Thermochemistry
 The study of energy changes in chemical reactions
Nature of Energy
Two types of energy
1) Kinetic energy- energy of motion; atoms and molecules possess KE
KE = ½ mv2
2) Potential energy- stored energy; found within the bonds of substances
**********potential energy can convert to kinetic energy***********
Energy is the ability to do work or produce heat
 Work -moving an object against a force
 Heat- energy transferred between object b/c of a temperature
difference
o Always moves from hot to cold
State Function
 A property of a substance that depends only on its present state
 Independent of the pathway taken or how you get from point A to
point B
 Examples: internal energy, enthalpy, temperature, pressure
 Non examples: distance traveled, work, heat
Think of the Universe being broken down into:
System
part you are concerned
with
Example: a reaction in a beaker
Surroundings
the rest
System: reactants(chemicals)
Surroundings: the beaker and everything beyond it
Types of Reactions
Endothermic Reactions
 System absorbs E form the surroundings
 Temperature of the surroundings lowers
 products are generally less stable (weaker bonds) than reactants
 ΔH is positive (+)
 phase changes of melting, vaporization and sublimation are examples
Exothermic Reactions
 System releases E into the surroundings
 Temperature of the surroundings raises
 products are generally more stable (stronger bonds) than reactants
 ΔH is negative (-)
 phase changes of freezing, condensation are examples
Every Energy (E) measurement has 3 parts:
1. A unit : Joules (J) or calories (cal))
 SI unit is the joule; 1 J = 1 kgm2/s2
 Usually express in kilojoules
 a calorie is equal to amount of heat needed to raise the
temperature of 1 gram of water by 1C; not a very common unit
used anymore
 Conversion factor between joules and calorie
4.184J= l cal
*********1 nutritional Calorie = 1000 cal or 1 kcal*********
2. A magnitude
3. A sign to tell direction
Three Laws of Thermodynamics
1. Law of Conservation of Energy (1st law of thermo)
 energy is can neither be created nor destroyed but converted from 1
form to another
 energy lost by the system is gained by the surroundings and vice versa
∆E=q+w
∆E =internal energy- sum of all PE and KE of system ( Ef – Ei)
q = heat (q is positive in endothermic reactions: heat added and negative in
exothermic reactions: heat removed)
w = work (w is positive if work is done on the system and negative if
the system does work)
W = - P∆V
units of liter-atm ( L· atm)
1L· atm = 101.3 J
Example 1
If a gas expands from 46 L to 64 L at a constant pressure of 15 atm. How
much work is done?
Example 2
What is the internal energy if 50 J of heat is added and 20 J of work are done
on the system
Example 3
Enthalpy (H)
 Focus on changes of H not H itself
 ΔH = ΔE + PΔV
 Remember ΔE= q +w so ΔE= q - pΔV rearranged q = ΔE + PΔV
So at constant pressure ΔH = qconstant pressure
(change in enthalpy of a system is due only to energy flow as heat
ΔH = H products - Hreactants
 ΔH = (+)  endothermic
ΔH (-)  exothermic
 Usually expressed as kilojoules per mole (kJ/mol)
Calorimetry
 process that measures the transfer of heat
 use a calorimeter( an insulated container): usually the change in
temperature of water is measured but often the heat capacity of the
calorimeter is known since the calorimeter also absorbs some heat as
well as the water in the device.
Heat Capacity (C)
 Amount of heat energy needed to raise temperature by 1º Celsius
C =__q___
∆T
Specific Heat Capacity (c) or (s)
 amount of energy needed to raise 1 gram of a substance by 1º Celsius
c=_ q
rearranges to q= mc∆T
m · ∆T
 Units: J/g·°C or J/g·K or cal/g·ºC or cal/g· K
 Each substance has its own specific heat
Example
The specific heat of graphite is .71 J/gºC. Calculate the energy needed to
raise the temperature of 75 Kg of graphite from 294 K to 348 K.
Molar Heat Capacity
 amount of heat required to raise 1 mole of substance by 1°C
c = ___q
rearranges to q = nc ∆T
n ·∆T
Coffee Cup Calorimetry (constant pressure)
 used to determine ΔH for reactions in solution
 uses an insulated cup full of water as the calorimeter
 use q= mc∆T to solve
 Water has a specific heat of 4.184 J/g·C
 Assume:
Heat released in a reaction = heat absorbed in water
Negative heat
positive heat
 qrxn + qsoln = 0
 qrxn = - qsoln if endothermic
 qsoln = - qrxn if exothermic
Example
A 46.2 g sample of copper is heated to 95.4 ºC and then placed in a
calorimeter containing 75.0 grams of water at 19.6 ºC. The final
temperature of both the water and the copper is 21.8 C. What is the
specific heat of copper?
Example
Suppose you place .500 g of magnesium chips into a coffee-cup calorimeter
and then add 100.0 ml of 1.00 M HCl. The reaction occurs is
Mg(s) + 2HCl (aq) --> H2(g) + MgCl2(aq)
The temperature of the solution increases from 22.2 to 44.8º C. What is the
enthalpy change per mole of magnesium. Assume specific heat capacity of
the solution is 4.20 J/g·K and the density of the HCl solution is 1.00 g/ml.
Example
Assume you mix 200. ml of .400 M HCl with 200. ml of .400M NaOH in a
coffee-cup calorimeter. The temperature of the solutions before mixing was
25.10 C; after mixing and allowing the reaction to occur, the temperature is
26.60C. What is the molar enthalpy of neutralization of the acid? Assume
the densities of the solution are 1.00g/ml and their specific heat capacities
are 4.20J/gK.
Bomb Calorimetry (constant volume)
 Material is put into a chamber with pure oxygen. Wires start
combustion; the container is put into a container of water
 The heat capacity of the calorimeter is known and tested
 Since volume cannot change, no work (-P ΔV) is done.
ΔE =q
 The heat capacity of the calorimeter MUST be known
qbomb = (Ccal) (ΔT)
qbomb + qwater = - qreaction
Example
A bomb calorimeter has a heat capacity of 11.3kJ/ºC. When a 1.50 grams
sample of methane(CH4) was burned in excess oxygen, the temperature
increased by 7.3 ºC. When a 1.15 g sample of hydrogen gas was burned in
excess oxygen, the temperature increased by 14.3 ºC. Calculate the Energy
of combustion per grams of hydrogen and methane.
Example
A 1.00 g sample of table sugar (C12H22O11) is burned in a combustion
calorimeter. The temperature of 1.50 x 103 grams of water in the calorimeter
rises from 25.00C to 27.32 C. If the heat capacity of the bomb is 837 J/K
and the heat capacity of the water is 4.184 J/gK, calculate the heat evolved
per gram of sucrose and the heat evolved per mole of sucrose?
Hess’s Law
 Is a state function- independent of the path
 A reaction can be carried in a single step or in a series of steps.
 If carried out in a series of steps, we can add equations to come up
with the desired final product and thus add ΔH for the overall reaction

One step:

Two step:
N2(g) + 2 O2(g)  2NO2(g)
ΔH = 68 kJ
N2(g) + O2(g)  2NO(g)
ΔH =180 kJ
2NO(g) + O2(g)  2NO2(g)
ΔH =-112 kJ
_________________________________________________
N2(g) + 2 O2(g)  2NO2(g)
ΔH = 68 kJ
Two Rules to Remember:
 If the reaction is reversed, the sign of ΔH is reversed
N2(g) + 2 O2(g)  2NO2(g)
ΔH = 68 kJ
2NO2(g)  N2(g) + 2 O2(g)
ΔH = -68 kJ
 The magnitude of ΔH is directly proportional to the moles of
reactants and products. If the coefficient of a reaction are multiplied
by and integer, the ΔH is also multiplied by the integer
******when using Hess’s law, work by adding the equations to make it look
like the answer, the other parts will cancel out******
Example C(s) + 2 H2(g) CH4(g)
ΔH = ?
Eq 1 C(s) + O2(g)  CO2(g)
ΔH1 = -393.5 kJ
Eq 2 H2(g) + ½ O2(g)  H2O(l)
ΔH2 = -285.8 kJ
Eq 3 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
ΔH3 = -890.3 kJ
Example
Given O2(g) + H2(g)  2 OH(g)
ΔºH = +77.9 kJ
O2(g) 2O(g)
ΔºH = +495 kJ
H2(g) 2H(g)
ΔºH = +435.9 kJ
Calculate the ΔºH for this reaction
O(g) + H(g)  OH (g)
Thermochemical Equations
 Used to calculate the enthalpy released or absorbed in a chemical
reaction
 2 H2(g) + O2(g)  2H2O(l)
ΔH = -572 kJ
Example
How much heat is evolved for the production of 1.00 mol of water?
Example
How much heat is evolved when 4.03 g of hydrogen is reacted with an
excess of oxygen?
Standard Enthalpies of Formation (Δ°Hf)
 Is the change in enthalpy that accompanies the formation of 1 mole of
a compound from its elements in their standard states
 standard conditions for a compound is 1atm for gases, 1M for
solutions
 standard conditions for an element is 1 atm and 25°C ( solids and
liquids in normal state)
 There is a table of standard heats of formation Hf for reference in
the back of your book
 Any element in their standard state has a Δ°Hf of 0.
To Calculate Standard Enthalpy
 need to be able to write the a formation equation showing the
formation of 1 mole of a compound from its elements in their standard
states
Example
What is the equation for the formation of NO2?
Example
Write the equation for the formation of methanol, CH3OH
The standard enthalpy change for any reaction can be found using this very
important equation
ΣΔ°Hf products - ΣΔ°Hf reactants = °Hreaction
Example
Using the standard enthalpies of formation, calculate the standard enthalpy
change for the overall reaction that occurs when ammonia is bi\burned in air
to form nitrogen dioxide and water.
4NH3(g) + 7 O2(g)  4 NO2(g) + 6 H2O(l)
Given the standard enthalpy of reaction , use the standard enthalpies of
formation to calculate the standard enthalpy of formation of CuO
CuO(s) + H2(g)  Cu(s) + H2O(l)
°Hreaction = -129.7 kJ
Bond Energies and Enthalpy
 Bond energy is required to break a bond.
 Since the breaking of a bond is an endothermic process, the bond
energy is always a positive number.
 When a bond is formed, energy equal to the bond energy is released.
Σ Bond energy of Reactants – Σ Bond energy of Products = Δ°Hrxn
Example:
Bond
H–H
O=O
O–H
Bond Energy(kj/mol)
436
499
463
Find the Δ°Hrxn for the following reaction
2 H2(g) + O2(g)  2H2O(g)