Download Thermochemistry and Measuring Energy Change Complete NOTES

Thermochemistry
Nearly every chemical reaction occurs with either the absorption or
release of energy. The study of such energy transfers is
thermochemistry.
System refers to what is being studied. In chemistry, the balanced
chemical equation indicates the chemical system.
Everything else in the universe is the surroundings. Often the chemical system is separated from
surroundings by a boundary, which prevents energy transfer and insulates the system (e.g. styrofoam).
 The flow of energy is controlled, allowing researchers to measure it.
Energy
Energy is defined as the ability to do work, is measured in joules (J). The first law of thermodynamics
states that energy cannot be created or destroyed but only change forms (e.g. thermal energy can be
converted to chemical energy).
Chemical energy describes the potential energy that is contained within a
compound. A more reactive substance contains greater chemical potential
to undergo a reaction. A substance that is very stable contains little
chemical potential energy.
In any type of change (i.e. physical, chemical, or nuclear), the energy
transfer can be determined by the amount of heat absorbed or released by
the system.
 This was initially measured in calories (cal) where 1 calorie is the
amount of energy required to raise 1 g of water 1oC.
1 cal = 4.184 J
Measuring Energy Changes: Calorimetry
Calorimetry is the technical process of measuring energy changes in a chemical system.
During a chemical change where energy is given off, the surroundings will be warmed. If energy is
absorbed, the surroundings will be cooled.
 the energy can be transferred to or absorbed from a particular amount of water causing its
temperature to rise
q = mcT
q is the amount of heat transferred (J)
m is the mass of substance heated (g)
T is the temperature change (oC)
c is the specific heat capacity (J/goC)
Specific heat capacity is a property of a substance that describes the amount of heat required to raise
the temperature of 1 g of the substance by 1oC.
 the specific heat capacity of water is 4.184 J/goC (arising from the value of 1 calorie)
Enthalpy

The chemical potential energy absorbed or released by a chemical system when it changes from
reactants to products is the enthalpy change.
 The enthalpy of reaction is the change in potential energy of a chemical system (sometimes
referred to as the “heat of reaction” or “change in heat”).

In an exothermic reaction, the enthalpy change for the system has a negative value, as chemical
potential energy is lost by the system.
 placed on products’ side of chemical equation

In an endothermic reaction, the enthalpy change for the system has a positive value, as chemical
potential energy is gained by the system.
 placed on reactants’ side of chemical equation
Molar Enthalpy (Hx)

The enthalpy change associated with one mole of a substance undergoing any change is called the
molar enthalpy.
 expressed in kilojoules per mole (kJ/mol)

If the enthalpy change is written with the balanced chemical equation, the value MUST correspond
to the molar coefficients in the equation (i.e. the value needs to be converted to molar enthalpy).

Hvap = 40.8 kJ/mol of H2O(l)
Evaporation of water (physical change)
1H2O(l)  1H2O(g)
H = 40.8 kJ
Incorporated into equation (on reactant side)
1H2O(l) + 40.8 kJ  1H2O(g)
2H2O(l) + 2(40.8 kJ)  2H2O(g)
Evaporation of 2 moles of water
2H2O(l)  2H2O(g)
H = 81.6 kJ
Condensation of water (reverse process)
2H2O(g)  2H2O(l)
1H2O(g)  1H2O(l)
H = -81.6 kJ
H = -40.8 kJ
Incorporated into equation (on product side)
1H2O(g)  1H2O(l) + 40.8 kJ
Combustion of methane gas
1CH4(g) + 2O2(g)  1CO2(g) + 2H2O(g)
H = -890 kJ
1CH4(g) + 2O2(g)  1CO2(g) + 2H2O(g) + 890 kJ
1CH4(g) + 2O2(g)  1CO2(g) + 2H2O(g)
H = -890 kJ (for 1 mol of CH4(g))
H = -890 kJ (for 2 mol of O2(g))
H = -890 kJ (for 1 mol of CO2(g))
H = -890 kJ (for 2 mol of H2O(g))
Combustion of 2 moles of methane gas
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) + 2(890 kJ)
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g)
H = -1780 kJ (for 2 mol of CH4(g))
Combustion of ½ mole of methane gas
1/2CH4(g) + 1O2(g)  1/2CO2(g) + 1H2O(g)
H = -445 kJ (for 0.5 mol of CH4(g))
Incorporated into equation (on product side)
1CH4(g) + 2O2(g)  1CO2(g) + 2H2O(g) + 890 kJ
1CO2(g) + 2H2O(g)  1CH4(g) + 2O2(g)
H = 890 kJ
A visual representation of chemical energy changes can be used to portray a chemical reaction.
 potential energy diagrams
Potential Energy Diagrams
Potential Energy Diagrams
products
stored
chemical
energy
is high
Energy is released
during the
chemical change.
products
Time
Energy
Energy
reactants
stored
chemical
energy is
low
stored
chemical
energy
is low
stored
chemical
energy is
high
Energy is absorbed
during the
chemical change.
reactants
Time
The value of the energy change shown in the diagram should reflect the molar coefficients of the
balanced chemical equation.
Standard Molar Enthalpy of Reaction


the enthalpy change associated with the reaction of one mole of a substance at 100 kPa and a
specified temperature of 25oC (SATP conditions)
use the symbol Hxo for standard enthalpy
H = nHxo
and sinceH = q (opposite signs)*
nHo= mcT
*However, we MUST indicate the correct sign depending on whether the change is endothermic or
exothermic.
standard heat of formation
(measured at standard conditions)
A + B
C
+ D
+ energy
if endothermic
2A + 2B
D + C
2C
B
H o = value of enthalpy change
+ energy
if exothermic
+ 2D
+ A
H o = 2(value of enthalpy change)
H o = - (value of enthalpy change)
Hess's Law of Heat Summation
A state function is any physical property
whose value does not depend on the system's
history (e.g. temperature, pressure, volume).
 enthalpy change (H) is a state function
Enthalpy change is determined only by the
enthalpies of the initial and final states of the
chemical system and not by the path taken.
Consider,
C(s) + O2(g)
Ho = -393.5 kJ
CO2(g)
This could be written as two steps:
C(s) +
0.5O 2(g)
CO (g)
Ho = -110.5 kJ
CO (g)+
0.5O 2(g)
CO 2(g)
Ho = -283.0 kJ
(-110.5 kJ) + (-283.0 kJ) = -393.5 kJ
Law of Heat Summation
For any reaction that can be written in steps, the standard heat of reaction is the same as
the sum of the standard heats for the steps.
C(s) + O 2(g)
Increasing H
H o = -110.5 kJ
CO (g)+
H o =
0.5O 2(g)
-393.5 kJ
H o =
-283.0 kJ
CO 2(g)
Enthalpy diagram forC(s) + O 2(g)
Total H o = (-110.5 kJ)
+ (-283.0 kJ) = -393.5 kJ
CO 2(g)
Standard Enthalpy of Formation
The standard enthalpy of formation of a compound Hfo is the amount of heat absorbed or evolved
when one mole of the compound is formed from its elements in their standard states.
H2(g) + 0.5O2(g)
H2O(l)
Hfo = -285.9 kJ/mol
The Ho for a reaction can be obtained by subtracting the sum of the standard
enthalpies of formation of the reactants from the sum of the standard enthalpies
of formation of the products.
nHof (products)
Ho =
aA + bB + ...
Ho =
[nHfo(N) + mHfo(M) + …]
-
 nHof (reactants)
nN + mM + ...
-
[aHfo(A) + bHfo(B) +…]
Standard Enthalpy of Combustion
The experimental determination of Hfo values is often difficult or impossible.
Combustion of organic compounds involves the reaction with molecular oxygen to form gaseous
carbon dioxide and liquid water, as seen in this example with sucrose:
C12H22O 11(s)+ 12O 2(g)
12CO2(g)+ 11H2O(l)
Hocombustion = -5640.9 kJ/mol
The enthalpy change for the combustion of one mole of a compound under standard conditions is
called the standard enthalpy of combustion. This value can be experimentally determined using a
bomb calorimeter (q = Ct, where the heat capacity of the bomb, C, must be known).
Because the Hfo for O2(g), CO2(g), and H2O(l) are known, the standard enthalpy of formation of sucrose
can be found.

(1mol)Hocomb = [(11 mol)Hfo(H2O) + (12 mol)Hfo(CO2)]
- [(1 mol)Hfo(C12H22O11) + (12 mol)Hfo(O2)]
(1 mol)(-5640.9 kJ/mol) = [(11 mol)(-285.8 kJ/mol) + (12 mol)(-393.5 kJ/mol)]
- [(1 mol)Hfo(C12H22O11) + ( 12 mol)(0 kJ/mol)]
-5640.9 kJ
= [(-3144 kJ) +(-4722kJ)] – [(1 mol)Hfo(C12H22O11]
(1 mol)Hfo(C12H22O11) =

Hfo(C12H22O11)
=
-2225 kJ
-2225 kJ/mol (units become “per mole” in the end)