Transcript Thermochemistry

Topic 6
Thermochemistry
Nearly all chemical reactions involve either the release
or the absorption of heat which is one form of energy.
Heat is involved in reactions by either going into or
out of the reaction system and is one aspect of a
reaction that can be monitored.
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Thermochemistry
Thermochemistry is the study of the energy effects that accompany
chemical reactions.
Why do chemical reactions occur? What is the driving force of rxn?
Answer: stability: reaction (rxn) wants to get to lower E. For a rxn
to take place spontaneously, the products of reaction must be more
stable (lower E) than the starting reactants. Nonspontaneous
reaction means the rxn will never happen by itself; the reaction can
happen, but it will need external help.
rxn must absorb E from external
source; therefore, rxn is nonspon
rxn itself releases E;
therefore, rxn is spon
P
R
E
P
reactants at higher E are less stable;
therefore, more reactive
E
R
reactants at lower E are more stable;
therefore, less reactive
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Energy
Energy is defined as the capacity to
move matter (“do work”). The concept of
energy gives us an alternative way to
describe motion.
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Energy
There are three broad concepts of
energy:
– Kinetic Energy, Ek, is the energy associated with
an object by virtue of its motion.
– Potential Energy, Ep is the energy an object has
by virtue of its position in a field of energy.
– Internal Energy, U, is the sum of the kinetic and
potential energies of the particles making up a
substance.
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Energy
The total energy of a system is the sum of its
kinetic energy (Ek), potential energy (Ep), and
internal energy (U).
Etot  Ek  Ep  U
Typically in a laboratory, the experimental vessel is at
rest making the kinetic and potential energies
contribution to the total energy zero.
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Energy
The Law of Conservation of Energy:
Energy may be converted from one form
to another, but the total quantities of
energy remain constant.
Energy may be transferred or
converted but not destroyed.
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Reaction Enthalpy
In chemical reactions, heat is often transferred from
the “system or reaction” to its “surroundings,” or
vice versa.
system is the substance or mixture of
substances under study in which a change
occurs.
surroundings are everything in the vicinity of
the thermodynamic system (room, flask,
solvent).
We tend to measure the
change in internal
energy of the system.
system or rxn
surroundings
+ into system
- out system
The sign (+ or -) for energy
indicates the direction of
flow meaning into or out of
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the system.
Heat of Reaction
Heat flow is defined as the energy that flows into
or out of a system. We follow heat flow by
watching the difference in temperature between
the system and its surroundings.
Often we follow the surroundings temp (solvent)
and must realize that the opposite is happening
to the system. If the system is absorbing heat (+)
from the surroundings than the temp of the
surroundings must be decreasing.
Tsystem (+)
Tsurr (-)
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Heat of Reaction
Heat flow or heat of reaction is denoted by the
symbol q and is the amount of heat required to return a
system to the given temperature at the completion of
the reaction.
For an endothermic rxn, the sign of q is positive;
heat is absorbed by the system from the
surroundings.
Tsurr
Surroundings
system absorbs heat from
surroundings; nonspon (endo)
P
E
R
+q
System
Tsystem
Dq > 0
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Heat of Reaction
For an exothermic rxn, the sign of q is negative;
heat is evolved (released) by the system to the
surroundings.
Tsurr
Surroundings
system releases heat to
surroundings; spon (exo)
R
-q
Tsystem
E
P
System
Dq < 0
HW 50
code: exo
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Enthalpy and Enthalpy Change
The heat absorbed or evolved by a reaction
depends on the conditions under which it
occurs. ex. constant pressure
Usually, a reaction takes place in an open
vessel, and therefore under the constant
pressure of the atmosphere.
heat of this type of reaction is denoted qp; this
heat at constant pressure is named enthalpy
and given the symbol H. H is the heat flow at
constant pressure.
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Enthalpy and Enthalpy Change
Enthalpy, denoted H, is an extensive property of a substance
that can be used to obtain the heat absorbed or evolved in a
chemical reaction at constant pressure.
Extensive property - depends on the quantity of substance (i.e.
mass, volume, heat); contrary to intensive property that does
not depend on amount of substance (i.e. color, boiling point).
Enthalpy is a state function. State functions are properties
associated with a system in equilibrium; for any given state, a
state function is uniquely defined (i.e. T, P, V, E). If a system
undergoes a change (from one equilibrium state to another),
and X is a state function, then DX (the change in the value of X)
is the same regardless of how the changed occurred. It only
depends on the final state (Xf) and the initial state (Xi) .
o
o
o
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Enthalpy and Enthalpy Change
The reaction enthalpy, DHrxn, for a
reaction carried out at constant pressure
is the enthalpy change associated with
the conversion of reactants to products
(at the same temperature).
DH  H (final)  H (initial )
DH  H (products)  H ( reactants)
-DH is associated with an exothermic rxn (releases heat)
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+DH is associated with an endothermic rxn (aborbs heat)
Enthalpy and Enthalpy Change
As we already stated the reaction enthalpy is equal to
the heat of reaction at constant pressure, qp. This
represents the entire change in internal energy (DU)
minus any expansion “work” done by the system;
therefore, we can define enthalpy and internal work by
the 1st law of thermodynamics:
In any process, the total change in energy of the
system, DU, is equal to the sum of the heat
absorbed, q, and the work, w, done by the
system.
DU = qp + w = DH + w
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Changes in E manifest themselves as exchanges
of energy between the system and surroundings.
These exchanges of energy are of two kinds:
heat and work that must be accounted.
Heat is energy that moves into or out of a system
because of a temperature difference between
system and surroundings.
Work is the energy exchange that results when a
force F moves an object through a distance d;
work (w) = Fd
In chemical systems, work is defined as a change
in volume at a given pressure, that is
w   PDV
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w   PDV
The negative sign is present in the equation to keep the sign
correct in terms of system. For expansion work, DV will be a
positive value; however, expansion involves the system doing
work on the surroundings and a decrease in internal energy
(negative sign keeps overall work term a negative value). For
contraction work, DV will be a negative value; however,
contraction involves the surroundings doing work on the system
and an increase in internal energy (negative sign keeps overall
work term a positive value).
Giving us the 1st law of thermo in its more useful form:
note:
system absorbs heat (+)
DU  DH  PDV
system releases heat (-)
surroundings doing work on system (+)
system doing work on surroundings (-)
HW 51
code: energy
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Thermochemical Equations
A thermochemical equation is the chemical
equation for a reaction (including physical states
{important}) in which the equation is given a molar
interpretation, and the reaction enthalpy for these
molar amounts is written directly after the equation.
Note: physical states are important because the DH for the formation of H2O (l) from H2 (g)
and O2 (g) is not the same as for the formation of H2O (g).
N 2 (g )  3H 2 (g )  2NH 3 (g ); DH  -91.8 kJ
Enthalpy is an extensive property. This means that the value of reaction
enthalpy is directly proportional to the amount of reactants and products
involved in the reaction. This equation basically gives a recipe of what and how
much to mix (1 mole N2 and 3 mols H2) to produce 2 mols NH3 and release 91.8
kJ of heat. Note that a negative value of DH indicates that the reaction is
exothermic; it releases heat. Since H is an extensive property, if the recipe is
varied, the amount of heat generated will also vary (i.e. only used ½ mole of N2
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gas, only ½ of -91.8 kJ of heat would be released).
Thermochemical Equations
Since enthalpy is an extensive property, we can
set up conversion factors based on the
coefficients and DH values in a thermochemical
equation and perform calculations similar to
stoichiometric calculations.
N 2 (g )  3H 2 (g )  2NH 3 (g ); DH  -91.8 kJ
−𝟗𝟏. 𝟖 𝒌𝑱
𝟏 𝒎𝒐𝒍 𝑵𝟐
−𝟗𝟏. 𝟖 𝒌𝑱
𝟑 𝒎𝒐𝒍 𝑯𝟐
−𝟗𝟏. 𝟖 𝒌𝑱
𝟐 𝒎𝒐𝒍 𝑵𝑯𝟑
Note: the minus sign is not a “negative sign”; it indicates heat is
being released by the system.
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Thermochemical Equations
If the DH for a reaction is x, then the DH for the reverse
reaction is –x. If a reaction is exothermic, its reverse
reaction is endothermic and vice versa; magnitude of
DH is the same, but the algebraic sign is reversed.
2 H 2 ( g )  O2 ( g )  2 H 2O( g ) ; DH o  - 483.7 kJ exo
equation reversed
2 H 2O( g )  2 H 2 ( g )  O2 ( g ) ; DH o   483.7 kJ endo
When a thermochemical equation is multiplied by
any factor, the value of DH for the new equation is
obtained by multiplying the DH in the original
equation by that same factor.
equation multiplied by
factor of 2
2 H 2 ( g )  O2 ( g )  2 H 2O( g ) ; DH o  - 483.7 kJ
4 H 2 ( g )  2O2 ( g )  4 H 2O( g ) ; DH o   967.4 kJ
Note: if DH has a degree symbol (o) like it does in the examples above
(DHo), it indicates thermo standard conditions at 25oC (298K) and 1 atm.
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Applying Stoichiometry and Heats of Reactions
Consider the reaction of methane, CH4, burning in the presence of oxygen
at constant pressure. Given the following equation, how much heat could
be obtained by the combustion of 10.0 g CH4? 10.0 g O2? 10.0 g of each?
CH 4 (g )  2O 2 (g )  CO 2 (g )  2H 2O(l );DH o  -890.3 kJ
We will determine the amount of heat by performing a stoichiometry calculation similar
to what we did in earlier sections except we will use a conversion factor involving heat.
We will convert mass to mols and then convert mols to kJ.
10.0 𝑔 𝐶𝐻4
1 𝑚𝑜𝑙 𝐶𝐻4
16.04 𝑔 𝐶𝐻4
−890.3 𝑘𝐽
1 𝑚𝑜𝑙 𝐶𝐻4
= -556 kJ
The second part of this problem will be conducted the same way but using O2 factors.
10.0 𝑔 𝑂2
1 𝑚𝑜𝑙 𝑂2
32.00 𝑔 𝑂2
−890.3 𝑘𝐽
2 𝑚𝑜𝑙 𝑂2
= -139 kJ
HW 52
code: stoich
The third part involves a limiting reagent problem. We must determine which
reactant will be exhausted first because it will dictate the overall heat involved in
the reaction. Since O2 is generating less kJ of heat, it is the limiting reagent
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meaning -139 kJ is also the answer to part 3.
Measuring Heats of Reaction
To determine how heats of reactions are
measured, we must examine the heat
required to raise the temperature of a
substance because a thermochemical
measurement is based on the
relationship between heat and
temperature change.
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Measuring Heats of Reaction
Heat Capacity and Specific Heat
– The heat capacity, C, of a sample of substance
is the quantity of heat required to raise the
temperature of the sample of substance one
degree Celsius (kJ/oC, cal/oC).
– Changing the temperature of the sample requires
heat equal to the following:
q  CDT
kJ =
𝒌𝑱
o
𝒐𝑪 x C
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A Problem to Consider
Suppose a piece of iron has a heat capacity of 6.70 J/oC.
The quantity of heat required to raise the temperature of
the piece of iron from 25.0oC to 35.0oC is:
Basically, we are trying to determine the heat associated with the
temperature increase of our iron system. We know the heat capacity
of iron in J/oC and the temperature change; therefore, we can
calculate the answer by following units or plugging into the previous
equation:
q  CDT
=
𝟔. 𝟕𝟎 𝑱
𝒐
𝑪
𝟑𝟓. 𝟎 − 𝟐𝟓. 𝟎 𝒐𝑪 = 𝟔𝟕. 𝟎 𝐉
DT = Tf - Ti
Note: The positive sign on heat indicates that the iron
(system in this problem) is absorbing the heat.
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Measuring Heats of Reaction
Heat capacities are also compared for one gram
amounts of substances. The specific heat
capacity, s, (or “specific heat”) is the heat
required to raise the temperature of one gram
𝑱
of a substance by one degree Celsius ( 𝑜 ).
𝒈 𝐶
To find the heat required you must multiply the
specific heat, s, of the substance times its mass in
grams, m, and the temperature change, DT.
q  s  m  DT
𝑱=
𝑱
𝒈 𝑜𝐶
𝒈 𝑻𝒇 − 𝑻𝒊 𝒐𝑪
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A Problem to Consider
Calculate the heat when the temperature of 15.0
grams of water is raised from 20.0oC to 50.0oC.
(The specific heat of water is 4.184 J/g.oC.)
This problem involves calculating the heat associated with a mass of
water being increased in temperature. You can follow the units on the
specific heat of water and cancel them to solve for J or use the equation
on the previous slide:
q  s  m  DT
𝒒=
𝟒. 𝟏𝟖𝟒 𝑱
𝒈 𝒐𝑪
Tf
- Ti
𝟏𝟓. 𝟎 𝒈 𝟓𝟎. 𝟎 − 𝟐𝟎. 𝟎 𝒐𝑪 = 𝟏. 𝟖𝟖 𝒙 𝟏𝟎𝟑 𝑱
Note: Positive heat indicates an endothermic
reaction; also note in this problem water is the
system.
HW 53
code: heat
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Heats of Reaction: Calorimetry
A calorimeter is a device used to measure heats of
reactions. The apparatus is designed with thermally
insulated walls.
If a temperature change of DT is observed, then the
heat flow that would be needed to reverse the
temperature change is (Ccal)(-DT), where Ccal is the
heat capacity and all of its contents (after the
reaction has occurred). Therefore:
𝒒𝒓𝒙𝒏 = −𝑪𝒄𝒂𝒍 ∆𝑻
Or another way to look at it is
𝒒𝒓𝒙𝒏 = −𝒒𝒄𝒂𝒍
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Heats of Reaction: Calorimetry
An important concept to remember is that we typically are
looking for the heat of a reaction (system) but often
measure data involving the surroundings/calorimeter. This
means we will need to change the sign of the heat value
calculated to determine the heat of the reaction (system).
𝒒𝒓𝒙𝒏 = −𝑪𝒄𝒂𝒍 ∆𝑻
𝒒𝒓𝒙𝒏 = −(𝒔 𝒎 ∆𝑻)𝒔𝒖𝒓𝒓
𝒒𝒓𝒙𝒏 = (𝒔 𝒎 ∆𝑻)𝒔𝒚𝒔𝒕𝒆𝒎
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A Problem to Consider
When 23.6 grams of calcium chloride, CaCl2, was dissolved in
water in a calorimeter, the temperature rose from 25.0oC to
38.7oC. If the heat capacity of the solution and the calorimeter is
1258 J/oC, what is the enthalpy change per mole of calcium
chloride?
First, this problem involves plugging the proper values into the equation
to determine the ratio of heat generated in this reaction:
𝑞𝑟𝑥𝑛 = −𝐶𝑐𝑎𝑙 ∆𝑇
𝑞𝑟𝑥𝑛
1258 𝐽
=− 𝑜
× 38.7 − 25. 0𝑜 𝐶 = −1.72 𝑥 104 𝐽
𝐶
𝟐𝟑. 𝟔 𝒈 𝑪𝒂𝑪𝒍𝟐
This is the amount of heat released when 23.6 g of calcium chloride are
dissolved, not per one mole of calcium chloride as requested in problem.
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A Problem to Consider
When 23.6 grams of calcium chloride, CaCl2, was dissolved in
water in a calorimeter, the temperature rose from 25.0oC to
38.7oC. If the heat capacity of the solution and the calorimeter is
1258 J/oC, what is the enthalpy change per mole of calcium
chloride?
This problem is looking for a ratio that will give us a conversion factor that
can be used to calculate the amount of heat associated with different
quantities of calcium chloride in future experiments. Basically, we will
convert g to moles giving us a conversion factor between heat and mols.
−1.72 𝑥 104 𝐽
23.6 𝑔 𝐶𝑎𝐶𝑙2
111.1 𝑔 𝐶𝑎𝐶𝑙2
−8.10 𝑥 104 𝐽
=
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
This factor can be use to calculate the heat released for any amount of
calcium chloride.
HW 54
code: calor
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Hess’ Law
Reaction enthalpies can be determined
using calorimetry only for reactions that go
to completion, are fast, and have no side
reactions. For reactions that do not meet
these criteria, the reaction enthalpy can be
determined indirectly using Hess’ Law.
Hess’ law states when a reaction can be
expressed as the algebraic sum of two or
more reactions, then the DH for the reaction
is the algebraic sum of the DH of the
reactions added.
Basically, R & P in individual steps can be added
like algebraic quantities in determining overall
equation and enthalpy change for that reaction.
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Given:
1) A + D  E + C
DH = X kJ
2) 2A + B  2C
DH = Y kJ
Determine the enthalpy change for the equation below
by using given equations.
2D  B + 2E
DH = ?
Typically, we look for a species in the desired equation that occurs in only one of the
given equations and manipulate the equation to place the species on the correct side
and correct number of moles.
If we examine the desired equation, we will notice that D occurs in only
equation 1. We ask ourselves two questions to determine how to manipulate
equation 1: First, is the species on the correct side of the equation? In this
case, D is on the correct side in equation 1 as compared to the desired
equation; therefore, we will use the equation just as written. Second, does the
given equation have the correct number of moles of the species as in the
desired equation? In this case, D needs 2 moles and it has one mole in
equation 1; therefore, we will multiply the whole reaction including DH by 2:
2A + 2D 
2E + 2C
DH = 2X kJ
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Given:
1) A + D  E + C
DH = X kJ
2) 2A + B  2C
DH = Y kJ
Determine the enthalpy change for the equation below
by using given equations.
2D  B + 2E
DH = ?
Now we look for a species in equation 2 that we can examine and do the same
process; we see that B only occurs in equation 2. We ask ourselves two questions to
determine how to manipulate equation 2: First, is the species on the correct side of
the equation? In this case, B is not on correct side in equation 2 as compared to the
desired equation; therefore, we must reverse the equation and change the sign of DH.
Second, does the given equation have the correct number of moles of the species as in
the desired equation? In this case, B needs 1 mole and it has only one mole in
equation 2; therefore, we do not need to multiply the equation by any factor:
2C

2A + B
DH = -Y kJ
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Given:
1) A + D  E + C
DH = X kJ
2) 2A + B  2C
DH = Y kJ
Determine the enthalpy change for the equation below
by using given equations.
2D  B + 2E
DH = ?
We then add the two equations to determine if we obtained the desired equation.
To obtain the DH for the desired equation, we add the manipulated DH’s. If we
do this, we obtain
2A + 2D  2E + 2C
DH = 2X kJ
2C  2A + B
DH = -Y kJ
_______________________________________
2D 
B + 2E
DH = 2X – Y kJ
Notice you can easily check the manipulation of the given equations. If
we did things correctly, the overall equation will be exactly the same as
the desired equation which is the case in this example.
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Hess’ Law
Let’s look at real chemistry example, suppose you are given
the following GIVEN data:
S(s )  O 2 (g )  SO 2 (g ); DH  -297 kJ x2
o
2SO 3 (g )  2SO 2 (g )  O 2 (g ); DH  198 kJ
o
use these data to obtain the enthalpy change for the following
reaction?
2S(s)  3O 2 (g )  2SO 3 (g ); DH  ?
o
First, we will look for a species that is present in reaction in question and
only in one of the given reaction. We find S (s) is such a species.
We next ask ourselves two questions:
1.) Is it on the same side in the question reaction as in the given equation?
The answer is yes (both on reactants side); therefore, we do not have to
reverse the reaction.
2.) Does it have the same number of moles in both reactions?
The answer is no; therefore, we will need to multiply the given equation
by a factor of 2 to make it the same as in the question reaction.
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Hess’ Law
Let’s look at real chemistry example, suppose you are given
the following GIVEN data:
S(s )  O 2 (g )  SO 2 (g ); DH  -297 kJ x2
o
2SO 3 (g )  2SO 2 (g )  O 2 (g ); DH  198reverse
kJ
o
use these data to obtain the enthalpy change for the following
reaction?
2S(s)  3O 2 (g )  2SO 3 (g ); DH  ?
o
Second, we will look for a different species that is present in the question reaction
and only in the given reaction remaining. We find SO3 (g) is such a species.
We next ask ourselves two questions:
1.) Is it on the same side in the question reaction as in the given equation?
The answer is no; therefore, we will have to reverse the reaction.
2.) Does it have the same number of moles in both reactions?
The answer is yes (both have 2 mols); therefore, we do not need use a multiplier.
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S(s )  O 2 (g )  SO 2 (g ); DH  -297 kJ
o
x2
2SO 3 (g )  2SO 2 (g )  O 2 (g ); DH  198 kJ
o
reverse
If we multiply the first equation by 2 and reverse
the second equation, they will sum together to
become the reaction in question.
2S(s )  2O 2 (g )  2SO 2 (g ); DH  (-297 kJ)  (2)
o
2SO 2 (g )  O 2 (g )  2SO 3 (g ); DH  (198 kJ)  (-1)
o
2S(s)  3O 2 (g )  2SO 3 (g ); DH o  -792 kJ
Notice the overall equation is exactly the same as the desired equation;
therefore, the sum of the DH equals the DH of the reaction.
HW 55
code: hess
note: HW may require the use of fractions as multipliers
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Standard Enthalpies of Formation
The standard enthalpy of formation of something,
denoted DHfo, is the enthalpy change when one mole
of that something at standard pressure (1 atm) and
298.15K (25oC) forms from its constituent elements
in their standard state. The something can be a
formula unit (molecular or ionic), an atom, or an ion.
An element is in its standard state if it is in its most
stable form at standard pressure and 298.15K.
i.e.
Ag (s) + ½ Cl2 (g)  AgCl (s)
DHfo AgCl
Note that the standard enthalpy of formation for a
pure element in its standard state and H+ are zero.
This means elements in their standard state have
DHfo = 0: metals - solids, diatomic gases, H+ ion.
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Standard Enthalpies of Formation
Another way to determine heat of
reaction is the the law of summation of
heats of formation which states that
the enthalpy of a reaction is equal to
the total formation energy of the
products minus that of the reactants.
DH   nDH (products )   nDH (reactants )
o
o
f
o
f
S is the mathematical symbol meaning “the sum
of”, and n is the coefficients of the substances in
the chemical equation.
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Here’s how you would set up a calculation using the law of
summations on the generic reaction below:
aA + bB  cC + dD
DH o   nDH of (products )   nDH of (reactants )
sum of nDHof of all products
sum of nDHof of all reactants
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A Problem to Consider
What is the standard reaction enthalpy , DHorxn,
for the following reaction?
4NH 3 (g )  5O 2 (g )  4NO(g )  6H 2O(g )
DH of :  45.9kJ / mol
0
90.3
 241.8
You will either be given or look up the DHof of the species in
question.
Note: that species in their standard state will be “0” and not
given or found in tables.
Next, we will use the law of summation and plug the values into
the calculation:
DH o   nDH of (products )   nDH of (reactants )
40
4NH 3 (g )  5O 2 (g )  4NO(g )  6H 2O(g )
DH of :  45.9kJ / mol
0
90.3
 241.8
Using the summation law:
o
o
o
DH   nDH f (products)   mDH f (reactants)
DH o  [4molNO (90.3kJ / molNO )  6molH 2O(241.8kJ / molH 2O)]
 [4molNH 3(45.9kJ / molNH 3)  5molO 2(0kJ / molO 2)] 
 [361.2kJ  (1450.8kJ )]  [(183.6kJ )  0kJ ] 
 1089.6kJ  183.6kJ  906kJ
DH  906 kJ
o
Be careful of arithmetic signs as they are a
likely source of mistakes.
HW 56
code: formation
41