 0.100 – 2(2.6  10–7) = 0.09999948 • Which rounds to 0.100 (3 decimal places) • [H2] = 2x = 2(2.6  10–7) = 5.2  10–7M Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  37. Calculations Simplifications: When Can You Ignore x In Binomial (Ci – x)? • If equilibrium law gives very complicated mathematical problems and if K is small • Then the change (x term) will also be small and we can assume it can be ignored when added or subtracted from the initial concentration, Ci. • How do we check that the assumption is correct? • If the calculated x is so small it does not change the initial concentration (e.g. 0.10 Minitial – 0.003 Mx-calc = 0.10) • Or if the answer achieved by using the assumption differs from the true value by less than five percent. This often occurs when Ci > 100 x Kc Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  38. Group Problem For the reaction 2A(g) B(g) given that Kp= 3.5 × 10–16 at 25 ˚C, and we place 0.2 atmA into the container, what will be the pressure of B at equilibrium? 2AB I 0.2 0 atm C –2x +x E 0.2 – 2xx ≈0.2 x = 1.4 × 10–17 [B]= 1.4 × 10–17atm Proof: 0.2 - 1.4 × 10–17 = 0.2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

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