Chapter 6

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Chapter 6 • Energy • Law of Conservation of Energy • Potential vs Kinetic • Heat – transfer of Energy because of temp. difference

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Chapter 6. Energy Law of Conservation of Energy Potential vs Kinetic Heat – transfer of Energy because of temp. difference. State function System vs Surroundings Exothermic Endothermic. Thermodynamics 1 st law – Energy of Universe is constant Internal energy  E = q + w w = -p V. - PowerPoint PPT Presentation

Transcript of Chapter 6

Page 1: Chapter 6

Chapter 6

• Energy

• Law of Conservation of Energy

• Potential vs Kinetic

• Heat – transfer of Energy because of temp. difference

Page 2: Chapter 6

• State function

• System vs Surroundings

• Exothermic

• Endothermic

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• Thermodynamics

• 1st law – Energy of Universe is constant

• Internal energy

• E = q + w

• w = -p V

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• EnthalpyH = qp

H = Hprod – Hreact

• Calorimetry

• Specific heat

• q= sm T

• Specific heat of water is 4.18 J/g0C

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• What is the specific heat of 15 g of an unkown at 45 0C if when mixed with 55 g of water at 25 0C the final temperature is 32 0C?

• - q hot = q cold

• - smTunk. = sm TH2O

• - s(15g)(32-45 0C) = (4.18J/g0C)(55g)(32-25 0C)

• s = - (4.18J/g0C)(55g)(7 0C)/((15g)(-130C))

• s = 8 J/g0C

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• Constant volume E = qv

• Bomb Calorimeter– q = C T C = heat capacity of the calorimeter

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• Hess’s LawHtotal = H1 + H2 + ….

• If reaction is reversed change the sign

• If you multiply the reaction’s coefficients you also multiply the H

• N2 + O2 2NO H1 = 180 kJ

• 2NO + O2 2NO2 H2 = -112 kJ

• -----------------------------------------------

• N2 + 2O2 2NO2 H = H1 + H2 = 68 kJ

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• Cgraph. + O2 CO2 H = -394 kJ

• Cdiam. + O2 CO2 H = -396 kJ

• Cgraph. + O2 CO2 H = -394 kJ

• CO2 O2 + Cdiam. H = +396 kJ

• ---------------------------------------------

• Cgraph. Cdiam. H = 2 kJ

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• Standard Enthalpies of Formation Hf

o

– Values for formation of 1 mole at standard conditions

– Reactants are all elements– Elements have a value of zero– From Appendix 4 on pg. A21

H rxn = nprodHfo prod - nreactHf

o react

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H rxn = nprodHfo prod - nreactHf

o react

• CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = ?

H = 1n(-394kJ/n)+2n(-286kJ/n) – 1n(-75kJ/n) – 0

H = -891 kJ