AP Chemistry

Thermochemistry



the branch of chemistry concerned with the quantities of HEAT evolved or absorbed during chemical reactions
 AP Chemistry

First Law of Thermodynamics


“Law of Conservation of Energy”

Energy can neither be

created or destroyed
– it is only transferred between the
system and the surroundings.
 AP Chemistry

Enthalpy (ΔH°)
• the thermodynamic function that accounts for heat flow.
• At constant pressure, the enthalpy is equal to the energy flow as
heat
ΔH = Hproducts – Hreactants
Energy (Heat) lost by the system to surroundings = Exothermic
Energy (Heat) gained from the surroundings to system=
Endothermic
ΔH = + for endothermic reactions (Energy of products > Reactants)
ΔH = - for exothermic reactions (Energy of reactants > products)
 AP Chemistry

Practice 1
A chemical reaction that gives off heat to its surroundings is said
to be endothermic/exothermic and has a positive/negative value
of ΔH.
 

As an ice cube melts under atmospheric pressure, is the process

endothermic or exothermic and what is the sign of ΔH?
 AP Chemistry

Practice 1- ANSWER
A chemical reaction that gives off heat to its surroundings is said
to be endothermic/exothermic and has a positive/negative
value of ΔH.
 

As an ice cube melts under atmospheric pressure, is the process

endothermic or exothermic and what is the sign of ΔH?
Melting (changing from a solid to a liquid) requires heat to be
absorbed by the ice from the surrounding and, therefore, is
endothermic. The sign of ΔH is positive.
 AP Chemistry

Potential Energy Diagrams

• PE Diagrams (or Enthalpy
Diagrams) can be used to
determine if a reaction is endo
or exothermic.
• By comparing the PE of the
reactants to the products, one
can determine the components
with the most potential energy
and determine is the net energy
is positive or negative – meaning
is energy as heat released or ΔH = Hproducts – Hreactants
absorbed by the system ΔH = 20kJ – 40kJ = -20kJ
Exothermic Reaction
 AP Chemistry

Practice 2
Given the following graph, is
the reaction endothermic or
exothermic? Justify your
answer with a calculation of
ΔH.
Indicate what value is
represented (a), (b), and (c).
 AP Chemistry

Practice 2- ANSWER
Given the following graph, is the
reaction endothermic or exothermic?
Justify your answer with a calculation
of ΔH.
Endothermic because the PE of the
products is greater than the
reactants so heat must be added to
get this reaction to work. ΔH =
300-100 = +200
Indicate what value is represented (a),
(b), and (c).
a- Activation Energy of the forward
reaction
b- Activation Energy of the reverse
reaction
c- Enthalpy of the forward reaction
(+200)
 AP Chemistry

Calculating Enthalpy values from Data

There are five ways to calculate Enthalpy:
• Stoichiometry
• Hess’s Law of Summation
• Summation Equation
• Using Bond Energies
• Calorimetry

Expect to see questions on ALL of these methods!

Note: This is probably not the order they are covered in the textbook.
 AP Chemistry

Units of Enthalpy
The SI unit for energy is the joule (J).
A joule (J) is defined as the kinetic energy of a 2kg mass moving at a speed
of 1m/s.
Ek= ½mv2 = ½(2kg)(1m/s)2 =1kgm2/s2 = 1J
A joule is not a large amount of energy, and we will often use kilojoules
(kJ).

Enthalpy is an extensive property meaning ΔH is proportional to the amount

of reactant consumed in the process. More reactant will generate more
heat.
Therefore, the ΔHrxn = ΔH/1mol of reaction and the units for ΔH are usually
kJ/mol.
 AP Chemistry

Calculating Enthalpy using Stoichiometry

For reactions that the enthalpy have been experimentally determined,
the value of the heat released or absorbed by a specific situation can be
calculated using stoichiometry. The stoichiometric ratios become
important here.
For the combustion of methane:
CH4 + 2O2→ CO2 + 2H2O ΔHrxn= -890kJ/mol
If 1 mole of CH4 and 2 moles of O2 are burned, then 890kJ of heat will be
released. (The reaction is exothermic because the ΔH is negative)
 
The equation could also be written:
CH4 + 2O2→ CO2 + 2H2O + 890kJ
This also shows the reaction is exothermic because heat is a product.
 AP Chemistry

Calculating Enthalpy using Stoichiometry

CH4 + 2O2→ CO2 + 2H2O ΔHrxn= -890kJ/mol
Given the reaction above, how much heat is released if 8.2g of
methane gas are burned in a constant pressure system?
8.2g CH4 x 1molCH4 x -890kJ = -460kJ
16.0g 1molCH4
or 460kJ of heat is released
 AP Chemistry

Calculating Enthalpy using Stoichiometry

CH4 + 2O2 → CO2 + 2H2O ΔHrxn= -890kJ/mol
How much heat is released if 12.3g of oxygen is burned in the presence
of methane gas?
12.3gO2 x 1mol O2 x -890kJ = -171kJ
32.0g 2molO2
(or 171kJ of heat is released)

*Watch the stoichiometry! The reaction requires 2 moles of O2 to

produce 890kJ of heat.
 AP Chemistry

Practice 3
2NO + O2 → 2NO2 ΔHrxn= -112kJ/mol

If 35.0g NO reacts with 42.0g of O2, how much heat will be

produced?
 AP Chemistry

Practice 3- ANSWER
2NO + O2 → 2NO2 ΔHrxn= -112kJ/mol

If 35.0g NO reacts with 42.0g of O2, how much heat will be produced?
Use the Limiting Reactant to determine the amount of heat produced!

35.0g NO x 1mol NO x -112kJ = -65.3kJ

30.01g 2mol NO
42.0g O2 x 1mol O2 x -112kJ = -147kJ
32.00g 1mol O2
Because the NO is the limiting reactant, only 65.3kJ of heat are produced.
The statement is produced because the reaction is exothermic.
 AP Chemistry

Hess’s Law of Summation

Hess’s Law states:
if a reaction is carried out in a series of steps, ΔHrxn for the overall reaction
equals the sum of the enthalpy changes for the individual steps.

The reaction of nitrogen and oxygen to produce nitrogen dioxide can be written in
one step:
N2 + 2O2 → 2NO2 ΔHrxn= 68kJ/mol
Or a series of steps:
N2 + O2 → 2NO ΔH1= 180kJ/mol
2NO + O2 → 2NO2 ΔH2= -112kJ/mol
Net reaction:
N2 + 2O2 → 2NO2 ΔHrxn= ΔH1 + ΔH2= 68kJ/mol
 AP Chemistry

Hess’s Law of Summation

To use Hess’s Law to compute enthalpy changes, it is important
to understand two characteristics of ΔHrxn:
• If a reaction is reversed, the sign of ΔHrxn is also reversed.
“FLIP”
• If an exothermic reaction is reversed, then it becomes endothermic –
hence the sign change or flip.
• The magnitude of ΔHrxn is directly proportional to the
quantities of reactants and products in a reaction.
• If the coefficients in a balanced reaction are multiplied, the value of
ΔHrxn is multiplied by the same value. “MULTIPLY”
 AP Chemistry

Hess’s Law of Summation

Calculate ΔHrxn for 2NO + O2 → N2O4 using the following information:

Reaction 1: N2O4 → 2NO2 ΔH1= +57.9kJ/mol

Reaction 2: 2NO + O2 → 2NO2 ΔH2= -112.0kJ/mol
 
Reaction 1: 2NO2 →N2O4 ΔH1= -57.9kJ/mol FLIP!
Reaction 1 must be flipped to make N2O4 a product and therefore the ΔH1 becomes -57.9kJ/mol.
Reaction 2: 2NO + O2 → 2NO2 ΔH2= -112.0kJ/mol
Reaction 2 has all reactants and products on the correct side with the correct coefficients.
When the two reactions are added together, the 2NO2 cancel out and you are left with the correct
overall reaction.
ΔHrxn = -57.9kJ/mol + -112.0kJ/mol = -169.9kJ/mol
The ΔHrxn for the overall reaction is -169.9kJ/mol.
 AP Chemistry

Practice 4
Calculate ΔH for the reaction:
2 C(s) + H2(g) → C2H2(g)
given the following chemical equations and their respective
enthalpy changes.
C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) ΔHrxn= -1299.6kJ/
mol
C(s) + O2 → CO2 ΔHrxn= -393.5kJ/mol
H2 (g) + ½ O2 (g) → H2O(l) ΔHrxn= -285.8kJ/mol
 AP Chemistry

Practice 4 - ANSWER
Calculate ΔH for the reaction:
2 C(s) + H2(g) → C2H2(g)
given the following chemical equations and their respective enthalpy changes.
C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O(l) ΔHrxn= -1299.6kJ/mol
C(s) + O2 → CO2 ΔHrxn= -393.5kJ/mol
H2 (g) + ½ O2 (g) → H2O(l) ΔHrxn= -285.8kJ/mol

2CO2(g) + H2O(l) →C2H2(g) + 5/2 O2(g) ) ΔHrxn= +1299.6kJ/mol FLIP

2C(s) + 2O2 → 2CO2 ΔHrxn= 2(-393.5kJ/mol) MULITPLY
H2 (g) + ½ O2 (g) → H2O(l) ΔHrxn= -285.8kJ/mol No change
2 C(s) + H2(g) → C2H2(g) ΔHrxn= +226kJ/mol
 AP Chemistry

Enthalpies of Formations (ΔH°f)

Tables of the experimentally determined values called the
enthalpy of formation (ΔH°f) can be used to determine the
enthalpy of a reaction.
It is important to note that a formation reaction is the formation
of 1 mol of a compound from its elements in their standard
states at standard conditions.
The “ ° ” on the (ΔH°f) indicates the standard conditions exist -
pressure is 1atm and temperature is 298K or 25°C.
By definition, the standard enthalpy of formation of the most
stable form of any element is ZERO.
 AP Chemistry

Formation Reactions
Example of a formation reaction:
C(s) + 2H2(g) + ½O2(g) → CH3OH(l) ΔH°f = -239kJ/mol
Note:
The elements are in there standard states:
C is solid at room temp
H2 is diatomic and a gas at room temp, etc
The equation is balanced for 1 mole of the compound which is
why the coefficient for the O2 is ½.
 AP Chemistry

Practice 5
Write the formation reaction for water.
 AP Chemistry

Practice 5 - ANSWER
Write the formation reaction for water.

2H2(g) + O2(g) → 2H2O(l)

 AP Chemistry

Calculating with

Enthalpies of Formations (ΔH°f)
The formation reactions can be used with Hess’s Law to
determine the enthalpy of a reaction or a simplified calculation
can be used:

ΔH◦rxn = ΣΔH°f(products) - ΣΔH°f(reactants)

The symbol Σ (sigma) means “the sum of” and the stoichiometric
coefficients cannot be forgotten.
 AP Chemistry

Calculating with

Enthalpies of Formations (ΔH°f)
Calculate the standard enthalpy change for the following reaction:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Use the equation given and looking up the enthalpy of formation (a table will usually
be given).
ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)
ΔH°rxn = [3(ΔH°f CO2(g)) +4(ΔH°f H2O(l))] – [1(ΔH°f C3H8(g)) + 5(ΔH°f O2(g))]
Substitute in the values from the table:
ΔH°rxn = [3(-393.5kJ) +4(-285.8kJ)] – [1(-103.85kJ) + 5(0kJ)]
ΔH°rxn = -2219.85kJ/mol

Note: the Stoichiometry coefficients are used, mathematical orders of operations are
followed and the Enthalpy of oxygen is 0 because it is an element!
 AP Chemistry

Practice 6
Calculate the standard enthalpy change for the combustion of 1
mol of benzene (C6H6) to CO2 and H2O(l).

Calculate the standard enthalpy change for the combustion of

ethanol.
 AP Chemistry

Practice 6 - ANSWERS
Calculate the standard enthalpy change for the combustion of 1 mol
of benzene (C6H6) to CO2 and H2O(l).
C6H6 + 15/2 O2 → 6CO2 + 3H2O(l)
ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)
ΔH°rxn = [6(ΔH°f CO2(g)) +3(ΔH°f H2O(l))] – [1(ΔH°f C6H6(g)) +
15/2(ΔH°f O2(g))]
ΔH°rxn = [6(-393.5kJ) +3(-285.8kJ)] – [1(+49.2kJ) + 15/2(0kJ)]
ΔH°rxn = -3267.6kJ/mol
 AP Chemistry

Practice 6 - ANSWERS
Calculate the standard enthalpy change for the combustion of ethanol
C2H5OH + 3O2 → 2CO2 + 3H2O(l)
ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants)
ΔH°rxn = [2(ΔH°f CO2(g)) +3(ΔH°f H2O(l))] – [1(ΔH°f C2H5OH(g)) +
3(ΔH°f O2(g))]
ΔH°rxn = [2(-393.5kJ) +3(-285.8kJ)] – [1(-278kJ) + 7(0kJ)]
ΔH°rxn = -1366.4kJ/mol
 AP Chemistry

Bond Energies and Enthalpy

Bond Energy values can be used to calculate the approximate enthalpies
for reaction.
In order to break a bond, energy must be added to the system
(endothermic). Why? Atoms are at their lowest potential energy levels
when bonds are formed and therefore more stable. To separate two
atoms (break a bond), energy must be added.
Conversely, when bonds are formed, atoms return to a stable state with
the lowest potential energy and the excess energy is release as heat
(exothermic).

Energy is required to break bonds, Energy is released when bonds

form.
 AP Chemistry

Bond Energies and Enthalpy Diagrams

This can be seen in the potential
energy diagram – the energy added
to break the bonds is represented
by the activation energy. When
products begin to form, energy is
released at atoms form more stable
relationships. The difference in the
energy absorbed to break the bonds
and the energy released when the
bonds are formed is the enthalpy of
the reaction.

Mathematically:
ΔH°rxn = ΣBond Energy (broken) – ΣBond Energy
 AP Chemistry

Calculating Enthalpy using Bond Energies

Calculate ΔH°rxn for the reaction:
H2(g) + F2(g) →2HF(g)
Given the following bond energies:
H-H F-F H-F
  432kJ/mol 154kJ/mol 565kJ/mol

ΔH°rxn = [1(432kJ/mol) + 1(154kJ/mol)] – [2(565kJ/mol)]

ΔH°rxn = -544kJ/mol
 AP Chemistry

Practice 7
Using bond energies, calculate the enthalpy change for the
combustion of ethanol.
 AP Chemistry

Practice 7 - ANSWER
Using bond energies, calculate the enthalpy change for the combustion of
ethanol.
C2H5OH + 3O2 → 2CO2 + 3H2O
ΔH°rxn = ΣBond Energy (broken) – ΣBond Energy (formed)
ΔH°rxn = [1(C-C) + 5(C-H) +1(C-O) + 1(O-H) + 3(O=O)]– [4(C=O) + 6(O-H)]
ΔH°rxn = [1(347) + 5(413) +1(358) + 1(467) + 3(498)]– [4(799) + 6(467)]
ΔH°rxn = [4731]– [5998] = -1267kJ/mol

*Note – this is the same as problem #6 so why are the answers different?
Bond Energies are measured for the GASEOUS state and so the water
produced is a gas. Heat is required to make the water for liquid to gas
(endothermic) therefore less heat is release from the reaction overall.
 AP Chemistry

Calorimetry
Calorimetry is the measure of heat flow. In the lab, this is one
way standard enthalpies are determined experimentally.

To raise the temperature of 1.00 gram of water 1.0°C, 4.184J

of heat is required (that the definition of one Calorie – what you
know of for food). This is referred to as the Heat Capacity or
Specific Heat.
 AP Chemistry

Calorimeter
The device used to measure heat flow
is a calorimeter.
In calorimetry, to measure heat flow,
we can measure the magnitude of the
temperature change as the heat flow
proceeds. There must be something in
the calorimeter to act as the
surroundings to absorb or release heat
to the system. Water is usually the
substance in a calorimeter.
Because the heat capacity of water is
known (C = 4.184J/g°C), the exchange
of heat between the system and the
surroundings can be determined.
 AP Chemistry

Measuring the heat flow

How is this used to determine standard enthalpy? Because the
heat is reactant dependent, if the amount of reactant is known
and the heat absorbed/released is measured, then the enthalpy
can be calculated.
qsolution = mCΔT
q = heat
m = mass of the solution
C = specific heat capacity of the solution
ΔT = Tfinal - Tinitial
 AP Chemistry

Calculating heat (q)

How much heat is needed to warm 300.g of water from 22°C to
98°C?

qsolution = mCΔT
qsolution = (300.g)(4.184J/g°C)(+76°C) = +95395.2J

Therefore, 95395.2J must be added to heat the water to 98°C.

 AP Chemistry

Using a Coffee Cup Calorimetry

Because the solution in the coffee cup is the “surroundings”, the
q of the “system” is equal in magnitude but opposite in sign
from the solution.
qsolution = -qrxn
Finally, the amount of heat released or absorbed by the reaction
is dependent of the limiting reactant and therefore ΔH°rxn =
qrxn /moles of Limiting Reactant.
ΔH°rxn = qrxn
mol LR
 AP Chemistry

Calculating using Calorimetry Data: 


A reaction
1.00L of 1.00M Barium Nitrate solution at 25.0°C is mixed with 1.00L of 1.00M
Sodium Sulfate solution at 25.0°C in a calorimeter. A solid forms and the
temperature of the mixture increases to 28.1°C.
Assuming the calorimeter absorbs only a negligible quantity of heat, that the
specific heat of the solution is 4.184J/g°C, and that the density of the solution
is 1.0g/ml, calculate the standard enthalpy change.
 
Determine the mass of the solution: 2.00L total x 1.0g/ml = 2000g
Calculate ΔT = 28.1°C-25.0°C = +3.1°C
Calculate qsolution = (2000g)(4.182J/g°C)(+3.1°C) = +25940.8J = +26000J
Determine qrxn (equal in magnitude to qsolution but opposite in sign) = -25940.8J
This makes sense because the temperature increased indicating an exothermic
reaction!
 AP Chemistry

Calculating using Calorimetry Data: 


A reaction (continued)
Now the limiting reactant must be determined:
Ba(NO3)2 + Na2SO4 → BaSO4 + 2NaNO3
1.00 mol of Barium Nitrate, 1.00 mol of Sodium Sulfate (either is
the LR)
Calculate ΔH°rxn = qrxn /moles of Limiting Reactant
ΔH°rxn = -25940.8J/1.00mol = -25940.8J/mol or -25.9408kJ/mol
ΔH°rxn = -26kJ/mol (exothermic!)
 AP Chemistry

Calculating using Calorimetry Data: 


Dissolving a Salt
When a 4.25g sample of solid ammonium nitrate dissolves in 60.0g of water in a
coffee-cup calorimeter, the temperature drops from 22.0°C to 16.9°C.
Calculate the enthalpy for the solution process. Assume that the specific heat
and density of the solution is the same as water.
 
Determine the mass of the solution: 4.25g +60.0 = 64.25g (the salt becomes
part of the solution)
Calculate ΔT = 16.9°C-22.0°C = -5.1°C
Calculate qsolution = (64.25g)(4.182J/g°C)(-5.1°C) = -1370.992J = -1400J
Determine qrxn (equal in magnitude to qsolution but opposite in sign) = +1370.992

This makes sense because the temperature decreased indicating an

endothermic reaction!
 
 AP Chemistry

Calculating using Calorimetry Data: 


Dissolving a Salt (continued)
The limiting reactant is the only reactant: Ammonium Nitrate
NH4NO3 (s) → NH4+(aq) + NO3-(aq)
4.25g NH4NO3 = .0531mol NH4NO3

Calculate ΔH°rxn = qrxn /moles of Limiting Reactant

ΔH°rxn = +1370.992J/.0531mol = +25826.27J/mol or
-25.82627kJ/mol
ΔH°rxn = +26kJ/mol (endothermic!)