Download Heat of Reaction

Unit #6
Thermochemistry
Thermochemistry The study of the heat changes in chemical
reactions and physical processes.
Examples of heat changes:
Fire
Chemical
Exo. Chemical potential to
heat/light via
combustion.
Friction
Physical
Exo. Kinetic to heat
Hair Drier
Physical
Exo. Electrical to heat
Air
Conditioner
Physical
?
Electrical to heat (high
pressure) to absorb heat
(low pressure).
Examples of heat changes con’t:
Ice
Melting
H2O
Evaporating
Physical
Endo.
Physical
Endo.
Low kinetic to
high kinetic
Low kinetic to high
kinetic.
All of the above result in a change in heat (energy).
Energy - The capacity to do work or supply heat.
Chemical potential energy Energy stored in chemicals because of their
composition.
Where is “this energy” stored ?
In the chemical
bonds.
The type of atoms and their arrangements
determine the amount energy stored in the bonds.
Heat A form of energy that ALWAYS flows from a
warmer object to a cooler object.
In physical sciences it is represented as a “q”
It is measured in “Joules”
What are the observable effects of adding heat ?
1. Increase of temperature.
2. Change of state (melting, boiling…)
3. Starting of a chemical reaction (combustion,
decomposition).
Temperature A measure of the average kinetic energy of the
molecules of an object.
Heat Capacity The amount of heat (energy) required to
change an object’s temperature by exactly 1C.
Measured in J/C or calories.
calorie The amount of heat required to increase the
temperature of 1 gram of water from 4C to 5C.
Notice this is a lower case word.
Calorie with a capital “C” is used when refering
to food. It equals 1000 calories (lower case “c”), or 1
Kcal (Kilocalorie).
Trivia:
A runner in a marathon (26.2 miles) could burn
1800 Calories during the run.
1800 Calories is about 4.5 servings of spaghetti.
1 Kcal = 4186 joules
Joule SI unit of heat and energy.
1 joule = 1 Kg x m2
s2
= 0.239 calories
Specific Heat Capacity The amount of heat required to raise 1 gram
of a substance 1C.
Measured in
J
g x C
Heat capacity (not specific heat) is dependant on
two things:
1. The mass of the substance.
The greater the mass, the greater the heat
capacity.
2. The chemical composition of the substance.
Example #1:
A single microwavable burrito lists its Calories
as 310. How many joules is this ?
310 Cal x
1000 calories
1 Cal
=
1,297,071.13 joules
Example #2:
Consider the following substances and their
specific heats:
H2O = 4.186 J/g xC
Al
= 0.90 J/g xC
Fe
= 0.44 J/g xC
Cu
= 0.38 J/g xC
Which heats
quicker??
cools slower
HCu
2O
Example #3:
A piece of silver has a heat capacity of 42.8
J/C. If the silver has a mass of 181 grams, what is
the specific heat ?
Specific heat =
Specific heat =
J
g x C
=
0.236 J/g xC
42.8 J
181 g x C
Example #4:
Calculate the heat (q) absorbed by 15 grams
of water to raise its temperature from 20C to 50C.
(assume constant pressure).
Specific heat =
4.186 J
g x C
Heat absorbed = 1881 J
=
XJ
15 g x 30C
Example #5:
You wish to heat water for coffee. How much
heat in Joules must be used to raise the temperature of
0.180 Kg of tap water (enough for 1 cup) from 15 C
to 96 C (the ideal brewing temp).
Specific heat =
Heat added =
4.186 J
g x C
61031.88 J
=
XJ
180 g x 81C
Example #6:
An iron skillet weighing 1.51 Kg is heated to
178C on the stove. It then was cooled to the room
temp of 21C . How much heat energy (in Joules) was
removed ?
Specific heat of Fe =
0.450 J
g x C
Heat removed = -106681.5 J
=
XJ
1510 g x -157C
Example #7:
Refer to the microwavable burrito in example #1,
in which 1,297,071.13 J of energy are stored. Assume
a 150 lb person (68.0 Kg) is 90% water by mass; what
is the theoretical change in temperature by eating this
burrito ?
Specific heat H2O =
4.186 J
g x C
Temp increase =
5.06C
HOMEWORK:
Combustion of propane yields 2043.96 KJ/mol
of heat. How much propane (in grams) must be
combusted to heat 50 gallon (189.3 L) of water from
50F (10C) to 140F (60C) ?
Specific heat H2O =
4.186 J
g x C
Assume pwater = 1 g/mL
The 0th Law of Thermodynamics
If “A” is in thermal equilibrium with “B”, and “B”
is in thermal equilibrium with “C”; then “A” is also in
equilibrium with “C”.
What is Thermal Equilibrium ?
The same
temperature
If A = B, & B = C, then A = C
This law justifies the use of a thermometer
Heat of Reaction The amount of energy involved in a chemical
change. It will be either Endo- or Exothermic.
To understand this concept, consider the
following 4 ideas:
1st idea)
The Total Energy of a substance = a + b + c
a = Kinetic energy =
1/2 mv2
b = Potential energy (energy due to position in a
force)
c = Internal energy =
ET = EK + E P + U
1st idea con’t)
ET = EK + EP + U
Normally in a lab (during testing) the substance
is at rest in a container.
Therefore: EK = 0
and its potential energy is constant.
Therefore: EP = 0
Then the ET (total energy) = U (internal energy).
1st idea con’t)
For example:
Water falling over the breast of a dam has
some amount of total energy. As it falls, potential
energy is converted to kinetic energy. But, some of
the kinetic energy may also be converted into random
molecular motion (U, internal energy) of the water.
Does this sound familiar ?
It should. It is the 1st law of Thermodynamics.
1st idea con’t)
1st Law of Thermodynamics
AKA
The Law of Conservation of Energy Energy is neither created or destroyed, it just
changes form.
2nd idea)
Thermodynamic system The substance or mixture of substances under
study in which a change occurs.
3rd idea)
Surroundings Everything in the vicinity of the thermodynamic
system.
4th idea)
Heat Remember an energy that ALWAYS flows from
a warmer object to a cooler object.
a.
Represented as “q”
b.
If heat is absorbed (added to) a system, “q” will
be a positive value.
c.
If heat is evolved (leaves) from a system, “q”
will be a negative value.
With the previously discussed 4 ideas we can
now consider the actual “Heat of Reaction”.
Heat of Reaction At a given temperature, (usually 298K,
24.85C), is the value of “q” required to return a
system to the given temperature at the end of the
reaction.
Example 1)
Combustion of methane.
CH4 + 2O2
CO2 + 2H2O
Will this be Endo or Exothermic ?
Exothermic
Do you expect “q” to be positive or negative ?
In fact it is negative, -890 KJ/mol
Example 2)
“Gasohol” contains Ethyl alcohol, CH3CH2OH,
which will combust. How much energy in KJ will be
produced by combusting 3.79 liter (1 gallon) of ETOH
? H = -1368 KJ/mol, p = 0.789 g/mL.
CH3CH2OH + 3O2
2 CO2 + 3H2O
Clues:
1.
Determine the mass of your sample.
2.
Determine how many moles of sample you have.
3.
Use the heat of combustion and a proportion
to solve.
ANSWER =
8.88 x 104 KJ
Example 3)
Ammonia combusts in the presence of a
platinum catalyst to produce Nitrogen monoxide,
water, and 295.5 KJ/mol.
Determine the balanced equation for this
reaction and the value of “q”.
ANSWER = q = -1170 KJ of energy
Example 4)
Ozone (O3) decomposing into oxygen is
exothermic.
What is the sign for “q” ?
- “negative”
What would the container feel like ?
Warm
Enthalpy & Enthalpy Change
Enthalpy A state function and an extensive property of a
substance that can be used to obtain the heat evolved
or absorbed in a chemical reaction.
Represented as “H”.
It is equal to the heat of the reaction at constant
pressure.
Extensive properties Properties that are dependant on the amount
of substance.
Examples:
Mass, Heat capacity, Volume
Intensive properties Properties that are independant on the amount
of substance.
Examples:
Boiling point, condensation point, melting point,
freezing point, Color, Odor, State, Density
State Function A property of a system that depends only on its
present state, which is determined by variables like
temperature or pressure, and is independent of any
previous history of the system.
Consider the following:
“Camp B”
5000 ft.
Path #1
Steep, short
“Camp A”
1200 ft.
Path #2
Not as steep, but longer
The final altitude is the same regardless of the
path chosen (previous history). This is analogous to
a “State Function”.
Enthalpy A state function and an extensive property of a
substance that can be used to obtain the heat evolved
or absorbed in a chemical reaction.
Represented as “H”.
It is equal to the heat of the reaction at constant
pressure.
Hess’ Law of Heat Summation For a chemical reaction that can be written as
the sum of two or more steps, the enthalpy change
for the overall reaction equals the sum of the
enthalpy changes for the individual steps.
Like the “Camps A & B”, no matter how you go from
reactants to products (whether in one step or
several) the enthalpy change is the same overall.
Another way to consider it:
H = Hfinal - Hinitial
OR
H = H(products) - H(reactants)
Application of Hess’ Law:
Given that:
H = -297 KJ
1.
S(s) + O2(g) -----> SO2(g)
2.
2SO2(g) + O2(g) -----> 2SO3(g) H = -198 KJ
The overall reaction is as follows:
2S(s) + 3O2(g) ----> 2SO3(g)
Calculate H for the above reaction.
Solution:
1.
2S(s) + 2O2(g) -----> 2SO2(g) H = 2(-297) KJ
2.
2SO2(g) + O2(g) -----> 2SO3(g) H = -198 KJ
2S(s) + 3O2(g) -----> 2SO3(g)
H = -792 KJ
Example #2
Hydrogen peroxide is a colorless liquid whose
solutions are used as a bleach and antiseptic. It can
be prepared as follows:
H2(g) + O2(g) ----> H2O2(l)
Given:
1. H2O2(l) ----> H2O(l) + 1/2 O2(g) H = -98.0 KJ
2. 2H2(g) + O2(g) ----> 2H2O(l)
H = -571.6 KJ
H2(g) + O2(g) ----> H2O2(l)
Calculate H.
Clues:
1.
You are permitted to rewrite an equation in
the opposite order, if necessary, so that a formula
will cancel out. But you must reverse the sign on H
if you do this.
2.
You may multiply or divide all coefficients in a
given equation, if necessary, so that a formula will
cancel out. But you must also multiply or divide the
value of H if you do this.
Example #3
Ammonia will burn in the presence of a platinum
catalyst to produce Nitrogen monoxide and water.
4NH3(g) + 5O2(g) ----> 4NO + 6H2O(g)
Given:
1. N2(g) + O2(g) ----> 2NO(g)
H = 180.6 KJ
2. N2(g) + 3H2(g) ----> 2NH3(g) H = -91.8 KJ
3. 2H2(g) + O2(g) ---> 2H2O(g) H = -483.7 KJ
Calculate H.
Clues:
1.
You are permitted to rewrite an equation in
the opposite order, if necessary, so that a formula
will cancel out. But you must reverse the sign on H
if you do this.
2.
You may multiply or divide all coefficients in a
given equation, if necessary, so that a formula will
cancel out. But you must also multiply or divide the
value of H if you do this.
Answer:
H = -906.3 KJ
You must provide the solution (work).
Example #4
Hydrazine, N2H4, is a colorless liquid used as a
rocket fuel. It can be synthesized with the following
reaction:
N2(g) + 2H2(g) ----> N2H4(l)
Given:
1. N2H4(l) + O2(g) ----> N2(g) + 2H2O(l)
H = -622.2 KJ
2. H2(g) + 1/2 O2(g) ----> H2O(l)
H = -285.8 KJ
N2(g) + 2H2(g) ----> N2H4(l)
Calculate H.
A derivative of Hess; Law is this:
H = Hf(products) - Hf(reactants)
H2O(l) = -285.8 KJ/mol
NaOH(s) = -426.8 KJ/mol
Na(s) and H2(g) = 0 KJ/mol
Use the above to calculate H for the
following reaction.
Sodium reacts with water producing Sodium
hydroxide and Hydrogen. Calculate H for this
reaction.
Answer: -282 KJ/mol
Example #2:
Silver sulfide reacts with water producing
Silver, Hydrogen sulfide, and Oxygen.
Using the list of “Thermochemical data”
provided to you, calculate the change in enthalpy for
this reaction.
Note: Hf(AgS) = -32.6 KJ/mol. All others on t”the
list”
Example #3:
Rust reacts with Carbon monoxide producing
Iron and Carbon dioxide.
Using the list of “Thermochemical data”
provided to you, calculate the change in enthalpy for
this reaction.
Answer:
-24.74 KJ
Example #4:
Methane reacts with Nitrogen yielding
Hydrogen cyanide and Ammonia.
Using the list of “Thermochemical data”
provided to you, calculate the change in enthalpy for
this reaction.
Answer:
163.8 KJ
Example #5:
What is the change in enthalpy of the
neutralization of 1 mole of Hydrochloric acid with 1
mole of Sodium hydroxide ?
Using the list of “Thermochemical data”
provided to you, calculate the change in enthalpy for
this reaction.
Answer:
-55.84 KJ
Cumulative skills problem:
How much heat (in Kilojoules) can be produced
via aerobic respiration of 4.0 grams of glucose ?
AND
How many “nutritional calories” (Calories) are
produced ?
Summary of applications of Hess’ law:
1.
You can “add” “reaction steps” to calculate the
H for a reaction that you
information on.
We have done several examples using this
concept. (The problems where you rearranged
equations so that they could cancel each other out,
leaving only the desired components.
Summary of applications of Hess’ law (con’t):
2.
You can calculate the HTotal from the Heats of
formation (Hf) of the various reactants and products.
We have done several examples using this
concept. (The problems where you utilized
H = Hf(products) - Hf(reactants) to solve.
Summary of applications of Hess’ law (con’t):
3.
You can calculate the H of phase transition
from Hf.
We will now do several examples using this
application.
Phase transition example #1:
When water evaporates, it absorbs heat from
it’s surroundings.
Determine how much propane, C3H8, (in grams)
must be combusted to evaporate 355 mL (about 20
oz.) of water.