Thermochemistry: Energy Flow and Chemical Reactions · PDF fileThermochemistry is a branch of...
Transcript of Thermochemistry: Energy Flow and Chemical Reactions · PDF fileThermochemistry is a branch of...
Thermochemistry:
Energy Flow
and
Chemical
Reactions
• thermodynamics
• internal energy – definition, first law
• enthalpy – definition, energy diagrams, calorimetry,
theoretical calculation (heats of formation and
bond energies), stoichiometry
• hess’s law
• energy from foods
Outline
Thermochemistry is a branch of thermodynamics
that deals withthe heat involved with chemical and
physical changes.
Thermodynamics is
the study of heat and
its transformations.
Thermodynamics
Internal Energy
What is the difference of the two pictures above?
Energy state
If we go from cold water to boiling water as our reaction:
H2O (solid) H2O (gas)
energy state 1 energy state 2
Internal Energy
we define E = a variable to denote the energy of our
reacting substances
= specifically, the internal energy of the substance
undergoing reaction
∆ E = change in energy
= E (boiling water) – E (cold water)
since Energy (gas) > energy (solid)
∆ E = positive , +
For the reaction to proceed, energy has to be supplied.
But where will it come from?
A chemical system and its surroundings.
the system
the surroundings
Internal Energy
System = the reaction itself, the object of our study
Surroundings = the rest of the universe
Internal Energy
DE = Efinal - Einitial = Eproducts - Ereactants
Energy diagrams for the transfer of internal energy (E) between a
system and its surroundings.
Internal Energy
Energy changes occurs because of the exchange of
energy between the system and the surroundings!
Internal Energy
Notice that all the equations/energies refer to the system!
But we can also define it from the point of view of the
surroundings.
Case 1: surroundings supplied energy to the system∆ E = E2, surr - E1, surr = negative
We have to specify which we are referring to: system or
surroundings
State 1: State 2:
E 1, surr
E 1, sys
E 2, surr
E 2, sys
E 2, sys - E1, sys = ∆ E system = positive, +
E 2, surr - E1, surr = ∆ E surroundings = negative, -
Internal Energy
So we have a discrepancy of notations and definitions.
As per convention, whenever we have E1, E2, etc., we are
always referring to the system. Henceforth, we are always
referring to the system in all our studies of energies.
Notice, however that∆ E system + ∆ E surr = 0
since: system + surroundings = universe
∆ E universe = 0 = E2, universe - E1, universe = 0
E2, universe = E1, universe
The energy of the universe is constant!
Internal Energy
First Law of Thermodynamics:
The energy of the universe is
constant.
Thus, energy can neither be
created nor destroyed. It can
be only be converted from
one form to another.
Internal Energy
Let’s define formally E = internal energy
= total energy of the system
= summation of all the kinetic and potential
energies of the system
Experimentally: ∆ E = q + w
q = heat = energy transferred from a hotter object to a colder one
w = work = energy used to cause an object to move against a force
That means, we can change the energy of the system by
applying heat and doing work. Or,
When energy is transferred from one object to another,
it appears as work and/or as heat.
A system transferring energy as heat only.
Internal Energy
A system losing energy as work only.
Energy, E
Zn(s) + 2H+(aq) + 2Cl-(aq)
H2(g) + Zn2+(aq) + 2Cl-(aq)
DE<0work done on
surroundings
Internal Energy
The Sign Conventions* for q, w and DE
q w+ = DE
+
+
-
-
-
-
+
+
+
-
depends on sizes of q and w
depends on sizes of q and w
* For q: + means system gains heat; - means system loses heat.
* For w: + means word done on system; - means work done by system.
Internal Energy
Internal Energy
DE = (sign), (number), (unit)
Sign = gives direction as to the transfer of energy
+ = E final > E initial = system gained energy from the surroundings
- = E final < E initial = system lost energy to the surroundings
Units of Energy
Joule (J)
calorie (cal)
British Thermal Unit
1 cal = 4.184 J
1 J = 1 kg*m2/s2
1 Btu = 1055 J
Internal Energy
Calorie (Cal) 1 Calorie = 1 000 calorie
Sample Problem
Hydrogen and oxygen reacts in a cylinder. As the reaction
occurs, the system loses 1150 J of heat to the
surroundings and the expanding gas product does 480 J of
work on the surroundings as it pushes against the
atmosphere. What is the change in the internal energy of
the system?
q = heat = - 1150 J
w = work = - 480 J
∆E = q + w= (-1150 J) + (- 480 J)
= - 1630 J
A system receives 425 J of heat and delivers 400 J of
work to its surroundings. What is the change in internal
energy of the system (in joules)?
Answer: ∆ E = q + w
= (+ 425) + (- 400 J)
= + 25 J
Sample Problem
Worksheet # 9-1
1. Complete combustion of 1.0 metric ton of coal
(assuming pure carbon) to gaseous carbon dioxide
releases 3.3 x 1010 J of heat. Convert this energy to (a)
kilojoules; (b) kilocalories and (c) british thermal unit.
2. A system conducts 225 cal of heat to the surroundings
while delivering 428 cal of work. What is the change in
internal energy of the system (in cal and J)?
3. What is the change in internal energy (in J) of a system
that releases 675 J of thermal energy to its
surroundings and has 525 calories of work done on it?
Worksheet # 9-1: Answers
1. Complete combustion of 1.0 metric ton of coal (assuming
pure carbon) to gaseous carbon dioxide releases 3.3 x 1010 J
of heat. Convert this energy to (a) kilojoules; (b) kilocalories
and (c) british thermal unit.
Answers: a) 3.3 x 107 kJ b) 7.9 x 106 kcal c) 3.1 x 107
Btu
2. A system conducts 225 cal of heat to the surroundings while
delivering 428 cal of work. What is the change in internal
energy of the system (in cal and J)? Answer: - 653 cal =
- 2732 J
3. What is the change in internal energy (in J) of a system that
releases 675 J of thermal energy to its surroundings and has
525 calories of work done on it? Answer: 1522 J
Some interesting
quantities of energy.
Two different paths for the energy change of a system.
Enthalpy
Pressure-volume
work.
Enthalpy
Enthalpy
There is another energy variable we call enthalpy:
H = E + PV
PV = work
∆H = ∆E + P ∆V
Formal definition: Enthalpy is the thermodynamic quantity
that is the sum of internal energy and the product of the
pressure-volume work.
The change in Enthalpy equals the heat gained or lost at
constant pressure.
Most chemical reactions occur at constant pressure, so ∆H
is more relevant than ∆E.
w = - PDV
DH = DE + PDV
qp = DE + PDV = DH
DH ≈ DE in
1. Reactions that do not
involve gases.
2. Reactions in which the
number of moles of gas does
not change.
3. Reactions in which the
number of moles of gas does
change but q is >>> PDV.
H = E + PV
where H is enthalpy
Enthalpy
Enthalpy
Let’s examine enthalpy and internal energy more closely:
We defined ∆E = E2 – E1
We can calculate ∆E from the first law of thermodynamics.
What about E2, E1 or the exact energy value of a system.
E, internal energy = summation of all the energies of the
system, including KE and PE
For a number of particles (1, 2, 3 particles) energy can be
calculated, but if we are dealing with macroscopic amounts (ex:
a mole of water, 6.02 x 1023 particles), the exact value is
incalculable and impossible!
But is it important?
Enthalpy
We do not need to calculate the energy values exactly. But
we assume that the energy of a system at a defined
condition is exact, fixed and unchanging.
Example:
1.0 mol of 2.5 L water at 90 C, 1.5 atm pressure has an
internal energy E and an enthaly H whenever this condition
is specified
Mathematically,
we call ∆E and ∆H = state functions
state function = property of a system that is determined by
specifying its conditions or its state (ex. T, P, etc.)
= the value of a state function does not depend on the
particular history of the sample, only on its present condition
Enthalpy
State function analogy: scaling a mountain
The height of the mountain is a state function. However, the
way you can climb the mountain is not.
H, E = state functions
q, w = not state functions, path dependent (value depends
on how the process took place)
Enthalpy
Exothermic Reactions: ∆H = H final – H initial = negative, -
energy is released by the system
Endothermic reactions: ∆H = H final – H initial = positive, +
energy is absorbed by the system
The sign of enthalpy (like internal energy) indicates the
direction of heat transfer during a process that occurs at
constant pressure.
Enthalpy diagrams for exothermic and endothermic processes.
Enth
alp
y, H
Enth
alp
y, H
CH4 + 2O2
CO2 + 2H2O
Hinitial
HinitialHfinal
Hfinal
H2O(l)
H2O(g)
heat out heat inDH < 0 DH > 0
A Exothermic process B Endothermic process
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)
Enthalpy
Enthalpy
Video: endothermic reaction (endo final.wmv)
Enthalpy
It turns out experimentally that:
∆E = energy transferred at constant volume conditions
∆H = energy transferred at constant pressure conditions
In most chemical reactions, we are mostly concerned with
enthalpy. As most chemical reactions occur at constant
pressure.
Thermochemical equations
2H2 (gas)+ O2 (gas)→ 2H2O (gas) ∆H = - 483.6 kJ
The enthalpy value is given at the end of the chemical
equation.
Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
heat of formation (DHf)
heat of fusion (DHfus)
heat of vaporization (DHvap)
C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)
K(s) + 1/2Br2(l) KBr(s)
NaCl(s) NaCl(l)
C6H6(l) C6H6(g)
Enthalpy
Write a balanced equation and draw an approximate
enthalpy diagram for each of the following:
a) Combustion of one mol of methane in oxygen
b) The evaporation of liquid water
Answer:
a) CH4 + 2O2 → CO2 + 2H2O, exothermic
b) H2O(l) → H2O(g), endothermic
Sample Problem
Enth
alp
y, H
CH4 + 2O2
CO2 + 2H2O Enth
alp
y, H
H2O(l)
H2O(g)
heat out heat in
Worksheet # 9-2
In each of the following cases, determine the sign of
enthalpy, state whether the reaction is exothermic or
endothermic and draw an enthalpy diagram.
1. H2 (g) + ½ O2 (g) → H2 O (l) + 285.8 kJ
2. CO2 (s) + 260 kJ → CO2 (g)
3. The heat released for the combustion of liquid ethanol
4. The vaporization of liquid ammonia
Worksheet # 9-2: AnswersE
nth
alp
y, H
Enth
alp
y, H
heat out
heat in
exothermic
endothermic
H2 (g) + ½ O2 (g)
H2 O (l) 2CO2 + 3H2O
C2H5OH + 3O2
CO2 (s)
CO2 (g)
+ 260 kJ
NH3 (l)
NH3 (g)
- 285.8 kJ
Enthalpy
Properties of enthalpy:
1. Enthalpy is an extensive property. Thus its magnitude is
directly proportional to the amount of substance reacted
or produced.
Implication: in a thermochemical equation, if the equation is multiplied by 2, the ∆H is also multiplied by 2. If the
equation is divided by 2, the ∆H is also divided by two.
Example: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ
2CH4(g) + 4O2 (g) → 2CO2 (g) + 4H2O(l) ∆H = 2(- 890) kJ
Enthalpy
2. ∆H reaction = - ∆H reverse reaction
CO2 (g) + 2H2O (l) → CH2 (g) + 2O2 (g) ∆H = 890 kJ
3. ∆H is depends on the state of the chemical species used in
the equation. The state of the substance has to be shown.
Example:
CO2 (g) + 2H2O (l) → CH4 (g) + 2O2 (g) ∆H = 890 kJ
CO2 (g) + 2H2O (g) → CH4 (g) + 2O2 (g) ∆H = 978 kJ
Enthalpy
Enthalpy and chemical stoichometry-The calculation of the enthalpy for a particular reaction and
its relation to the chemical equation is governed property 1
of enthalpy – it is an extensive property.
Example:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ
relationships: 1 mol CH4 is to ∆H = - 890 kJ
2 mol O2 (g) is to ∆H = - 890 kJ
1 mol CO2 is to ∆H = - 890 kJ
2 mol H2O is to ∆H = - 890 k
Sample Problem
Consider the equation:
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ∆H = - 890 kJ
1. How many moles of methane, CH4 (g) has to be burned
to generate 1500 kJ of energy?
2. How many grams of carbon dioxide is produced if the
enthalpy generated in the combustion is 1830 kJ?
Consider the following balanced thermochemical equation
for a reaction sometimes used for H2S production:
1/8 S8 + H2 → H2S ΔH = - 20.2 kJ
a) Is this an exothermic or endothermic reaction?
b) What is the ΔHrxn for the reverse reaction?
c) What is ΔH when 3.2 mol of S8 reacts?
d) What is ΔH when 20.0 g of S8 reacts?
Answers:
a) Exothermic
b) 20.2 kJ
c) -517 kJ
d) -12.6 kJ
Sample Problem
AMOUNT (mol)
of compound A
AMOUNT (mol)
of compound B
HEAT (kJ)
gained or lost
molar ratio from
balanced equation
DHrxn (kJ/mol)
Summary of the relationship
between amount (mol) of
substance and the heat (kJ)
transferred during a reaction.
Enthalpy
Worksheet # 9 - 3
1. Consider the following balanced thermochemical equation for the
decomposition of the mineral magnesite:
MgCO3 → MgO + CO2 ∆H = 117.3 kJ
(a) is heat absorbed or released in the reaction?(b) what is the ∆H for the reverse reaction?(c) what is the ∆H when 5.35 mol of CO2 reacts with excess MgO?(d) What is the ∆H when 2.36 mol of magnesite reacts?(e) what is the ∆H when 0.15 mol of MgO is produced from the reaction?
2. When 1.0 mol of NO forms from its elements, 90.29 kJ of heat is absorbed.(a) write a balanced thermochemical equation for this reaction.(b) how much heat is involved when 1.50 mol of NO decomposes to its elements?(c) how much heat is involved when 1.75 grams of NO decomposes to its elements?(d) how much heat is involved when 3.50 grams of N2 is produced?
Worksheet # 9 – 3: Answers
1. Consider the following balanced thermochemical equation for the
decomposition of the mineral magnesite:
MgCO3 → MgO + CO2 ∆H = 117.3 kJ
(a) is heat absorbed or released in the reaction? Absorbed(b) what is the ∆H for the reverse reaction? -117.3 kJ(c) what is the ∆H when 5.35 mol of CO2 reacts with excess MgO?-627.6 kJ(d) What is the ∆H when 2.36 mol of magnesite reacts? + 276.8 kJ(e) what is the ∆H when 0.15 mol of MgO is produced from the reaction? + 17.6 kJ
2. When 1.0 mol of NO forms from its elements, 90.29 kJ of heat is absorbed?(a) write a balanced thermochemical equation for this reaction?
½ N2 + ½ O2 → NO ∆H = 90.29 kJ(b) how much heat is involved when 1.50 mol of NO decomposes to its elements?(c) how much heat is involved when 1.75 grams of NO decomposes to its elements?(d) how much heat is involved when 3.50 grams of N2 is produced?
b) -135.4 kJ c) -5.3 kJ d) -22.6 kJ
Enthalpy
Calorimetry = measurement of heat flow
Calorimeter = apparatus that measures heat flow
∆H reaction can be determined experimentally using a calorimeter.
Coffee-cup calorimeter.
Enthalpy
Enthalpy
What did the
thermometer say to
the graduated
cylinder?
"You may have
graduated but I've got
many degrees."
Enthalpy
In chemical equations, energy changes can be calculated by
monitoring the temperature changes in the reaction.
q is proportional to ∆T
In equation form: q α ∆T
q = C ∆T
where C = constant of proportionality
= heat capacity
= amount of heat required to raise the temperature of
a substance by 1.0 Kelvin
The larger the heat capacity the greater the heat required to
produce a given rise in temperature.
For 1.0 mol substance , C = C (mol)
Lets look at another experiment:
1.0 L water and 10.0 ml water are heated separately such that each would
have its temperature increase by 10.0 C. What would be the result.
The 10.0 ml water sample will have a greater increase in energy as some of
its molecules can turn into gas.
Implication: mass is also a factor in measuring enthalpies.
q = C ∆T
As there is no mass in the equation, it can be in the heat capacity factor.
C = m c
where m = mass of the substance
c = specific heat of the substance
Thus, q = m c ∆T
Enthalpy
Heat Capacity C = q / ΔT
Specific heat capacity (c) = q / (mass x ΔT)
unit: J/K
unit: J/gK
Molar heat capacity (C mol) = q / (moles x ΔT)
unit: J/mol K
Enthalpy
Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials
Specific Heat
Capacity (J/g*K)
SubstanceSpecific Heat
Capacity (J/g*K)
Substance
Compounds
water, H2O(l)
ethyl alcohol, C2H5OH(l)
ethylene glycol, (CH2OH)2(l)
carbon tetrachloride, CCl4(l)
4.184
2.46
2.42
0.864
Elements
aluminum, Al
graphite,C
iron, Fe
copper, Cu
gold, Au
0.900
0.711
0.450
0.387
0.129
wood
cement
glass
granite
steel
Materials
1.76
0.88
0.84
0.79
0.45
Enthalpy
Relation of enthalpy and q (from calorimetry experiments)
q = m c ∆T = q solution
q solution = - q reaction = ∆H reaction
∆H reaction = - m c ∆T
A 295 g aluminum engine part at an initial temperature of
3.0 C absorbs 85.0 kJ of heat. What is the final
temperature of the part (specific heat of aluminum = 0.900
J/g K)?
Answer: Tf = 323 C
Sample Problem
q = m c ∆T85 000 J = 295 g Al (0.9 J/gK) (Tf – 3.0 C)
When 165 ml of water at 22 C is mixed with 85 mL of water
at 82 C, what is the final temperature? Assume that no
heat is lost to the surroundings (density of water is 1.0
g/mL).
(m c ∆T) water 1 = (m c ∆T) water 2
165 ml ( Tf – 22 C ) = 85 ml (Tf – 82 C)
Answer: Tf = 42 C
Sample Problem
Worksheet # 9 - 4
1. Calculate q when 12.0 grams of water is heated from 20
degrees to 100 degrees? ( c water = 4.184 J/g C)
2. The specific heat capacity of silver is 0.24 J/g C.
a) Calculate the energy required to raise the temperature of
150.0 grams Ag from 273 K to 298 K.
b) Calculate the energy required to raise the temperature of
1.0 mol Ag by 1.0 C
c) It takes 1.25 kJ of energy to heat a sample of pure silver
from 12.0 C to 15.2 C. Calculate the mass of the sample
of silver.
3. A 30.0 g of water at 280 K is mixed with a 50.0 g water at
330 K. Calculate the final temperature of the mixture
assuming no heat loss to the surroundings.
Worksheet # 9 – 4: Answers
1. Calculate q when 12.0 grams of water is heated from 20
degrees to 100 degrees? Ans = 4016.6 J
2. The specific heat capacity of silver is 0.24 J/g C.
a) Calculate the energy required to raise the temperature of
150.0 grams Ag from 273 K to 298 K. Ans = 900 J
b) Calculate the energy required to raise the temperature of
1.0 mol Ag by 1.0 C. Ans = 25.9 J
c) It takes 1.25 kJ of energy to heat a sample of pure silver
from 12.0 C to 15.2 C. Calculate the mass of the sample
of silver. Ans = 1627.6 g
3. A 30.0 g of water at 280 K is mixed with a 50.0 g water at
330 K. Calculate the final temperature of the mixture
assuming no heat loss to the surroundings. Ans = 311 K