11
Relations between ∆E and ∆H – Thermodynamic Processes
We can also calculate ∆E for any of the processes in the previous lecture.
Let’s assume that all products are either solids or ideal gases.
∆E = ∆H - ∆(PV). All changes in volume associated with work terms are caused
by gases. So ∆(PV) = ∆(nRT). Let’s assume that the process occurs at constant T. So all
contributions to the change in the PV product come from changes in the numbers of
moles of gaseous products and reactants.
∆(nRT) = RT∆n = RT(moles of gaseous products – moles of gaseous reactants)
Consider C(s) + ½O2(g) → CO(g), reaction 1. One mole of gaseous products, ½ moles
of gaseous reactants. ∆n = 1 – ½ = ½
∆E1
= ∆H1 – (½ mole) (8.31 J mole-1 K-1)(298 K) (1 kJ/1000 J)
= -110.5 – 1.24 kJ = -111.7 kJ mol-1. The correction is small, but significant.
Let’s add the reaction shown in the thermodynamic cycle shown last time:
∆H1
CO(g)
½O2(g) ∆H2
½O2(g)
C(s) + ½O2(g)
CO2(g)
∆H3
C(s) + ½O2(g)
→ CO(g)
CO(g) + ½O2(g)
→ CO2(g)
_____________________________
C(s) + O2(g)
→ CO2(g)
(1)
∆H1 = -110.5 kJ/mol
(2)
∆H2 = -283.0 kJ/mol
(3)
∆H3 = -393.5 kJ/mol
Reaction (3) is the sum of reactions (1) and (2), and the enthalpies add correspondingly.
The additivity of enthalpies implied by the independence of path of ∆H is called Hess’
Law.
12
So, we have three reactions, but we only need two independent changes in enthalpy to
specify all reaction chemistry.
We do not need to tabulate the heats of reaction for every conceivable combination of
reactants and products. Instead, we can consider a special thermodynamic cycle that
involves reactions that form specific chemical compounds from the elements in their
standard states.
First of all, let’s define the standard state of a substance: The standard state is the state of
matter of a given substance at a specified temperature and pressure.
A common set of temperature and pressure is 25 °C (298.15 K) and 1 atm. Under these
conditions, the standard state of elemental carbon is graphite. The standard state of H2O
under these conditions is liquid water.
Now elemental carbon may exist in other forms, including diamond and Buckyball.
Neither of these allotropic forms of carbon is the standard state, however.
A common mistake: people confuse standard state with specific values of
temperature and pressure. The standard state is not 298.15 K and 1 atm. It is
the state of matter of a particular substance at those (or perhaps other)
temperatures and pressures.
Let’s consider the following reaction:
Elements in their standard states → substance in its standard state ∆H°rxn for this
reaction is the standard enthalpy (or heat) of formation of the substance:
Example:
C(s) + 2 H2(g) + ½O2(g) → CH3OH( l )
∆H°rxn = -239.5 kJ/mol
This reaction tells us that the standard enthalpy of formation, ∆H°f of liquid methanol is
–239.5 kJ/mol
Information of this type is readily available at the National Institute of Standards and
Technology, via this web site:
http://webbook.nist.gov/chemistry/
DEMO this web site.
If we decompose a substance into elements at their standard state, then ∆H°rxn = - ∆H°f
Enthalpies of the elements in their standard states are zero.
13
So now, let’s go back to this cycle:
Elements
− ∑ (∆H )
0
f reac tan ts
∑ (∆H
Reactants
0
f products
)
Products
∆Hrxn0
The left leg of the upper cycle is just the reverse of the sums of the heats of formation of
the reactants. The right leg of the cycle corresponds to the cumulative heats of formation
of the products.
∆Hrxn0 =
∑ (∆H )
0
f products
− ∑ (∆H 0f ) reac tan ts
a couple of graphs that are useful in visualization:
Atoms
Elements
Sum of reactant bond
energies
Sum of product bond energies
-(Sum of reactant heats of
formation)
Sum of product heats of
formation
Reactants
∆Hf °
Products
14
We can see that a chemical reaction is exothermic if the sum of the heats of formation of
products is more negative (or less positive) than the sums of the heats of formation of the
reactants.
We also see that by taking the reactants and products to the limit of free atoms, we break
all of the bonds in the reactants and form all of the bonds in the products. The products
of exothermic reactions have stronger bonds than the reactants. Exothermic reactions are
driven by the formation of stronger bonds.
Note: actually measuring the heat released in the formation of a substance from the
elements is often very difficult, if not impossible. However, heat release in combustion is
a process that can be probed quite easily.
An example:
Suppose we have a hydrocarbon CxHy, and we wish to measure its enthalpy of formation.
Let’s combust this molecule, to the products CO2(g) and H2O( l )
Start with a balanced chemical equation:
CxHy + (x + y/4)O2(g) → xCO2(g) + ½y H2O( l )
∆H°combustion = x[∆Hf ° (CO2(g)] + ½y[ ∆Hf °(H2O( l )] - ∆Hf °(CxHy)
We measure ∆H°combustion, and heats of formation for CO2(g) and H2O( l ) are known. So,
the only unknown quantity is ∆Hf °(CxHy),
15
Thermodynamic Processes
Let’s put our knowledge of heat and work together in a cyclic process that interconverts
heat into work. This problem is a variation on Example 9.2. 2 moles of an ideal monatomic gas
are confined as indicated in the diagram.
A, 122 K
B, 366 K
2
X
X = positive
heat
absorbed
X
P, atm
1
C, 183 K
D, 61 K
10
V, L
30
16
Let’s calculate the work and the heat absorbed by the system for this cyclic process. First the
work. From A to B, the gas expands against a constant pressure.
A, 122 K
B, 366 K
2
X
X = positive
heat
absorbed
X
P, atm
1
C, 183 K
D, 61 K
10
V, L
30
WAB = -Pext ∆V = -2 atm × (20 L) = -40 L-atm × 101.3 J/( L-atm ) = -4.05 × 103 J
WBC = 0
WCD = --Pext ∆V = -1 atm (-20 L) = +20 L-atm × 101.3 J/( L-atm ) = + 2.03 × 103 J
WDA = 0
So, the net work done on the system is negative. The net work done by the system is
positive. Graphically, the net work done on the system is the negative of the inscribed area.
Now, to calculate positive heat absorbed, we need to know the temperatures at each
point.
TA = PAVA/nR = (2)(10)/[2(.0821)] = 122 K
17
TB = PBVB/nR = (2)(30)/[2(.0821)] = 366 K
TC = PCVC/nR = (1)(30)/[2(.0821)] = 183 K
TD = PDVD/nR = (1)(10)/[2(.0821)] = 61 K
The gas is heated on legs AB and DA.
qAB involves heating the gas from 122 K to 366 K at constant pressure
qDA involves heating the gas from 61 K to 122 K at constant volume
5
2
3
QDA = nCV∆T = 2( R)(122 - 61 ) = 3(8.31)(61) = 1521 J
2
QAB = nCP∆T = 2( R)(366 – 122 ) = 5(8.31)(244) = 10140 J
So, the next expenditure of energy in terms of heat absorbed is 11,661 J.
The work done by the system in this cyclic process is 2030 J.
Efficiency = work done by the system/heat absorbed by the system
= -(work done on the system)/heat absorbed
= (030/11661) × 100 = 17.4 %