1 / 61

Topic 5. Chemical Energetics

Topic 5. Chemical Energetics. IB Topic 5: Energetics 5.1: Exothermic and Endothermic Reactions. 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction ( Δ H o ). 5.1.2 State that combustion and neutralization are exothermic processes.

dwalden
Download Presentation

Topic 5. Chemical Energetics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 5. Chemical Energetics

  2. IB Topic 5: Energetics5.1: Exothermic and Endothermic Reactions 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (ΔHo). 5.1.2 State that combustion and neutralization are exothermic processes. 5.1.3 Apply the relationship between temperature change, enthalpy change and the classification of a reaction as endothermic or exothermic. 5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of enthalpy change for the reaction.

  3. Heat and Temperature Heatis energy that is transferred from one object to another due to a difference in temperature Temperature is a measure of the average kinetic energy of a body Heat is always transferred from objects at a higher temperature to those at a lower temperature 3

  4. 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (ΔHo). Exothermic Reaction: A process that releases heat to its surroundings. Products have less energy than the reactants Endothermic Reaction : A process that absorbs heat from the surroundings. Products have more energy than the reactants.

  5. 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (ΔHo). Standard Enthalpy Change of Reaction (∆H): The heat energy exchanged with the surroundings when a reaction happens under standard conditions (NOT STP… see below). Since the enthalpy change for any given reaction will vary with the conditions, esp. concentration of chemicals, ΔH are measured under standard conditions: pressure = 101.3 kPa temperature = 25ºC = 298 K Concentrations of 1 mol dm-3 The most thermodynamically stable allotrope (which in the case of carbon is graphite) Only ΔH can be measured, not H for the initial or final state of a system.

  6. 5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (ΔHo). Pseudonyms (other names) for H • Heat of Reaction: Hrxn heat produced in a chemical reaction • Heat of Combustion: Hcomb heat produced by a combustion reaction • Heat of Neutralization: heat produced in a neutralization reaction (when an acid and base are mixed to get water, pH = 7) • Heat of solution: Hsol heat produced by when something dissolves • Heat of Fusion: Hfus heat produced when something melts • Heat of Vaporization: Hvap heat produced when something evaporates • Heat of Sublimation: Hsub heat produced when something sublimes • Heat of formation: Hf change in enthalpy that accompanies the formation of 1 mole of compound from it’s elements (this has special uses in chemistry…)

  7. 5.1.2 State that combustion and neutralization are exothermic processes. Combustion Exothermic reaction General Combustion Reaction Formula: Compound (usually hydrocarbon) + O2 CO2 + H2O + energy CH4 + 2O2 CO2 + 2H2O + 890kJ ∆H = -890kJ Neutralization Exothermic reaction Acid + Base  Salt + Water + energy HCl + NaOH  NaCl + H2O + 57.3 kJ ∆H = -57.3kJ

  8. 5.1.3 Apply the relationship between temperature change, enthalpy change and the classification of a reaction as endothermic or exothermic. Exothermic Heat flows out of the system Surroundings heat up Heat change (ΔH) < 0 (negative) C8H18+ 12½O2 8CO2 + 9H2O ΔH = -5512 kJ mol-1 H2 + ½O2  H2O ΔH = -286 kJ mol-1 Endothermic Heat flows into the system Surroundings cool down Heat change (ΔH) > 0 (positive) H2O(s)  H2O(l) ΔH = +6.01 kJ mol-1 ½N2 + O2 NO2 ΔH = +33.9 kJ mol-1

  9. 2H2(g) + O2(g) 2H2O (l) + energy H2O (g) H2O (l) + energy energy + 2HgO (s) 2Hg (l) + O2(g) energy + H2O (s) H2O (l) Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings. Endothermic process is any process in which heat has to be supplied to the system from the surroundings. 6.2

  10. 6.5

  11. 5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of enthalpy change for the reaction. Exothermic Reactions Products more stable than reactants (lower energy). ΔH = Hproducts – Hreactants Since the products have less energy than the reactants, the ΔH value is negative. Endothermic Reactions Products less stable than reactants (higher energy) ΔH = Hproducts – Hreactants Since the products have more energy than the reactants, the ΔH value is positive.

  12. Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. DH = H (products) – H (reactants) DH = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants Hproducts > Hreactants DH < 0 DH > 0 6.4

  13. 5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of enthalpy change for the reaction.

  14. REVIEW

  15. REVIEW

  16. IB Topic 5: Energetics5.2: Calculation of Enthalpy Changes 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. 5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water 5.2.4 Evaluate the results of experiments to determine enthalpy changes.

  17. Factors Affecting Heat Quantities The amount of heat contained by an object depends primarily on three factors: The mass of material The temperature The kind of material and its ability to absorb or retain heat. 17

  18. Heat Quantities The heat required to raise the temperature of 1.00 g of water 1 oC is known as a calorie Calorie (with a capital “C”): dietary measurement of heat. Food has potential energy stored in the chemical bonds of food. 1 Cal = 1 kcal = 1000 cal The SI unit for heat is the joule. It is based on the mechanical energy requirements. 1.00 calorie = 4.184 Joules 18

  19. 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. Heat Energy Change q = m x c x ΔT q = heat (joules or calories) m = mass (g) c = specific heat (J g-1oC-1) The amount of heat required to raise the temperature of 1 g of a substance 1 oC. Specific heat of water = 4.184 Joules / ΔT = change in temperature

  20. 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. How much heat in joules will be absorbed when 32.0 g of water is heated from 25.0 oC to 80.0 oC? q = m x c x ΔT q = ? m = 32.0 g c = 4.18 J g-1oC-1 ΔT = 80.0-25.0 = 55.0 oC q = 32.0 x 4.18 x 55.0 = 7,360 J When 435 J of heat is added to 3.4 g of olive oil at 21 oC, the temperature increases to 85 oC. What is the specific heat of olive oil? q = m x c x ΔT q = 435 J m = 3.4 g c = ? ΔT = 85-21 = 64 oC 435 = 3.4 x c x 64 = 2.0 J g-1oC-1

  21. Heat Transfer Problem 1 Calculate the heat gained in an aluminum cooking pan whose mass is 400 grams, from 20oC to 200oC. The specific heat of aluminum is 0.902 J g-1oC-1. Solution DQ = mCDT = (400 g) (0.902 J g-1oC-1)(200oC – 20oC) = 64,944 J = 60,000J OR 6 x 104 J OR 60 kJ 21

  22. 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. How much heat in joules is required to raise the temperature of 250 g of mercury 52 oC? A 1.55 g piece of stainless steel absorbs 141 J of heat when its temperature increases by 178 oC. What is the specific heat of the stainless steel?

  23. 5.2.1 Calculate the heat energy change when the temperature of a pure substance is changed. How much heat in joules is required to raise the temperature of 250 g of mercury 52 oC? q = m x c x ΔT q = ? m = 250 g c = 0.14 J g-1oC-1 (Table 11.2) ΔT = 52 oC q = 250 x 0.14 x 52 = 1800 J A 1.55 g piece of stainless steel absorbs 141 J of heat when its temperature increases by 178 oC. What is the specific heat of the stainless steel? q = m x c x ΔT q = 141 J m = 1.55 g c = ? ΔT = 178 oC 141 = 1.55 x c x 178 = 0.511 J g-1oC-1

  24. 5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions. • Calorimeter: Reactions used to heat up an external source of water. Temperature change of water, mass of material and mass of water are measured. Use q = m x c x ΔT to solve for q then find the heat of reaction in kJ/mol of reacting substance.

  25. Calorimetry Calorimetry involves the measurement of heat changes that occur in chemical processes or reactions. Determines the ΔH by measuring temp Δ's created from the rxn The heat change that occurs when a substance absorbs or releases energy is really a function of three quantities: The mass The temperature change The heat capacity of the material 25

  26. Heat Capacity and Specific Heat The ability of a substance to absorb or retain heat varies widely. The heat capacity depends on the nature of the material. The specific heat of a material is the amount of heat required to raise the temperature of 1 gram of a substance 1 oC (or Kelvin) 26

  27. Specific Heat values for Some Common Substances 27

  28. 5.2.2 Design suitable experimental procedures for measuring the heat energy changes of reactions. Heat Energy Change q = -q HUH?!?

  29. Heat Transfer Problem 2 What is the final temperature when 50 grams of water at 20oC is added to 80 grams water at 60oC? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g-1 oC-1 Solution:DQ (Cold) = DQ (hot) mCDT= mCDT Let T = final temperature (50 g) (4.184 J g-1oC-1)(T- 20oC) = (80 g) (4.184 J g-1oC-1)(60oC- T) (50 g)(T- 20oC) = (80 g)(60oC- T) 50T -1000 = 4800 – 80T 130T =5800 T = 44.6 oC 29

  30. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Using Thermochemical Equations Calcium oxide combines with water to produce calcium hydroxide and heat (exothermic reaction). • CaO(s) + H2O(l)  Ca(OH)2(s) + 65.2 kJ OR • CaO(s) + H2O(l)  Ca(OH)2(s) H = -65.2 kJ • These H values assume 1 mole of each compound (based on the coefficients) How many kJ of heat are produced when 7.23 g of CaO react? • Write out and balance equation: already balanced • Determine the number of moles: 7.23 g/56.01 gmol-1 = 0.129 mol • Multiply: 0.129 mol CaO x 65.2 kJ mol-1 * Notice that mol / mol cancel out and you’re left with kJ • Solve = 8.41 kJ

  31. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Using Thermochemical Equations Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to sodium carbonate, water, and carbon dioxide. How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)? • Write out and balance the equation • Determine the number of moles of NaHCO3(s) • Set up the ratio • Solve for x • State, with justification, whether the reaction is endothermic or exothermic.

  32. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Using Thermochemical Equations Sodium hydrogen carbonate absorbs 129 kJ of energy and decomposes to sodium carbonate, water, and carbon dioxide. • 2NaHCO3(s) + 129 kJ  Na2CO3(s) + H2O(g) + CO2(g) OR • 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g) H = 129 kJ How many kJ of heat are needed to decompose 2.24 mol NaHCO3(s)? • Balance equation • Moles NaHCO3(s) = 2.24 mol • Ratio: x kJ = 129 kJ 2.24 NaHCO3(s) 2 mol • Solve for x = 144 kJ • The reaction is endothermic because energy is being absorbed / the H is positive.

  33. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Using Experimental Data In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup calorimeter. At the start, the solutions and the calorimeter are all at 25.0 oC. During the reaction, the highest temperature observed is 32.0 oC. Calculate the heat (in kJ) released during this reaction. Assume the densities of the solutions are 1.00 g mL-1.

  34. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Using Experimental Data In a neutralization reaction, 25.0 mL of water containing 0.025 mol HCl is added to 25.0 mL of water containing 0.025 mol NaOH in a foam cup calorimeter. At the start, the solutions and the calorimeter are all at 25.0 oC. During the reaction, the highest temperature observed is 32.0 oC. Calculate the heat (in kJ) released during this reaction. Assume the densities of the solutions are 1.00 g mL-1. Use q = m x c x ΔT m = mass of solution = 50.0 mL x 1.00 g mL-1 = 50.0 g C = 4.18 j g-1oC-1 ΔT = 32.0 – 25.0 = 7.0 oC q = 50.0 g x 4.18 J g-1oC-1 x 7.0 oC = 1463 J = 1.5 kJ

  35. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Using Thermodynamic Quantities (Standard Heats of Formation) • Heat of reaction can be found by: sum the heats of formation of all the products – sum of heats of formation of all the reactants Hrxn=Hf products – Hf reactants • Hf= standard enthalpy of formation. Energy required to form a compound from its elements. • “standard” is a term used a lot in chemistry. It usually means that the values are experimentally determined and compared to an agreed upon reference value • Since the Hf is given per mole, we must multiply by coefficients

  36. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Using Thermodynamic Quantities (Standard Heats of Formation) Using the table of thermodynamic quantities, calculate the heat of reaction for 2SO2(g) + O2(g)  2SO3(g) Heat of reaction = Hf products – Hf of reactants Heat of products: Hf (SO3) = -395.2 kJ/mol x 2 = -790.4 kJ Heat of reactants = Hf (SO2) + Hf (O2) (-296.9 kJ/mol x 2) + (0) = -593.8 KJ Heat of reaction = -790.4 kJ – (-593.8 kJ) = -196.6 kJ

  37. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. N2O4 + 3 CO  N2O + 3CO2 Hf products = 1(81 kJ mol-1) + 3(-393 kJ mol-1) = -1098 kJ/mol Hf reactants = 1(-9.7 kJ mol-1)+ 3(-110 kJ mol-1) = -320.3 kJ mol-1 Hf products – Hf reactants = (-1098 kJ mol-1) – (-320.3 kJ mol-1) = -778 kJ mol-1 Hrxn = -778 kJ mol-1 Therefore it is exothermic

  38. 5.2.3 Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water. Ca(OH)2(s) + CO2 (g)  H2O(g) + CaCO3 (s) Hf products = 1(-241.8 kJ mol-1) + 1(-1206.9 kJ mol-1) = -1448.7 kJ/mol Hf reactants = 1(-986.1 kJ mol-1)+ 1(-393.5 kJ mol-1) = -1379.6 kJ mol-1 Hf products – Hf reactants = (-1448.7 kJ mol-1) – (-1379.6 kJ mol-1) = -69.1 kJ mol-1 Hrxn = -69.1 kJ mol-1 Therefore it is exothermic

  39. IB Topic 5: Energetics5.3 Hess’s Law & 5.4 Bond Enthalpies 5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes. 5.4.1 Define the term average bond enthalpy 5.4.2 Explain, in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic

  40. 5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes. Hess’s Law: Reactions can be added together in order to determine heats of reactions that can’t be measured in the lab. C(diamond)  C(graphite) This reaction is too slow to be measured in the lab. Two reactions can be used that can be measured in the lab: a) C(graph) + O2(g)  CO2(g) H = -393.5 kJ b) C(diam) + O2(g)  CO2(g) H = -395.4 kJ Since C(graphite) is a product, write equation a) in reverse to give: c) CO2(g)  C(graph) + O2(g) H = 393.5 kJ Now add equations b) and c) together: C(diam) + O2(g) + CO2(g)  C(graph) + O2(g) + CO2(g) H = -395.4 kJ + 393.5 kJ = -1.9 kJ Final equation: C(diamond)  C(graphite) H = - 1.9 kJ

  41. 5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes. Given the following thermochemical equations, calculate the heat of reaction for: C2H4(g) + H2O(l)  C2H5OH(l) a) C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) H = -1367 kJ b) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1411 kJ

  42. 5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes. Given the following thermochemical equations, calculate the heat of reaction for C2H4(g) + H2O(l)  C2H5OH(l) a) C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) H = -1367 kJ b) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1411 kJ Since C2H5OH(l) is a product, write equation a) in reverse order: c) 2CO2(g) + 2H2O(l)  C2H5OH(l) + 3O2(g) H = 1367 kJ Add equations b) & c) together, cancelling out substances on opposite sides of the arrow. Add the heat values to obtain the heat of reaction. C2H4(g) + H2O(l)  C2H5OH(l) H = -44 kJ

  43. 5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes. Given the following thermochemical equations, calculate the heat of reaction for: C(s) + 2H2(g)  CH4(g) a) C(s) + O2(g)  CO2(g) H = -393 kJ mol-1 b) H2(g) + ½ O2(g)  H2O(l) H = -286 kJ mol-1 c) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ mol-1

  44. 5.3.1 Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes. Given the following thermochemical equations, calculate the heat of reaction for: C(s) + 2H2(g)  CH4(g) C(s) + O2(g)  CO2(g) H = -393 kJ 2(H2(g) + ½ O2(g)  H2O(l)) H = 2(-286 kJ mol-1) 2H2(g) + O2(g)  2H2O(l)) H = -572 kJ CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ mol-1 CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) H = 890 C(s) + 2H2(g)  CH4(g)H = -75 kJ

  45. 5.4.1 Define the term average bond enthalpy In a chemical reaction Chemical bonds are broken Atoms are rearranged New chemical bonds are formed These processes always involve energy changes 45

  46. Energy Changes Breaking chemical bonds requires energy  Endothermic Forming new chemical bonds releases energy  Exothermic 46

  47. 5.4.1 Define the term average bond enthalpy Enthalpy changes of reactions are the result of bonds breaking and new bonds being formed. Remember… Breaking bonds requires energy Forming new bonds releases energy Bond enthalpy is the energy required to break one mole of a certain type of bond in the gaseous state averaged across a variety of compounds. FYI: Bond enthalpies for unlike atoms will be affected by surrounding bonds and will be slightly different in different compounds so average bond enthalpies are used.

  48. 5.3.2 Using bond enthalpy to determine enthalpy change of a reaction The average bond enthalpies for several types of chemical bonds are shown in the table below: 48

  49. 5.3.2 Using bond enthalpy to determine enthalpy change of a reaction • H = ∑ (energy required – ∑(energy released to break bonds) when bonds are formed) OR • H = ∑ (bond enthalpy – ∑(bond enthalpy of reactants) of products)

  50. 5.3.2 Using bond enthalpy to determine enthalpy change of a reaction H = ∑ (bond enthalpy – ∑(bond enthalpy of reactants) of products)

More Related