Transcript Heat capacity - Chemistry & Physics Armstrong State

Chapter 5
Principles of Chemical Reactivity
Basic Principles
Thermodynamics: The science of heat and
work
Energy: the capacity to do work
-chemical, mechanical, thermal, electrical,
radiant, sound, nuclear
-affects matter by raising its temperature,
eventually causing a state change
-All physical changes and chemical changes
involve energy
Potential Energy:
energy that results from an object’s position
-gravitational, chemical, electrostatic
Kinetic Energy:
energy of motion
Basic Principles
Law of Energy Conservation:
Energy can neither be created nor destroyed
-a.k.a. The first law of thermodynamics
-The total energy of the universe is constant
Temperature vs. Heat:
– Temperature is the measure of an object’s
heat energy
– Heat ≠ temperature
The Measurement of Heat
Thermal Energy
depends on temperature and the amount
(mass or volume) of the object
-More thermal energy a substances has the
greater the motion its atoms/molecules have
-Total thermal energy of an object is the sum of
the individual energies of all
atoms/molecules/ions that make up that object
SI unit: Joule (J)
1 calorie = 4.184 J
English unit = BTU
Converting Calories to Joules
Convert 60.1 cal to joules
Basic Principles
System: object or collection of objects being
studied
– In lab, the system is the chemicals inside the beaker
Surroundings: everything outside of the system
that can exchange energy with the system
– The surroundings are outside the beaker
Universe: system plus surroundings
Exothermic: heat transferred from the system to
the surroundings
Endothermic: heat transferred from the
surroundings to the system
Specific Heat Capacity (C)
amount of heat required to raise the
temperature of 1gram of a substance by 1
kelvin
SI Units: Specific heat capacity = J /g.K
J
Specific heat of water = 4.184 g.K
Heat Transfer
Heat transfer equation
used to calculate amounts of heat (q) in a
substance
J
q  m  C  T
J
g.K
g
q1 + q2 + q3 … = 0
K
or
qsystem + qsurroundings = 0
Heat Transfer
Calculate the amount of heat to raise the
temperature of 200 g of water from 10.0 oC to
55.0 oC
Heat Transfer
Calculate the amount of heat energy (in joules)
needed to raise the temperature of 7.40 g of
water from 29.0°C to 46.0°C
Heat Transfer
A 1.6 g sample of metal that appears to be gold requires
5.8 J to raise the temperature from 23°C to 41°C.
Is the metal pure gold?
J
Specific heat of gold is 0.13 .
gK
Therefore the metal cannot be pure gold.
Changes of State
occurs when enough energy is put into a
substance to over come molecular interactions
Solid-liquid:
molecules in a solid when heated move about vigorously
enough to break solid-solid molecular interactions to
become a liquid
Liquid-gas:
molecules in a liquid when heated move about more
vigorously enough to break liquid-liquid molecular
interactions to become a gas
Note: This happens in reverse by removing heat energy
Energy and Changes of State
Heat of fusion:
heat needed to convert a substance from a solid
to a liquid (at its melting/freezing point)
333 J/g for water
Heat of vaporization:
heat needed to convert a substance from a
liquid to a gas (at its boiling/condensation point)
2256 J/g for water
Example: Calculate the amount of heat involved to
convert 100.0 g of ice at -50.0°C to steam at
200.0°C.
The First Law of Thermodynamics
This law can be stated as, “The combined
amount of energy in the universe is constant”
Also called-The Law of Conservation of Energy:
– Energy is neither created nor destroyed in
chemical reactions and physical changes.
The First Law of Thermodynamics
There are two basic ideas for thermodynamic
systems:
1. Chemical systems tend toward a state of
minimum potential energy
Some examples of this include:
-H2O flows downhill
-Objects fall when dropped
Epotential = mg(h)
State Function
The value of a state function is independent of
pathway
-An analogy to a state function is the energy required to climb
a mountain taking two different paths:
E1 = energy at the bottom of the mountain
E1 = mgh1
E2 = energy at the top of the mountain
E2 = mgh2
E = E2-E1 = mgh2 – mgh1 = mg(h)
Examples of state functions:
Temperature, Pressure, Volume, Energy, Entropy, and enthalpy
Examples of non-state functions:
Number of moles, heat, work
The First Law of Thermodynamics
2. Chemical systems tend toward a state of
maximum disorder
Common examples of this are:
- A mirror shatters when dropped and does not
reform
- It is easy to scramble an egg and difficult to
unscramble it
- Food dye when dropped into water disperses
The First Law of Thermodynamics
Thermodynamic questions:
1. Will these substances react when they are
mixed under specified conditions?
2. If they do react, what energy changes and
transfers are associated with their reaction?
3. If a reaction occurs, to what extent does it
occur?
Changes in Internal Energy (E or U)
all of the energy contained within a
substance
– all forms of energy such as kinetic, potential,
gravitational, electromagnetic, etc.
First Law of Thermodynamics states that the
change in internal energy, E, is determined
by the heat flow (q) and the work (w)
E = q + w
book: U = q + w
Changes in Internal Energy (E)
E = Eproducts – Ereactants
E = q + w
at constant pressure:
w = -P x V
q > 0 if heat is absorbed by the system
q < 0 if heat is absorbed by the surroundings
w > 0 if surroundings do work on the system
w < 0 if system does work on the surroundings
Changes in Internal Energy (E)
If 1200 joules of heat are added to a system, and
the system does 800 joules of work on the
surroundings, what is the:
1. energy change for the system, Esys?
2. energy change of the surroundings, Esurr?
Changes in Internal Energy (E)
E is negative when energy is released by
a system
-Energy can be written as a product of the
process
C5 H12(  )  8 O 2(g)  5 CO2(g)  6 H 2O(  )  3.516  103 kJ
E  - 3.516  10 kJ
3
Changes in Internal Energy (E)
E is positive when energy is absorbed by a
system undergoing a chemical or physical
change
– Energy can be written as a reactant of the
process
5 CO2(g)  6 H 2O(  )  3.516  10 kJ  C5 H12(  )  8 O 2(g)
3
E   3.516  103 kJ
Enthalpy Changes for Chemical Reactions
Exothermic reactions:
release energy in the form of heat to the
surroundings (H < 0)
-heat is transferred from a system to the surroundings
Endothermic reactions:
gain energy in the form of heat from the
surroundings (H > 0)
-heat is transferred from the surroundings to the system
For example, the combustion of propane:
Combustion of butane:
Enthalpy Change (H)
Heat content of a substance at constant
pressure
- Chemistry is commonly done in open beakers on a
desk top at atmospheric pressure
- Therefore enthalpy change (H) is the change in
heat content: H = qp at constant pressure
E = qv
at constant volume
If E and H < 0:
energy is transferred to the surroundings
If E and H > 0:
energy is transferred to the system
- Enthalpy and energy differ by the amount of work
E = H + w
and w = -PV
Enthalpy Changes for Chemical Reactions
Exothermic reactions generate specific amounts of heat
– Because the potential energies of the products are lower
than the potential energies of the reactants
Endothermic reactions consume specific amounts of heat
– Potential energies of
the reactants are lower
than the products
H for the reverse reaction
is equal, but has the
opposite sign to the forward
reaction
Thermochemical Equations
balanced chemical reaction with the H value
for the reaction
C5H12( )  8 O2(g)  5 CO2(g)  6 H2O()
Horxn  - 3523kJ
H < 0 designates an exothermic reaction:
heat is a product, the container feels hot
H > 0 designates an endothermic reaction:
heat is a reactant, the container feels cold
Hess’s Law
If a reaction is the sum of two or more other
reactions, H for the overall process is the
sum of the H for the component reactions
– Hess’s Law is true because H is a state
function
If we know the following H values:
Target:
1 4 FeO(s)  O 2(g)  2 Fe2O3(s) H o  560
? kJ

2.3 4 Fe(s)  3 O 2(g)  2 Fe 2 O 3(s)
1. 2 2 Fe(s)  O 2(g)  2 FeO(g)
H o  544 kJ
H  1648 kJ
o
Hess’s Law
We can calculate the H for the reaction by
properly adding (or subtracting) the H for
reactions 1 and 2
– Notice that the target reaction has FeO and O2 as
reactants and Fe2O3 as a product
– Arrange reactions 1 and 2 so that they also have FeO
and O2 as reactants and Fe2O3 as a product
• Each reaction can be doubled, tripled, or multiplied by a half,
etc.
• H values are then doubled, tripled, etc.
• If a reaction is reversed the H value is changed to the
opposite sign
4 FeO
4 Fe
2 O2
2 x [1.
2] 2(2 FeO  s   2 Fe  s   O 2 g  )
+
2. 3 4 Fe  s   3 O 2 g   2 Fe 2 O 3 s 
1
1 4 FeO s   O 2 g   2 Fe 2 O 3
H 0
2( 544 ) kJ
+1088 kJ
 1648 kJ
 560 kJ
Hess’s Law
Given the following equations and H
o
values
H  kJ 
[1] 2 N 2 g   O 2 g   2 N 2 O  g 
164.1
[2] N 2 g  + O 2 g   2 NO  g 
180.5
[3] N 2 g  + 2 O 2 g   2 NO 2  g 
66.4
o
calculate
for 2the
reaction
below:
N 2 O g  H
+ NO

3
NO

H
?
g 
g 
You do it!
Hess’s Law
-The + sign of the H value tells us that the reaction is
endothermic.
-The reverse reaction is exothermic, i.e.
3 NO  g   N 2 O  g  + NO 2  g 
H
o
= -155.5 kJ
Standard Enthalpy of Formation
Thermochemical standard state conditions
The thermochemical standard T = 298.15 K
The thermochemical standard P = 1.0000 atm
- Be careful not to confuse these values with STP
Thermochemical standard states of matter
For pure substances in their liquid or solid phase the
standard state is the pure liquid or solid
- For aqueous solutions the standard state is 1.00 M
concentration
For gases the standard state is the gas at 1.00 atm of
pressure
- For gaseous mixtures the partial pressure must be
1.00 atm
Standard Molar Enthalpies of
Formation, Hfo
The enthalpy change for the formation of one
mole of a substance formed directly from its
constituent elements in their standard states
– The symbol for standard molar enthalpy of formation is
Hfo (KJ/mol)
The standard molar enthalpy of formation for MgCl2
is:
Standard Molar Enthalpies of
Formation (Hfo)
Standard molar enthalpies of formation have been
determined for many substances and are tabulated in
Appendix L
Standard molar enthalpies of elements in their most
stable forms at 298.15 K and 1.000 atm are zero.
Example: The standard molar enthalpy of formation for
phosphoric acid is -1279 kJ/mol. Write the equation for the
reaction for which Hof = -1279 kJ
Note: P in standard state is P4 (s)
Phosphoric acid in standard state is H3PO4(s)
Standard Molar Enthalpies of
Formation
Calculate the enthalpy change for the reaction of
one mole of H2(g) with one mole of F2(g) to form
two moles of HF(g) at 25oC and one atmosphere.
Standard Molar Enthalpies of
Formation
Calculate the enthalpy change for the reaction in
which 15.0 g of aluminum reacts with oxygen to
form Al2O3 at 25oC and one atmosphere.
You do it!
Hess’s Law
Version II
For chemical reaction at standard conditions:
– the standard enthalpy change is the sum of the
standard molar enthalpies of formation of the
products minus the sum for the reactants
- each enthalpy of formation is multiplied by its coefficient in
the balanced chemical equation
H
0
rxn
  n H
0
f products
  n H
n
n
n  stoichiome tric coefficien ts
Final - Initial
0
f reactants
Hess’s Law
Calculate the H°f for the following reaction from
the data in appendix L:
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
Hess’s Law
Given the following information, calculate Hfo for
H2S(g)
2 H2Sg + 3 O2g 2 SO2g +2 H2Ol
Hof
?
0
-296.8 -285.8
(kJ / mol)
You do it!
Ho298 = -1124 kJ
Calorimetry
An experimental technique that measures the heat
transfer during a chemical or physical process
Constant pressure calorimetry:
A styrofoam coffee-cup calorimeter is
used to measure the amount of heat
produced (or absorbed) in a reaction
– This is one method to measure
qP (called H) for reactions in solution
qreaction + qsolution = 0
Note: Assuming no heat transfer to the surroundings
Calorimetry
If an exothermic reaction is performed in a calorimeter,
the heat evolved by the reaction is determined from
the temperature rise of the solution
– This requires a two part calculation
When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00
mL of 0.600 M CH3COOH already in the calorimeter at the
same temperature, the resulting temperature is observed to
be 25.947oC. Determine heat of reaction and then calculate
the change in enthalpy (as KJ/mol) for the production of
NaCH3COO.
CH3COOH(aq) + NaOH(aq)  NaCH3COO(aq) + H2O(l)
Calorimetry
Constant volume calorimetry:
Or Bomb calorimetry measures measure the amount of heat
produced (or absorbed) in a chemical reaction
-this method is used for measuring qv (E)
qreaction + qbomb + qwater = 0