Download Week 1 - University of Guelph

THERMODYNAMICS
• science of the interconversion of different forms of ENERGY
e.g. heat and work
• an essential foundation for understanding physical sciences.
• at the centre of any physical description
Key Questions:
1. Interconversion of energy – how, why, etc.
2. Is a particular reaction possible?
3. How much useful work can be obtained from a reaction?
4. Can we understand chemical equilibria?
Chemical reactions involve a transfer of energy between
the system and the surroundings.
DEFINITIONS:
SYSTEM – the part of the universe under observation
e.g., engine, electrical cell, reaction in a flask, etc.
The system can be:
• “Open” – both mass and energy may leave and enter.
• “Closed” – energy can be exchanged but no mass can enter or leave.
• “Isolated” – neither mass or energy may enter or leave.
SURROUNDINGS – the rest of the universe.
Therefore, the “UNIVERSE” is made up of the “SYSTEM” and the “SURROUNDINGS”.
The “STATE” of a system is characterised by a set of variables, e.g., P, V, T, n, etc.
These are STATE VARIABLES also known as STATE FUNCTIONS.
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P, V, T, n and mass can be classified as either:
Intensive
– independent of size.
e.g. P, T
Extensive
– dependent on size.
e.g. m,V, n
ENERGY TRANSFER
• We can transfer energy in and out of a system, whereby the energy of the
system may change.
• Internal Energy (U) is the sum of the kinetic and potential energy of system.
• Two ways U can change:
1. q – heat flow
2. w – work
ENERGY TRANSFER
SYSTEM
WORK (w)
HEAT (q)
Thermal
Surroundings
INTERNAL
ENERGY (U)
Mechanical
Surroundings
We can ONLY measure changes in U, not absolute values,
so for any reaction or process, i.e., Ufinal – Uinitial = ΔU
We can change the internal energy, U, by any combination of q and w.
∴ ΔU = q + w
This is a statement of the conservation of energy:
1st LAW OF THERMODYNAMICS
The Law of Conservation of Energy states energy may be converted from
one form to another, but the total quantity of energy remains constant.
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HEAT ENERGY, HEAT TRANSFER and TEMPERATURE
ΔU = q + w tells us we cannot create or destroy energy, but it can be
transferred in and out of the system!
T 1 > T2
T1
T2
Object 1
Object 2
TEMPERATURE – an indication of thermal energy.
• objects at same T do not exchange heat energy.
• if T’s are different, energy (heat) flows from high T (hot) to low T (cold).
•
THERMOCHEMISTRY (Chapter 6)
– a subset of thermodynamics
– study of the heat given off or absorbed in a chemical reaction.
CHEMICAL REACTIONS AND ENERGY CHANGES
All chemical reactions are accompanied by an ENERGY CHANGE
Chemical reactions may release heat to the surroundings, e.g.,
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) (burning or combustion of natural gas)
If heat is released (given off), the reaction is exothermic.
This arises due to differences in the energies of reactants and products.
ALL combustion reactions are EXOTHERMIC.
If heat is absorbed (taken in), the reaction is endothermic.
e.g., 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) (photosynthesis)
Note: The amount of heat depends on the size of the system.
(This is an EXTENSIVE property, like mass, V and n.)
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SIGN CONVENTION FOR HEAT PROCESSES
PROCESS
SIGN
Heat absorbed by the system from the surroundings (endothermic)
+
Heat absorbed from the system by the surroundings (exothermic)
–
Most reactions are run under constant pressure conditions. The heat given off (or
absorbed) at constant pressure is known as the ENTHALPY change, and is
denoted by the symbol, ΔH
ENTHALPY - is a measure of the difference in heat content, H,
between final and initial states (e.g., products and reactants).
i.e., ΔHRxn = Σ ΔHproducts – Σ ΔHreactants = qp = heat flow
EXOTHERMIC and ENDOTHERMIC PROCESSES
Reactants
Enthalpy
ΔHRxn < 0
(heat released)
EXOTHERMIC
Products
Products
ΔHRxn > 0
(heat absorbed)
ENDOTHERMIC
Enthalpy
Reactants
Remember:
• Heat changes at constant P are equal to the change in ENTHALPY
• If heat is given off, the rxn is EXOTHERMIC (ΔH is negative)
• If heat is absorbed, the rxn is ENDOTHERMIC (ΔH is positive)
We can measure this heat change through the science of calorimetry.
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CALORIMETRY (Refer to page 238 in text )
– measurement of the quantities of heat in bodies.
Experimentally, we need to:
• measure temperature change during a reaction, then
• convert that change in temperature to change in energy.
How?
HEAT CAPACITY (C) – energy required to raise the temperature of a substance
by 1C° or 1 K unit, i.e., C = q/ΔT
YOU should distinguish between:
A) Heat Capacity (C) – energy required to raise a substance by 1C° or
1 K unit. Units: J K–1
B) Molar Heat Capacity (c) – energy needed to raise 1 mole through 1C°or
1 K unit (or the heat capacity per mole). Units: J K–1 mol–1
C) Specific Heat (s, CSP or CS) – amount of energy to raise 1 g through 1C°
or 1 K unit (or the heat capacity per gram of substance).
Units: J g–1 K–1
e.g., H2O(l) Specific Heat = 4.184 J g–1 deg–1
Hence, q = C × ΔT = n × c × ΔT = mass × s × ΔT
In calorimetry· the idea is to ISOLATE the SYSTEM from the SURROUNDINGS.
e.g. "Coffee–Cup" Calorimeter – refer to Figure 6.12 in text
To measure heat given off (or absorbed) by the chemical reaction:
• q = CΔT (calculate this usually)
• C – know this for the vessel and water (sometimes given together,
sometimes separately).
• ΔT – measure this
Note: Heat released in the reaction is equal to the heat gained by the surroundings,
i.e., qSystem + qSurroundings = 0
therefore, qSystem = –qSurroundings = qRxn
qRxn = –(qsolution + qcalorimeter)
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HEAT OF SOLUTION
- enthalpy change associated with dissolution of 1 mole solid:
i) take known mass of water at temperature, T1
ii) add mass of solid, dissolve and measure temperature T2
iii) calculate the change in temperature: ΔT = T2 – T1 = TFinal – TInitial
Example – Dissolution of a solid (KF) in water:
Exactly 1.23 g KF (s = 0.84 J g–1 deg–1) is dissolved in 10.0 g H2O (s = 4.184 J g–1
deg–1). The H2O temp rises from 23.2 to 33.1°C. What is heat of solution per mole
for KF? (MM KF = 58.1 g mol–1)
KF dissolves in the water and liberates heat which warms the solution.
Energy = Heat gained by water + Heat gained by KF
Energy =
q(H2O)
+
q(KF)
Energy =
Note the size of q(KF) versus q(H2O)!
In many cases, we will just assume the specific hear of water is very close
to that of the solution and simplify this calculation to “Energy = q(H2O)”.
MM for KF = 58.1 g/mol, therefore # moles of KF added =
1 mole of KF would therefore evolve _______________________ of heat.
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Bomb Calorimeter (refer to Figure 6.13)
•
•
•
carries out combustion reactions at constant volume conditions
the heat produced raises the temperature of the bomb and the
surrounding liquid (the bomb is usually placed in a water bath).
Sometimes we evaluate the heat capacity of the bomb and the water
separately (i.e., qTotal = qBomb + qwater bath) and sometimes the heat
capacity for the entire calorimeter (bomb + water bath) has been
determined experimentally. You must read the question carefully to
determine which case you have.
Example:
Exactly 1.00 g of (CH3)2N2H2 (MM = 60.1 g/mol) is burned in a bomb calorimeter
with a heat capacity = 1840 J deg–1. A 5.00 kg water bath rises in temperature from
24.62 to 26.07°C. Calculate the energy released when 1 mole is burned.
Total heat produced =
warms bomb
+
warms water bath
1 mole of (CH3)2N2H2 evolves ____________________________ kJ of heat energy.
EXTRA PROBLEMS TO TRY
Problem 1:
50.0 mL dilute AgNO3 is added to a solution with OH– ions in a coffee cup calorimeter.
Ag2O precipitates and the temperature of the liquid goes from 23.78 to 25.19°C. Assume
that the mixture has the same specific heat as water. Calculate "qsurr" for a mass of 150. g. Is
the reaction exothermic or endothermic?
Answer: qSurroundings = 885 Joules
Rxn exothermic (temp rose).
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Problem 2:
When ice at 0°C melts to liquid H2O, it absorbs 334 J per gram. The specific heat of water
is 4.184 J g–1 deg–1. Suppose the heat needed to melt a 35.0g ice cube is absorbed from
water (0.210 kg) in a calorimeter at 21.0°C. What is the final water temperature? [This is
the kind of calculation that all experienced party givers do before throwing a party!]
Answer: TFinal = 6.6°C
WHAT ABOUT WORK?
The first law of thermodynamics states that the change in the internal energy is
dependent upon the flow of heat, but is also depend on WORK (refer to sec. 18.1)
WORK – the energy exchange that results when a force moves an
object through a distance.
EXAMPLES OF WORK
Mechanical: w = force × distance = F × Δx = m × a × Δx
Units: (mass) × (acceleration) × (distance) = kg•m2•s-2 = N•m = Joule (J)
Electrical:
w = charge × potential difference = ΔQ × V
Units: (coulombs) × (volts) = Joule
Note: We will look at this kind of work when we discuss electrochemistry later on in the course.
COMPRESSION/EXPANSION or P-V WORK
(see Figures 6.9 & 18.2 in the text)
At a CONSTANT external pressure, Pext., the work (in Joules) done BY the
surroundings ON a gas (i.e., system) is given by
w = force × distance = F × Δx
w = F/A × A × Δx = Pext. × A × Δx
where the gas is compressed a distance Δx by a pressure, Pext..
But, A × Δx can be represented by the volume decrease, or,
w = Pext. ΔV
where ΔV = V2 – V1
Compression: work is done ON the system BY the surroundings
Our definition says work is to be positive because for a gas to be
compressed, energy is added to the system, i.e., the surroundings
does work on the system.
So our relationship must be:
w = – Pext.ΔV
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Expansion work is negative since energy leaves the system as the
system does work on the surroundings.
Pressure-Volume (or P-V) work is the most common type of work
encountered in chemical systems.
UNITS FOR ENERGY
1 J = 1 kg•m2•s-2 (SI definition) i.e., mass × velocity2
Note: 1 Pa = 1 N•m-2 = 1 kg•m-1•s-2 ∴1 J = Pa•m3
Hence PΔV → Pa•m3 = Joule
An older unit still in use is the “calorie”, where 1 cal = 4.184 J (by definition)
EXAMPLE: Calculate the work at 25°C and 1.00 atm for
a) CO2(g) + NaOH(s) → NaHCO3(s)
b) 2O3(g) → 3O2(g)
Consider:
– only PV work,
– that the gases behave ideally (i.e., follow the ideal gas law) and
– the equations represent the amounts in moles.
In a), we move from 1.00 mole of gaseous CO2 on the reactant side to 0 moles of
gas on the product side. Work is done ___ the system ____ the surroundings. This is
an example of a __________________. Work done is then ____________ (+/–).
First determine the volume of the CO2(g), then determine the work done.
Note: The gas constant R = 8.314 J K-1 mol-1 = 0.0821 L•atm K–1•mol–1
In b), we move from 2 moles of gaseous O3 on the reactant side to 3 moles of O2 on
the product side. Work is done ___ the system ____ the surroundings. This is an
example of a __________________. Work done is then _______________(+/–).
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QUESTIONS:
1) A gas sample is heated in a cylinder using 550 kJ of heat. A piston
compresses the gas, doing 700 kJ of work. What is the change in internal
energy of the system?
2) Calculate the work done when 6300 L of an ideal gas (initially at 1.00 atm)
is compressed (at constant T) by a constant pressure of 1.40 atm to a final
volume of 45L.
SIGN CONVENTION FOR WORK and HEAT PROCESSES
PROCESS
SIGN
Work done ____ the system ____ the surroundings (____________)
–
Work done ____ the system ____ the surroundings (____________)
+
Heat absorbed ____ the system ____ the surroundings (____________)
+
Heat absorbed ____ the system ____ the surroundings (____________)
–
•
•
if w and q are positive, they tend to increase U of the system
work done by the system is a reduction in U, hence negative.
INTERNAL ENERGY & ISOTHERMAL EXPANSIONS
•
•
•
•
•
Internal energy of an ideal gas depends only on its temperature.
To change the temperature of a gas, the internal energy must be changed by
doing work on it or transferring energy to it, i.e., ΔU = q + w.
When a gas expands, the forces due to pressure push the container walls
back, so the gas is doing work. Change in w is negative.
For an isothermal (i.e., ΔT = 0) expansion of an ideal gas, ΔU = 0 because
the kinetic energy of the gas is related to its temperature (which here is
unchanged) and the potential energy is related to bonding and structure
(which here is unchanged).
Since ΔU = q + w = 0, ∴ w = –q. The system does w joules of PV-work
and absorbs an equivalent amount of heat q = –w.
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HEAT CHANGES IN REACTIONS
a) Reactions at Constant Volume, i.e., a sealed container
System cannot expand or compress (ΔV = 0), so the work is zero, i.e., w = 0
Using the First Law of Thermodynamics, ΔU = q + w;
∴At constant volume, qV = ΔU (e.g., in a bomb calorimeter)
i.e., at constant volume (say in a sealed, “bomb” calorimeter), the
increase/decrease in internal energy corresponds to the heat
supplied/absorbed by the surroundings. Therefore the molar heat capacity ,
c, becomes: cV = qV/ΔT = ΔU/ΔT
b) Reactions at Constant Pressure, i.e., open to the atmosphere
The heat will cause the volume to change, i.e., ΔV ≠ 0
Some work will be done, e.g., pushing back the atmosphere
i.e., w = –Pext.ΔV
Hence, ΔU = q – Pext.ΔV
OR
qP = ΔU + Pext.ΔV = ΔH
∴At constant pressure, qP = ΔΗ
Most reactions run under constant pressure (including biological processes).
We can measure this heat change with a coffee-cup calorimeter (refer to
page 239). When dealing with calorimetry measurements at constant
pressure, the molar heat capacity becomes: cP = qP/ΔT = ΔH/ΔT
Therefore, heat capacities give the rate of change of U and H with T.
ΔU and ΔH are STATE functions. (q and w are not.)
X
Final
Height
X
Initial
ΔHEIGHT = HEIGHTFinal – HEIGHTInitial
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A state function for a system depends on the differences between the final
and initial state ONLY, NOT on the pathway. This will become important
later when we discuss Hess’ Law.
HOW DO ΔH and ΔU COMPARE FOR A REACTION?
ΔH = ΔU + Δ(PV)
ΔH and ΔU differ only in the PV product.
For reactions at constant P, the above equation becomes: ΔH = ΔU + PΔV OR
ΔH – ΔU = –w
OR ΔU – ΔH = w
i.e., the difference between the change in internal energy and the enthalpy is the
amount of work that needs to be done to make room for the products of the reaction.
For reactions at constant pressure, the volume change of solids and liquids is very
small. Hence, for these cases ΔH is similar to ΔU.
For an ideal gas at constant T and P, the equation PV = nRT will let us calculate the
volume change, i.e.,
Δ(PV) = Δ(nRT)
PΔV = ΔnRT
i.e., Δn = moles of gaseous products minus moles of gaseous reactants.
∴ ΔH = ΔU + ΔnRT
Example:
At 298 K, CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔHRxn = –890.6 kJ mol–1
What is ΔURxn?
At constant pressure, q =
ΔURxn =
Could calculate the actual volume change from PV = nRT.
Not necessary (in this case), since we can use PΔV = ΔnRT.
Δn =
ΔU =
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Net result is a shrinking volume (compression), therefore work done by
____________________ on _________________________.
At 25°C, RT = 8.314 JK-1 mol–1 × 298K ≈ 2.5 kJ mol-1, therefore, ΔHRxn and ΔURxn
are not that far apart in value.
PROBLEM:
When an ideal gas expands at constant temperature (i.e., isothermal expansion), ΔU is zero.
Consider 1.00 L of an ideal gas initially at 9.00 atm and 15°C.
(a) Calculate q and w if the gas expands isothermally against a pressure of 1.00 atm.
(b) Calculate q and w if the gas expands isothermally first against a pressure of 3.00 atm.
And then against 1.00 atm.
(c) Calculate q and w if the gas expands isothermally first against a pressure of 3.00 atm,
then 2.00 atm and then against 1.00 atm.
(d) Comment on the results.
Answers:
a) q = –w = +0.810 kJ
b) w1 + w2 = –1.216 kJ
q1 + q2 = +1.216 kJ
c) w = w1 + w2 + w3 = –1.368 kJ
q = q1 + q2 + q3 = +1.368 kJ
d) The conclusion is that you would get
the maximum amount of work done
on the surroundings by doing the
expansion in an infinite number of
very small steps.
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