#### Transcript Heat

```Homework Problems
Chapter 5 Homework Problems: 1, 12, 16, 18, 20, 28, 30, 36, 38, 40,
48, 50, 60, 62, 66, 87, 96, 108, 122
CHAPTER 5
Thermochemistry
Thermochemistry, Work, and Heat
Thermochemistry is the study of energy flow in chemical
systems.
Work (w) is (force) . (displacement)
Heat, or thermal energy, (q) is the energy that moves from a hot
object to a cold object when placed in contact with each other.
T1 > T3 > T2
heat flows from hot
to cold block
Energy
Energy is the capacity to do work or transfer heat. In principal
any kind of energy can be converted into an equivalent amount of work
or heat.
We divide energy into two general types:
kinetic energy (due to motion in a particular direction)
EK = 1/2 mv2
potential energy (due to position or composition) Examples chemical potential energy, gravitational potential energy.
The total amount of all the different kinds of energy for a system
is called the internal energy (U).
Units of Energy
All forms of energy can be expressed in the same units. To find
the MKS unit for energy, it is convenient to use the equation for kinetic
energy.
EK = 1/2mv2 So units are (kg) (m/s)2 = kg.m2 = 1 Joule = 1 J
s2
Since 1 J is a small amount of energy, we often express energy in terms
of kJ (kilojoule). 1 kJ = 1000. J
Other common units for energy include:
calorie (cal) Amount of heat needed to raise the temperature of 1 g of
water by 1 C
1 cal = 4.184 J (exact)
1 kcal = 1 food calorie = 1000 cal = 4184. J (exact)
Heat and Chemical Reactions
Some chemical reactions release heat into their surroundings.
Other reactions require that the surroundings provide heat so that the
reaction can proceed.
An exothermic reaction is a reaction that releases heat to the
surroundings. Combustion reactions are one common type of exothermic reaction
2 H2(g) + O2(g)  2 H2O(l)
releases 572. kJ of heat
per mole of reaction
An endothermic reaction takes up heat from the surroundings.
2 HgO(s)  2 Hg(l) + O2(g)
takes up 181. kJ of heat
per mole of reaction
By convention, q < 0 for an exothermic reaction, and q > 0 for an
endothermic reaction.
System, Surroundings, and Universe
The system is a part of the universe that we have separated off for
study.
The surroundings are everything not included in the system.
The universe is everything - system + surroundings.
While in principle the surroundings are everything not included
in the system, in practice we can usually focus on that part of the
surroundings in the immediate vicinity of the system.
Open, Closed, and Isolated Systems
Systems can be divided into various categories based on how
they interact with their surroundings.
Open system - A system that can exchange both energy and mass
with its surroundings.
Closed system - A system that can exchange energy with its
surroundings, but which cannot exchange mass with its surroundings.
Isolated system - A system that cannot exchange either energy or
mass with its surroundings.
State Function
A state function is any function whose change in value depends
only on the initial and final state of the system, but which is independent
of the pathway used to go between them. The value for something that is
not a state function depends not only on the initial and final state but also
on the pathway used to travel between them.
Altitude is a state function, distance traveled is not a state function. In
thermodynamics, U is a state function, q and w are not state functions (in
fact, they are not functions at all!)
Mechanical Work
Mechanical work is the work associated with the change in the
volume of a system.
Consider the expansion of a gas
Derivation of the Expression For Mechanical Work
w = F . d (from the definition of work)
But p = F/A, and so F = p . A
So |w| = p . A . d However, from the diagram d = h
So |w| = p . A . h
But A . d = V where V is the change in volume.
So |w| = p . V (absolute value)
But notice that when a gas expands
its volume change is in a direction opposite
to that of the applied pressure. This introduces a negative sign into the expression for
mechanical work. Therefore
w = - p . V
p
d
First Law of Thermodynamics
The first law of thermodynamics gives a relationship between
internal energy, work, and heat.
U = q + w
U = Uf - Ui = change in internal energy
q = heat
w = work
For now, we limit ourselves to mechanical work, where w = - p V.
Note that the first law is based on experimental observation and is
not derived from some other law or principle. The first law relates the
change in a state function (U) to the sum of two things that are not state
functions (q and w).
Sign Convention for Heat and Work
We use the following sign convention for heat and work.
q > 0 heat flows from the surroundings into the system (endothermic)
q < 0 heat flows from the system to the surroundings (exothermic)
w > 0 work is done on the system (system is compressed)
w < 0 work is done by the system (system expands)
Note that qsur = - qsys, and wsur = - wsys.
Use of the First Law
A gas is confined inside of a cylinder. The applied pressure is
1.000 atm, and the initial volume occupied by the gas is 2.000 L. 8000. J
of heat is added to the gas under conditions of constant applied pressure.
The final volume occupied by the gas is 6.000 L. What are q, w, and U
for the process?
A gas is confined inside of a cylinder. The applied pressure is
1.000 atm, and the initial volume occupied by the gas is 2.000 L. 8000. J
of heat is added to the gas under conditions of constant applied pressure.
The final volume occupied by the gas is 6.000 L. What are q, w, and U
for the process?
From the first law, U = q + w.
Heat is added to the system, so q = + 8000. J
For constant pressure w = - p V
= - (1.00 atm) (6.000 L – 2.000 L)
= - 4.00 L.atm 101.3 J = - 405. J *
1 L.atm
Finally, U = q + w = 8000. J + (- 405. J) = 7595. J
*Note: To find the conversion between L.atm and J we do the following
# J = 1 L.atm 1 m3
1.013 x 105 N/m2 = 101.3 N.m = 101.3 J
1000 L
1 atm
Constant Volume Processes
Consider some general process taking place at constant volume.
From the first law
U = q + w
But for mechanical work, w = - pV
So at constant volume, V = 0, so w = 0
Therefore, for a process carried out at constant volume
U = qV (V = constant)
What does this mean? For a process carried out at constant
volume, q is a state function, and so no information is needed concerning
path. This makes it far easier to calculate and keep track of heat flow for
these kinds of processes.
Enthalpy
Most processes in the laboratory are carried out at constant
pressure instead of constant volume. It would be nice to have a state
function whose change in value was equal to q for constant pressure
processes. Enthalpy is such a function.
We define enthalpy, H, as follows
H = U + pV
Since U, p, and V are state functions, it follows that enthalpy is
also a state function.
Constant Pressure Processes and Enthalpy
Consider some general process taking place at constant pressure.
From the definition of enthalpy
H = U + pV
H = U + (pV) = U + (pfVf - piVi)
Now, if pressure is held constant, then pf = pi = p, and so
H = U + p(Vf - Vi) = U + pV
Now, from the first law
U = q + w
If we only have mechanical work
U = q - p V
If we now substitute into the expression for enthalpy, we get
H = U + pV = (q - p V) + p V
or (finally!)
H = qp (p = constant)
What does this mean? For a process carried out at constant
pressure, q is a state function, and so no information is needed
concerning path. This makes it far easier to calculate and keep track of
heat flow for these kinds of processes.
To summarize
U = qV (constant volume processes)
H = qp (constant pressure processes)
Since q is something that can be measured experimentally, we now have
a way to relate this information to changes in state functions (internal
energy or enthalpy).
Enthalpy of Reaction (Hrxn)
Because of the importance of enthalpy, we define enthalpy
changes for specific kinds of processes.
Hrxn (enthalpy of reaction) - The enthalpy change when one
mole of a specific reaction is carried out at p = 1.00 atm (); equals qp for
the process.
Example:
CaCO3(s)  CaO(s) + CO2(g)
Hrxn = + 178.3 kJ/mol
2 CaCO3(s)  2 CaO(s) + 2 CO2(g)
Hrxn = + 356.6 kJ/mol
CaO(s) + CO2(g)  CaCO3(s)
Hrxn = - 178.3 kJ/mol
Notice that multiplying a reaction by a constant multiplies the
value for Hrxn by the same constant. Changing the direction of a
reaction changes the sign for Hrxn .
We can see the last result as follows:
CaCO3(s)  CaO(s) + CO2(g)
Hrxn = + 178.3 kJ/mol
CaO(s) + CO2(g)  CaCO3(s)
Hrxn =
?
__________________________________________________________
CaCO3(s)  CaCO3(s)
Htotal = 0.0 kJ/mol
Carrying out the first process followed by the second process
means that our final state is the same as our initial state. There is no
change in the system, and so Htotal = 0.0 kJ/mol. Therefore, the change
in enthalpy for the second process must be the same as the change in
enthalpy for the first process, except for the sign.
The Physical State of the Reactants
and Products
The value for the enthalpy change for a process depends not only
on the identity and amounts of the reactants and products but also their
state.
Consider the following example:
2 H2(g) + O2(g)  2 H2O(l)
Hrxn = - 571.6 kJ/mol
2 H2(g) + O2(g)  2 H2O(g)
Hrxn = - 483.6 kJ/mol
The difference in is due to the fact that heat has to be added to liquid
water to convert it into a gas. In fact, at T = 25.0 °C
H2O(l)  H2O(g)
Hrxn = + 44.0 kJ/mol
We can check this as follows. If we add the first two reactions
below we get the third reaction as our final result.
2 H2(g) + O2(g)  2 H2O(l)
2 H2O(l)  2 H2O(g)
Hrxn = - 571.6 kJ/mol
Hrxn = + 88.0 kJ/mol
__________________________________________________________
2 H2(g) + O2(g)  2 H2O(g)
Hrxn = - 483.6 kJ/mol
Since enthalpy is a state function (path independent) the change in
enthalpy for the combination of the first two processes has to be the same
as the change in enthalpy for the third process. This is a simple example
of a general principle called Hess’ law.
Relationship Between Hrxn and Urxn
For a process carried out at constant pressure
H = U + pV
If
V > 0 then H > U
V = 0 then H = U
V < 0 then H < U
For chemical reactions it is usually true that
|U| >> |pV|
and so
Urxn  Hrxn
For precise work we can apply a correction to correctly convert
between Urxn and Hrxn. (using H = U + pV).
Heat Capacity
Consider adding a known amount of heat to a substance. One
would expect that the temperature of the substance would increase.
Heat capacity (C) is a measure of how much heat is required to
cause the temperature of a substance to change by a given amount.
Heat capacity is defined as
C = q/T
where q = amount of heat added
T = Tf - Ti = change in temperature
Notes on Heat Capacity
1) Since C = q/T, the units for C are (energy)/(temperature).
The derived MKS units for C are J/K, though it is often given in J/C.
2) Note that the numerical value for C when expressed in J/K or
J/C is identical. That is because we are using the change in temperature.
Since the size of a degree Kelvin and a degree Celsius is the same, the
numerical value for T is the same whether expressed in K or C.
Example: Ti = 15.0 C (288.2 K) and Tf = 21.5 C (294.7 K)
T = Tf – Ti = 21.5 C – 15.0 C = 6.5 C
= 294.7 K - 288.2 K = 6.5 K
3) C is not in general a state function, since q is not a state
function. However, if we restrict ourselves to processes occurring under
conditions of constant pressure then C is a state function.
Specific Heat and Molar Heat Capacity
There are two quantities that are closely related to heat capacity.
Specific heat, (s) - The heat capacity per gram of substance.
Molar heat capacity, (Cm) - The heat capacity per mole of
substance.
The relationships between C, s, and (heat capacity, specific heat,
and molar heat capacity) are as follows:
s = C/m
where m= mass of substance
Cm = C/n
where n = moles of substance
So s has units of J/gC and Cm has units of J/molC.
Note that C is an extensive property, but s and Cm are intensive
properties.
Heat Capacity Problem
The initial temperature of a metal block of mass 38.44 g is Ti =
18.3 °C. After the addition of 147. J of heat the temperature of the metal
increases to a final value Tf = 25.7 °C. What are C and s for the metal
block?
The initial temperature of a metal block of mass 38.44 g is Ti =
18.3 °C. After the addition of 132. J of heat the temperature of the metal
increases to a final value Tf = 25.7 °C. What are C and s for the metal
block?
From the definition of heat capacity
C=
heat absorbed
= q =
132. J
= 18. J/°C
change in temperature T
(25.7 °C - 18.3 °C)
s= C=
m
18. J/°C = 0.46 J/gC
38.44 g
Note that C is an extensive property but s is an intensive property
of the metal.
Specific Heat For Common Substances
Substance
s (J/g°C)
Al(s)
0.900
Au(s)
0.129
C(graphite)
0.720
C(diamond)
0.502
Cu(s)
0.385
Fe(s)
0.444
Hg(l)
0.139
H2O(l)
4.184
Pb(s)
0.128
Sn(s)
0.227
Zn(s)
0.389
The above table is at T = 25. C. Note that water has an unusually large value for heat capacity, which acts to moderate temperatures for
cities surrounded by large bodies of water.
Calorimetry
Calorimetry is the experimental method used to measure the heat
produced or taken up in a chemical reaction or physical process. The
device used in these measurements is called a calorimeter.
If the heat capacity and temperature change for the calorimeter
are known, we can find the value for q for the process
If the process is carried out at constant pressure, then
qsys = Hrxn
If the process is carried out at constant volume, then
qsys = Urxn
Constant Pressure Calorimetry
In a coffee cup calorimeter a chemical reaction or other process is
carried out under conditions of constant pressure.
Example
A metal block of mass 14.486 g at an initial temperature of 74.25
°C is placed inside a coffee cup calorimeter containing 100.00 g of water
at a temperature of 17.42 °C. After equilibrium is reached the final
temperature of the water is 20.84 °C. What is s, the specific heat, of the
metal? (Note: s = 4.184 J/g°C for water.)
A metal block of mass 14.486 g at an initial temperature of 74.25
°C is placed inside a coffee cup calorimeter containing 100.00 g of water
at a temperature of 17.42 °C. After equilibrium is reached the final
temperature of the water is 20.84 °C. What is s, the specific heat, of the
metal? (Note: s = 4.184 J/g°C for water.)
qwater = - qmetal
swater mwater Twater = - smetal mmetal Tmetal
smetal = - swater (mwater/mmetal)( Twater/Tmetal )
So
smetal = - (4.184 J/g°C) 100.00 g (20.84 °C - 17.42 °C)
14.486 g (20.84 °C - 74.25 °C)
= 0.462 J/g°C
Note we did not take into account the heat capacity of the calorimeter
itself, but assumed it was small. For more precise work the heat capacity
of the calorimeter would need to be included in the calculations.
Bomb Calorimetry
In bomb calorimetry a measured mass of a chemical substance
reacts with excess oxygen to form combustion products.
Bomb Calorimetry Calculations
If the heat capacity of the calorimeter apparatus is known (and
this can be determined experimentally) then
q = - C T
C = heat capacity of calorimeter
T = Tf - Ti = change in temperature
measuring the value of q for the surroundings, and qsys = - qsur.
Since the combustion process occurs under conditions of constant
volume, q = U, the change in internal energy.
Example: 1.412 g of carbon (M = 12.01 g/mol) are burned in a
bomb calorimeter (C = 10325. J/C). The temperature of the calorimeter
increases by 4.48 C. What are U (change in internal energy) and Um
(change in internal energy per mole of carbon) for the process.
Example: 1.412 g of carbon (M = 12.01 g/mol) are burned in a
bomb calorimeter (C = 10325. J/C). The temperature of the calorimeter
changes by 4.48 C. What are U (change in internal energy) and Um
(change in internal energy per mole of carbon) for the process.
q = - C T = - (10325. J/C) (4.48 C) = - 46300. J
For a process carried out at constant volume q = Um, so
U = - 46300. J
Finally, the change in internal energy per mole of carbon is
Um = U/n
n = 1.412 g 1 mol = 0.1176 mol
12.01 g
Um = - 46300. J = - 393000 J/mol = - 393. kJ/mol
0.1176 mol
Note that Hm  Um = - 393. kJ/mol, as previously noted.
Standard Conditions and Standard State
It is convenient to report thermodynamic data for a particular set
of conditions. These conditions are called the thermodynamic standard
conditions, and are usually (but not always) taken to be p = 1.00 atm*, T
= 25.0 C = 298.2 K.
The standard state for an element is the most stable form of the
element for standard conditions.
carbon C(s)
nitrogen N2(g)
bromine Br2()
oxygen O2(g)
iron Fe(s)
argon Ar(g)
mercury Hg ()
chlorine Cl2(g)
sulfur S(s)
__________________________________________________________
* Technically, standard pressure is now taken to be 1.00 bar = 0.987 atm. This makes only a
minor difference in values for thermodynamic quantities.
Enthalpy of Formation (Hf)
The formation reaction for a substance is defined as the reaction
that produces one mole of a single product out of elements in their
standard state. Because of the way we have defined the formation
reaction, we may have to use fractional stoichiometric coefficients for
some or all of the reactants. The enthalpy change for this reaction is
defined as the enthalpy of formation for the substance.
H2(g) + 1/2 O2(g)  H2O(l)
Hf(H2O(l)) = - 285.8 kJ/mol
H2(g) + 1/2 O2(g)  H2O(g)
Hf(H2O(g)) = - 241.8 kJ/mol
C(s) + O2(g)  CO2(g)
Hf(CO2(g)) = - 393.5 kJ/mol
Pb(s) + C(s) + 3/2 O2(g)  PbCO3(s) Hf(PbCO3(s)) = - 699.1 kJ/mol
3/
2
O2(g)  O3(g)
Hf(O3(g)) = + 143.0 kJ/mol
Data for formation enthalpies are given in Appendix 2 of Burdge (with
separate tables for inorganic and organic substances).
Note that based on the above definition it follows that the
enthalpy of formation of an element in its standard state is 0.0 kJ/mol.
We may see this as follows:
The formation reaction for N2(g), by definition (formation of one
mole of a single product out of elements in their standard state), is
N2(g)  N2(g)
But nothing happens in the above process, and so it follows that
Hrxn = Hf(N2(g) ) = 0.0 kJ/mol
This will be true for any element in its standard (thermodynamically
most stable) state.
Example: Write the formation reaction for carbon tetrachloride
(CCl4(l)), acetone (CH3COCH3(l)), and nitrous oxide (N2O(g)).
Example: Write the formation reaction for carbon tetrachloride
(CCl4(l)), acetone (CH3COCH3(l)), and nitrous oxide (N2O(g)).
C(s) + 2 Cl2(g)  CCl4(l)
3 C(s) + 3 H2(g) + ½ O2(g)  CH3COCH3(l)
N2(g) + ½ O2(g)  N2O(g)
Note that the enthalpy change when 1 mole of any of the above reactions
is carried out corresponds to the enthalpy of formation for the substance.
Enthalpy of Combustion (Hc)
The combustion reaction for a substance is defined as the reaction
of one mole of a single substance with O2(g) to form combustion
products. Because of the way in which we have defined the combustion
reaction we may have to use fractional coefficients for some of the
reactants and products. The enthalpy change for this reaction is defined
as the enthalpy of combustion for the substance.
Combustion products for common elements are
C  CO2(g)
H  H2O(l) N  N2(g)
S  SO2(g)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Hc(CH4(g)) = - 890.3 kJ/mol
C6H6() + 15/2 O2(g)  6 CO2(g) + 3 H2O(l)
Hc(C6H6()) = - 3267.4 kJ/mol
Writing Formation and Combustion Reactions
We may use the definition of formation reaction and combustion
reaction to write own the balanced chemical equations corresponding to
these reactions.
Example: Hexane (C6H14(l)) is a hydrocarbon often used as a
solvent in organic reactions. Write the formation reaction and the
combustion reaction for hexane.
CH3CH2CH2CH2CH2CH3
Example: Hexane (C6H14(l)) is a hydrocarbon often used as a
solvent in organic reactions. Write the formation reaction and the
combustion reaction for hexane.
Formation - One mole of a single product out of elements in their
standard state.
?
 C6H14(l)
6 C(s) + 7 H2(g)  C6H14(l)
(balanced)
Combustion - One mole of a single reactant, plus oxygen, to form
combustion products.
C6H14(l) + ? O2(g) 
?
C6H14(l) + ? O2(g)  6 CO2(g) + 7 H2O(l)
C6H14(l) + 19/2 O2(g)  6 CO2(g) + 7 H2O(l)
(balanced)
Enthalpy Change For Phase Transitions
There are three phase transitions that can occur by adding heat to
a substance.
s  l fusion (melting)
Hfus
l  g vaporization
Hvap
s  g sublimation
Hsub
The enthalpy change when one mole of a substance undergoes the
transition at the normal transition temperature is defined as the enthalpy
change for the phase transition. It represents the amount of heat required
to convert one mole of the substance from the initial phase to the final
phase.
H2O(l)  H2O(g)
Hvap(H2O) = 40.7 kJ/mol at T = 100.0 C
Unlike other processes, the temperature for a phase transition is
usually taken to be the temperature for which the two phases exist at
equilibrium when p = 1.0 atm.
Hess’ Law
We previously noted that the change in the value for a state
function depends only on initial and final state and is independent of the
path used to travel between the two states. We may put this in a more
formal manner in terms of Hess’ law.
Hess’ law – The change in value for any state function will be the
same for any process or combination of processes that have the same
initial and final state.
We are particularly interested in applying Hess’ law to chemical
reactions.
Hess’ Law For a Chemical Reaction
Let us use Hess’ law to find the value for (Hrxn) for the
following chemical reaction:
CaCO3(s)  CaO(s) + CO2(g)
Hrxn
We may obtain this same reaction as follows:
Ca(s) + ½ O2(g)  CaO(s)
Hf(CaO(s))
C(s) + O2(g)  CO2(g)
Hf(CO2(g))
CaCO3(s)  Ca(s) + C(s) + 3/2 O2(g)
- Hf(CaCO3(s))
CaCO3(s)  CaO(s) + CO2(g)
Hrxn
So by Hess’ law,
Hrxn = Hf(CaO(s)) + Hf(CO2(g)) - Hf(CaCO3(s))
We can get Hrxn using only data on enthalpies of formation!
General Method For Finding Hrxn
Based on the same procedure used in the previous example the
following general relationship can be derived
Hrxn = [  Hf(products) ] - [  Hf(reactants) ]
Notice what this means. If we have a table for formation
enthalpies we can find the value for Hrxn for any chemical reaction. In
fact, the same general procedure can be used to find the values for the
change in any state function.
Also note that when we use the superscript  this indicates not
only standard conditions but also standard concentrations for reactants
and products.
gases p = 1.0 atm
solutes [M] = 1.0 mol/L
solids, liquids, solvents must be present in the system
Example: Find Hrxn for the conversion of acetylene into dichloroethane
C2H2(g) + 2 HCl(g)  CH2ClCH2Cl(l)
Example: Find Hrxn for the conversion of acetylene into dichloroethane
C2H2(g) + 2 HCl(g)  CH2ClCH2Cl(l)
Hrxn = [Hf(CH2ClCH2Cl(l)) ] - [Hf(C2H2(g)) + 2 Hf(HCl(g))]
= [ (- 165.2 kJ/mol) ] - [ (226.7 kJ/mol) + 2 ( - 92.3 kJ/mol) ]
= - 207.3 kJ/mol
Thermodynamic data are given in the Appendix 2 of Burdge, with
separate tables for inorganic and organic substances.
Other Uses of Hess’ Law
Hess’ law can be used to find Hrxn for any process that can be
written as a combination of other chemical reactions whose values for
Hrxn are known.
Example: Consider the following formation reaction
5 C(s) + 6 H2(g)  C5H12(l)
Hrxn = ?
Using the following information find Hrxn for this reaction
(1) C5H12(l) + 8 O2(g)  5 CO2(g) + 6 H2O(g) Hrxn = - 3505.8 kJ/mol
(2) C(s) + O2(g)  CO2(g)
Hrxn = - 393.5 kJ/mol
(3) 2 H2(g) + O2(g)  2 H2O(g)
Hrxn = - 483.5 kJ/mol
Note all of the above are combustion reactions, which are particularly
easy to carry out experimentally.
Example: Consider the following formation reaction
5 C(s) + 6 H2(g)  C5H12(l)
Hrxn = ?
Using the following information find Hrxn for this reaction
(1) C5H12(l) + 8 O2(g)  5 CO2(g) + 6 H2O(g) Hrxn = - 3505.8 kJ/mol
(2) C(s) + O2(g)  CO2(g)
Hrxn = - 393.5 kJ/mol
(3) 2 H2(g) + O2(g)  2 H2O(g)
Hrxn = - 483.5 kJ/mol
reverse 1st reaction
5 CO2(g) + 6 H2O(g)  C5H12(l) + 8 O2(g) Hrxn = + 3505.8 kJ/mol
5 times the 2nd reaction
5 C(s) + 5 O2(g)  5 CO2(g)
Hrxn = - 1967.5 kJ/mol
3 times 3rd reaction
6 H2(g) + 3 O2(g)  6 H2O(g)
5 C(s) + 6 H2(g)  C5H12(l)
Hrxn = - 1450.5 kJ/mol
Hrxn = + 87.8 kJ/mol
End of Chapter 5
“In this house we obey the laws of thermodynamics!”
- Homer Simpson
“...the Dutch physicist Heike Kamerlingh Onnes gave H the name
enthalpy, from the Greek  (in) and  (heat), or from the single
Greek word  (enthalpos), to warm within.” K. J. Laidler, The
World of Physical Chemistry
“[Thermodynamics] is the only physical theory of universal
content which, within the framework of the applicability of its basic
concepts, I am convinced will never be overthrown.” Albert Einstein
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