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UNIT 5 THERMOCHEMISTRY

UNIT 5 THERMOCHEMISTRY. THERMOCHEMISTRY. Thermochemistry is the study of heat flow or heat energy in a chemical reaction. Chemical energy is transformed into Heat Energy during a chemical reaction. HEAT ENERGY.

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UNIT 5 THERMOCHEMISTRY

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  1. UNIT 5 THERMOCHEMISTRY

  2. THERMOCHEMISTRY • Thermochemistry is the study of heat flow or heat energy in a chemical reaction. • Chemical energy is transformed into Heat Energy during a chemical reaction.

  3. HEAT ENERGY • Heat energy flows into or out of a system because of a difference in temperature between it and its surroundings.

  4. The system is a part of the universe on which we focus our attention. • The surrounding is the rest of the universe.

  5. Heat is expressed by the symbol “q”. • “q” is expressed in Joules or kJ. It can also be expressed in calories. • 1 calorie = 4.184 Joules.

  6. FIRST LAW OF THERMODYNAMICS • Energy can neither be created nor destroyed. • Any energy lost by the system must be gained by the surroundings and vice-a-versa

  7. Internal Energy The internal Energy is the sum of all of the Kinetic and Potential Energies of the System. It is designated by E

  8. ∆E The change in internal heat = ∆E = Efinal - Einitial

  9. ∆E • Internal Energy is a State Function. • State Functions depend only on the present state of the system and not on the path that got it there. • ∆E is also a State Function.

  10. ∆E ∆E is the sum of all of the heat added or taken away from the system, and the work done by the system. ∆E = q + w

  11. A gas is placed under constant pressure. When .750 kJ of heat is added to the gas, it expands and does 250.J of work on its surroundings. Calculate ∆E- ∆E = q + w 750. Joules + 250. Joules ∆E = 1000 Joules or 1.00 kJ

  12. ENDOTHERMIC • When the heat flows from the surroundings into the system, the process is called endothermic. • Heat is taken up or absorbed during the reaction. • ∆Ewill be +

  13. EXOTHERMIC • When heat flows from the system into the surroundings, the process is called exothermic. • ∆Ewill be negative.

  14. SPECIFIC HEAT • C is the Specific Heat Capacity. • Specific heat is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree C. • The above equation is often written as q = m c ∆t

  15. The specific heat of iron is C = 0.466 J/g(°C). How much heat is given off by a 50.0 g sample of Fe when it cools from 80.0 °C to 50.0 °C?

  16. Cr has a specific heat of 0.449 J/g(°C).5.00 g of Cr at 23°C absorbs 60.5 J of heat. Calculate the final temperature.

  17. ENTHALPY • Enthalpy is a chemical energy represented by the symbol H. Enthalpy (H) is expressed in joules (J) or kilojoules.

  18. ∆H is Change in ENTHALPY • So ∆H for a chemical reaction is expressed as: ∆H = HProducts - HReactants

  19. Thermochemical Equation • Consider the chemical reaction: 2 H2 (g) + O2 (g)  2 H2O (g) ∆H = - 483.6 kJ • The chemical equation is balanced and it shows the enthalpy change associated with it. • The negative value of ∆H indicates that the reaction is exothermic.

  20. Consider the reaction: 2 H2O (g) + CO2 (g)  CH2 (g) + 2 O2 (g) ∆H = 890 kJ • The positive value of ∆H indicates that the thermochemical reaction is an endothermic reaction.

  21. Enthalpies of Reaction Enthalpy of reaction or Heat of Reaction can be written as: ∆Hrxn

  22. ENTHALPY RULES

  23. Enthalpy Rule #1 • The magnitude of ∆H is proportional to the amount of products and reactants.”

  24. Consider the chemical reaction: • 2 H2 (g) + O2 (g)  2 H2O (g) ∆H = - 483.6 kJ • According to this reaction: 2 moles of H2 reacts with 1 mole of O2 to produce 2 moles of H2O and the amount of energy released by this reaction is 483.6 kJ. • For 1 mole of H2O its

  25. Consider the following: 2 H2 (g) + O2 (g)  2 H2O (g) ∆H = - 483.6 kJ • Calculate ∆H when 1.00g H2 reacts

  26. ENTHALPY Rule #2 ∆H for a reaction is equal in magnitude, but opposite in sign, for the reverse reaction.

  27. Consider this reaction: • H2 (g) + Cl2 (g)  2 HCl (g) ∆H = -185 kJ • In this reaction: 1 mole of H2 reacts with 1 mole of CI2 to produce 2 moles of HCI and the amount of energy released is 185 kJ. • So in: 2 HCl (g)  H2 (g) + Cl2 (g) ∆H = +185 kJ • In this reaction: 2 moles of HCI break down to form 1 mole of H2 and 1 mole of CI2. Energy absorbed is 185 kJ.

  28. Consider the following equation: H2 (g) + Cl2 (g)  2 HCl (g) ∆H = -185 kJ • Calculate ∆H for: HCl (g)  ½ H2 (g) + ½ Cl2 (g) • Reverse and apply Rule # 2. 2 HCl (g)  H2 (g) + Cl2 (g) ∆H = Then calculate ∆H for one mole of HCI, use Rule #1. ∆H =

  29. Enthalpy Rule #3 The Enthalpy Change is affected by the states of the reactants and the products. Each state has their own quantity of Enthalpy Change. Therefore, the states of everything must be expressed.

  30. 2Mg (s) + O2 (g) 2MgO (s) ∆H= -1204 kJ Calculate the amount of heat transferred when 24.3 grams of Magnesium reacts? It is an Exothermic reaction and ∆H =

  31. 2Mg (s) + O2(g) 2MgO(s) ∆H= -1204 kJ • How many kilojoules of heat are absorbed when 161.2 g of Magnesium Oxide Decomposes?

  32. C6H6 (l) 3C2H2 (g) ∆H = +630 kJ 1- What is the Enthalpy change of the reverse reaction? 2- Find the ∆H for the formation of 1 mole of Acetylene? 3- How many grams of Acetylene will be produced when +1260 kJ of heat are absorbed by Benzene? 1 = kJ 2 = kJ 3 = g

  33. CALORIMETRY • A calorimeter is used for the measurement of heat. It is thermally isolated from the surroundings  q = 0 for the whole calorimeter (no heat escapes or enters the calorimeter).

  34. Within a CALORIMETER • Whatever is lost by a material (Exothermic) will be gained by the water. • Therefore if you calculate the heat energy change of the water, you’ll know what is lost by the material.

  35. When 2.00 g of MgCl2 is added to 65.0 g of water in a calorimeter, it dissolves according to: MgCl2(s) Mg +2(aq) + 2 Cl -(aq) The temperature rises from 25.00 °C to 29.87 °C. Calculate the heat of dissolution of MgCl2. Express with the correct Sig Figs. Joules

  36. HESS’s LAW • A chemical reaction can be represented as a combination of several steps. • The ∆Ho for the overall reaction is the same whether the chemical reaction occurs in one step or several steps.

  37. Consider a chemical reaction that occurs in two steps: • Step 1: C (s) + ½ O2(g) CO (g) ∆Ho = - 110.5 kJ • Step 2: CO (g) + ½ O2(g) CO2(g) ∆Ho = - 283.0 kJ • The Overall reaction is: C (s) + O2(g) CO2(g) ∆Ho = -393.5 kJ

  38. HESS’s LAW • Since Overall reaction = Step 1 reaction + Step 2 reaction, then, • ∆Ho Overall = ∆Ho Step 1 + ∆Ho Step 2 • This relationship is called Hess’s law.

  39. HESS’s LAW • Given: Step #1 S (s) + O2(g) SO2(g) ∆Ho = - 297 kJ Step #2 2 SO3(g) 2 SO2(g) + O2(g) ∆Ho = 198 kJ • Calculate ∆Ho for the overall reaction: 2 S(s) + 3 O2(g) 2 SO3(g)

  40. HESS’s LAW • S is a reactant, but in the Overall reaction, there are 2 moles of S whereas in Step 1 there is 1. • So Multiply Step 1 by 2. 2 S(s) + 2 O2(g)  2 SO2(g) ∆H0 = - 594 kJ • Ignore O2 since it appears in both steps.

  41. HESS’s LAW • In the overall equation, SO3 appears as a product whereas in Step 2, it appears as a reactant. There are 2 moles of SO3 in the Overall reaction as well as in Step 2. • Reverse the reaction. 2 SO2(g) + O2(g) 2 SO3(g) ∆H0 = -198 kJ

  42. HESS’s LAW • ∆H0 = (-594 kJ) + (-198 kJ) = -792 kJ • Add reactants and products of both steps, this would result in the Overall reaction. 2 S (s) + 3 O2(g) 2 SO3 (g) • Now apply Hess’s law to calculate ∆HO

  43. Here are some more HESS’s LAW problems….

  44. 2CO (g) + 2NO (g) 2CO2 (g) + N2 (g) Use the following to determine ∆H for the above reaction: 2CO (g) + O2(g) 2CO2(g) ∆H = -566.0 kJ N2 (g)+O2(g) 2NO (g)∆H = 180.6 kJ ∆H =

  45. 4Al (s) + 3MnO2(s) 2Al2O3 (g) + 3Mn(s) Use the following to determine ∆H for the above reaction: 4Al (s)+ 3O2(g) 2Al2O3(s)∆H = -3352 kJ Mn(s)+O2(g) MnO2(s)∆H = -521 kJ ∆H =

  46. 2H2O2(l) 2H2O(l) + O2 (g) Use the following to determine ∆H for the above reaction: 2H2(g) + O2(g)  2H2O ∆H = -572 kJ H2 (g)+O2(g) H2O2(l)∆H = -188 kJ ∆H =

  47. ENTHALPIES of FORMATION • Enthalpy of Formation- ΔHf • Standard Enthalpy Change- ΔH° • Standard Enthalpy of Formation- ΔHf° All substances that make it up are in their standard states. These include 1 mole at 298 Kelvin and they are expressed in kJ/mole

  48. ENTHALPIES of FORMATION *********************************For the most stable state of the ELEMENT, the Standard Enthalpy of Formation is 0. \*********************************

  49. Using these enthalpies of formation and Hess' Law the enthalpy change of any reaction involving the compounds with a known enthalpy of formation can be determined.

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