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Chapter 5:. Thermochemistry. Thermochemistry:. Energy Kinetic & Potential First Law of Thermo internal energy, heat & work endothermic & exothermic processes state functions Enthalpy Enthalpies of Reaction. Calorimetry heat capacity and specific heat constant-pressure calorimetry

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slide1

Chapter 5:

Thermochemistry

thermochemistry
Thermochemistry:
  • Energy
    • Kinetic & Potential
  • First Law of Thermo
    • internal energy, heat & work
    • endothermic & exothermic processes
    • state functions
  • Enthalpy
  • Enthalpies of Reaction
slide3
Calorimetry
    • heat capacity and specific heat
    • constant-pressure calorimetry
    • bomb calorimetry (constant-volume calorimetry)
  • Hess’s Law
  • Enthalpies of Formation
    • for calculation of enthalpies of reaction
  • Foods and Fuels
slide4
Energy
  • work is a form of energy w = F x d
  • energy is the capacity to do work or transfer heat
  • Kinetic Energy
    • energy of motion E = ½ mv2
  • potential energy
    • energy of position
    • applies to electrostatic energy
    • applies to chemical energy (energy of bonds)
slide5
energy units
    • one joule = energy of a 2 kg mass moving at 1 m/s
    • E = ½ mv2 (½)(2 kg) (m/s)2 = kg m2/s2 = 1 J
    • 1 cal = 4.184 J 1 kcal = 1 food calorie (Cal)

Systems & Surroundings

  • system -- chemicals in the reaction
  • surroundings -- container & all outside environment
  • closed system can exchange energy (but not matter) with its surroundings
slide6

Closed System

energy(as heat or work)

2H2(g) + O2(g)

¯

2H2O(l)+ energy

(system)

no exchg of matterwith surroundings

slide7
First Law of Thermo.
  • Energy is always conserved
    • any energy lost by system, must be gained by surroundings
  • Internal Energy -- total energy of system
    • combination of all potential and kinetic energy of system
    • incl. motions & interactions of of all components
    • we measure the changes in energy E = Efinal - Einitial
slide8
+ D E = Efinal > Einitial system has gained E from surroundings
  • - D E = Efinal < Einitial system has lost E to surroundings
  • Relating D E to heat and work
    • D E = q + w q is positive if heat goes from surroundings to system w is positive if work is done on system by surroundings
slide9

+q +w

- q - w

surroundings

surroundings

system

system

heat

heat

work

work

slide10
Endothermic
  • system absorbs heat or heat flows into the system

Exothermic

  • system gives off heat or heat flows out of the system

State Function

  • a property of a system that is determined by specifying its condition or state (T, P, etc.)
  • internal energy is a state function, \DE depends only on Efinal & Einitial
slide11
Enthalpy
  • for most reactions, most of the energy exchanged is in the form of heat, that heat transfer is called enthalpy, H
  • enthalpy is a state function
  • like internal energy, we can only measure the change in enthalpy, DH
    • DH = qp when the process occurs under constant pressureDH = Hfinal - Hinitial = qp
  • - DH Þ exothermic process
  • +DH Þ endothermic process
slide12

system

DH > 0

surroundings

system

surroundings

DH < 0

slide13

Enthalpies of Reaction

    • D Hrxn = Hprod - Hreact
  • enthalpy is an extensive property
    • magnitude of D H depends directly on the amount of reactant
    • C(s) + 2H2(g)® CH4(g)D H = -74.8 kJ/mol
    • 2C(s) + 4H2(g)® 2CH4(g)D H = -149.6 kJ/2mol
slide14
enthalpy change for forward rxn is equal in magnitude but opposite in sign for the reverse rxn
    • CH4(g) ® C(s) + 2H2(g)D H = +74.8 kJ/mol
    • C(s) + 2H2(g)® CH4(g) D H = - 74.8 kJ/mol
  • enthalpy change for a reaction depends on the state of the reactants and products
    • C(g) + 2H2(g)® CH4(g)D H = -793.2 kJ/mol
    • 2H2(g) + O2(g)® 2H2O(g)D H = -486.6 kJ/mol
    • 2H2(g)+ O2(g)® 2H2O(l)D H = -571.7 kJ/mol
slide15

DH = Hfinal - Hinitial

H2O(g)

-241.8 kJ

44 kJ

-

+

Enthalpy

H2O(l)

-285.8 kJ

slide16
Practice Ex. 5.2:
  • Hydrogen peroxide can decompose to water and oxygen . Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.
    • 2H2O2(l)® 2H2O(l) + O2(g)D H = -196 kJ
    • 5.00 g H2O2(l) x 1 mol = 0.147 mol H2O2(l) 34.0 g H2O2(l)
    • 0.147 mol H2O2(l) x -196 kJ H2O2(l) = -14.4 kJ 2 mol
slide17
Calorimetry
  • experimental determination of D H using heat flow

heat capacity

  • measures the energy absorbed using temperature change
  • the amount of heat required to raise its temp. by 1 K
  • molar heat capacity -- heat capacity of 1 mol of substance
slide18
specific heat
  • heat energy required to raise some mass of a substance to some different temp.
  • specific heat = quantity of heat trans. (g substance) (temp. change)
  • = q . m DT
  • S.H. = joule g K
  • q = (S.H.) (g substance) (D T)

remember: this is change in temp.

slide19
Practice Ex. 5.3:
  • Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temp. increases by 12.0 °C if the specific heat of the rocks is 0.82 J/gK.
    • S.H. x g x DT = joules
    • What unit should be in the solution?
    • joules -- quantity of heat
    • 0.82 J x 50.0 x 103 g x 12.0 K = 4.9 x 105 Jg K
slide20
Constant-Pressure Calorimetry
  • D H = qp at constant pressure as in coffee cup calorimeter
    • heat gained by solution = qsoln
    • \qsoln = (S.H.soln)(gsoln)(DT)
    • heat gained by solution must that which is given off by reaction
    • \qrxn = - qsoln = - (S.H.soln)(gsoln)(DT)

must be opposite in sign

if DT is positive then qrxn is exothermic

slide21
Practice Ex. 5.4:
  • When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from 22.30°C to 23.11°C. Calculate D H for this reaction, assuming that the combined solution has a mass of 100.0 g and a S.H. = 4.18 J/g °C.
    • AgNO3(aq) + HCl(aq) ® AgCl(s) + HNO3(aq)
    • qsoln = 4.18 J x 100.0 g soln x 0.81°C = 3.39 x 102 J g °C
    • qrxn = - qsoln = - 3.39 x 102 J = - 68,000 J or 0.00500 mol - 68 kJ/mol
slide23
Bomb Calorimetry (Constant-Volume)
  • bomb calorimeter has a pre-determined heat capacity
  • sample is combusted in the calorimeter and D T is used to determine the heat change of the reaction
  • qrxn = - Ccalorimeter x D T

heat capacity of calorimeter

because rxn is exothermic

slide24

thermometer

insulation

water

rxn

slide25
Practice Ex. 5.5:
  • A 0.5865 g sample of lactic acid, HC3H5O3, is burned in a calorimeter with C = 4.812 kJ/°C. Temp. increases from 23.10°C to 24.95°C. Calculate heat of combustion per gram and per mole. D T = +1.85°C
    • qrxn = - (4.812 kJ/°C) (1.85°C) = - 8.90 kJ per 0.5865 g lactic acid-8.90 kJ = - 15.2 kJ/g 0.5865 g
    • - 15.2 kJ x 90 .1 g = - 1370 kJ/mol 1 g 1 mol
slide26
Hess’s Law
  • rxns in one step or multiple steps are additive because they are state functions
    • eg.
    • CH4(g) + 2O2(g)® CO2(g) + 2H2O(g)D H = - 802 kJ
    • CH4(g) + 2O2(g)® CO2(g) + 2H2O(l)D H = - 890 kJ

2H2O(g)® 2H2O(l)D H = - 88 kJ

slide27
Practice Ex. 5.6:
  • Calculate D H for the conversion of graphite to diamond:
    • Cgraphite® Cdiamond
    • Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ
    • Cdiamond + O2(g)® CO2(g)D H = -395.4 kJ
    • Cgraphite + O2(g)® CO2(g)D H = -393.5 kJ
    • CO2(g)® Cdiamond + O2(g)D H = 395.4 kJ

Cgraphite® CdiamondD H = + 1.9 kJ

slide28
Enthalpies of Formation
  • enthalpies are tabulated for many processes
    • vaporization, fusion, formation, etc.
  • enthalpy of formation describes the change in heat when a compound is formed from its constituent elements, DHf
  • standard enthalpy of formation, DHfo, are values for a rxn that forms 1 mol of the compound from its elements under standard conditions, 298 K, 1 atm
slide29
For elemental forms:
    • eg. C(s) graphite, Ag(s) , H2(g) , O2(g) , etc.
    • DHfo, for any element is = 0
  • used for calculation of enthalpies of reaction, DHrxn
  • DHrxn = S DHfo prod - S DHfo react
slide30
Practice Ex. 5.9:
  • Given this standard enthalpy of reaction, use the standard enthalpies of formation to calculate the standard enthalpy of formation of CuO(s)(
    • CuO(a) + H2(g)® Cu(s) + H2O(l)DHo = -130.6 kJ
    • DHrxn = S DH f o prod - S DH f o react
    • -130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)]
    • DHfo CuO = -155.2 kJ/mol
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