THERMOCHEMISTRY or Thermodynamics

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THERMOCHEMISTRY- A
By Dr. Hisham E Abdellatef
2011-2012
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THERMOCHEMISTRY
The study of heat released or required by
chemical reactions or heat changes caused by
chemical reactions
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SURROUNDINGS

• Surrounding the part of the universe which
  surround the system.

System part of universe selected for the
thermodynamic study.
HEAT
HEAT
HEAT
HEAT
SYSTEM
SYSTEM
EXOTHERMIC
ENDOTHERMIC
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Heat of Reaction

• Heat of reaction the change in
  energy which accompanies a chemical reaction. ?
  H
• Endothermic reaction the reaction which is
  supplied by heat (absorb heat and ? H is ve).
• Exothermic reaction the reaction which is
  accompanied with evolution of heat (evolve heat
  and ? H is - ve).

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System and Surroundings

• In chemical reactions, heat is often transferred
  from the system to its surroundings, or vice
  versa.

• The substance or mixture of substances under
  study in which a change occurs is called the
  thermodynamic system (or simply the system.)
• The surroundings are everything outside of the
  thermodynamic system.

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Exothermic
If ?E lt 0, Efinal lt Einitial
cellular respiration of glucose
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Endothermic
If ?E gt 0, Efinal gt Einitial
Photosynthesis is an endothermic reaction
(requires energy input from sun)
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What is Energy?
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Specific HEAT Jg-1C-1

• The amount of heat which required to raise the
  temperature of 1gof substance by 1C.
• Molar heat capacity
• The amount of heat which required to raise the
  temperature of one mole of substance by 1C
• molar heat capacity specific heat x M.wt.

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• Heat of reaction
• The change in energy which accompanies a chemical
  reaction.
• Heat of combustion
• The change in enthalpy (heat evolved) when one
  mole of the substance is completely burned in
  presence of excess oxygen.
• Heat of formation
• The change in enthalpy (heat evolved) when one
  mole of the substance is formed from its elements
  their standard state. (t 25 C and P 1 atm).

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HEAT CAPACITY

• JC-1 Specific Heat x mass
• The amount of heat which required to raise the
  temperature of the substance by 1C.

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How do we relate change in temp. to the energy
transferred? Heat capacity (J/oC) heat
supplied (J)
temperature (oC)
Heat Capacity heat required to raise temp. of
an object by 1oC
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Specific heat capacity is the quantity of energy
required to change the temperature of a 1g sample
of something by 1oC
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Specific Heat Capacity

• How much energy is transferred due to T
  difference?
• The heat (q) lost or gained is related to
• a) sample mass
• b) change in T and
• c) specific heat capacity

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Specific Heat Capacity

• Substance Spec. Heat (J/gK)
• H2O 4.184
• Ethylene glycol 2.39
• Al 0.897
• glass 0.84

Aluminum
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Specific Heat Capacity

• If 25.0 g of Al cool from 310 oC to 37 oC, how
  many joules of heat energy are lost by the Al?

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Specific Heat Capacity

• If 25.0 g of Al cool from 310 oC to 37 oC, how
  many joules of heat energy are lost by the Al?

where ?T Tfinal - Tinitial q (0.897
J/gK)(25.0 g)(37 - 310)K q - 6120 J
Notice that the negative sign on q signals heat
lost by or transferred OUT of Al.
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UNITS OF ENERGY

• 1 calorie heat required to raise temp. of 1.00
  g of H2O by 1.0 oC.
• 1000 cal 1 kilocalorie 1 kcal
• But we use the unit called the JOULE
• 1 cal 4.184 joules

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FIRST LAW OF THERMODYNAMICS

• ?E q w

Energy is conserved!
?????????
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SYSTEM
?E q w
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• 1st Law of Thermodynamics
• Energy can neither created not destroyed, only
  transformed from one form to another

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Exothermic
Endothermic
6.2
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Enthalpy Diagrams

• Values of DH are measured experimentally.
• Negative values indicate exothermic reactions.
• Positive values indicate endothermic reactions.

An increase in enthalpy during the reaction DH
is positive.
A decrease in enthalpy during the reaction DH is
negative.
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Exothermic Examples

• Oxidation wooden splint burning (giving off
  light, heat, CO2, H2O
• Burning H2 in air,
• body reactions,
• dissolving metals in acid,
• mixing acid and water,
• sugar dehydration

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Endothermic Examples

• Electrolysis (breaking water down into H2 and O2
  by running electricity in it)
• Photosynthesis, pasteurization, canning
  vegetables
• 2 H2 O2 ? 2H2O energy
• 4 g 32 g ? 36 g 136 600 cal
• 2H2O energy ? 2H2 O2
• 36 g 136 600 cal ? 4g 32 g

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Changes in Internal Energy

• If ?E gt 0, Efinal gt Einitial
• Therefore, the system absorbed energy from the
  surroundings.
• This energy change is called endergonic.

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Changes in Internal Energy

• If ?E lt 0, Efinal lt Einitial
• Therefore, the system released energy to the
  surroundings.
• This energy change is called exergonic.

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Thermochemical equations

• factors which affects the quantity of heat
  evolved or absorbed during a physical or a
  chemical transformation.
• Amount of the reactants and products
• Physical state of the reactants and products
• Temperature
• Pressure

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Thermochemical equations

• It must essentially
• be balanced
• give the value of ?E or ?H corresponding to the
  quantities of substances given by the equation
• mention the physical states of the reactants and
  products . The physical states are represented by
  the symbols (s), (L), (g) and (aq) for solid,
  liquid, gas and aqueous states respectively.

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Example of thermochemical equation

• H2 ½ O2 ? H2O ?H -68.32 Kcal
• 1 mole of hydrogen reacts with 0.5 mole of
  oxygen, one mole of water is formed and 68.32
  Kcal of heats evolved at constant pressure.
• But, not specify whether water is in the form of
  steam or liquid
• H2 (g) ½ O2 (g)? H2O (L) ?H -68.32
  Kcal
• H2 (g) ½ O2 (g)? H2O (g) ?H -57.80
  Kcal.

Effect of temperature ?????
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Standard enthalpy change ?H.

• The heat change at
• 298 K and
• one atmosphere pressure
• is called the standard heat change or standard
  enthalpy change. It is denoted by ?H.

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Enthalpy (H) of the reaction(Comes from Greek
for heat inside)

• the sum of internal energy and the product of
  this pressure and volume.
• H E PV
• E is the internal energy,
• P is the pressure and
• V is the volume of the system.
• It is also called heat content.
• ?H H product H reactants Hp Hr

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Calculation of ?H from ?E

• When the system changes at constant pressure, the
  change in enthalpy, ?H, is
• ?H ?(E PV)
• ?H ?E P?V
• At constant pressure and temperature
• The enthalpy of a chemical is measured in
  kilojoules per mole (kJmol-1).

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?H ?E P?V
For solid and liquid
?H ?E

• At constant volume

Liquids and solids are neglected in calculation
of ?n ?n moles of gaseous products - moles of
gaseous reactants.
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• In case of gases
• ?H ?E P?V (I)
• ?V ?n x V
• ?n no of moles of products - no of moles of
  reactants
• Px?V PVx?n (II)
• But PV RT (for one mole of gas)
• Putting RT in place of PV in equation (II) we get
• P?V RT?n
• Substituting the value of P AV in equation (I) we
  get
• ?H ?E ?n RT

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• R 1.987 cal. (2 cal.)
• or
• 8.314 joules

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Example 1

• Calculate ? E for the following reaction
• 2 CO(g) O2(g) ? 2 CO2(g)
• Where ? H - 135272 cal. At 25C
• Solution
• 
  ? n 2 -3 -1?H ?E ?nRT
  T(K) 25 273
• - 135272 cal. ?E (- 1x2x298)
• ?E - 135272 596 - 134676 cal.

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Example 2

• Calculate ?E and ?H for vaporization of water at
  100C and 1 atm., the specific heat of
  vaporization of water under these conditions is
  540 cal/g.
• Solution
• H2O(l) ? H2O(vap)
• Sp. heat of vaporization 540 cal/g. 1 mole H2O
  18g.
• ?H molar heat of vaporization 540 x 18 at
  constant pressure
• Endothermic reaction since ?H 9720 cal.
• ?n 1 - 0 1 ? H ?E ?nRT
• 9720 ?E 1 x 2 x 373 ? E 8974 cal.

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Example 3

• 6.4g of naphthalene C10H8 when burned under
  constant volume gave 123 KJ at 20C, calculate ?E
  and ?H.
• Solution at constant volume the heat of
  combustion is equal to ?E gave means exothermic
• C10H8(s) 12 O2(g) ? 10CO2(g) 4H2O(l)
• 6.4g at constant volume ? -123KJ
• 128g at constant volume ??E
• ?E -2460KJ - 2460 x 103 J
  
• ?H ?E ?nRT
• - 2460 x 103 (- 2 x 8.3 x 293)
• - 2460 X10-3 - 4863.8 J
• - 2464863 J
• - 2464.863 KJ

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Example

• The heat of combustion of ethylene at 17 C and
  at constant volume is -332.19 kcal. Calculate the
  heat of combustion at constant pressure
  considering water to be in liquid state (R 2
  cal.).

The chemical equation for the combustion of
ethylene is C2H4 3 O2 2CO2(g) 2H2O
(1) 1 mole 3 moles 2moles
negligible volume No. of moles of the
products 2 No. of moles of the reactants
4 ?n (2-4) -2
?H ?E ?n RT ?H -332.19 2 x
I0-3x -2 x 290 -333.3 kcal
Given that ?E-332. 19 kcal. T 27317
290k R2cal2xlO-3kcals.
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Example

• The heat of combustion of carbon monoxide at
  constant volume and at 17 C is -283.3 Kj.
  Calculate its heat of combustion at constant
  pressure(R 8.314 J degree-1 mole-1).

CO(g) ½ O2(g) ?CO2(g) 1 mole ½ mole 1
mole No. of mles of products 1 No. of moles of
reactants 1.5 n No. of moles of products - No.
of moles of reactants 1-1.5 -0.5 ?H ?E
?n x RT ?H -283.3 (-0.5x (8.314x10-3) x 290
- 283.3-1.20 -284.5 KJ Heat of combustion of CO
at constant pressure is -284.5 kJ.
Given that ?E -283.3 kJ T (27317) 290
K. R 8.314 J or 8.314x10-3 KJ
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Heat of combustion

• The change in enthalpy (heat evolved) when 1 mole
  of a substance is completely burnt in presence of
  excess oxygen.
• Organic compound O2(g) ? CO2(g) H2O(l)
• The thermochemical equation must be balanced
  firstly.

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Example 4

• ? H for the combustion of liquid heptane C7H16
  into CO2(g) and H2O(l) is - 1151 kcal/mole at
  20C calculate ? E.
• Solution
• C7H16(l) 11O2(g) ? 7CO2(g) 8H2O(i)
• ?n 7-11 -4
• From thermochemical equation
• ? H ? E ? nRT
• ?E ? H - A nRT - 1151 x 10-3 - (- 4x2x293)
• - 1151 x 103 - (-2344) cal./mol.
• -1151 x 103 2344
• -1148.656 x103 cal/mo

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Example 5

• The heat of combustion of benzoic acid C6H5COOH,
  into CO2(g) and H2O(i) at constant pressure is -
  78.2kJ/mole at 27C calculate heat of combustion
  at constant volume?
• Solution C6H5COOH(S) 7.5 O2(g) ? 7CO2(g)
  3H2O(i)
• ? H - 78.2 x 103 J
• ?n 7-7.5 -0.5
• ? H ? E ?nRT ? E ? H - ?nRT
• ? E - 73200 - (-0.5 x 8.3 x (27 273))
  -81935 J

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• Try to solve 3.2 g of naphthalene C10H8 solid
  when burnt in excess O2 gas into CO2(g) and
  H2O(l) under constant volume gives 1423 KJ at
  20C calculate ? E and ? H? (M.Wt 128).

Calculate ?H of reaction?
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Heat of Formation (Enthalpies of Formation)
?Hf

• is the enthalpy change for the formation of one
  mole of the substance from its elements, at
  standard pressure (1 atm) and a specified
  temperature (25C)
• H2(g) ½ O2(g) ? H2O(l) ? Hf -285.8
  KJ
• C (graphite) 2H2(g) ? CH4(g) ? Ho f
  -74.9 KJ

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Example 6

• Calculate the ? H of the reaction
• Fe2O3(s) 3CO(g) ? 2Fe(s) 3CO2(g)
• given that ?Hf CO2(g) - 393.5 KJ/mol
• ? Hf Fe2O3 -822.2 KJ/mol
• ?Hf CO(g) - 110.5KJ/mol
• Solution ? Ho (products) - (reactants).
• (3X-393.5) (2x0) - (lx-822.2)
  (3x-110.5)J
• -1180.5 1153.7 -26.8 KJ.

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Example 7

• Give the following data
• B2H6(g) 6 H2O(l) ? 2H3BO3(s) 6 H2(g)
• ? H - 493.4 KJ
• Hf of H3BO3(S) is - 1088.7 KJ/mol, and
• ? Hf of H2O(l) is - 285.9 KJ/mol
• Calculate the standard enthalpy of formation of
  B2H6

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• Solution
• 2? Hf H3BO3 - ? Hf B2H6 6? Hf H2O
• - 493.4 KJ 2x- 1088.7 KJ- 1x ?Hf B2H6 6X
  -285.9KJ
• - 2177.4KJ - ?Hf B2H6 - 1715.4KJ
• - 493.4KJ - 462.0KJ - ? HofB2H6
• ?HfB2H6 31.4KJ/mol

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Variation of heat (or enthalpy) of reaction with
temperature Kirchoff's equations.
internal energies of the reactants and products.

• 1. At constant volume,
• ?EE2-E1
• Differentiating this equation with respect to
  temperature at constant volume, we get

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Kirchoff's equations.

• But we have already seen that

heat capacities
heat capacities of the products and reactants
Integrating between temperature T l and T2 ,
we have
?E2- ?E1 ? E2 -?E1 ? Cv
(T2-T1) (3)
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Kirchoff's equations.

• 2. At constant pressure

  ?HH2-H1
And finely . ?H2- ?H1 ?Cp (T2-T1) (6)
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Example 3 3

• The heat of reaction
• ½ H2 ½ Cl2 HCl at 27 C is -22.1
  Kcal.
• Calculate the heat of reaction at 77C. The
  molar heat capacities at constant pressure at 27
  C for hydrogen, chlorine and HCl are 6.82, 7.70
  and 6.80 cal mol-1, respectively.

Here ½ H2 ½Cl- ? HCl ? H -22.1 Kcal ? Cp
Heat capacities of products - Heat capacities of
reactants 6.80-½(6.82) ½ (7.70) 6.80 -
7.26 -0.46 x 10-3 Kcal ?H2- ?H1 ?Cp
(T2-T1) ?H2 - (-22.1) (-0.46 x 10 -3) x 50
-21.123
T2 273 77 350 K T1 273 27 300 K
T2-T1 (350-300)K50K
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Example 3 4

• Calculate the standard heat of formation of
  propane (C3H8) if its heat of combustion is
  2220.2 KJ mol-1. The heats of formation of CO2
  (g) and H2O (L) are - 393.5 and -285.8 KJ mol-1
  respectively.

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• C3H8 (g) 5 O2 (g) ? 3CO2 (g) 4 H2O (L)
  ? Hc 2220.2 KJ
• C (s) O2 (g) ? CO2(g)
  ?Hf- 393.5 KJ
• H2(g)½O2(g) ?H2O (L)
  ?Hf-285.8KJ
• We should manipulate these equations in a way so
  as to get the required equation
• 3 C (s) 4 H2 (g) ? C3H8 (g) ?H?

Multiplying equation (ii) by 3 and equation
(iii) by 4 and adding up we get 3C(s) 3O2 (g) ?
3 CO2 (g) ?H -1180.8 KJ4H2 (g) 2O2(g) ?
4H20(L) ?H -1143.2 KJ 3C(s) 4H2 (g)
502(g) ? 3CO2(g) 4H2O(L) ?H - 2323.7
KJ subtracting equation (i) from equation (iv),
we have3C(s)4H2(g) ? C3H8(g) ? H -103.5 KJ
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Heat of transition

• the change in enthalpy which occurs when one mole
  of an element changes from one allotropic form to
  another.

• P white ? P red
  ?H -1.028 kcal
• S monoclipic ? S rhombic ?H-0016
  Kcal

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A ?B q1 ?H1 q1
B ? C q2 ?H2 q2
C ? Z q3 ?H3 q3
A ?z Q1 ?H1 -Q1 The total evolution of heat
q1 q2 q3 Q2 According to Hess's law Q1 Q2
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We can find DH(a) by subtracting DH(b) from DH(c)
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Hesss Law Energy Level Diagrams

• Forming CO2 can occur in a single step or in a
  two steps.
• ?Htotal is the same no matter which path is
  followed.

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Hesss Law Energy Level Diagrams
Forming H2O can occur in a single step or in a
two steps. ?Htotal is the same no matter which
path is followed.
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Illustrations of Hess's law

• (1) Burning of carbon to CO2

• 1st way
• C(S) 02(g) ? CO2 (g) ?H -94.05kcal
  (-393.50 KJ)
• 2nd way
• C(s)½O2(g) ?CO(g) ?H -26.42 kcal and
• CO2 (g) ½ O2 (g) ? CO2 (g) ?H -67.71
  kcal
• Overall change C (s) O2 (g) ?CO2 (g) ?H
  -94.13 kcal

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Illustrations of Hess's law

• (2) Formation of sodium by Hydroxide from Na

• 1st way
• 2Na (s) ½ O2 (g) ? Na2O (s)
  ?H -100 kcal
• Na2O H2O (L) ?2 NaOH (aq)
  ? H -56 kcal
• 2 Na (s) H2O (L) ½ O2 (g) ? 2 NaOH (aq) ?H
  - 156 Kcal
• 2nd way
• 2 Na (s) 2 H2O (L) ? 2 NaOH (aq) H2 (g)
  ?H - 88 Kcal
• H2 (g) ½ O2 (g) ? H2O (g)
  ?H - 68.5 Kcal
• 2 Na (s) H2O (L) ½ O2 (g) ? 2 NaOH (aq) ?H
  - 156 Kcal

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Example 8
Calculate the heat of hydrogenation of
acetylene 1) C2H2(g) 2 H2(g) ? C2H6(g)
?H1 ? At 25C and 1 atm., given that 2)
2C2H2(g) 5 O2(g) ? 4CO2 2H2O(l) ?H2 -
2602 KJ 3) 2C2H6(g) 7O2(g) ? 4CO2(g) 6H2O(I)
?H3 -3123KJ 4) H2(g) ½O2(g) ?H2O(l)
?H4 -286KJ
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Solution 5) C2H2(g) 5/2 O2(g) ? 2CO2(g)
H2O(l) ????? (2) ??? 2 ?H5
-2602/2 -1301KJ 6) 2H2(g) O2(g) ? 2H2O(I)
???? 4 ?? 2 ? H6
2 (-286KJ) - 572 KJ 7) 2CO2(g) 3H2O (l)
? 2C2H6(g) 7/2 O2(g) ??? ? ???? (3) ??? 2 ?H7
3123/2 1561KJ
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Adding equation 5,6 and 7 gives C2H2(g) 2H2(g)
7/2O2(g) 2CO2 3H2O(I) ? 2CO2(g) 3H2O()
C2H6(g) 7/2 O2(g) Canceling elements that are
the same on both sides, we get C2H2(g)
2H2(g) ? C2H6(g) ??H1 ? H5 ?H6 ? H7
(-1301 KJ) (-572 KJ) (1561 KJ) -312 KJ  
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Calorimetry

• We measure heat flow using calorimetry.
• A calorimeter is a device used to make this
  measurement.
• A coffee cup calorimeter may be used for
  measuring heat involving solutions.

A bomb calorimeter is used to find heat of
combustion the bomb contains oxygen and a
sample of the material to be burned.
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Bomb Calorimeter
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Coffee-cup calorimeter.
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