Chemistry 30
Unit 2 – Thermochemistry
Law of Conservation of Energy – energy cannot be created or destroyed, only converted from one form to
another
potential energy (Ep) – stored energy (in intermolecular and intramolecular bonds)
kinetic energy (Ek) – energy in motion (in the movement of molecules & atoms)
energy – measured in Joules (J) – the ability to do work
temperature – the average Ek of the molecules (C or K) – no indication of the amount of energy STORED in
the molecules
heat – the transfer of thermal energy from the hot to cold object
specific heat capacity (c) – the amount of heat energy required to raise the temperature of 1 g of a substance by
1C (J/gC)
- substances are all different depending on thow they store energy -  intermolecular forces,  c
- water has a very high c – helps moderate temperatures on earth, in our bodies
- to increase temp, must increase motion (vibrational, rotational and/or translational)
- may also be measured using different units
- heat capacity (of a system) – for a specific mass & composition (J/C)
- molar heat capacity – amount of energy per mole (J/molC)
Heat calculation
Q=mcT
*** just make sure that the units you want are the units you get
(carry and cancel your units!)
Some additional definitions
system – includes reactants, products and solvents only
surroundings – everything else
enthalpy (H) – the heat content of a chemical system (cannot be measured directly)
enthalpy of reaction (H) – the change in the heat content of a system during a reaction
Types of Energy Changes & Measuring those Changes
1. Nuclear Changes
decreasing energy involved
2. Chemical Changes
3. Physical Changes
Phase Changes
Hsub = +6.03kJ/mol + 40.8 kJ/mol = +46.8 kJ/mol
endothermic
Hfus = +6.03 kJ/mol
Hsol = -6.03 kJ/mol
Hvap = +40.8 kJ/mol
Hcond = -40.8 kJ/mol
Hdep = -6.03kJ/mol - 40.8 kJ/mol = -46.8 kJ/mol
Chemistry 30 Notes – Unit 2
exothermic
page 1
The Effect of Heat on the Temperature of Water
Q=Hvap*n (Hvap=+40.8 kJ/mol)
Temp (C)
100
Q=mcT (c=4.19 J/gC), Ek
0
Q=mcT
(c=4.19 J/gC)
Ek
time
Q=Hfus*n (Hfus=+6.03 kJ/mol)
Q=mcT (c=2.01 J/gC), Ek
the graph above tracks changes in temperature (Ek) – can also track changes in Ep
-
H2O(g)
Ep (kJ)
H (kJ)
H=+40.8 kJ/mol
H=+6.03 kJ/mol + 40.8 kJ/mol
= + 46.8 kJ/mol
H2O(l)
H=+6.03 kJ/mol
H2O(s)
time
Calculations Involving Phase Change & mcT
- these calculations involve from 1 to 5 steps depending on how many phase changes are encountered
ex: calculate how much energy is released when 45.7 g of H2O(g) becomes a liquid with no change in
temperature
1
2
3
use only part 2
4
5
Q  Hfus  n
 40.8kJ 1mol
Q

 45.7g
mol
18.02g
Q  103kJ
ex: how much energy is required to completely vaporize 350g of ice initially at –35.0C?
4
2
5
3
use parts 1-4
1
Q3  mc T
(350g)( 4.19J)(100C)
Q3 
gC
Q3  1.47  10 2 kJ
Chemistry 30 Notes – Unit 2
Q1  mc T
Q2  Hfus  n
(350g)( 2.01J)(35C)
 6.03kJ 1mol
Q1 
Q2 

 350g
gC
mol
18.02g
Q1  24.6J
Q2  117kJ
Q4  Hvap  n
 40.8kJ 1mol
Q4 

 350g
mol
18.02g
Q4  792kJ
QT   Qn
QT  1.08  10 2 kJ
page 2
Other Changes in Enthalpy
- all chemical reactions may be discussed in terms of changes in enthalpy.
- Some “special ones”
Hfus = the amount of energy required to melt one mole of a substance (no change in temperature)
Hsol = the amount of energy released when one mole of a substance solidifies
Hvap = the amount of energy required to vapourized one mole of a substance
Hcomb = the amount of energy released when one mole of a substance is burned completely
H ….
the most important…
H f  the amount of energy released/consumed when one mole of a substance is formed in its standard state
from its elements in their standard state
standard state – the state (s, l, g) of an element/compound at 25C and standard pressure (see pp.6-7in the
data booklet for heats of formation)
ex: the heat of formation of NH4Cl(s)
½ N2(g) + ½ Cl2(g) + H2(g)  NH4Cl(s)
-
H= -314.4 kJ/mol
heats of formation are always described as kJ/mol, so you have to balance the equation with 1 mol of the
product (use fractions)
Hess’ Law (part 1)
An application of the “Law of Conservation of Energy”, it doesn’t matter how you get from reactants to products,
the change in enthalpy of reaction depends only on the products and reactants
ex: H2O(s)  H2O(g)
H=?
two steps:
H2O(s)  H2O(l)
H2O(l)  H2O(g)
H=+6.03 kJ/mol
H=+40.8 kJ/mol
H2O(s) H2O(g)
H=+46.8 kJ/mol
if you know the H for the steps to get to where you want to go, you can add the Hs together to get the final
H, as long as every chemical species you don’t want in the final equation cancel out.
ex: Calculate the H for the reaction 3 C2H2(g)  C6H6(l), given
1. C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(g)
2. 6 CO2(g) + 3 H2O(l)  C6H6(l)
-
H= -1299 kJ
H= +3267 kJ
since ethyne is present in the first equation and we want it as a reactant, copy the first equation as is…
but, since we need 3 ethynes, multiply the first equation (all reactants, products and H by 3)
since we want the second equation as is, copy it below
cancel anything that appears on reactant and product sides…
add the equations
3 C2H2(g) + 15/2 O2(g)  6 CO2(g) + 3 H2O(g)
6 CO2(g) + 3 H2O(l)  C6H6(l)
H= (-1299 kJ)3
H= +3267 kJ
3 C2H2(g)  C6H6(l)
H= -1299x3 + 3267
H= -630 kJ
Chemistry 30 Notes – Unit 2
page 3
ex: Calculate the heat of reaction (i.e. balance the equation using the smallest whole number ratios) for the
reaction 3 N2H4(l) + 4 ClF3(g)  3 N2(g) + 12 HF(g) + Cl2(g)
given:
1. N2H4(l) + O2(g)  N2(g) + 2 H2O(l)
2. 2 ClF3(g) + 2 NH3(g)  N2(g) + 6 HF(g) + Cl2(g)
3. 4 NH3(g) + 3 O2(g)  2 N2(g) + 6 H2O(l)
-
H=-622 kJ
H= -1196 kJ
H= +530 kJ
start with equation #1 (N2H4 appears in both the equation wanted and #1, only once), multiply by 3 (to match
the coefficient in the desired equation)
multipy equation #2 by 2 (to match the coefficient in the desired equation)
reverse equation #3, to enable the cancelling of the H2O(l) and the O2(g) (multiply the H by –1 (change the
sign) for the reversed equation)
3 N2H4(l) + 3 O2(g)  3 N2(g) + 6 H2O(l)
4 ClF3(g) + 4 NH3(g)  2 N2(g) + 12 HF(g) + 2 Cl2(g)
2 N2(g) + 6 H2O(l)  4 NH3(g) + 3 O2(g)
H=(-622 kJ)*3
H=( -1196 kJ)*2
H= -530 kJ
3 N2H4(l) + 4 ClF3(g)  3 N2(g) + 12 HF(g) + Cl2(g)
H= -2728 kJ
Hess’ Law Part 2
Since the number of steps don’t matter, as long as you have the reactants and the products and their Hf, then
you can determine the heat of reaction…
ex: Calculate the heat of combustion for octane (C 8H18(l))
- start with the balanced desired equation: C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g)
if you want to use all of the Hf reactions,
8 C(s) + 9 H2(g)  C8H18(l)
O2(g)  O2(g)
C(s) + O2(g)  CO2(g)
H2(g) + O2(g)  H2O(g)
Hf = -250.0 kJ/mol
Hf = 0 kJ/mol (all elements…)
Hf = -393.5 kJ/mol
Hf = -241.8 kJ/mol
switching around & multiplying….
C8H18(l)  8 C(s) + 9 H2(g)
25/2 O2(g)  25/2 O2(g)
8 C(s) + 8 O2(g)  8 CO2(g)
9 H2(g) + 9 O2(g)  9 H2O(g)
H = +250.0 kJ/mol
H = 0 kJ/mol*25/2
Hf = -393.5 kJ/mol * 8
Hf = -241.8 kJ/mol * 9
C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g)
Hcomb = [+250 + 25/2(0) –393.5*8 – 241.8*9 ] kJ/mol
Hcomb = -5074.2 kJ/mol
Is there an easier way than writing this out all of the time?
H= nHfp - nHfr (the change in enthalpy of the reaction is equal to the sum of the heats of formation of the
products multiplied by their coefficients from the balanced equation minus the sum of the
heats of formation of the reactants multiplied by their coefficients from the balanced
reaction)
* this makes sense, the H for any reaction is comparable to any other  (T = Tf – Ti, H = Hfinal-Hinitial)
Chemistry 30 Notes – Unit 2
page 4
ex: Calculate the amount of energy released when one mole of propane (C 3H8(g)) is burned completely.
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
Hcomb
-
= nHfp - nHfr
= [(3*-393.5 kJ) + (4*-241.8 kJ)] – [1(-103.8 kJ) + 5(0 kJ)]
= [-2147.7 kJ] – [ -103.8 kJ]
= -2043.9 kJ/mol C3H8(g)
from this information, a potential energy diagram may be drawn – the heats of the reactants are where
the reaction begins, the heats of the products are where the reaction ends. The H is the difference
between.
Ep
(kJ)
-103.8
rxn mechanism
C3H8(g) + 5 O2(g)
H =
-2043.9 kJ
-2147.7
3 CO2(g) + 4 H2O(g)
ex: Calculate the heat of formation of a nonane (C 9H20(l)) if the combustion of one mole released 5327.5 kJ of
energy.
C9H20(l) + O2(g)  CO2(g) + H2O(g)
balance the equation
C9H20(l) + 19/2 O2(g)  9 CO2(g) + 10 H2O(g)
(use fractions for the O2 )
use Hess’ Law – but the unknown isn’t Hcomb, it’s the Hformation of the fuel nonane.
H= nHfp - nHfr
-5327.5 kJ = [(9*-393.5 kJ) + (10*-241.8 kJ)] – [( Hf + (19/2*0 kJ)]
-5327.5 kJ = -5959.5 kJ - Hf
-632.0 kJ = Hf (nonane)
Chemistry 30 Notes – Unit 2
page 5
Determining Heats of Reaction Experimentally (Calorimetry)
- how are the heats of reaction determined?
- a method to measure the amount of energy consumed or released when a reaction occurs must be
devised
Calorimetry
the measurement of heat changes in reactions
uses a device called a calorimeter – a device that will absorb or release heat from/to the reaction
types of calorimeters:
1. Simple (solution) calorimeter
- made from a styrofoam cup (or two stacked) filled with H2O(l)
- placing a lid on the top will improve the “isolatedness” of the system
- must assume any heat from the reaction is absorbed by the water – not by the cup
- as long as there is not too much T, the results are fairly accurate
- can also be constructed from a tin (or aluminum) can –since metals readily absorb and
transfer heat energy, you can’t assume no heat is absorbed by the calorimeter, and its
specific heat capacity and energy absorbed/released MUST be included in the calculation
- used for heats of solution, fusion, neutralization, reaction (double replacement, single replacement),
specific heat capacity determination
2. Flame calorimeter
- used to measure heast of combustion (the amount of energy released when a mole of a
substance is burned)
- also used for “heats of combustion” for foods – “calorie content”
- the fuel is burned under a can, and heats the water in the can (** the can cannot be
excluded from the calorimeter – the amount of heat absorbed by the calorimeter = the
amount of heat absorbed by the water + the amount of heat absorbed by the can)
3. Bomb calorimeter
- provides as close to an isolated system as possible, containing all reactants and products in a sealed
container that is well insulated from the outside environment
- usually used to measure heats of combustion, but could also be used for other heats of reaction
4. Human calorimeter
- used to determine the amount of energy expended in physical exercise (used to predict calorie
consumption so it can be included with exercise equipment, etc.)
- in a closed, insulated room, have a person perform activities, measure the change in temperature
Calorimetry Problems
All calorimetry problems apply the Law of Conservation of Energy
Heat Lost = - Heat Gained
*** notice the negative sign (since heat lost + heat gained = 0, one is equal to the negative of the other)
to do calormetry problems, always divide them into the two parts
reaction side
H – combustion, reaction, solution, fusion…
mc T – for specific heat capacity or mixture problems
calorimeter side
ALWAYS mc T (or c T if the capacity of the
calorimeter doesn’t include the mass)
** may have two steps if the calorimeter is made of
metal
Chemistry 30 Notes – Unit 2
page 6
-
the only connection between the two sides of the calorimeter is the amount of heat transferred.
i.e. – if the reaction side lost 1.9 kJ (Q= -1.9 kJ), the calorimeter side must have gained 1.9 kJ (Q=+1.9
kJ)
*** PLEASE DON’T FORGET THAT IF THE REACTION IS ENDOTHERMIC (gains energy), IT HAD TO GET
THAT ENERGY FROM THE CALORIMETER (so the calorimeter loses energy – is EXOTHERMIC)
ex: (calculating the heat of fusion) A 70.0 g sample of cesium is sealed in a glass vial and is lowered into 250
mL of water at 90.00C. When the cesium had melted, the temperature of water had fallen to 88.98C.
Determine the molar heat of fusion (Hfus) for cesium.
-
separate the parts of the calorimeter and reaction
find Q for whichever side of the question you have enough information for
determine what is asked for in the question
reaction side (cesium)
m = 70.0 g
Hfus=?
molar mass = 132.91 g/mol
calorimeter side (water only)
m=250 mL (250 g)
c = 4.19 J/gC
T = -1.02C
You have enough info
to solve the calorimeter
side first
Q  mc T
Q=+1.07kJ (gained)
4.19 J
)( 1.09 C)
gC
Q  1.07kJ(lost )
Q  (250 g)(
Hfus  n  Q
Q
n
132 .91g
Hfus  ( 1.07kJ)(
)( 70 .0g)
mol
Hfus  2.03kJ / mol
Hfus 
Once you have Q, change the sign
(heat lost by the calorimeter is
gained by the reaction) and move it
to the reaction side.
ex: What is the heat of combustion of an unknown fuel (in J/g) if burning 0.75 g of fuel increased the
temperature of 175 mL of water in a 12.0 g metal calorimeter (c=0.700 J/gC) from 10C to 45C?
reaction side (combustion)
m = 0.75 g
Hcomb=?
Q=-26kJ (lost)
If the water gained energy
(endothermic), then the reaction
must be exothermic
Hcomb  m  Q
Q
m
Hcomb  ( 26kJ)(
Hcomb 
1
)
0.75 g
Hcomb  35kJ / g
Chemistry 30 Notes – Unit 2
Heat of combustion is
usually measured in kJ/mol,
but if the fuel is unknown
(or for complex foods),
comb is usually in kJ/g
or kJ/serving
calorimeter side (water & metal)
m=175 mL (175 g)
c = 4.19 J/gC
T = 35C
m=12.0 g
c = 0.700 J/gC
T = 35C
Q  mc T  mc T
4.19 J
0.700 J
Q  (175 g)(
)( 35 C)  (12 .0g)(
)( 35 C)
gC
gC
Q  26kJ(gained )
You have enough info to solve the
calorimeter side first – the
calorimeter consists of the water &
the metal can
page 7
ex: (Finding the final temperature of a mixture) If 75 mL of water with an initial temperature of 100C is mixed
with 135 mL of water with an initial temperature of 30C, find the final temperature
reaction side (water - mcT)
m = 75 mL (75 g)
c = 4.19 J/gC
Ti= 100C
You don’t have enough information
to solve for Q on either the
calorimeter or the reaction side, so
you have to use the original equation
Q lost = - Q gained
If the two parts of the mixture
are the same substance, the
specific heat capacities(c)
cancel making the math much
simpler!
calorimeter side (water - mcT)
m=135 mL (135 g)
c = 4.19 J/gC
Ti = 30C
Qlost  Qgained
mc (Tf  Ti )  mc (Tf  Ti )
4.19 J
4.19 J
(75 g)(
)( Tf  100 C)  (135 g)(
)( Tf  30 C)
gC
gC
75(Tf  100 C)  135 (Tf  30 C)
75 Tf  7500 C  135 Tf  4050 C
210 Tf  11550 C
Tf  55 C
Be careful with your signs –
if you get a final
temperature that doesn’t
make sense, check that you
didn’t miss the negative
sign!
ex: When solutions of an acid and a base are mixed, heat is released. Using a coffee cup calorimeter, 100 mL
of 1.00 mol/L hydrochloric acid are mixed with 100 mL of 1.00 mol/L sodium hydroxide solution. The initial
temperature of the solutions was 22.6C, and the temperature after mixing was 29.5C. Calculate the molar
enthalpy of solution for the reaction:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
reaction side (neutralization)
Hneut=?
Q=-5.78kJ (lost)
If the water gained energy
(endothermic), then the reaction
must be exothermic
calorimeter side (water)
m=200mL (200g)
c = 4.19 J/gC
T = 6.9C
Hneut  n  Q
Q  mc T
Q
Hneut 
n
Q  (200 g)(
1L
)( 0.100L )
1.00mol
Hneut  57 .8kJ / mol
Hneut  ( 5.78kJ)(
The mass of the water is the mass of the
solutions – a solution is made of two
parts – the solvent (the calorimeter) and
the solutes – the reactants
4.19 J
)( 6.9C)
gC
Q  5.78kJ(gained )
Use the concentration and
volume of the solution to
find moles – the molar
enthalpy for this reaction
will be per mol of NaOH (or
equally, per mol of HCl)
Chemistry 30 Notes – Unit 2
page 8