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Thermochemistry

Thermochemistry. Chapter 16 . Thermochemistry. The study of energy involved during chemical reactions. Energy sources: ~ chemical ~ nuclear ~ solar ~ geothermal ~ wind/water. Heat: the energy of motion of molecules

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Thermochemistry

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  1. Thermochemistry Chapter 16

  2. Thermochemistry • The study of energy involved during chemical reactions. Energy sources: ~ chemical ~ nuclear ~ solar ~ geothermal ~ wind/water

  3. Heat: the energy of motion of molecules • Kinetic molecular theory: substances are composed of particles that are continually moving and colliding with other particles. • Kinetic energy: energy of motion • Potential energy: stored energy

  4. Temperature: • transfer of heat to a substance because of faster molecular movement (as long as there is no phase change) • Measured with a thermometer

  5. A temperature change is explained as a change in kinetic energy • Temperature depends on the quantity of heat (q) flowing out or in of the substance.

  6. Heat (q) q=mc ∆t • q=heat • m=mass • ∆t=change in temperature (tf-ti) • c=specific heat capacity (J/(g oC) Specific heat capacity is the quantity of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.

  7. Table of Specific Heats

  8. Law of conservation of energy • ∆E universe = O • The total energy of the universe is constant, it is not created or destroyed, however it can be transferred from one substance to another. • ∆E universe = ∆E system + ∆E surroundings

  9. First Law of thermodynamics • Any change in energy of a system is equivalent by an opposite change in energy of the surroundings. • ∆E system = - ∆E surroundings • According to this law, any energy released or absorbed by a system will have a transfer of heat, q. • So, q system = -q surroundings

  10. Sample Problem • 15 g of ice was added to 60.0g of water. The Ti of water was 26.5 oC, the final temperature of the mixture was 9.70 oC. How much heat was lost by the water? • q=mc ∆t q=(60.0g) (4.18 J/g oC) (9.70-26.5 oC) q= - 4213.44 J =-4.21 kJ

  11. Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 56oC. q = m x c x (Tf - Ti)q = 250g x 4.18 J/g oC x (56oC - 20oC)q = 37 620 J = 38 kJ

  12. Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25oC to 60oC. • q = m x c x (Tf - Ti)q = 204.75 Jm = 15gTi = 25 oCTf = 60 oC 204.75 J = 15 g x c x (60oC - 25oC)204.75 J= 15 g x c x 35oC204.75 J= 525 g oC x cc = 204.75 J ÷ 525 g oC = 0.39 J/ g oC

  13. Watch flash about heat flow http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/enm1s3_4.swf

  14. Homework: • Handout

  15. Enthalpy (H) • Total kinetic and potential energy (internal energy) of a system under constant pressure. • System: area where reaction takes place • Its particles average motion define its properties. • Surroundings: outside of the system

  16. Open-system • both matter and energy can freely cross from the system to the surroundings and back. • Ex: an open test tube

  17. Closed-System • energy can cross the boundary, but matter cannot. • Ex: a sealed test tube

  18. Isolated-System • neither matter nor energy can cross between the system and the surroundings. • Ex: The universe • there are no surroundings to exchange matter or energy with (as far as we know!)

  19. The internal energy of a reactant or product cannot be measured, but their change in enthalpy (heat of reaction) can. ∆ H = Hproducts – Hreactants A change in enthalpy occurs during phase changes, chemical reactions and nuclear reactions. ∆ H system = q surroundings

  20. If energy is released out of the system then it is considered to be an exothermic enthalpy change (exit). If energy is absorbed into the system then it is considered to be an endothermic enthalpy change (enter).

  21. Endothermic Reactions

  22. Endothermic Reactions Method 2: 2 HgO (s) 2 Hg (l) + O2(g)ΔH=181.67 kJ Method 3: 2 HgO (s) + 181.67 kJ 2 Hg (l) + O2(g)

  23. Exothermic Reactions

  24. Exothermic Reactions Method 2: 4Al(s)+ 3O2(g) 2 Al2O3(g)ΔH=-1675.7 kJ Method 3: 4 Al(s) + 3O2(g) 2 Al2O3(g) +1675.7 kJ

  25. Calorimeter • Instrument used to measure amount of energy involved in a chemical reaction. • It is equivalent to an isolated or closed system. (nothing may enter or exit the system) • The energy change is not measured within the system, but the energy transferred to its surroundings.

  26. A basic calorimeter • Two styrofoam cups nestled within one another (insulation), then filled with a specific quantity of water. • A chemical reaction or phase change takes place inside and a thermometer is placed within to measure any change in temperature that occurs to the system.

  27. Assumptions • It is an isolated or closed system and there is no heat transfer between the calorimeter and its surroundings. • The amount of heat absorbed or released by the calorimeter itself is too small to influence calculations. • Any dilute solutions involved in the reaction are treated as if they are water.

  28. Bomb calorimeter • Cannot use basic design for combustion reactions. • Used in research for H for fuels, oils, food, explosives… • Larger and more sophisticated • The reaction container is strong enough to with stand an explosion, hence the name “bomb”

  29. Have fixed components, like volume of water, thermometer… • Heavily insulated or vacuum insulated so no convection or conduction can occur affecting the enthalpy of the system. • Use the equation: q=CΔt

  30. A Bomb Calorimeter

  31. Problems • Pg 638 # 5-7 • Handout Questions 1-10

  32. Standard Molar Enthalpy of Formation • Quantity of energy released (-) or absorbed (+) when one mole of a compound is formed directly from its elements at standard temperature and pressure. • We use a table to find them. • Unit for ΔHf: kJ/mol • Watch your states!

  33. Calculating enthalpy changes • Amount of a substance reacting matters, so can use q= nΔH. • Remember n=amount of moles. • If you are given a mass (g) and molar mass (g/mol), then you can solve for n by dividing mass by molar mass. (review from chem 11 stoichiometry section)

  34. Sample Problem: • Show the formation reaction of methanol. • If I had 10.5 g of CH3OH(l), how much energy would be released? C(s) + 2H2(g) + 1/2O2(g)  CH3OH(l) ΔHf = -239 kJ/mol mm= (12.01) +(4 x 1.01) + (16.00) = 32.05 g/mol q = nΔH = (10.5 g / 32.05 g/mol) (-239 kJ/mol) = -78.3 KJ

  35. On Your Own: • I have 125.6 g of NaOH(s) formed at standard temperature and pressure, how much energy will be released? Answer: - 1336.4 kJ

  36. Standard Molar Enthalpy of Combustion • Energy changes involved with combustion reactions of one mole of a substance. • Remember that these reactions are only measured once cooled to 25oC • Combustion is a reaction with oxygen as a reactant

  37. Practice Problem: • Write the combustion reaction for methane (CH4) • ΔHcomb for CH4 is -965.1 kJ/mol • How much energy is released when 12.8 g of CH4 is combusted? CH4(g) + 2O2 (g)  2H2O(l) + CO2 (g) ΔHcomb = -965.1 kJ/mol q=nΔH = (12.8 g/ 16.05 g/mol) (-965.1 kJ/mol) = - 769.7 kJ

  38. Combustion reaction: • Always form water and carbon dioxide. • Ex: CH4 + 2O2 CO2 + 2H2O • Remember your alkanes: CnH2n+2 • Meth: C=1, Eth: C=2, Prop: C=3, But: C=4, Pent: C=5, Hex: C=6…

  39. Practice • Pg 643 # 15-17 • Pg 645 # 19,21,23

  40. Enthalpy and changes of state A change in matter without any change in chemical composition of the system *** Always involves energy changes, but not temperature changes. There is constant temperature during a phase change. Why?

  41. Because… No change in kinetic energy, however we know that energy is entering the system because the bonds that are holding the molecules together are being broken or altered. This increases Ep of the molecules

  42. Just think! • We know that there are energy changes during phase changes, because when we sweat our body is trying to cool itself down using evaporation!

  43. Phase changes: (surroundings)

  44. Latent Heat of Phase ChangeΔH melt = - ΔH freez Molar Heat of Fusion or melting The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point. • Molar Heat of Freezing or solidifying • The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

  45. ΔH vap = - ΔH cond Molar Heat of Vaporization The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point. • Molar Heat of Condensation • The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.

  46. Molar Enthalpy for changes of state • You can find tables of these values listed • Unit: kJ/mol

  47. Latent Heat – Sample Problem Problem: The molar heat of fusion of water is 6.009 kJ/mol. How much energy is needed to convert 60g of ice at 0C to liquid water at 0C? Mass of ice Heat of fusion Molar Mass of water

  48. Chemical reactions: • If a reaction is exothermic, energy is released to the surroundings (temp increases) and chemical potential energy of the system decreases • ∆ H is a negative number

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