Transcript Thermochemistry - Virginia State University

Thermochemistry
Chapter 6
Dr. Victor H. Vilchiz
Stoichiometry and Enthalpy
• Consider the reaction of methane,
CH4, burning in the presence of
oxygen at constant pressure. Given
the following equation, how much
heat would be produced by the
combustion of 10.0 grams CH4?
CH4 (g )  2O 2 (g )  CO2 (g )  2H 2O(l ); Ho  -890.3 kJ
10.0 g CH 4 
1 mol CH 4
16.0 g

890.3 kJ
1 mol CH 4
 556 kJ
Heats of Reaction
• In general the heat exchanged in a
reaction is called the heat of reaction
HRxn; however, there are some
important reaction types that they
get their own H name.
– Combustion reactions HRxn= Hcomb
– Formation reactions HRxn= Hf
– Fusion reactions HRxn=Hfus
– Vaporization Reactions HRxn = Hvap
– Sublimation Reactions HRxn= Hsub
Heat of Rxn
• Where does the heat of reaction
come from?
– During a chemical reaction bonds must
be broken.
• Breaking bonds requires energy.
– As the products are formed bonds are
formed.
• Forming Bonds releases energy.
HRxn is just a book keeping method
Measuring Heats of Reaction
• To see how heats of reactions are
measured, we must look at the heat
required to raise the temperature of
a substance, because a
thermochemical measurement is
based on the relationship between
heat and temperature change.
• The branch of science that studies
heats of reaction is Calorimetry
The heat required to raise the temperature
of a substance is its heat capacity.
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.
Changing the temperature of the sample
requires heat equal to:
q  CT
A Problem to Consider
• Suppose a piece of iron requires
6.70 J of heat to raise its
temperature by one degree Celsius.
The quantity of heat required to
raise the temperature of the piece of
iron from 25.0 oC to 35.0 oC is:
q  CT  (6.70 J / C)  ( 35.0 C  25.0 C)
o
q  67.0 J
o
o
Measuring Heats of Reaction
Heat capacities are also compared for one
gram amounts of substances. The specific
heat capacity (or “specific heat”) is the heat
required to raise the temperature of one
gram of a substance by one degree Celsius.
To find the heat required you must multiply the
specific heat, s (c), of the substance times its
mass in grams, m, and the temperature change,
T.
q  s  m  T
A Problem to Consider
• Calculate the heat absorbed when
the temperature of 15.0 grams of
water is raised from 20.0 oC to 50.0
oC. (The specific heat of water is
4.184 J/g.oC.)
q  s  m  T
q
o
J
(4.184 go C )  (15.0g )  (50.0  20.0 C)
q  1.88  10 J
3
Heats of Reaction: Calorimetry
• A calorimeter is a device used to
measure the heat absorbed or evolved
during a physical or chemical change.
(coffee cup and bomb calorimeters)
The heat absorbed by the calorimeter and its
contents is the negative of the heat of reaction.
qcal  qrxn
qrxn  nH Rxn
qcal  Ccal T
H Rxn
 Ccal T

n
A Problem to Consider
• When 23.6 grams of calcium chloride,
CaCl2, was dissolved in water in a
calorimeter, the temperature rose from
25.0 oC to 38.7 oC.
If the heat capacity of the solution and the
calorimeter, Ccal, is 1258 J/oC, what is the
enthalpy change per mole of calcium chloride?
Heats of Reaction: Calorimetry
First, let us calculate the heat
absorbed by the calorimeter.
qcal  Ccal T  (1258
J
o
C
)  (38.7 C  25.0 C )
o
o
qcal  1.72  10 J
4
Now we must calculate the heat per mole
of calcium chloride.
Heats of Reaction: Calorimetry
Calcium chloride has a molecular
mass of 111.1 g, so
(1 mol CaCl2 )
( 23.6 g CaCl2 ) 
 0.212 mol CaCl2
111.1 g
Now we can calculate the heat per mole
of calcium chloride.
qrxn
 17.2 kJ
H 

 81.1 kJ / mol
mol CaCl2 0.212 mol
Calorimeter Calculations
• There are three different formulas
used to calculate heat transfer. Each
is used in different situations.
– q=msT; used when only the temperature
changes. (Heating and Cooling)
– q=CcalT; used when a calorimeter’s heat
capacity is not negligible.
– q=mH (nH); used when there is a
chemical or physical change. Note: H are
sometimes provided in kJ/mol or J/g hence
the two variations of the formula.
Heating/Cooling Curve
Hess’ Law
• Hess’ law of heat summation states
that for a chemical equation that can
be written as the sum of two or more
steps, the enthalpy change for the
overall equation is the sum of the
enthalpy changes for the individual
steps.
• Hess’ law works because H is a
state function.
Hess’ Law
• For example, suppose you are given
the following data:
S(s)  O 2 (g )  SO 2 (g ); H  -297 kJ
o
2SO 3 (g )  2SO 2 (g )  O 2 (g ); H  198 kJ
o
Could you use these data to obtain the
enthalpy change for the following reaction?
2S(s )  3O 2 (g )  2SO 3 (g ); H o  ?
Hess’ Law
• If we multiply the first equation by 2
and reverse the second equation,
they will sum together to become
the third.
2S(s)  2O 2 (g )  2SO 2 (g ); H  (-297 kJ)  (2)
o
2SO 2 (g )  O 2 (g )  2SO 3 (g ); H o  (198 kJ)  (-1)
2S(s )  3O 2 (g )  2SO 3 (g ); H o  -792 kJ
Standard Enthalpies of Formation
• The term standard state refers to
the standard thermodynamic
conditions chosen for substances
when listing or comparing
thermodynamic data: 1 atmosphere
pressure and the specified
temperature (usually 25 oC).
The enthalpy change for a reaction in which
reactants are in their standard states is denoted
Ho (“delta H zero” or “delta H naught”).
Standard Conditions
Compound
 For
a gas, pressure is exactly 1 atm
 For a solution, Molarity is exactly 1M.
 Pure substance (liquid or solid), it is the
pure liquid or solid.
Element
 The
form [N2(g), K(s)] in which it exists at
1 atm and 25°C.
Standard Enthalpies of Formation
• The standard enthalpy of formation of a
substance, denoted Hfo, is the
enthalpy change for the formation of
one mole of a substance in its standard
state from its component elements in
their standard state.
Note that the standard enthalpy of formation
for a pure element in its standard state is zero.
Standard Enthalpies of Formation
• The law of summation of heats of
formation states that the enthalpy of a
reaction is equal to the total formation
energy of the products minus that of the
reactants.
H  
o
o
nH f (products ) 

o
mH f (reactants )
m and n are the coefficients of the
substances in the chemical equation.
A Problem to Consider
• Large quantities of ammonia are
used to prepare nitric acid according
to the following equation:
4NH 3 (g )  5O 2 (g )  4NO(g )  6H 2O(g )
What is the standard enthalpy change for this
reaction? Use Table 6.2 for data.
A Problem to Consider
• You record the values of Hfo under
the formulas in the equation,
multiplying them by the coefficients
in the equation.
4NH 3 (g )  5O 2 (g )  4NO(g )  6H 2O(g )
4( 45.9)
5(0)
4(90.3)
6( 241.8)
You can calculate Ho by subtracting the values
for the reactants from the values for the
products.
A Problem to Consider
• Using the summation law:
H o   nH of (products )   mH of (reactants )
H o  [4(90.3)  6( 241.8)] kJ  [4( 45.9)  5(0)] kJ
H  906 kJ
o
Be careful of arithmetic signs as they are a
likely source of mistakes.
Operational Skills
• Writing thermochemical equations.
• Manipulating thermochemical equations.
• Calculating the heat of reaction from the
stoichiometry.
• Relating heat and specific heat.
• Calculating H from calorimetric data.
• Applying Hess’s law.
• Calculating the enthalpy of reaction from
standard enthalpies of formation.
Specific Heat Capacities
Return to Lecture
Figure 6.11: Coffee-cup calorimeter.
Return to Lecture
Bomb Calorimeter
Return to Lecture
Enthalpy Calculations Guide
Return to Lecture
Heats of Formation
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