tjBCNetwork
Thermochemistry Lecture
Jennifer Fang
1. Enthalpy
2. Entropy
3. Gibbs Free Energy
4. q
5. Hess’ Law
6. Laws of Thermodynamics
ENTHALPY – total energy in all its forms; made up of the kinetic energy of the motion of the
molecules
-
The symbol for enthalpy is H
-
We cannot measure H directly, but we can measure changes in enthalpy, represented by
the symbol ΔH where delta(Δ) = change
-
If change in enthalpy is positive then the reaction absorbed heat (endothermic)
-
If change in enthalpy is negative then the reaction released heat (exothermic)
-
Change in enthalpy is directly dependent on number of moles; for example, the change in
enthalpy for the reaction of 2O  O2 is twice as large as the change in enthalpy for the
reaction O  ½ O2
-
ΔH°f is enthalpy of formation; ie the amount of enthalpy required to form a pure
compound or molecule from its elements
-
Ex: C + 4H = CH4 has a ΔH°f of -74.9 kJ/mol
1. Given the equation 3 O2(g)→2 O3(g) ∆H = +285.4 kJ, calculate ∆H for the following
reaction. 3/2 O2(g) → O3(g)
 Since 3/2 O2(g) → O3(g) is ½ of 3 O2(g)→2 O3(g) the enthalpy of the reaction will
be ½ the enthalpy of the reaction 3 O2(g)→2 O3(g)
tjBCNetwork
 ½ (+285.4kJ) = +142.7kJ
ENTROPY – commonly referred to as disorder or randomness
-
The symbol for entropy is S, and the symbol for change in entropy is ΔS
-
Entropy of gases is greater than entropy of liquids and solids
-
Entropy is higher as temperature increases
-
Entropy of larger molecules is greater than entropy of smaller molecules
-
Δ Suniverse = ΔSsys + ΔSsurr where ΔSuniverse is the change in the entropy of the
universe and ΔSsys is the change in entropy of the reaction and ΔSsurr is the change in
entropy of the surrounding of the reaction
-
ΔSsurr of a spontaneous process is positive (entropy of the surroundings increases =
entropically favored = spontaneous)
-
ΔSsurr of a nonspontaneous process is negative
-
Most, but not all, exothermic processes are spontaneous giving them a positive value of S
-
ΔS°f is entropy; ie the amount of free energy required to form a pure compound or
molecule from its elements
-
Ex: C + 4H = CH4 has a ΔS°f of 186 J/mol
2. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Given:
S°NH3 = 193 J/K·mol
S°O2 = 205 J/K·mol
S°NO = 211 J/K·mol
S°H2O = 189 J/K·mol
ANSWER
The change in the standard molar entropy of a reaction can be found by the difference
between the sum of the molar entropies of the products and the sum of the molar
entropies of the reactants.
ΔS°reaction = Δ S°products - Δ S°reactants
ΔS°reaction = (4 S°NO + 6 S°H2O) - (4 S°NH3 + 5 S°O2)
ΔS°reaction = (4(211 J/K·K) + 6(189 J/K·mol)) - (4(193 J/K·mol) + 5(205 J/
tjBCNetwork
K·mol))
ΔS°reaction = (844 J/K·K + 1134 J/K·mol) - (772 J/K·mol + 1025 J/K·mol)
ΔS°reaction = 1978 J/K·mol - 1797 J/K·mol)
ΔS°reaction = 181 J/K·mol
GIBBS FREE ENERGY – the amount of work obtainable from a thermodynamic system at a
set temperature and pressure
-
The formula to determine change in Gibbs Free energy is ΔG = ΔH - T ΔS where delta H
is change in enthalpy, T is temperature in Kelvins, and delta S is change in entropy
-
G = H – TS is the formula to determine Gibbs Free Energy
-
ΔG° is the value of change in Gibbs Free energy at STP (standard temperature and
pressure of 25 Celsius and 1 atm)
-
A positive ΔG means that a reaction is spontaneous
-
The following table demonstrates the relation ship between the signs of ΔH and ΔS on
ΔG
ΔH
+
ΔS
+
+
-
+
-
ΔG
There will be a temperature above which entropy will win and make
the process spontaneous
Always positive
Always negative
There is a temperature below which enthalpy will win and make the
process spontaneous
-
ΔG°f is Gibbs free energy of formation; ie the amount of free energy required to form a
pure compound or molecule from its elements
-
Ex: C + 4H = CH4 has a ΔG°f of 50.6 kJ/mol
3. Calculate H° and S° and ΔG° for the following reaction at 298 K and
determine if it is spontaneous or nonspontaneous.
tjBCNetwork
N2(g) + 3 H2(g)
2 NH3(g)
ANSWER
Using a standard-state enthalpy of formation and absolute entropy data table,we
find the following information:
Compound
Hfo(kJ/mol)
S°(J/mol-K)
N2(g)
0
191.61
H2(g)
0
130.68
NH3(g)
-46.11
192.45
The reaction is exothermic ( H° < 0), which means that the enthalpy of
reaction favors the products of the reaction:
Ho = Δ Hfo(products) - Δ Hfo(reactants)
= [2 mol NH3 x 46.11 kJ/mol] - [1 mol N2 x 0 kJ/mol + 3 mol H2 x 0 kJ/
mol]
= -92.22 kJ
So = Δ So(products) - ΔSo(reactants)
= [2 mol NH3 x 192.45 J/mol-K] - [1 mol N2 x 191.61 J/mol-K + 3 mol
H2 x 130.68 J/mol-K]
= -198.75 J/K
ΔG° = ΔH - T ΔS = -92.22 kJ – (298K) (-198.75 J/K)
 BE CAREFUL: the units of change in entropy are commonly in joules while the units
of change in enthalpy are usually kilojoules, you have to convert one or the other to J/
kJ to be able to subtract one from the other
ΔG° = ΔH - T ΔS = -92.22 kJ – (298K) (-0.19875 kJ/K) = -32.9925 kJ which
means that this reaction is spontaneous
Q
4. Q of a reaction is equal to heat
5. The formula for q is q = mc∆T where m = mass, c = specific heat, and ∆T is the change
in temperature
6. Usually you solve for q as heat when you are determining how many calories or
kilocalories it takes to raise a certain substance a certain number of degrees
tjBCNetwork
7. C as specific heat is defined as the amount of heat per unit mass required to raise the
temperature by one degree Celsius
8. Usually c is express in units of calories/g *degrees Celsius
9. Some textbooks may also call specific heat Cp
4. How much heat, in calories and kilocalories, does it take to raise the temperature of
814g of water from 18.0 C to 100 C?
Answer
q = mc∆T = (814g)(1cal/g C)(100C – 18C) = 6.67x10^4 cal = 66.7 kcal
∆T in this case = +82 C because you are trying to raise the temperature that many degrees
HESS’ LAW – Hess's law states that the enthalpy or heat change accompanying a chemical
reaction is independent of the pathway between the initial and final states.
In other words, if a chemical change takes place by several different routes, the overall enthalpy
change is the same, regardless of the route by which the chemical change occurs (provided the
initial and final condition are the same).
10. Basically, reactions can be added up to find the total enthalpy change of a reaction that
we cannot measure the independent parts of
11. In a laboratory, it is impossible to find the change in enthalpy for the reaction C2H4(g) +
H2(g)  C2H6(g), but we can measure the enthalpy changes for the reactions involving
intermediates of those reactions and add those reactions up to find the total enthalpy
change for C2H4(g) + H2(g)  C2H6(g
5. Calculate the enthalpy change for the reaction
C2H4(g) + H2(g)  C2H6(g) ∆H = ?
Given:
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) ∆H = -1410.9kJ
2 C2H6(g) + 7O2(g)  4CO2(g) + 6 H2O (l) ∆H = -3119.4kJ
2H2(g) + O2(g)  2 H2O (l) ∆H = -571.6kJ
ANSWER:
tjBCNetwork
To figure this out, we need to rearrange all the equations and play with them a little bit to
make the intermediates cross out so that all we have left are C2H4, H2, and C2H6.
Our intermediates are O2, CO2, and H2O. Somehow we need to flip some equations or
multiply equations by certain coefficients to cross out all those intermediates.
Remember, when you flip an equation, delta H is multiplied by -1. When you multiply an
equation by a coefficient, delta H is multiplied by that coefficient.
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) ∆H = -1410.9kJ
(keep as is so we have exactly 1 C2H4 on the reactants side)
2 C2H6(g) + 7O2(g)  4CO2(g) + 6 H2O (l) ∆H = -3119.4kJ
(multiply by -1 and divide by two so we get exactly 1 C2H6 on the products side)
2H2(g) + O2(g)  2 H2O (l) ∆H = -571.6kJ
(divide by two to get exactly 1 H2 on the reactants side)
NEW REACTIONS:
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) ∆H = -1410.9kJ
2CO2(g) + 3H2O (l)  C2H6(g) + 7/2 O2(g) [∆H = -3119.4kJ TIMES -1 AND DIVIDED BY 2]
H2(g) + ½ O2(g)  H2O (l) [∆H = -571.6kJ DIVIDED BY 2]
There are 7/2 O2 on each side so we can cross that out, there are 3H2O on either side so we can
cross that out, and there are 2CO2 on each side so we can cross that out
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) ∆H = -1410.9kJ
2CO2(g) + 3H2O (l)  C2H6(g) + 7/2 O2(g) [∆H = -3119.4kJ TIMES -1 AND DIVIDED BY 2]
H2(g) + ½ O2(g)  H2O (l) [∆H = -571.6kJ DIVIDED BY 2]
Resulting equation = C2H4(g) + H2(g)  C2H6(g)
∆H = -1410.9kJ + 1559.7 kJ + -285.8 kJ = -137.0kJ
LAWS OF THERMODYNAMICS
1. The first law of thermodynamics states that energy cannot be created or destroyed.
tjBCNetwork
2. The second law of thermodynamics states that for any spontaneous process there is an
increase of entropy in the universe; and the entropy of the universe is always increasing.
3. The third law of thermodynamics states that the entropy of a perfect crystal at 0K is 0.