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7/8/14
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Lecture 4. THE FIRST LAW OF THERMODYNAMICS
THERMODYNAMICS: Basic Concepts Thermodynamics: (from the Greek therme,
meaning "heat" and, dynamis, meaning "power") is the study of energy conversion between heat and mechanical work, and subsequently the macroscopic variables such as temperature, volume and pressure
Energy: capacity to do work Work: motion against an opposing force System: part of the world which we have interest or
being investigated. Surrounding: is where we make our observations
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Definition of Work: • Work is motion against opposing force. • Work is defined as a force acting through a displacement x, the
displacement being in the direction of the force.
xfw Δ=
Consider a Gas Expansion Work: Initial State Final State
P1, V1, T P2, V2, T
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Consider a Gas Expansion Work:
:is pressure, opposing external The ,
)( 12
exP
hmgwhhmgw
Δ−=
−−=
VPwVVPhAPw
AmgP
ex
exex
ex
Δ−=
−−=Δ−=
=
)( 12
or
Consider a Gas Expansion Work:
2
1
1
2
2
1
lnln
,2
1
2
1
PPnRT
VVnRTw
VdVnRTw
VnRTP
PdVPw
PP
dVPw
v
v
in
in
v
v in
ex
v
v ex
−=−=
−=
=
−=
−=
∫
∫
∫
and
gas, the of pressure theis
than greater mallyinfinitesi onlyis instant, every at
volume in increase ssimalinfiniteti against done work a For
in
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Consider a Gas Expansion Work: q the external pressure is infinitetissimally smaller than the internal pressure at all stages of the expansion - reversible process. A reversible change in thermodynamics is a change that can be reversed by an infinitesimal modification of a variable. q in reversible processes, there is maximum amount of work done that could possible extracted from a process
Gas Expansion Work: Sample Problem A sample of 4.50 g of methane occupies 12.7 L at 310 K. (A) Calculate the work done when the gas expands isothermally against a constant external
pressure of 200 Torr until its volume has increased by 3.3 L. (B) Calculate the work that would be done if the same expansion occurred reversibly.
J 87.97
atm L 1J101.3
atm L 0.87
atm L 0.87
L) torr760
atm 1 torr
process) ble(irreversi formula the use we constantis pressure opposing since A.
:SOLUTION
:
−=
×−=
−=
×−=
Δ−=
w
w
w
w
VPw ex
3.3)(200(
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Consider a Gas Expansion Work:
J -
KmolK J 1-mol g
g
:expansiongas reversible a for B.:SOLUTION
1-1-
4.1677.12)7.123.3(ln310)3145.8()
165.4(
ln
=
+×××−=
−=
wLLw
VV
nRTwi
f
Definition of HEAT:
Heat is the transfer of energy between two bodies that are at different temperatures. Heat appears at the boundary of the system. 0th Law: heat is transferred from the hotter object to the colder one. Heat is path dependent.
heat specificsetemperatur in change T mass,m
=
=Δ=
Δ= Tmsq
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THE FIRST LAW OF THERMODYNAMICS:
“Energy can neither be created nor destroyed but can be converted from one form to another.” Law of Conservation of Energy. ENERGY OF THE UNIVERSE:
surrsys
surrsysuniv
surrsysuniv
EEEEE
EEE
Δ−=Δ
=Δ+Δ=Δ
+=
therefore
system, given a forenergies inchanges the and0
If a system undergoes an energy change, the surroundings must also undergo a change equal in magnitude but opposite in direction
Total ENERGY of a system; Internal Energy, U
TOTAL ENERGY:
UE
UPEKEE
total
total
=
++=
0 are PE and KE rest, at system a for
Internal Energy , U: q energy associated with the chemical system q depends on the thermodynamic parameters such as T, P, V,
composition, etc. q consists of translational, rotational, vibrational, electronic energies,
as well as intermolecular interactions
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Internal Energy, U:
For a given system, we do not know the exact nature of U, we will be only interested in the changes in U for a particular process that the system undergoes.
12 UUU −=Δ
Mathematical expression of the first law of thermodynamics:
wqU +=Δ
or for an infinitesimal change:
dwdqdU +=
The change in the internal energy of a system in a given process is the sum of the heat exchange, q, between the system and its surroundings and the work done on or by the system.
HEAT and WORK sign conventions:
Process Sign
Work done by the system on the surroundings
-
Work done on the system by the surroundings
+
Heat absorbed by the system from the surroundings
+
Heat absorbed by the surroundings from the system
-
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Constant volume adiabatic bomb calorimeter
Adiabatic means no heat exchange with the surroundings
Sample inside an oxygen filled container is ignited by an electrical discharge. Heat released is measured by the increase in the temperature of the water.
v
v
v
qU
VPqwqU
V
=Δ
Δ−=
+=Δ
=Δ 0 VP constant at ,
,
ENTHALPHY In processes carried out under constant P,
PVUHEnthalpyPVUPVUq
VVPqUUorVPwqU
p
p
v
+=
+−+=
−−=−
Δ=+=Δ
or
,
)()(
),(,
1122
1212
Total enthalpy of a system can not be measured directly, so the change in enthalpy, ΔH, is a more useful value than H itself.
For a constant P process;
VPdUdH
VPUH
Δ+=
Δ+Δ=Δ
change, simalinfiniteti for
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Entalphy, ΔH vs. Internal Energy, ΔU
Consider the following reaction:
2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
Heat evolved by the reaction is 367.5 kJ, reaction takes place at constant pressure,
qp= ΔH = -367.5KJ.
Internal Energy ΔU:
The volume of the H2(1 mole) generated by the reaction occupies 24.5 L, therefore –PΔV = -24.5 L at or -2.5 kJ Finally,
ΔU = -367.5 kJ – 2.5 kJ
ΔU = -370 kJ.
- Difference is due to expansion work.
VPHU Δ−Δ=Δ
SAMPLE PROBLEM 4.02
In which of the following processes is ΔH = ΔE? A. 2.0 moles of NH3 are colled from 325oC to 300oC at 1.2 atm B. 1.0 g of water is vaporized at 100oC and 1.0 atm C. 2.0 moles of hydrogen iodide gas react to form hydrogen gas and iodine gas
in a 40 L container D. Calcium carbonate is heated to form calcium oxide and carbon dioxide in a
container with variable volume E. 1.0 mole of carbon dioxide sublimes to gas phase.
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SAMPLE PROBLEM 4.03
For which of the following reactions will ΔH be equal to ΔE?
A. H2(g) + Br2(g) 2HBr(g)
B. H2O(l) H2O(g)
C. CaCO3 CaO(s) + CO2(g)
D. 2H2(g) + O2(g) 2H2O(l)
E. CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
Constant Pressure Calorimeter:
A constant-pressure calorimeter made of two plastic cups. The outer cup helps insulate the reacting mixture from the surroundings. Two solutions of known volume containing the reactants at the same temperature are carefully mixed in the calorimeter. The heat produced or absorbed by the reaction can be determined by the temperature change, the quantities and specific heats of the solutions used, and the heat capacity of the calorimeter.
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HEAT CAPACITIES: Calorimetry: measurement of heat changes in chemical and physical processes.
When heat is added to the system, the corresponding temperature rise will depend on a) The amount of heat delivered, b) The amount of the substance, c) The chemical nature and the physical state of the substance, d) The conditions at which the energy is added to the system.
HEAT CAPACITIES:
Thus for a given amount of a substance, change in temperature is related to the heat added:
capacity heat molaris C mol K J
capacity heatis C K J
1-1-
1-
,
,
,
TnqC
nC
TqC
orTCq
Δ==
Δ=
Δ=
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HEAT CAPACITY at constant volume
TCnTTCdTCU
dTCdUTU
TqC
vv
T
T v
v
vv
Δ=−==Δ
=Δ
Δ=
Δ=
=Δ
∫ )( 122
1
or
q U volume, constant at v
HEAT CAPACITY at constant pressure processes
TCnTTCdTCH
dtCdHTH
Tq
C
pp
T
T p
p
pp
Δ=−==Δ
=Δ
Δ=
Δ=
=Δ
∫ )( 122
1
or
q H process, pressure constant at p
for gases, Cp > Cv because of the work to be done to the surroundings in constant pressure processes for condensed phases, Cp and Cv are identical for most purposes for ideal gases, RCC vp =−
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Ch 45.5 First Long Exam 8:30 – 9:25 July 14 Multiple Choice July 16 Problem Solving
Coverage: Properties of Gases; Kinetic Molecular Theory of Gases, First Law of Thermodynamics and Thermochemistry.
Calculate the values of ΔE and ΔH for heating of 55.40 g of Xenon from 300 K to 400 K in a closed vessel. Assume ideal gas behavior and the heat capacities are independent of temperature. Xenon is a monoatomic gas. Its Cv is equal to 12.47 J K-1 mol-1.
SAMPLE PROBLEM 4.04
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When 229 J of energy is supplied as heat at constant pressure to 3.00 mole CO2(g), the temperature of the sample increases by 2.06 K. Calculate the molar heat capacities at constant volume (Cv) and constant pressure (Cp) of the gas.
SAMPLE PROBLEM 4.05
Use the Cp and Cv values from the previous problem to calculate the change in a) molar enthalpy b) molar internal energy when carbon dioxide is heated from 15oC (temperature when air is inhaled) to 37oC (blood temperature, temperature in our lungs).
SAMPLE PROBLEM 4.06
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A 50.0 mL sample of 0.400 M Copper (II) sulfate solution at 23.35oC is mixed with 50 mL of 0.600M sodium Hydroxide solution, also at 23.35oC in a coffee cup calorimeter. After the reaction occurs the temperature of the resulting mixture is measured to be 25.23oC. The density of the final solution is 1.02 g/mL. Calculate the amount of heat evolved. Specific heat of solution is 4.184 J/g oC
SAMPLE PROBLEM 4.07
GAS EXPANSION: ISOTHERMAL EXPANSION
q- w thatfollows it therefore 0U ,then
T, on dependentis energy internal the gases, ideal For.maintainedis mequilibriu thermal the that rate slow a
such at happen must typically system the of out or into transfer heat The constant. heldis etemperatur :process Isothermal
==Δ
0 H then T, constant at constantis PV and 0, U ncesi
PVUH
=Δ
=Δ
Δ+Δ=Δ
For isothermal expansion by an ideal gas, the heat absorbed is equal to the work done by the ideal gas on its surroundings.
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GAS EXPANSION: ADIABATIC EXPANSION
VPVP or
CC where
PP
VV
,0qgssurroundin and system between exchange heat no :process diabaticA
2211
v
p
1
2
2
1
γγ
γ
=
=γ=⎟⎟⎠
⎞⎜⎜⎝
⎛
=
57 that so
R57C and R2
5C gases, diatomic for3
5 that so
,R25C and R2
3C gases, monoatomic for
pv
pv
=γ
==
=γ
==
Calculate the values of ΔU, and ΔH for the reversible adiabatic expansion of 1.0 mole of a monoatomic ideal gas from 2 L to 6 L. The temperature of the gas is initially 298 K.
SAMPLE PROBLEM 4.08
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THERMOCHEMISTRY:
Heat of reaction is the heat change in the transformation of reactants at a given temperature and pressure to products at the same conditions.
For constant pressure processes, the heat of reaction qp is equal to enthalpy change of the reaction, ΔH.
positive.is H gs.surroundin the from heatabsorbs system :reactions cEndothermi
negative.is H gs.surroundin to heat offgives system :reactions rmicExothe
r
r
Δ
Δ
THERMOCHEMISTRY:
2H2(g) + O2(g) 2HO2(g)
241.8 kJ of heat is given off. The enthalpy change for this process is called standard enthalphy of reaction, ΔHo
In general, standard enthalpy change of a chemical reaction is the total enthalpy of the products minus the total enthalpy of the reactants
tcoefficien tricstoichiomeis v enthalpy; molar standard theis H
)reactants(Hv)products(HvHo
ooor ∑∑ −=Δ
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THERMOCHEMICAL EQUATIONS:
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) + 1367 kJ
The coefficient in the balanced thermochemical equations refer to the number of moles of reactants and products. The numerical value of ΔH refers to the number of moles of substances specified by the equation The physical states of the species are important and must be specified. The value of ΔH does not change significantly with moderate changes in temperature.
When 2.61 g of dimethyl ether, CH3OCH3, is burned at constant pressure, 82.5 kJ of heat is given off and producing only CO2 and water. Find the ΔH for the reaction.
SAMPLE PROBLEM 4.09
SAMPLE PROBLEM 4.10
Write the thermochemical equation for for the reaction in sample problem 4.07.
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When aluminum metal is exposed to atmospheric oxygen, (as in aluminum doors and windows), it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation of 24.2 grams of aluminum at 25oC and 1.0 atm. (ΔH = - 3352 kJ)
SAMPLE PROBLEM 4.11
STANDARD STATES AND STANDARD ENTHALPY CHANGES
The thermodynamic standard state of a substance is the most stable pure form under standard pressure (1.0 atm) and at standard temperature (25oC) Thermochemical Standard States:
For pure substances in the liquid or solid states, the standard state is the pure liquid or solid
For gases, the standard state is the gas at P = 1.0 atm and T at 25oC. For substances in solution, standard state refers to 1.0 M
concentration. The standard enthalpy change, refers to the ΔH when a specified number of moles of reactants, all at standard states are converted completely to a specified number of moles of products.
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STANDARD MOLAR ENTHALPIES OF FORMATION
The standard molar enthalpy of formation of a substance is the enthalpy change for the reaction in which one mole of the substance in a specified state is formed from its elements in their standard states. By convention the ΔH of formation value for any element in its standard state is zero.
SAMPLE PROBLEM 4.11
Write the standard formation reaction for ethanol a) C2H5OH(l) B) ammonium perchlorate
THERMOCHEMISTRY:
∑∑ Δ−Δ=Δ
+−+=Δ
+→+
)tstanreac(Hv)products(HvH
general; in or))B(Hb)A(Ha()D(Hd)C(HcH
:is reaction the of enthalpy standard thedDcCbB Aa
of
of
or
ooooor
formation. of enthalpy molar standard theis H ofΔ
- is the enthalpy change when 1 mole of a compound is formed from its constituent elements at 1 bar and 298 K.
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THERMOCHEMISTRY:
0))g(O(H ;0))g(H(H
0H :etemperatur particular a atforms allotropic stable most their inelements for
:H
2o
f2o
f
of
of
=Δ=Δ
=Δ
Δ
2H2(g) + O2(g) H2O(g)
kJ/mole 8.241H of −=Δ
MEASUREMENTS OF ΔH of reaction
DIRECT METHOD: measure ΔfHo of direct synthesis from their elements:
2H2(g) + O2(g) H2O(g)
[ ]8.241H
))g(O(H))g(H(H2))g(OH(H2H
)tstanreac(Hv)products(HvH
or
2o
f2o
f2o
fo
r
of
of
or
−=Δ
Δ+Δ−Δ=Δ
Δ−Δ=Δ ∑∑
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Calculate the reaction enthalpy of the following reaction:
3HN3(l) + 2NO(g) H2O2(l) + 4N2(g) Using their standard enthalpies of formation. ΔfHo values: kJ/mol HN3(l) 264.0 NO(l) 90.25 H2O2(l) -187.78
SAMPLE PROBLEM 4.12
SAMPLE PROBLEM 4.13
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SAMPLE PROBLEM 4.14
SAMPLE PROBLEM 4.15
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MEASUREMENTS OF ΔH of reaction
INDIRECT METHOD: HESS’S LAW: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
ofHΔ for carbon monoxide:
C(graphite) + 1/2O2(g) CO(g)
C(graphite) + O2(g) CO2(g)
CO + 1/2O2(g) CO2(g)
kJ/mole 5.393H or −=Δ
kJ/mole 0.283H or −=Δ
THERMOCHEMICAL EQUATIONS
Enthalpy changes are additive. • the ΔH of the reverse reaction will have the opposite sign • multiplying a reaction by a factor, you also multiply the ΔH by the same factor.
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MEASUREMENTS OF ofHΔ
ofHΔ for carbon monoxide:
C(graphite) + O2(g) CO2(g)
CO2(g) 1/2O2(g) + CO(g)
kJ/mole 5.393H or −=Δ
kJ/mole 0.283H or =Δ
C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or −=Δ
MEASUREMENTS OF ofHΔ
C(graphite) + O2(g) CO2(g) + 393.5kJ
kJ/mole 5.393−=Δ orHC(graphite) + O2(g) CO2(g)
Hess’s Law can be rephrased as: ‘ The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which the reaction might be divided.’
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Calculate the standard molar enthalpy of formation of acetylene (C2H2) from its elements.
kJ/mole 8.2598H or −=Δ
kJ/mole 5.393H or −=Δ
2C(graphite) + H2(g) C2H2(g)
The equations for combustion and the corresponding enthalpy changes are:
(1) C(graphite) + 1/2O2(g) CO2(g)
(2) H2(g) + 1/2O2(g) H2O(l)
(3) 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
kJ/mole 8.285H or −=Δ
SAMPLE PROBLEM 4.16
SAMPLE PROBLEM: o
fHΔ for carbon monoxide:
C(graphite) + O2(g) CO2(g)
CO2(g) 1/2O2(g) + CO(g)
kJ/mole 5.393H or −=Δ
kJ/mole 0.283H or =Δ
C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or −=Δ
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SAMPLE PROBLEM:
Use the thermochemical equation shown below to determine the enthalpy of the reaction:
H2(g) + 1/2O2(g) H2O(g)
CO2(g) C(s) + O2(g) ΔfHo = +590.2 kJ/mol
C2H6(g) 2C(s) + 3H2(g) = 127.0 kJ/mol
C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O (l) =-2340.7 kJ/mol
SAMPLE PROBLEM 4.17
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SAMPLE PROBLEM 4.18
SAMPLE PROBLEM 4.18
From the following enthalpies of reaction, CaCO3(s) CaO(s) + CO2(g) ΔH = -178.1 kJ CaO(s) + H2O(l) Ca(OH)2(s) ΔH = -65.3 kJ Ca(OH)2(s) Ca+2(aq) + 2OH-(aq) ΔH = -16.2 kJ Calculate ΔHrxn for: Ca+2(aq) + 2OH-(aq) + CO2(g) CaCO3(s) + H2O(l)
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TEMPERATURE DEPENDENCE OF ENTHALPY CHANGE
KIRCHHOFFS’s LAW: The difference in enthalpies of a reaction at two different temperatures, T1 and T2, is just the difference in the enthalpies of heating the products and reactants from T1 to T2.
)( 1212 TTCHH pr −Δ=Δ−Δr
Where ΔCp is the difference in molar heat capacities between the products and reactants
The standard enthalpy change for the reaction: 3O2(g) 2O3(g) Is given by ΔrHo = 285.4 kJ/mol at 298 K and 1 bar. Calculate the value of ΔrHo at 380 K. Assume that the Cp values are all independent of temperature.
SAMPLE PROBLEM 4.18
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Bond Dissociation Enthalpy:
Bond Enthalpy: change in enthalpy when bonds form or break in diatomic molecule Bond Dissociation Enthalpy: average change in enthalpy in polyatomic molecules when individual bonds form or break.
N2(g) 2N(g)
H2O(g) H(g) + OH(g) ΔrHo = 502 kJ/mol OH(g) H(g) + O(g) ΔrHo = 427 kJ/mol
SAMPLE PROBLEM:
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ENTHALPY CHANGE and BOND ENERGIES:
The Enthalpy of a Reaction can be estimated by the enthalpies of the total number of bonds broken and formed in the reaction.
released energy total- input energy total (products))(reactantsr
=
−=Δ ∑ ∑BEBEH o
Bond Enthalpy Changes
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Estimate the enthalpy of combustion of methane:
CH4(g) + O2(g) CO2(g) + H2O(g) Using bond enthalpies in Table 4.4. Compare your result with that calculated from the enthalpies of formation of products and reactants. C-H 410 kJ/mole O=O 494 kJ/mole C=O 563 kJ/mole H-O 460 kJ/mole
SAMPLE PROBLEM 4.19
SAMPLE PROBLEM 4.20
Ethylamine undergoes an endothermic gas phase dissociation to produce ethylene and ammonia. CH3CH2NH2 H2C=CH2 + NH3 ΔH = +53.6 kJ/mol The following average bond energies are given:
C-H = 413 kJ C-C = 346 kJ C=C = 602 kJ N-H = 391 kJ
Calculate the C-N bond energy in ethylamine.
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End of Chapter
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