Chem 161-2010 Final exam review session

Chem 161-2010 Final exam review session - CHEM 161-2010...

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CHEM 161-2010 FINAL EXAM REVIEW SESSION Coverage: Chapter 6 – Thermochemistry Enthalpy Calorimetry and heat capacity Chapter 11 – States of matter and intermolecular forces Phase changes and phase diagrams Intermolecular forces (dipole-dipole, London, etc.) Structures and properties of solids (crystallography) Chapter 10 – Bonding theory and molecular structure (VSEPR) Geometric isomerism Chapter 9 – Chemical bonds Alkenes and alkynes Chapter 7 – Light and atomic structure Quantum mechanics/wave-particle duality/Planck’s equation/Einstein’s photoelectric effect/DeBroglie theorem Chapter 5 – Gases Partial pressures Chapter 4 – Chemical reactions in aqueous solutions Oxidation-reduction reactions Chem 161-2010 Lecture 22 1
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Chapter 6 – Thermochemistry Enthalpy Calorimetry and heat capacity CHAPTER 6 RELEVANT EQUATIONS ∆E = ΔU = q + w 1 Latm = 101.3 J Pressure-volume work: w = -PΔV; therefore, ∆E = q –P∆V Non-pressure-volume work: w = F x d (note, using a piston and cylinder, the piston has to move); F = m x a; w = m x a x d = m x g x d Units of non-pressure-volume work: m in kg, g = 9.81 m x s -2 , d in m; Units: kg x ms -2 x m = kgm 2 s -2 = J Heat Capacity = C = q/ΔT molar heat capacity = C/mol = q/(ΔT x mol) Specific Heat Capacity = SHC = SH = C/g = (q/ΔT)/g = q/ΔTg q = SHC x ΔT x g SHC H2O = 4.18J/g o C First law of thermodynamics: Energy is neither created nor destroyed. Therefore, heat lost by one = the heat gained by other. -q 1 = +q 2 Hess’s Law : ΔH is a state function, so path not important. Rules for changing enthalpy equations: o If the enthalpy equation is reversed, sign of ΔH is reversed. o If the enthalpy equation is halved, quantity of ΔH is halved. o If the enthalpy equation is doubled, quantity of ΔH is doubled. Enthalpy of Formation (more correctly, “Standard enthalpy of formation”) = ΔH o f o is for formation of one mole of product o from its elements o all reactants (i.e., elements) and products in their standard states (1 atm & 298K) o all elements in their most stable form, e.g., oxygen = O 2(g) ; mercury = Hg (l) ; sodium = Na (s) Chem 161-2010 Lecture 22 2
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o ΔH o f of elements = 0 ET: Discuss extensive and intensive properties; Use 100 mL H 2 O → 200 mL H 2 O as an example for mass change being extensive, and T and d change being intensive; ΔH is extensive ENTHALPY OF REACTION 1 Chem 161-2005 Hourly Exam II Chapter 6 – Thermodynamics Enthalpy H 2 (g) + Br 2 (g) 2HBr(g) H = -104 kJ The information given in the equation above means that: A. 104 kJ is released for each mole of HBr formed B. 104 kJ is absorbed for each mole of HBr formed C . 52 kJ is released for each mole of HBr formed D. 52 kJ is absorbed for each mole of HBr formed Ε. ∆ H f for HBr(g) = -104 kJ A negative H means that the reaction is exothermic, i.e., heat is released. H of -104 kJ means that 104 kJ of heat is released when 1 mol of H 2 reacts with 1 mol of Br 2 to form 2 moles of HBr. Therefore, 52 kJ of heat is released when 1 mol of HBr is formed.
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Chem 161-2010 Final exam review session - CHEM 161-2010...

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