Unit 2: Energy &
Rates of Reaction
Thermochemistry
is the study of heat change in chemical
reactions.
Chapter 5
5.1: Energy: is the capacity to do work (J)
•
Potential energy is the energy available by virtue of an object’s
position
•
Kinetic energy is the energy of an object due to its motion
•
Thermal energy is the energy associated with the random motion
of atoms and molecules
•
Chemical energy is the energy stored within the bonds of
chemical substances
•
Nuclear energy is the energy stored within the collection of
neutrons and protons in the atom
5.1
Energy Changes in Chemical Reactions
Heat (q) is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature (T) is a measure of the average kinetic energy of
the particles in a sample
Thermal energy: the total quantity of kinetic and potential energy
in a substance
Temperature ≠ Thermal Energy
90℃
40℃
greater thermal energy
5.1
Law of Conservation of Energy: Energy cannot be created
or destroyed
The system is the specific part of the universe that is of interest
in the study.
SYSTEM
SURROUNDINGS
open
Exchange: mass & energy
closed
isolated
energy
nothing
5.1
Systems & Surroundings
• Energy cannot be created or destroyed…but it can change
forms and often it is ‘LOST’…as heat
• The total energy of a system and its surroundings
remains the same→LAW OF CONSERVATION OF ENERGY
• Chemical System: the reactants and the products
being studied
• Surroundings: all matter not included within the
system (reaction). Usually only consider anything
nearby that is capable of exchanging energy or
matter.
Types of systems:
• Open: both energy and matter can be transferred in and out of the
system (ex: BBQ)
• Closed: energy can enter and leave the system, but matter cannot
(ex: glow stick)
• Isolated: neither matter nor energy can move in or out of the
system.
• (Bond making is an) Exothermic process: is any process that
gives off heat – transfers thermal energy from the system to the
surroundings.
• (bonds broken < bonds made).
• It takes more energy to make bonds
• System loses potential energy; products have a lower potential
energy (and stronger bonds on average) than the reactants
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
5.1
Endothermic process is any process in which heat has to
be supplied to the system from the surroundings.
• (bonds broken > bonds made)
• takes more energy to break bonds
• System gains potential energy; products have a higher
potential energy (on average) and weaker bonds
than the reactants
energy + 2HgO (s) → 2Hg (l) + O2 (g)
energy + H2O (s )→ H2O (l)
Diagrams from p. 288-289
•
•
•
•
•
Nuclear Energy
All nuclear reactions are exothermic.
Per unit of mass, nuclear reactions release much more energy
than exothermic chemical reactions.
Two main types of Reactions: Fusion and Fission
A fusion reaction occurs when nuclei of small atomic mass
combine to form larger molecule.
2 H + 3 H → 4 He + 1 n
1
1
2
0
In fission, large nuclei with high atomic mass are split to smaller,
lighter mass by collision with a neutron:
235
92U +
1
0n →
92
36Kr +
141
1 n + 1.9 x 10 10 kJ/mol
Ba
+
3
56
0
See p. 290 Table 1
5.1
Nuclear Reactions
For each of the following situations state whether the:
• process is endothermic or exothermic
• system or surroundings gain energy
a) An electric current converts water into H2(g) and O2(g).
Endothermic or exothermic
system or surrounding gain
b) Wood burning in a fireplace is converted into gases and leaves ashes.
Endothermic or exothermic
system or surrounding gain
c) When ammonium chloride, a salt, is dissolved in water, the solution
and beaker become cold.
Endothermic or exothermic
system or surrounding gain
d) Iodine vapour changes directly to a solid.
Endothermic or exothermic
system or surrounding gain
e) A hydrogen molecule which is composed of two atoms held together
by electric forces is split into two separate atoms.
Endothermic or exothermic
system or surrounding gain
f) Positively charged ions coalesce with negatively charged ions and
form a solid ionic crystal.
Endothermic or exothermic
system or surrounding gain
h) Solid carbon dioxide (dry ice) is converted directly into the gaseous
phase.
Endothermic or exothermic
system or surrounding gain
Enthalpy and Enthalpy Change
What is enthalpy? Total amount of thermal energy
(kinetic + potential energy) in a substance (kJ)
• entHalpy… “Heat” of reaction
Enthalpy Change (∆H):
• Enthalpy cannot be measured, however energy changes
in physical and chemical Reactions
• The enthalpy change of the chemical system is equal to
the flow of thermal energy in and out of the system,
• ∆𝐻𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 > 0, Endothermic reaction
• ∆𝐻𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 < 0, Exothermic reaction
Energy Changes during Exothermic & Endothermic Reaction
in an Open System
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
5.2
Molar Enthalphy Change
Molar enthalpy, ∆𝐻𝑟𝑥𝑛 , the enthalpy change associated with a physical,
chemical, or nuclear change involving one mole of a substance.
Types of molar Enthalpies
Dissolution (∆Hsol)
NaBr(s) →Na+(aq) + Br-(aq)
Combustion (∆Hcomb) CH4(g) + 2O2(g) → CO2(g) + H2O(l)
Vapourization (∆Hvap) CH3OH(l) → CH3OH(g)
Freezing (∆Hfr)
H2O(l) → H2O(s)
Neutralization (∆Hneut) 2NaOH(aq) + H2SO4(aq) →Na2SO4(aq) + H2O(l)
Formation (∆Hf) C(s) +2H2(g) + 1/2O2(g) → CH3OH(l)
∆𝑯
𝒒
∆𝑯𝒓𝒙𝒏 =
∆𝑯𝒓𝒙𝒏 =
𝒎𝒐𝒍
𝒎𝒐𝒍
Where n is the amount in mols and ∆Hr is the molar enthalpy change of
the reaction
5.2
5.2
Molar Enthalpy Calculations
Calculate ∆H for Vaporization Reactions
Ethanol, CH3CH2OH(l), is used to disinfect the skin prior to an injection.
If a 1.00 g sample of ethanol is spread across the skin and evaporated,
what is the expected enthalpy change? The molar enthalpy of
vaporization of ethanol is 38.6 kJ/mol.
Given:
methanol=1.00 g; ∆Hvap=38.6 kJ/mol MMethanol= 24 +6 +16
46g/mol
Required = ∆H = q
Analysis= q = n∆Hvap
n=
1.00𝑔
46𝑔/𝑚𝑜𝑙
= 0.022 mol
q = 0.022 mol x 38.6 kJ/mol
= 0.849 kJ
5.2
Representing Molar Enthalpy Changes
The equations we use to represent energy changes are called
thermochemical equations.
• Endothermic enthalpy changes are reported as positive values;
H2O(l) → H2 + ½ O2(g) ∆Hdecomp = + 285.8 KJ/mol H2O
• Exothermic enthalpy changes are reported as negative values
H2(l) + ½ O2 (g) → H2O(l) ∆Hcomb = -285.8 KJ/mol H2
Energy changes can be communicated in four different
ways.
Three of them are thermochemical equations and one uses diagram.
1. By including an energy value as a term in the thermochemical
equation
e.g., CH3OH(l) + 3/2 O2 → CO2(g) + 2H2O(g) + 726 KJ
(Exothermic Reaction)
2. By writing a chemical equation and stating its enthalpy change
e.g., CH3OH(l) + 3/2O2(g)→CO2(g)+ 2H2O(g)
∆H = -726 KJ
(Exothermic Reaction)
3. By stating the molar enthalpy of a specific reaction
e.g., ∆Hcombustion or ∆Hc = - 726 KJ/mol CH3OH
5.2
4. Drawing a chemical potential energy diagram
• Exothermic
• Endothermic
Potential Energy Diagram
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every
1 mole of ice that melts at 00C and 1 atm.
Thermochemical Equations
H2O (s)
H2O (l)
DH = 6.01 kJ
5.2
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane that is
combusted at 250C and 1 atm.
Thermochemical Equations
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l)
DH = -890.4 kJ
5.2
Calculations in Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l)
•
H2O (l)
H2O (s)
DH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
5.2
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
H2O (l)
DH = 6.01 kJ
H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burns in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
DHc = -3013 kJ/mol
3013 kJ
= 6470 kJ
1 mol P4
5.2
Stoichiometry and Thermochemical Calculations
• Enthalpy is dependent on the quantity of the products.
Example: Aluminum reacts readily with chlorine gas to produce
aluminum chloride. The enthalpy of this reaction is −1408 𝑘𝐽. What
amount of heat is transferred when 1.0 kg of Al reacts completely
with excess Cl2?
2Al(s) +
3Cl2(g) 
2AlCl3 (s) +
1408 kJ
∆𝐻
𝑛
=
−1408 𝑘𝐽 2 𝑚𝑜𝑙 𝐴𝑙
𝑛𝐴𝑙 =
1000 𝑔
= 37 𝑚𝑜𝑙
26.98 𝑔/𝑚𝑜𝑙
∆𝐻
37 𝑚𝑜𝑙
=
−1408 𝑘𝐽 2 𝑚𝑜𝑙 𝐴𝑙
∆𝐻 = −26 048 𝑘𝐽
−1408 𝑘𝐽 refers to every 2 moles of Al(s)
𝑞 = ∆𝐻 = 𝑛∆𝐻𝑟
∆𝐻𝑟 =
∆𝐻
𝑛
−1408 𝑘𝐽
𝑘𝐽
∆𝐻𝑟 =
= −704
2 𝑚𝑜𝑙 𝐴𝑙
𝑚𝑜𝑙
𝑞 = 𝑛𝐴𝑙 ∆𝐻𝑟
𝑘𝐽
𝑞 = 37 𝑚𝑜𝑙 × −704
𝑚𝑜𝑙
𝑞 = −26 048 𝑘𝐽
The amount of heat transferred is 26 048 𝑘𝐽
5.2
5.2: Calorimetry and Enthalpy
The heat capacity (c) of a substance is the amount of heat (q)
required to raise the temperature of a given quantity (m) of the
substance by one degree Celsius.
The specific heat (c) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the substance
by one degree Celsius.
Heat (q) absorbed or released:
q = mcDt
Dt = tfinal - tinitial
5.2
Calorimetry: the measurement of heat into or out of a
system for chemical and physical processes.
the heat released = the heat absorbed
qsurroundings = - qsystem
The device used to measure the absorption or
release of heat in chemical or physical processes is
called a “Calorimeter”; measure enthalpy changes
for reactions.
It works by insulating the system form the
surroundings (ex. Styrofoam cup).
5.2
a “Calorimeter”
Calorimetry Calculations
When analyzing data obtained using a calorimeter, make
these assumptions:
• Any thermal energy transferred from the calorimeter to
the outside environment is negligible.
• Any thermal energy absorbed by the calorimeter itself is
negligible.
• All dilute, aqueous solutions have the same density (1.00
g/mL) and specific heat capacity (4.184 J/(g∙°C)) as water
• thermal energy absorbed or released by a
chemical system is given the symbol q
. qwater = mcDt
5.2
Calculations Involving Thermal Energy Transfer
Thermal Energy transfer by metal
How much heat is given off when an
869 g iron bar cools from 940C to
50C?
Solution:
Given:
c of Fe = 0.444 J/g • 0C
m = 869 g
Dt = tfinal – tinitial = 50C – 940C = -890C
q = mcDt
= 869 g x 0.444 J/g • 0C x (-890C)
= -34,000 J
No heat enters or leaves! = -34 kJ
5.2
Calculations of metal in water
A student places 50.0 mL of liquid water at 21.00 °C into a coffee-cup
calorimeter. She places a sample of gold at 100 °C into the
calorimeter. The final temperature of the water is 21.33 °C. The
specific heat capacity of water is 4.18 J/g∙°C and the density of water,
d, is 1.00 g/mL. Calculate the quantity of thermal energy, q ,
absorbed by the water in the calorimeter.
SOLUTION:
Given: VH2O = 50.0 mL; c=4.18 J/(g∙°C); dH2O=1.00 g/mL;
Tinitial=21.00 °C;Tfinal=21.33 °C
Required: Quantity of thermal energy absorbed, q
𝑚
𝐷=
𝑣
m= vd, m = 50mL x 1.00g/mL
= 50 g
∆t = 21.33oC – 21.00oC
5.2
q = mc∆t
=50 g x 4.18 J x 0.33oC
g. oC
= 69 J
Determining specific heat capacity of a substance
Using the value of q above, calculate the specific heat capacity, c, of
gold if its mass is 6.77 g. Final temperature of the gold is same as that
of water in the calorimeter.
Given:mAu =6.77 g; qsys = -69 J; t initial =100 oC; t final = 21.33oC
∆t =21.33 oC – 100oC
= - 78.67 oC
𝑞
𝑐=
Required: cAu
𝑚∆𝑇
−69𝐽
q = mc∆t
𝑐=
Rearrange and solve for c
(6.77 𝑔)(−78.67℃)
𝐽
𝑐 = 0.129
𝑔℃
Thermal Energy Transfer during a Neutralization reaction
A 50.0 mL sample of a 1.0 mol/L aqueous solution of hydrochloric acid, HCl(aq),
was mixed with 50.0 mL of a 1.0 mol/L aqueous solution of sodium hydroxide,
NaOH(aq), at 25.0 °C in a calorimeter. After the solutions were mixed by stirring,
the temperature was 31.9 °C.
(a) Determine the quantity of thermal energy transferred by the reaction to the
water (surroundings), q , Assume that the specific heat capacity and density of
both solutions is the same as that of liquid water
(b) State whether the reaction was endothermic or exothermic.
Given: Total volume = 100 mL; c water= 4.18J/g. oC
dwater = m =1g/mL
v
m = vd; msolution = 100 g
∆t = 31.9 oC – 25.0 oC
= 6.9 oC;
q surr= mc∆t
=100 g x 4.18 J x 6.9oC
g• oC
= 2.9 x103 J
= + 2.9 kJ
q surr= -qsystem
+2.9 kJ = -2.9 kJ
Exo.