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Chapter 6. Thermochemistry. Jozsef Devenyi Department of Chemistry, UTM. The Nature of Energy. Recall: force:. a push or pull on an object. work:. the product of force applied to an object over a distance. energy:. the work done to move an object against a force. - PowerPoint PPT Presentation
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1 Chapter 5
Chapter 6
Thermochemistry
Jozsef DevenyiDepartment of Chemistry, UTM
2 Chapter 5
• kinetic energy is the energy of motion:
The Nature of EnergyRecall: force:
work:
energy:
a push or pull on an object
the product of force applied to an object over a distance
the work done to move an object against a force(energy is the capacity to do work or transfer heat)
3 Chapter 5
• potential energy is the energy an object possesses by virtue of its position
• potential energy can be converted into kinetic energy
The Nature of Energy
example: a ball is falling from a balcony
4 Chapter 6
We sometimes use the calorie instead of the joule:1 cal = 4.184 J (exactly)
Units of Energy
SI Unit for energy is the joule, J:
The Nature of Energy
for example, an object with a mass of 2 kg that moves at a speed of 1 m/s; its kinetic energy is
5 Chapter 6
• system: part of the universe we are interested in
Systems and Surroundings
• surroundings: the rest of the universe
The Nature of Energy
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First Law of Thermodynamics Internal Energy• Internal energy: total energy of a system
(cannot measure absolute internal energy)
• Change in internal energy:
Chapter 6
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Relating DE to Heat and Work:
• internal energy of a system:
• Energy cannot be created or destroyed.
energy of (system + surroundings) is constant
• Any energy transferred from a system must be transferred to the surroundings (and vice versa).
when a system undergoes a physical or chemical change, the change in internal energy is given by the heat released or absorbed by the system plus the work done on or by the system
First Law of Thermodynamics
first law of thermodynamics:
Chapter 6
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Relating DE to Heat and Work:
First Law of Thermodynamics
Chapter 6
First Law of Thermodynamics
Chapter 6 9
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Exothermic and Endothermic Processes
• endothermic: absorbs heat from the surroundings
example: (an endothermic reaction feels cold)
First Law of Thermodynamics
Chapter 6
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Exothermic and Endothermic Processes• exothermic: transfers heat to the surroundings
(an exothermic reaction feels warm/hot)example:
First Law of Thermodynamics
Chapter 6
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State FunctionsFirst Law of ThermodynamicsDE is a state function; that is, the value of DE depends only on the initial andfinal states of system, not onhow change occurred
Chapter 6
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we can measure the change in enthalpy:
Enthalpy
enthalpy is astate function
DH = Hfinal – Hinitial = qp
Chapter 6
enthalpy (H): heat transferred between the system and its surroundings while pressure is constant;
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Enthalpies of ReactionFor a reaction:enthalpy is an extensive property (magnitude DH isdirectly proportional to amount):
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g) DH = - 802 kJ
2 CH4 (g) + 4 O2 (g) 2 CO2 (g) + 4 H2O(g) DH =
Chapter 6
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Enthalpies of ReactionWhen reaction is reversed the sign of DH is reversed:
CO2 (g) + 2 H2O(g) CH4 (g) + 2 O2 (g) DH =
Change in enthalpy also depends on physical state:
H2O(g) H2O(l) DH = - 88 kJ
Chapter 6
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Example:Enthalpies of Reaction
Chapter 6
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Example:Enthalpies of Reaction
Chapter 6
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Calorimetry
calorimeter = apparatus that measures heat flow by measuring the change in temperature
heat capacity (C) = the amount of energy required to raise the temperature of an object by one degree Celsius (J / oC)
calorimetry = measurement of heat flow
Chapter 6
Heat Capacity and Specific Heat
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Calorimetry
molar heat capacity = heat capacity of 1 mol of a substance[J/(mol . oC)]
Chapter 6
Heat Capacity and Specific Heat
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specific heat (s) = specific heat capacity = heat capacity of 1 g of a substance unit: J/(g . oC)
Calorimetry Heat Capacity and Specific Heat
Chapter 6
that is, the amount of energy required to raise the temperature of 1 g of substance by one degree Celsius (J /g oC)
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Calorimetry Heat Capacity and Specific Heat
heat released/absorbed:
q = (specific heat) x (grams of substance) x Dt =
= s.h. X m x Dt where Dt = tfinal - tinitial
Chapter 6
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Examples:A)Calorimetry
Chapter 6
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Examples:CalorimetryB)
Chapter 6
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Constant-Pressure Calorimetry
DH = qp
• at constant atmosphericpressure:
• in such system, we assume that no heat is lost to surroundings
Calorimetry
Chapter 6
qrxn + qwater = 0
qrxn = - qwater
therefore
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Example:Calorimetry
Chapter 6
When a student mixes 50.0 mL of 1.0 M NaOH solutionand 50.0 mL of 1.0 M HCl solution in a coffee cupcalorimeter, the temperature of the resultant solutionincreases from 21.3 oC to 27.8 oC. Calculate the enthalpy change for this neutralization reaction.
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Example:Calorimetry
Chapter 6
Assume that i. the calorimeter loses only negligible quantity of heat,
ii. the total volume of the solution is 100 mL,
iii. the density and the specific heat of the solution are the same as those of water, 1.00 g/mL and 4.184 J/g oC, respectively.
(these assumptions are usually true, unless stated otherwise)
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Example:Calorimetry
Chapter 6
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Example:Calorimetry
Chapter 6
29 Chapter 5
Constant-Volume Calorimetry (Bomb Calorimetry)• rxn carried out under constant volume• uses a device called bomb calorimeter• usually used to study combustion rxns
Note: since pressure is not constant under these conditions, q measured this way is not equal to DH.
qwater + qcal = - qrxn
Calorimetry
qrxn + qwater + qcal = 0
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Example:Calorimetry
Chapter 6
In a laboratory test 9.20 g of ethanol, C2H5OH was burnedin a bomb calorimeter that contained 500.0 g of water andthe heat capacity of the calorimeter is 4.821 kJ/oC. Thetemperature increased from 22.9 oC to 24.85 oC. A) Calculate the heat of combustion per gram ethanol.
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Calorimetry
Chapter 6
Example:
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B)
Calorimetry
Chapter 6
Example:
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For example:CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O(g) DH = - 802 kJ
Hess’s LawHess’s law: if a reaction is carried out in a number of steps, DH for the overall reaction is the sum of DH for each individual step.
2 H2O(g) 2 H2O(l) DH = - 88 kJ
Chapter 6
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DH1 = DH2 + DH3Hess’s Law Note that:
Chapter 6
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Hess’s Law
N2 (g) + 2 O2 (g) 2 NO2 (g) DH1 = + 67.6 kJ
calculate the heat of reaction (enthalpy of reaction, DHrxn) for the following rxn: N2 (g) + O2 (g) 2 NO(g) .
2 NO(g) + O2 (g) 2 NO2 (g) DH2 = - 521 kJ
From the two reactions:
Chapter 6
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Hess’s Law
Chapter 6
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Hess’s Law
Chapter 6
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• If 1 mol of compound is formed from its constituent elements, then the enthalpy change for the
reaction is called the enthalpy of formation, DHf .
Enthalpies of Formation
• Standard enthalpy, DHo, is the enthalpy measured when everything is in its standard state (every component).
• Standard conditions (standard state): 1 atm and 25 oC (298 K).
Chapter 6
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Enthalpies of Formation• Molar enthalpy of formation: 1 mol of compound is formed from substances in their standard states.
• If there is more than one state for a substance under standard conditions, the more stable one is used.
Chapter 6
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• Standard enthalpy of formation of the most stable formof an elementis zero.
Enthalpies of Formation
Chapter 6
See Table 6.5
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Enthalpies of Formation
Chapter 6
42 Chapter 6
Using a set of DHof values to calculate DHrxn .
We use Hess’ Law to calculate the enthalpy of any reaction using the enthalpies from the table of enthalpies of formation.
Enthalpies of Formation
where - “S“ means “the sum”- n and m are stoichiometric coefficients for the each product and reactant, respectively
Enthalpies of FormationExample:Consider the following combustions reaction of methane:
Using Hess’ Law and the relevant standard enthalpies of
formation, calculate DHrxn for this reaction.
CH4 (g) + O2 (g) CO2 (g) + H2O(g) DHrxn = ??
Chapter 6 43
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Enthalpies of Formation
Chapter 6
Example:
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Example:Enthalpies of Formation
Chapter 6
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Example:Enthalpies of Formation
Chapter 6
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Example:Enthalpies of Formation
Chapter 6
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End of Chapter 6
Thermochemistry
Chapter 6
