### Q4 U2 Energy and Chemical Reactions

```Q4U2


Law of Conservation of Energy: energy can
not be created or destroyed, only changed
from one form to another.
Two Main Forms of Energy
◦ Kinetic energy is motion energy.
 Defined as the energy of a moving object.
 A thrown football, a speeding automobile, a waterfall, or a
rock falling from a cliff are examples of objects that have
kinetic energy.
◦ Potential energy is energy stored in matter.
 Potential energy appears in many different forms
 Defined as the energy in matter due to its position or
the arrangement of its parts.
•Rubber bands or Gasoline both have potential
energy
•The energy stored in molecules is called chemical
potential energy.
•This energy is stored in bonds
•The chemical makeup (arrangement of molecules) of gasoline makes
it a good fuel source (source of potential energy).
•The energy stored in gasoline is released by burning it
(combustion).
•The airplane motor uses this released energy to turn a propeller.
•During combustion, chemical bonds are broken
and reformed creating new products and
releasing energy.


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

Is Related to heat lost or gained in chemical
reactions
Heat is released, given off, in an exothermic
reaction
Heat is absorbed, taken in, in an endothermic
reaction
The reaction between H2 and O2 is highly
exothermic
The energy from it powers cars
The heat content of a chemical system is called the enthalpy
(symbol: H)
 The enthalpy change (ΔH) is the amount of heat released or
absorbed when a chemical reaction occurs at constant
pressure.
ΔH is specified per mole of substance
Standard units are:

◦ kJ mol-1 (kJ/mol) or
◦ kcal mol-1 (kcal/mol)

A calorie is the amount of heat needed to raise the temp of 1g
of water by 1 ˚C
1 calorie (1 cal) = 4.184 joules (4.184 J)
1kilocalorie = 1 Calorie =1000 calories

The Bond Enthalpy is the energy required to break a chemical bond. It
is usually expressed in units of kJ mol-1, measured at 298 K.
◦ The Standard unit for energy is the joule
 Energy changes are measured under standard laboratory conditions (STP)



Bond Enthalpy is the energy required to break
a chemical bond. It is usually expressed in
units of kJ/mol
ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)
Add up all the energies of the broken bonds;
add up all the energies of the bonds that are
reformed and subtract one from the other.


1.
Find ΔH for the following reaction given the following bond energies:
ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)
Bond Bond Energy(kJ/mol)
H-H
436
O=O
499
O-H
463
Figure out which bonds are broken and which bonds are formed.
2 H-H bonds are broken.
1 O=O bond is broken
◦ 2 O-H bonds are formed per water molecule, and there are 2 water molecules
formed, therefore 4 O-H bonds are formed
2.
Substitute the values given into the equation:
Calculate the enthalpy of the combustion of Methane
CH4 + O2  CO2 + H2O
1.
Balance the equation
CH4 + 2O2  CO2 + 2H2O
2.
Write the Lewis Dot Structure
H
H-C-H, O=O, O=O  O=C=O + H-O-H, H-O-H
H
3.
Count bonds broken and bonds Formed
Bonds broken (reactants)
Bonds Formed(Products)
4 C-H 413 kJ/mol each
2 C=O 799 kJ/mol each
2 O=O 495 kJ/mol each
4 O-H 463 kJ/mol each

4.
5.
Calculate the sum of the bond enthalpies of products and
reactants
∑ ΔH(bonds broken) = (4 X 413) + (2 X 495) = 2642
∑ ΔH(bonds formed) = (2 X 799)+ ( 4 X 463) = 3450
Calculate the enthalpy of reaction
ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)
2642 – 3450 = -808 kJ/mole
Solutions
1. H-C-C-C-H + O=O  O=C=O + H-O-H

Broken bonds(ΔH each) Formed bonds(ΔH each)
8 C-H 413 kJ/mol
6 C=O 799 kJ/mol
2 C-C 348 kJ/mol
8 O-H 463 kJ/mol
5 O=O 495 kJ/mol
∑ΔH(bonds broken)=(8 X413)+(2X348)+(5X495)=6475
∑ ΔH(bonds formed) = (6 X 799)+ ( 8 X 463) = 8498
ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)
6475 – 8498 = -2023 kJ/mole
Solutions
2. H-C-C-O-H + O=O  O=C=O + H-O-H
Broken bonds(ΔH each) Formed bonds(ΔH each)
5 C-H 413 kJ/mol
4 C=O 799 kJ/mol
1 C-C 348 kJ/mol
6 O-H 463 kJ/mol
1 C-O 358 kJ/mol
1 O-H 463 kJ/mol
3 O=O 495 kJ/mol

∑ΔH(bonds broken)=(5 X413)+348+358+463+(3X495)=4719 kJ/mol
∑ ΔH(bonds formed) = (4 X 799)+ ( 6 X 463) = 5974 kJ/mol
ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)
4719 – 5974 = -1255 kJ/mole
Solutions
H-C=C-H + O=O  O=C=O + H-O-H
Broken bonds(ΔH each) Formed bonds(ΔH each)
4 C-H 413 kJ/mol
8 C=O 799 kJ/mol
2 C=C 839 kJ/mol
4 O-H 463 kJ/mol
5 O=O 495 kJ/mol

∑ΔH(bonds broken)=(4 X413)+(2X 839)+(5X495)= 5805kJ/mol
∑ ΔH(bonds formed) = (8 X 799)+ ( 4 X 463) = 8744kJ/mol
ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)
5805 – 8744 = - 2939kJ/mole
N2 + 3H2  2NH3
Broken bonds(ΔH each)
1 N=N 941 kJ/mol
3 H-H 436 kJ/mol

Formed bonds(ΔH each)
6 N-H 391 kJ/mol
∑ΔH(bonds broken)=941+(3X 436)= 2249kJ/mol
∑ ΔH(bonds formed) = (6 X 391) = 2346kJ/mol
ΔH = ∑ ΔH(bonds broken) - ∑ ΔH(bonds formed)
2249– 2346 = -97 kJ/mole
Exothermic Reaction
Endothermic Reaction
Energy is released.
Energy is absorbed.
Energy is a product of the
reaction.
Energy is a reactant of the
reaction.
Energy of the reactants is
greater than
the energy of the products
ΔH(reactants) >ΔH(products)
Energy of the reactants is
less than
the energy of the products
ΔH(reactants) < ΔH(products)
ΔH =ƸΔH(products) - ƸΔH(reactants)
ΔH = Negative number
ΔH = Positive number

Heat of Reaction: The difference in energy
between the Energy of the Products and the
Energy of the Reactants.
Δ H = H Prod - H Rxts
•This is an exothermic reaction, the energy of the
products is less than the energy of the reactants
•The energy was lost to the environment as heat!
•The Products have a higher energy level than reactants
•Reactants are located on the flat portion to the left of
the peak.
•Products are located on the flat portion to the right of
the peak.
•The activated complex is located at the peak of the
reaction diagram.
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An activated complex is an intermediate state
that is formed during the conversion of
reactants into products.
It is the structure that results at the
maximum energy point along the reaction
path.
The activation energy of a chemical reaction
is the difference between the energy of the
activated complex and the energy of the
reactants.
The lower the level of energy in a system the
more stable it is(less likely to react).
•The amount of energy required to start the reaction (Ea)
•The difference between the energy level of the reactants
and the maximum energy in the reaction progress equals
the activation energy


Catalysts lower the activation energy of a reaction
allowing it to start more easily
They do not effect the amount or type of product
formed!

Entropy is a measure of how much of the energy of a
system is potentially available to do work and how
much of it is potentially manifest as heat.(heat is
often lost to the environment)
 Manufacturing equipment: heats the room because of heat
loss to surroundings

Example of ice melting
 the difference in temp. between a warm room (the
surroundings) and a cold glass of ice water (the system) is
equalized as heat from the warm surroundings are
transferred to the cooler ice and water mixture.

In an isolated system such as the room and ice
water together, the dispersal of energy from
warmer to cooler regions always results in a
net increase in entropy
 The more disorder the higher the entropy!
Complex
(higher energy)
(lower entropy)
exothermic
endothermic
(higher entropy)
Chaos
(lower energy)
A.
B.
C.
D.
E.
F.
Does the graph represent an endothermic or exothermic
reaction?
Label the position of the reactants, products, and
activated complex.
Determine the heat of reaction, ΔH, (enthalpy change) for
this reaction.
Determine the activation energy, Ea for this reaction.
How much energy is released or absorbed during the
reaction?
How much energy is required for this reaction to occur?
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A) The reaction is endothermic
B) Reactants are located on the flat portion to
the left of the peak. Products are located on
the flat portion to the right of the peak. The
activated complex is located at the peak of
the reaction diagram.
C) The enthalpy change is +50 kJ
D) The activation energy is 200 kJ
E) 50 kJ is absorbed in this reaction
F) An input of 200 kJ is required for this
reaction to occur



Consider a general reversible reaction such as:
A+B↔C+D
Given the following potential energy diagram
for this reaction, determine ΔH and Ea for both
the forward and reverse directions.
Is the forward reaction endothermic or
exothermic?
•Activation Energy (Ea) is the difference in energy between
the activated complex and the reactants
•Large energy barriers separates the reactants and the
products so that only very energetic molecules can pass over
this barrier.
•The forward reaction is exothermic.

When 2 moles of H2 react with 1 mole of O2 to
produce 2 moles of H2O (g), heat is released.
2 H2 (g) + O2 (g)  2 H2O(g) + 484kj

The heat needed to form a compound from its
elements is called the energy of formation (∆Hf)
◦ This information is found in tables, for 1 mole of
different compounds


If the ∆Hf of each reactant and each product is
known, a balanced equation can be used to
calculate the energy released or absorbed during
a reaction.
The difference between the sum of the energies
of the products and the sum of the energies of
the reactants is the energy absorbed(+) or
released(-) by the reaction
∆Hf(reaction) = ∑∆Hf(products) - ∑∆Hf(reactants)
∑ stands for the sum

How much energy is produced by the reaction of 16.0g of Fe2O3
with excess aluminum metal according to the equation:
Fe2O3(s) + 2Al(s)  Al2O3(s) + 2Fe + 852 kj
Steps to Solve
1.
Begin with balanced equation above
2.
Convert grams of Fe2O3(s) to moles
16.0g Fe2O3 1 mol Fe2O3
160g Fe2O3
3.
From the equation, determine the number of moles and the
16.0g Fe2O3 1 mol Fe2O3
160g Fe2O3
852 kj
= 85.2 kj produced
1 mol Fe2O3
1.
A rocket fuel is prepared by reacting
hydrazine and dinitrogen tetroxide according
to the equation:
2N2H4 (l) + N2O4 (l)  3N2(g) +4H2O(g) + 2400 kj
2.
calculate the heat released when 3200 grams
of hydrazine are consumed in the rocket
engine.
The dissociation of ammonia into its elements
is an endothermic reaction, absorbing 92.2kj
of energy according to the equation:
2NH3(g) + 92.2 kj  3H2(g) + N2(g)
How much energy will be required to
decompose 85.0g of ammonia?
1.
2.
3200.0g N2H4 1 mol N2H4 1200 kj
32.0 g N2H4 1 mol N2H4
85.0gNH3 1 mol NH3
17.0 g NH3
= 120,000kj produced
46.1 kj
= 231 kj produced
1 mol NH3

Chemistry Connections: Specific Heat
Capacity and Calorimetry
◦ http://player.discoveryeducation.com/?blnPreviewO
nly=1&guidAssetId=928367f4-9cf5-4ed6-bae5eded91b7dac9


The heat required to change the temperature of a
substance depends upon the amount and nature of
the substance as well as the extent of the
temperature change
For example:
◦ 1g of water requires 4.18joules of energy to change the
temp1 ̊C.
◦ It takes only 0.987j to raise the temp of AlF3 1 ̊C


Energy is transferred between a reaction and its
surroundings.
The amount transferred can be calculated:
heat gained/lost= (mass)(change in temp)(specific heat)
q=(m)(∆T)(Cp)
q is heat gained/lost
m is mass(g)
∆T is change in temp
Cp is the specific heat


A physical property of matter.
The temperature of matter is a direct
measure of the motion of the molecules
◦ The greater the motion the higher the temperature:

Motion requires energy: The more energy
matter has the higher temperature it will also
have.
◦ Typically this energy is supplied by heat.
◦ Heat loss or gain by matter is equivalent to energy
loss or gain

How much will the temperature of an object
increase or decrease by the gain or loss of
heat energy?
◦ The answer is given by the specific heat (S) of the
object.



Heat is gained or lost during chemical and
physical changes
Calorimetry is a technique used to measure
those changes in heat.
Important definitions:
◦ Specific heat of water (Shw )=4.184 J / ( g oC )
 This is the amount of heat required to raise the temp of
1 gram of water, 1 degree Celsius
◦ Heat of fusion of ice (Hfi )=334 J / g
 This is the amount of heat required to melt1 gram of
ice at 0.00 degree Celsius (solid to liquid)
◦ Heat of vaporization of water (Hvw)=2256 J / g
 This is the amount of heat required to VAPORIZE 1
gram of water at 100.00 degrees Celsius (liquid to gas)

How much heat is needed to melt a 60.0g ice
cube at 0.0°C?
◦ Heat of fusion of ice (Hfi )=334 J / g
◦ Remember this is 1 gram of ice, we have 60.0g!
◦ 334 J
g
60.0g = 20040 J
1

How much heat is needed to melt a 45.0g ice
cube at 0.0°C?
◦ 334 J
g

45.0g = 15030 J
1
How much heat is needed to vaporize 16.3g
of water at 100.00 degrees Celsius?
◦ (Hvw)=2256 J / g
◦ 16.3 g 2256 J
= 36772.8 J
1
g



Take an object of mass m, put in x amount of
heat and carefully note the temperature rise, then
S is given by
In this definition mass is usually in either grams
or kilograms and temperature is either in kelvin
or degrees Celsius.
Note that the specific heat is "per unit mass".
◦ This means the volume does not matter



11 grams of substance is heated from 20.0˚C
to 30.0˚C. The substance absorbed 4253 J of
energy. What is the specific heat of the
substance?
S= X/(mass)(∆T)
S = 4253J
(11.0g)(10˚C)
=38.6 J/g ˚C
How much heat is required to raise the
temperature of 68.0g of AlF3 from 25.0˚C to
80.0˚C ?
 Steps to Solve
1. Find specific heat of AlF3 from table
Cp of AlF3 = 0.8948J/g ֹ C˚
2. q=(m)(∆T)(Cp)
q=(68.0g) (80.0 – 25.0˚C) (0.8948J/g ˚C)
q= 3350 J


A Calorimeter containing water is used to measure the
heat absorbed or released in a chemical reaction.
◦ The temperature change in the water is used to measure the
amount of heat absorbed or released by the reaction

Law of conservation of energy says energy is neither
created or destroyed.
◦ In an insulated system, any heat lost by one quantity of
matter must be gained by another.

Energy flows from the warmer material to the cooler
material until they are both the same temperature.
Heat lost = Heat gained
m(∆T) Cp= m(∆T) Cp

The heat evolved from the reaction changes the
temperature of a working substance (usually water) with
a known heat capacity.
◦ A measurement of the temperature rise in the surroundings
(calorimeter body) allows a determination of the heat crossing
the boundary between the system (where the reaction takes
place) and the surroundings (where the temperature change is
measured).
A Calorimeter is an instrument for measuring the heat of a reaction
during a well defined process
A constant Volume calorimeter
•A much simpler, but less accurate
calorimeter, which, by its
construction, is necessarily
constant pressure
Suppose a piece of lead with a mass of 14.9g at a temp of 92.5˚C is
dropped into an insulated container of water. The mass of water is
165g and its temperature before adding the lead is 20.0˚C. What is
the final temperature of the system? Cp lead =0.1276 J/g ֹ ˚C
Remember Heat lost = Heat gained
m(∆T) Cp= m(∆T) Cp (The lead will lose energy)
a) The heat lost by lead is q= m(∆T) Cp = 14.9 g X (92.5 ˚C – Tf ) X 0.1276 J/gֹ ˚C
b) The heat gained by water is q= m(∆T) Cp = 165g X (Tf - 20.0˚C) X 4.18 J/gֹ ˚C
c) The heat gained must equal the heat lost:
165g X(Tf - 20.0˚C)X 4.18 J/gֹ ˚C=14.9 g X(92.5 ˚C– Tf )X 0.1276 J/gֹ ˚C
(Tf - 20.0˚C) X 690. J/˚C = (92.5 ˚C – Tf ) X 1.90J/˚C
(690. J/˚C )(Tf ) – 13800 J = 176 J – (1.90J/˚C )(Tf )
(690. J/˚C )(Tf ) + (1.90J/˚C )(Tf ) = 176 J + 13800 J
(690. J/˚C + 1.90J/˚C )(Tf ) = 14000 J
Tf = 20.2 ˚C
1.
2.
3.
How much heat is absorbed by a reaction that
lowers the temperature of 500.0g of water in a
calorimeter by 1.10˚C ?
Aluminum reacts with iron(III) oxide to yield
aluminum oxide and iron. Calculate the heat
given off in the reaction if the temperature of
1.00 kg of water in the calorimeter increases by
3.0˚C.
Burning 1.00g of a fuel in a calorimeter raises
the temperature of 1.0 kg of water from 20.0˚C
to 28.05˚C. Calculate the heat given off in this
reaction. How much heat would one mole of
fuel give off, assuming the molar mass of
65.8g/mol?
1.
qw = m(∆T) (Cw )
= 500.g(1.10˚C)(4.184J/g˚C)
= 2300 J
= 2.30 kJ
2.
qw = m(∆T) (Cw )
= (1.0 x 103g)(3.00˚C)(4.184J/g˚C)
= 12600 J
= 12.6 Kj
3.
∆T = 28.05 ˚C – 20.0 ˚C = 8.05 ˚C
for 1g: qw = m(∆T) (Cw )
= (1. 0 x 103g)(8.05˚C)(4.184J/g˚C)
= 33700 J
= 33.7 kJ for 1.0g
for 1 mol: qw = (33.7 kj/1g)(65.8g/1mol)
= 2220kJ/mol
An equation which shows both mass and heat
relationships between products and reactants is
called a thermochemical equation.
Example
2 H2(g) + O2(g) ----> 2 H2O(l) ΔH = -571.6 kJ



The magnitude of ΔH is directly proportional to
the amount of reactants or products.
A + 2 B ----> C
ΔH = -100 kJ
1/2 A + B ----> 1/2 C
ΔH = -50 kJ
ΔH for a reaction is equal in magnitude but
opposite in sign for the reverse reaction.
A + 2 B ----> C
ΔH = -100 kJ
C ----> A + 2 B
ΔH = +100 kJ

Hess' Law states that the heat evolved or

constant heat summation.
Hess's law can be applied to calculate
enthalpies of reactions that are difficult to
measure.
absorbed in a chemical process is the same
whether the process takes place in one or in
several steps. This is also known as the law of

For a chemical equation that can be written as
the sum of two or more steps, the enthalpy
change for the overall equation equals the
sum of the enthalpy changes for the
individual steps.
◦ No matter how you go from given reactants to
products (whether in one step or several), the
enthalpy change for the overall chemical change is
the same.

The heat transferred in a given change is the
same whether the change takes place in a
single step or in several steps.
In order to use Hess's Law:
1) If a rxn is reversed, the sign of ΔH is reversed.
S(s) + O2 (g) ---> SO2 (g) ΔH = -296 kJ
SO2 (g) ---> S(s)+ O2 (g) ΔH = +296 kJ

2) If the coefficients in a balanced equation are
multiplied by some number, the value of ΔH must be
multiplied by that same number.
S(s) + O2 (g) ---> SO2 (g) ΔH = -296 kJ
2 S(s) + 2O2 (g)--->2 SO2 (g) ΔH = (2)(-296 kJ) = -592 kJ
Example
S(s) + O2 (g) ---> SO2 (g) ΔH = -296 kJ
SO2 (g) + 1/2 O2 (g) ---> SO3 (g) ΔH = -98.9 kJ
____________________________________________
S(s) + 1 1/2 O2 (g) ---> SO3 (g) ΔH = -394.9 kJ
```