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Chapter 10

Chapter 10. Thermodymanics. Thermodynamics. Thermodynamics is a branch of science that focuses on energy changes that accompany chemical and physical changes. 10.1 Heat Capacity. Objective: To calculate heat capacity. Energy As Heat.

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Chapter 10

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  1. Chapter 10 Thermodymanics

  2. Thermodynamics • Thermodynamics is a branch of science that focuses on energy changes that accompany chemical and physical changes.

  3. 10.1 Heat Capacity • Objective: To calculate heat capacity.

  4. Energy As Heat • Heat (q): The energy transferred between objects that are at different temperatures. • Unit: joules (J)

  5. Molar Heat Capacity • Molar Heat Capacity: Energy (heat) needed to increase the temperature of 1 mol of substance by 1 K. q = nC∆T q = heat n = # of moles C = molar heat capacity ∆T = change in temperature

  6. Molar heat capacity: Applications • The molar heat capacity of water is larger than the molar heat capacity of land. This means that water does not heat up as easily as land does. As a result, oceans can help keep coastal areas cool during the summer. • The filling of a fruit pie has a larger heat capacity than the crust. This means that fruit filling will retain heat better and the crust will cool much quicker. As a result, eating the fruit filling can cause burns (even though it may appear that the pie is cool).

  7. Example • Determine the energy (heat) needed to increase the temperature of 10.0 mol of Hg by 7.5 K. The value of C for Hg is 27.8 J/K۰mol.

  8. Example • Determine the energy (heat) needed to increase the temperature of 10.0 mol of Hg by 7.5 K. The value of C for Hg is 27.8 J/K۰mol. q = ? n = 10.0 mol C = 27.8 J/K۰mol ∆T = 7.5 K q = nC∆T q = (10.0 mol)(27.8 J/K۰mol)(7.5 K) q= 2.1 x 103 J

  9. Practice #1 • The molar heat capacity of tungsten is 24.2 J/K ۰mol. Calculate the energy as heat needed to increase the temperature of 0.404 mol of W by 10.0 K.

  10. Practice #1 • The molar heat capacity of tungsten is 24.2 J/K ۰mol. Calculate the energy as heat needed to increase the temperature of 0.404 mol of W by 10.0 K. q = ? n = 0.404 mol C = 24.2 J/K۰mol ∆T = 10.0 K q = nC∆T q = (0.404 mol)(24.2 J/K۰mol)(10.0 K) q= 97.8J

  11. Practice #2 • Suppose a sample of NaCl increased in temperature by 2.5 K when the sample absorbed 1.7 x 102 J energy (heat). Calculate the number of moles of NaCl if the molar heat capacity is 50.5 J/K۰mol.

  12. Practice #2 • Suppose a sample of NaCl increased in temperature by 2.5 K when the sample absorbed 1.7 x 102 J energy (heat). Calculate the number of moles of NaCl if the molar heat capacity is 50.5 J/K۰mol. q = 1.7 x 102 J n = ? C = 50.5 J/K۰mol ∆T = 2.5 K q = nC∆T 1.7 x 102 J = n(50.5J/K۰mol)(2.5 K) n= 1.3 mol

  13. Practice #3 • Calculate the energy as heat needed to increase the temperature of 0.80 mol of nitrogen, N2, by 9.5 K. The molar heat capacity of nitrogen is 29.1 J/K۰mol.

  14. Practice #3 • Calculate the energy as heat needed to increase the temperature of 0.80 mol of nitrogen, N2, by 9.5 K. The molar heat capacity of nitrogen is 29.1 J/K۰mol. q = ? n = 0.80 mol C = 29.1 J/K۰mol ∆T = 9.5 K q = nC∆T q= (0.80 mol)(29.1J/K۰mol)(9.5 K) q= 2.2 x 102 J

  15. Practice #4 • A 0.07 mol sample of octane, C8H18, absorbed 3.5 x 103 J of energy. Calculate the temperature increase of octane if the molar heat capacity of octane is 254.0 J/K۰mol.

  16. Practice #4 • A 0.07 mol sample of octane, C8H18, absorbed 3.5 x 103 J of energy. Calculate the temperature increase of octane if the molar heat capacity of octane is 254.0 J/K۰mol. q = 3.5 x 103 J n = 0.07 mol C = 254.0 J/K۰mol ∆T = ? q = nC∆T 3.5 x 103 J = (0.07 mol)(254.0J/K۰mol) ∆T ∆T = 2.0 x 102 K

  17. 10.2 Enthalpy • Objective: To calculate the change in enthalpy.

  18. Enthalpy • The total energy of a system is impossible to measure. • However, we can measure the change in enthalpy of a system. • Enthalpy: The total energy content of a sample. • H = Enthalpy • ∆H = Change in Enthalpy

  19. Calorimeter • ∆H can be measured with a calorimeter. • A calorimeter is used to measure the heat absorbed or released in a chemical or physical change.

  20. Enthalpy • Enthalpy changes can be used to determine if a process is endothermic or exothermic. • Exothermic Reaction: Negative Enthalpy Change • Endothermic Reaction: Positive Enthalpy Changes

  21. Standard enthalpy of formation • The standard enthalpy of formation ( ) is the enthalpy change in forming 1 mol of a substance from elements in their standard states. • Note: The value for the standard enthalpy of formation for an element is 0. • The values for the standard enthalpies of formation can be found using a table.

  22. Calculating a reaction’s change in enthalpy • Step 1: Determine for each compound using enthalpy table. • Step 2: Multiply by the coefficients from the balanced equation (# of moles). • Step 3: Set up ΔH equation. • Step 4: calculate. ΔHreaction = ΔH products - ΔHreactants

  23. Calculating a reaction’s change in enthalpy: Example Calculate ΔH for the following reaction and determine if the reaction is exothermic or endothermic. SO2(g) + NO2(g)  SO3(g) + NO(g) ΔHreaction = ΔH products - ΔHreactants

  24. Calculating a reaction’s change in enthalpy SO2(g) + NO2(g)  SO3(g) + NO(g) -296.8(1) 33.1(1) -395.8(1)90.3(1) kJ/mol kJ/mol kJ/mol kJ/mol ΔH= [(-395.8 kJ/mol)(1) + (90.3 kJ/mol)(1)] – [(296.8 kJ/mol)(1) + (33.1 kJ/mol)(1)] Δ H = -305.5 kJ – 329.9 kJ Δ H = -635.4 kJ The reaction is exothermic. ΔHreaction = ΔH products - ΔHreactants

  25. Practice #1 • Calculate the enthalpy change for the following reaction: 2C2H6(g) + 7O2(g)  4 CO2(g) + 6H2O(g)

  26. Practice #2 • Calculate ΔH for the following reaction: CaO(s) + H2O(l)  Ca(OH)2(s)

  27. Practice #3 • Calculate the enthalpy change for the combustion of methane gas. CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

  28. 10.3 Entropy • Objective: To calculate the change in entropy.

  29. Entropy • A reaction is more likely to occur if enthalpy (ΔH) is negative. • However, some endothermic reactions can occur easily. Why? Entropy! • Entropy (ΔS): A measure of the randomness or disorder of a system. • A process if more likely to occur if there is an increase in entropy (or if ΔS is positive).

  30. Factors that affect entropy • Entropy is increased by the following factors: • Diffusion (process of dispersion) • Dilution of a solution • Decreasing the pressure of a gas • Increasing temperature • The number of moles of product is greater than the number of moles of reactant • Increasing the total number of particles in a system • When a reaction produces more gas particles (opposed to liquid or solids)

  31. Calculating change in entropy ΔSreaction = ΔSproducts - ΔSreactants

  32. Calculating the change in entropy: example #1 • Find the change in entropy for the following reaction: 2Na (s) + 2HCl (g)  2NaCl (s) + H2(g)

  33. Calculating the change in entropy: example #2 • Find the change in entropy for the following reaction: 2Na (s) + 2H2O (l)  2NaOH (s) + H2(g)

  34. 10.4 gibbs energy • Objectives: • To calculate the change in Gibbs energy. • To determine if a reaction is spontaneous or nonspontaneous.

  35. Gibbs energy or free energy • Gibbs Energy: the energy in a system that is available to do useful work. • Gibbs Energy is also called Free Energy. • ΔG = Change in Gibbs Energy

  36. Calculating the change in Gibbs energy (ΔG) ΔGreaction = ΔGproducts - ΔGreactants

  37. Calculating ΔG: example • Calculate ΔG for the following water-gas reaction: C(s) + H2O (g)  CO (g) + H2 (g)

  38. Calculating ΔG: Practice #1 • Calculate the Gibbs energy change that accompanies the following reaction: C(s) + O2 (g)  CO2 (g)

  39. Calculating ΔG: practice #2 • Calculate the Gibbs energy change that accompanies the following reaction: CaCO3 (s)  CaO (s) + CO2 (g)

  40. Gibbs energy determines spontaneity • If ΔG is negative, the forward reaction is spontaneous. • If ΔG is 0, the system is at equilibrium. • If ΔG is positive, the forward reaction is nonspontaneous. • NOTE: In this case, the reaction is spontaneous in the reverse direction.

  41. Calculating ΔG from ΔH and ΔS ΔG = ΔH - TΔS • Step 1: Organize the information. • Step 2: Change the units. • Step 3: Step up ΔG equation. • Step 4: Calculate.

  42. Calculating ΔG from ΔH and ΔS: Example • Given the changes in enthalpy and entropy are -139 kJ and 277 J/K respectively for a reaction at 25⁰C, calculate the change in Gibbs energy.

  43. Calculating ΔG from ΔH and ΔS: Example • Given the changes in enthalpy and entropy are -139 kJ and 277 J/K respectively for a reaction at 25⁰C, calculate the change in Gibbs energy. ΔH = -139 kJ ΔS = 277 J/K / (1000 J/kJ) = 0.277 kJ/K T = 25 ⁰C + 273 = 298 K ΔG = ?

  44. Calculating ΔG from ΔH and ΔS: Example • Given the changes in enthalpy and entropy are -139 kJ and 277 J/K respectively for a reaction at 25⁰C, calculate the change in Gibbs energy. ΔH = -139 kJ ΔS = 277 J/K / (1000 J/kJ) = 0.277 kJ/K T = 25 ⁰C + 273 = 298 K ΔG = ? ΔG = ΔH - T ΔS ΔG = (-139 kJ) – (298K)(0.277 kJ/K) ΔG = (-139 kJ) – (82.546 kJ) ΔG = -221.55 kJ The reaction is spontaneous.

  45. Calculating ΔG from ΔH and ΔS: practice #1 • A reaction has a ΔH of -76 kJ and a ΔS of -117 J/K. Calculate the ΔG at 298 K. Is the reaction spontaneous?

  46. Calculating ΔG from ΔH and ΔS: practice #2 • A reaction has a ΔH of 11 kJ and a ΔS of 49 J/K. Calculate ΔG at 298 K. Is the reaction spontaneous?

  47. 10.5 phase changes: Melting & Boiling Points • Objective: To calculate the melting and boiling point of a substance.

  48. Calculating Melting and boiling points • Tmp: melting point temperature • Tbp: boiling point temperature • ΔHfus: molar enthalpy of fusion • ΔSfus: molar entropy of fusion • ΔHvap: molar enthalpy of vaporization • ΔSvap: molar entropy of vaporization

  49. Calculating melting and Boiling points • Step 1: Organize the information. • Step 2: Change the units. • Step 3: Step up the equation. • Step 4: Calculate.

  50. Calculating Melting and boiling points: example • The enthalpy of fusion of Hg is 2.295 kJ/mol, and the molar entropy of fusion is 9.79 J/mol·K. The enthalpy of vaporization at the boiling point if 59.2 kJ/mol, and the molar entropy of vaporization is 93.8 J/mol·K. Calculate the melting point and boiling point of Hg.

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