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Outline of Chapter 2
• Basic Concepts: Heat, Work, Energy• First Law of Thermodynamics• Expansion work• Measurement of Heat• Enthalpy changes• Adiabatic changes (Cooling)• Thermochemistry• Inexact Differntials
System, Surroundings, Universe
• System- Part of the world of interest
• Surroundings- Region outside the system
• Open system- Allows matter and energy to pass
• Closed system- Cannot allow matter to pass
• Isolated system- Cannot allow matter or energy to pass
Heat, and Heat Transfer
Heat can be transferred byConvectionConductionRadiation
Heat- is the transfer of energy from one body to another by thermal contact
Internal Energy: Energy that exist internal to the system
U = q + wsystem
First Law: The internal energy of an isolated system is a constant; Energy cannot be created or destroyed.
ΔU= change in internal energyq = heat (released or absorbed)w = work done on or by the system
Internal energy can have different forms:
Translational Kinetic EnergyRotational Kinetic energyVibrational Kinetic energy
*Energy within the atom
U = U 0total system surroundingsU
Internal Energy of a monoatomic gas
60
2m mU T U RT
50
2m mU T U RT
Linear molecule
Non-Linear molecule
30
2m mU T U RT
Monoatomic gas
RT
RT
RT
RT RT
Expression for WorkWork (W) = force x distance
W = - F Z
Change equation to relate pressure and volume
Force FP = =
Area AF = P A
W = - P A ΔZ
Volume change (ΔV) = A ΔZ
W = - P ΔV
True expression for Work
dW = - P dVexW= -
f
i
V
exVP dV
Types of Expansion Work• Free Expansion- work done by
expansion against zero opposing force or zero pressure
f
i
V
exVw P dV
0
0exP
W
U q
Types of Expansion Work• Expansion against a constant external pressure
(irreversible)
f
i
V
exVw P dV
If external pressure does, not vary, this expression can be integrated
f
1
V
ex Vw = -P dV
ex f iw = -P V -V
exw = -P ΔV
Types of Expansion Work
• Expansion is reversiblef
i
V
exVw P dV
In this case, the external pressure Pex
exactly equals the internal pressure P at each stage of the expansion.
f
1
V
Vw = - P dV
f
i
V
V
nRTw = - dV
V
exP = P
The integral cannot be evaluated becausetemperature is not a constant throughout the integration.
Types of Expansion Work
• Expansion is Isothermal, reversiblef
i
V
exVw P dV
The system is kept at a constant temperature.The external pressure Pex equals the internalpressure P at each stage of the expansion.
f f
i i
V V
V V
nRT 1w = - dV = -nRT dV
V V
exP = P
Moles is always assumed
constant unless otherwise noted.
f
i
Vw = -nRT Ln
VΔT = 0
Problems: Calculate the work needed for a 65 kg person to climb up 4.0 meters on (a) the earth, (b) the moon (g=1.60 m/s2)
Problem 1: A chemical reaction takes place in a container of cross sectional area of 50.0 cm2. As a result of the reaction, a piston is pushed out through 15 cm against an constant external pressure of 121 kPa. Calculate the work (in Joules) done by the system
Problem 2: A sample consisting of 2.00 mol of He is expanded at 22ºC from 22.8 dm3 to 31.7 dm3. Calculate q, w and ΔU (in Joules) when the expansion occurs under the following conditions: (a) Freely (b) Isothermal, and constant external pressure equal to the final pressure of the gas (c) Isothermal, Reversibly
Reversible, NonspontaneousReversible Process: The thermodynamic process in which a system can be changed from its initial state to its final state then back to its initialstate leaving all thermodynamic variables for the universe (system + surroundings) unchanged.
A truly reversible change will:
- Occur in an infinite amount of time - All variables must be in equilibrium with each other at every stage of the change
IrreversibleReversible
Irreversible, SpontaneousIrreversible Process: The thermodynamic process in which a system that is changed from its initial state to its final state then back to its initialstate will change some thermodynamic variables of the universe.
A truly irreversible change will:
- Occur in an finite amount of time - All variables will not be in equilibrium with each other at every stage of the change
IrreversibleReversible
Heat exchange can be measured in a Bomb Calorimeter
U = q + w
First Law: The internal energy of an isolated system is a constant.
Constant Volume: no work done
exΔU = q - P ΔV
vΔU = q ΔV = 0
Vq = heat at a constant volume
Calorimetry is the study of heat transfer during a chemical or physical process
vΔU = q
Heat
q = C ΔT
Heat is measured by the change in temperature of the water surrounding thecalorimeter
C = Calorimeter constant
The calorimeter constant is the heat capacity of the system
Heat Capacity (Cv) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant volume.
v
VV
ΔU = q
ΔU dU UC = = =
ΔT dT T
VC T
Heat Capacity at a constant volume
We need partial derivatives because the change holds the volume constant
v
VV
ΔU = q
ΔU U dUC = =
ΔT T dT
VC T
Heat Capacity at a constant volumeWe need partial derivatives becausethe change holds the volume constant
VV
U dU ΔUC =
T dT ΔT
How we can evaluate the Heat capacity equation
If the heat capacity (Cv) does not vary withtemperature during the range of temperaturesused, then the equation can be brokendown to:
VΔU = C ΔT
V Vq = C ΔT
Because at constant volume, change in internalenergy is equal to the heat (qv).
Problem 2.4: A sample consisting of 1.00 mol of a perfect gas, for which Cv,m = (3/2)R, initially at P = 1.00 atm and T = 300 K, is heated reversibly to 400 K at a constant volume. Calculate the final pressure, ΔU, q, and w
We cannot measure internal energy when the volume is not
constant
U = q + w
Typical reactions occur a constant external pressure
ΔU = q - PΔV
Heat exchange can be measured in a Differential Scanning Calorimeter at a
constant pressure
U = q + w
First Law: The internal energy of an isolated system is a constant.
Lets define Enthalpy as:
ΔU = q - PΔV
q = ΔU + PΔV
PΔH = q ΔP = 0
Pq = heat at a constant pressure
H = U + PV
Heat Capacity (Cp) is the amount of heat needed to change the temperature of a system by 1 *C at either a constant pressure.
ΔH = q
ΔH dHC = = =
ΔT dT T
P P
PP
C T
H
Heat Capacity at a constant pressure
We need partial derivatives becausethe change holds the pressure constant
dH ΔHC =
T dT ΔTPP
H
How we can evaluate the Heat capacity equation
If the heat capacity (CP) does not vary withtemperature during the range of temperaturesused, then the equation can be brokendown to:
ΔH = C ΔTP
q = C ΔTP P
Because at constant volume, change ininternal energy is equal to the heat (qv).
Problem 2.20: When 2.25 mg of anthracene, C14H10 (s) was burned in a bomb calorimeter, the temperature rose by 1.35 K. Calculate the calorimeter constant for the system. By how much will the temperature rise if 135 mg of phenol (C6H5OH) is burned in the calorimeter under the same conditions?
(ΔcHΘ(C14H10) = -7061 kJ/mol)
Adiabatic Expansion: work is done but no heat enters the system
• When a gas expands adiabatically, work is done but no heat enters or leaves the system.
• The internal energy falls and the temperature also falls.
q = 0U = q + w
VΔU = w = C ΔT
ad V= U = C ΔTw
Adiabatic changes to an Ideal Gas can be related by the
following equations1c
if i
f
VT T
V
1
if i
f
VP P
V
C = CV/nR
γ = Cp/CV
The derivation of these equations is quite long and follows the next slide
Adiabatic Expansion of an Ideal Gas (Derivation)
VC dT = -P dV
U = q + wFor an adiabatic change: q = 0
addU = w
nRTP =
VV
nRTC dT = - dV
V
V
1 nRC dT = - dV
T V
f f
i i
T V
VT V
1 nRC dT = - dV
T V
f f
i i
T V
V T V
1 1C dT = -nR dV
T V
Assume the heat capacity does notvary during the temperature interval
f fV
i i
T VC Ln = - nR Ln
T V
V f f
i i
C T V Ln = - Ln
nR T V
f i
i f
T Vc Ln = Ln
T VVCc = nR
f i
i f
T V=
T V
c
Problem 2.1- A 3.75 mole sample of an ideal gas with Cv,m = 3R/2 initially at a temperature Ti=298 K, and Pi= 1.00 bar is enclosed in an adiabatic piston and cylinder assembly. The gas is compressed by placing a 725 kg mass on the piston of diameter 25.4 cm. Calculate the work done in this process and the distance the piston travels. Assume the mass of the piston is negligible.
Thermochemistry
The study of energy transferred as heat during the course of chemical reactions.
In thermochemistry, we rely on calorimetry to measurethe internal energy (∆U) enthalpy (∆H).
Changes in matter are:
Physical ChangesSolid Liquid Melting (Fusion)Liquid Gas Boiling (Vaporization)Solid Gas Sublimation
Chemical ChangesFormation reactionsCombustion reactionsOther reactions
The Thermodynamic Standard State
3CO2(g) + 4H2O(g)C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)C3H8(g) + 5O2(g) ∆H = -2220 kJ
Thermodynamic Standard State: Most stable form of a substance at 1 bar and at a specified temperature, usually 25 °C (for non-solutions).
3CO2(g) + 4H2O(g)C3H8(g) + 5O2(g) ∆H = -2044 kJ
∆H° = -2044 kJ
The standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. The International Union of Pure and Applied Chemistry (IUPAC) recommends using a standard pressure po = 1 bar (100 kilopascals).
State FunctionsInternal energy and enthalpyare is a state function.
f
iΔU= dU
f
iΔH= dH
Work and Heat are pathfunction. This means we cannever write Δq or Δw
Example of Hess’s Law
Solve for the reaction below
N2H4(g)2H2(g) + N2(g)
Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.
2NH3(g)3H2(g) + N2(g) ∆H°1 = -92.2 kJ
∆H°3 = ?
2NH3(g)N2H4(g) + H2(g) ∆H°2 = -187.6 kJ
Germain Hess
Example 1: The two reactions below are known methods to produce ammonia by theHaber process. Use the information to solve for the enthalpy change in thereaction below.
Solve for the reaction below
N2H4(g)2H2(g) + N2(g)
2NH3(g)3H2(g) + N2(g) ∆H°1 = -92.2 kJ
∆H°3 =
2NH3(g)N2H4(g) + H2(g) ∆H°2 = -187.6 kJ
Example 1: The two reactions below are known methods to produce ammonia bythe Haber process. Use the information to solve for the enthalpy changein the reaction below.
Steps to solve the problem - Find known reactions that will add up to the desired reaction
- Add up the reactions and add up the ΔH, ΔU, ΔG or other state functions
Θ ΘΔH A B = -ΔH B A
3 2 4 22NH N H (g) + H (g) 2NH3(g)3H2(g) + N2(g)
∆H°2 = 187.6 kJ∆H°1 = -92.2 kJ
2
-92.2 kJ + (187.6 kJ) = 95.4 kJ
Standard Heats of FormationStandard Heat of Formation (∆fH° ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states.
∆H°f = -74.8 kJCH4(g)C(s) + 2H2(g)
Standard states
1 mol of 1 substance
Standard Heats of Formation are used to determine reaction enthalpies (ΔrHº)
cC + dDaA + bB
ReactantsProducts
∆rH° = {c ∆fH°(C) + d ∆fH°(D)} – {a ∆fHº(A) + b ∆fH°(B)}
Reaction enthalpies (ΔrHº) can be determined by the difference of the product enthalpies of formation and the reactant enthalpies of formation.
0 0 0r f f
Products Reactants
Δ H = v Δ H - v Δ H Example:
Standard Heats of Formation
Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O.
C6H12O6(s) + 6*O2(g)6*CO2(g) + 6*H2O(l) ∆rH° = ?
ΔrH° = [∆H°f (C6H12O6)] - [6*∆H°f (CO2) + 6*∆H°f (H2O)]
[(6 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)]
∆rH° = [(1 mol)(-1273.3 kJ/mol)] -
= 2802.5 kJ
Answer
Products Reactants
Standard State (ΔfHº)= 0
Heats of Combustion (ΔcHº) is a special type of reaction enthalpy
CO2(g) + H2O(l)CH4(g) + O2(g)
= -890.3 kJ
∆cH°= [∆fH°(CO2) + 2 ∆fH°(H2O)] - [∆fH°(CH4)]
[(1 mol)(-74.8 kJ/mol)]
= [(1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol)] -
Calculate the standard enthalpy of combustion for methane (CH4(g)).
Answer
ΔfHº (CH4) = -74.8 kJ/mol
ΔfHº (CO2) = -393 kJ/mol
ΔfHº (H2O) = -285 kJ/mol
Combustion reactions are always balanced with 1 as the coefficient of the hydrocarbon
2 2
Temperature dependence of reaction enthalpy (ΔrHº)
• Often standard reaction enthalpies need to be corrected for temperature differences.
o o or p p,m p,m
Products Reactants
Δ C = v C - v C
2
1
T
1 2 pTH T = H T + C dT
2
1
To o or 1 r 2 pT
Δ H T = Δ H T + C dTr
If Cp is independent of temperature during the temperature range, the integral can be evaluated.
The equation doesn’t work if there is a phase transitionthrough the temperature range.
Problem: Balance the reactions and determine ΔrHΘ and ΔrUΘ
23 2g g g gNH NO N H O
NO: 91.3 kJ/molNH3: -45.9 kJ/molH2O: -241.8 kJ/mol
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