CH1121
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The capacity to do work or transfer heat
Law of conservation of energy states that
energy cannot be create nor destroyed but
can change forms
◦ Energy lost by a system must be gained by the
surroundings and vice versa
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Radiant energy
◦ From electromagnetic radiation
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Thermal energy
◦ From heat (movement of particles)
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Chemical energy
◦ Stored in the bonds of chemical compounds,
released in reactions
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Heat is the energy used to cause the
temperature of an object to increase
Thermodynamics is the study of energy and
its transformations
Thermochemistry is a portion of
thermodynamics that studies chemical
reactions and energy changes that involve
heat
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System is the portion of the universe we
single out for study
◦ In chemistry the reactants and products are the
system
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Surroundings are everything else besides the
system
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Open systems can exchange matter and
energy with the surroundings
◦ Uncovered pot of boiling water on the stove
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Closed systems can exchange energy but not
matter with the surroundings
◦ Piston with H2 and O2 reacting to produce H2O and
energy in the form of heat and work
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Exothermic process – the system loses heat
to the surroundings (spontaneous)
◦ Combustion of gasoline
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Endothermic process – the system gains heat
from the surroundings (not spontaneous)
◦ Melting ice cubes
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Internal energy plus the product of pressure and
volume of the system
𝐻 = 𝐸 + 𝑃𝑉
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Heat evolved (absorbed or released) by a reaction
is the change in enthalpy (ΔH)
∆𝐻 = 𝑞𝑃 = ∆𝐸 + 𝑃∆𝑉

qP = heat or energy evolved from a system at a
constant pressure
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“Heat of reaction”
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Enthalpy change associated with a chemical reaction
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-ΔH = exothermic = gives off heat
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+ΔH = endothermic = absorbs heat
◦ Unit are in Joules (J)/kilojoules (kJ)
𝑞
∆𝐻 =
𝑛
𝑞𝑟𝑥𝑛 = −𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
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In your data sheets booklet these equations
are given
We must incorporate mol calculations into
these questions
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Consider the following reaction:
2Mg(s) + O2(g)  2MgO(s)
ΔH=-1204kJ
1.
2.
3.
4.
Is the reaction exothermic or endothermic?
Calculate the amount of heat transferred when 3.55g of
Mg(s) reacts at a constant pressure.
How many grams of MgO(s) are produced during an
enthalpy change of -234kJ?
How many kJ of heat are absorbed when 40.3g of MgO(s)
is decomposed into Mg(s) and O2(g) at constant pressure?
1.
2.
3.
4.
Exothermic (Δ𝐻 = −)
1𝑚𝑜𝑙 𝑀𝑔
24.3050𝑔 𝑀𝑔
3.55𝑔 𝑀𝑔
−234kJ
2𝑚𝑜𝑙 𝑀𝑔𝑂
−1204𝑘𝐽
40.3𝑔 𝑀𝑔𝑂
−1204𝑘𝐽
2𝑚𝑜𝑙 𝑀𝑔
40.3044𝑔 𝑀𝑔𝑂
1𝑚𝑜𝑙 𝑀𝑔𝑂
1𝑚𝑜𝑙 𝑀𝑔𝑂
40.3044𝑔 𝑀𝑔𝑂
= −87.9𝑘𝐽
= 15.7𝑔 𝑀𝑔𝑂
1204𝑘𝐽
2𝑚𝑜𝑙 𝑀𝑔𝑂
= +602𝑘𝐽
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Measurement of heat flow
◦ Carried out in/with a calorimeter by measuring the
magnitude of temperature change the heat flow
produces
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In this course we do calculations using a
specific type of calorimeter (coffee cup
calorimeter)
𝑞 = 𝑚𝐶∆𝑇
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Amount of heat required to raise its
temperature by 1K (or 1℃)
Greater C means more heat required top
produce a temperature increase
Sand vs. water
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Heat capacity for pure substances is usually
for a given amount
Specific heat is the amount of energy needed
to raise the temperature of 1g of a substance
by 1K.
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Calorimeters are vessels used to carry out
experiments that allow us to analyze
temperature changes
This allows us to see the energy being
created by a reaction
Bomb Calorimeter
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Used to study
combustion
Coffee Cup Calorimeter
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Used to study
principles of
calorimetry
◦ Mostly solutions
Constant volume
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Constant pressure
Bomb Calorimeter
Coffee Cup Calorimeter
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For calculations we use:
q = heat
m = mass
C = specific heat (or heat capacity)
ΔT = change in temperature

The specific heat of octane (C8H18) is
2.22J/g•K
1. How many joules of heat are needed to raise the
temperature of 80.0g of octane from 10.0 to
25.0℃?
2. Which will require more heat, increasing the
temperature of 1mol of C8H18 by a certain amount
or increasing the temperature of 1mol of H2O by
the same amount?
1.
2.
𝑞 = 𝑚𝐶∆𝑇 = 80.0𝑔
1𝑚𝑜𝑙 𝐶8𝐻18
1𝑚𝑜𝑙 𝐻2𝑂
2.22𝐽
𝑔∙𝐾
15𝐾 = 2.66 × 103 𝐽
114.22852𝑔 𝐶8𝐻18
1𝑚𝑜𝑙 𝐶8𝐻18
18.01528𝑔 𝐻2𝑂
1𝑚𝑜𝑙 𝐻2𝑂
2.22𝐽
𝑔∙𝐾
4.184𝐽
𝑔∙𝐾
=
=
253.59𝐽
𝐾
75.376𝐽
𝐾
It will take more energy to increase one mol
of C8H18 that one mol of H2O

When 50mL of 1.0M HCl and 50mL of 1.0M
NaOH is combine in a coffee cup calorimeter
the temperature of the resultant solution
increases from 21.0℃ to 27.5℃. Calculate the
enthalpy change for the reaction in kJ/mol
HCl, assuming that the calorimeter loses only
negligible quantity of heat, that the total
volume of the solution is 100mL, that its
density is 1.0g/mL, and that its specific heat
is 4.18J/g•K.
𝑚 = 100𝑚𝐿
1.0𝑔
= 100𝑔
1𝑚𝐿
∆𝑇 = 27.5℃ − 21.0℃ = 6.5℃ = 6.5𝐾
4.18𝐽
𝑞𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 𝑚𝐶∆𝑇 = 100𝑔
6.5𝐾
𝑔∙𝐾
=2717𝐽=2.7𝑘𝐽=-ΔH (exothermic)
0.050𝐿 𝐻𝐶𝑙
∆𝐻
𝑚𝑜𝑙
=
1.0𝑚𝑜𝑙 𝐻𝐶𝑙
1𝐿
−2.7𝑘𝐽
0.050𝑚𝑜𝑙
= 0.050𝑚𝑜𝑙𝐻𝐶𝑙
= −54𝑘𝐽/𝑚𝑜𝑙
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Used to give a visual representation of energy
changes throughout a reaction
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Things to remember
◦ Enthalpy is an extensive property
 Changes with the number of mols being reacted
◦ Enthalpy stays the same magnitude for the reverse
reaction but changes the sign
◦ Enthalpy change depends on the states of the reactants
and products
 Enthalpy of gases>liquids>solids
◦ +ΔH = endothermic (products higher than reactants)
◦ -ΔH = exothermic (products lower than reactants)
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Draw the enthalpy diagram for the following
equation:
2H2(g) + O2(g) 2H2O(g) ΔH = -483.6kJ
Remember to include:
◦
◦
◦
◦
◦
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Direction of increasing enthalpy
Products (with proper coefficients and states)
Reactants (with proper coefficients and states)
Direction of reaction
ΔH (heat of reaction) with proper sign
“Exothermic” or “endothermic”
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No matter how many steps a reaction has the
total enthalpy change is equal to the sum of
all enthalpy changes
◦ ΔHtotal rxn = ΔHall steps
Calculate ΔH for the reaction:
2C(s) + H2(g)  C2H2(g)
given the following chemical equations and their
respective enthalpy changes:

C2H2(g) +
5
O
2 2(g)
 2CO2(g) + H2O(l) ΔH=-1299.6kJ
C(s) + O2(g)  CO2(g)
H2(g) +
1
2
O2(g)  H2O(l)
ΔH=-393.5kJ
ΔH=-285.8kJ
2CO2(g) + H2O(l)  C2H2(g) +
2C(s) + 2O2(g)  2CO2(g)
H2(g) +
1
O2(g)
2
 H2O(l)
5
O
2 2(g)
ΔH=1299.6kJ
ΔH=-787.0kJ
ΔH=-285.8kJ
--------------------------------------2C(s) + H2(g)  C2H2(g)
ΔH=226.8kJ
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Things to remember
◦ Products and reactants should be on same side in
component equations as in net equation
◦ Mol ratio should add up to be the same in component
and net equations
◦ Reversing an equation changes the sign of ΔH for that
equation
◦ Add ΔH’s with appropriate signs
◦ When multiplying an equation by any number multiply
the ΔH as well