First Law of Thermodynamics
The Energy of the Universe is a Constant
Lect-12
If energy is the ability to do work, then work must be energy expended. Given 100% efficiency between energy expended
and work performed, energy expended would be equal to work performed. Work is defined as force times distance, and
has the same units as does energy. Work, in thermodynamics, is defined as all forms of energy, except heat. The
question to be addressed is: "How does a chemical reaction system do work?"
The answer depends on whether any substances are present in the GAS state. Consider the following development:
work = ( force ) * ( change in distance )
so:
and
pressure = ( force / area )
work = ( pressure ) * ( area ) * ( change in distance )
but : (area * change in distance ) = ( change in volume ) =
∆ V = ( Vf - Vi )
so work = P
∆ V = P( Vf - Vi )
When Vf > Vi, the system gas has expanded, ∆ V will be positive, and the system has done work on the surroundings.
It takes energy to do this, so the system lost energy in expanding the gas against the surroundings. The system lost
energy so work( system ) will have a negative sign.
When Vf < Vi, the system gas has been compressed, ∆ V will be negative, and the surroundings have done work on the
system. It takes energy to do this, so the surroundings lost energy in compressing the gas in the system. The system
gained this energy so work( system ) will have a positive sign.
Notice that the algebraic sign of work sys is opposite that of
∆ V; in an expansion ∆ V is positive, but work( system ) is
negative; and conversely for a compression. It is to account for this difference in sign behavior that chemical work is
work( system ) = − P∆V( system )
defined as follows:
The first law of thermodynamics deals with the conservation of energy and defines the total energy of a system (E) as
composed of a heat term (q), and all other forms of energy grouped together as work (w), in the expression E = q + w.
As before, absolute values of E cannot be measured, only differences in E's, i.e., so we use
∆E( system ) = q + w
∆E .
This expression is the mathematical form of the First Law of Thermodynamics.
The following development shows how the First Law of Thermodynamics is applied to practical matters.
Chemical work only occurs when there is a change in gas volumes. Therefore no chemical work can occur in processes
carried out under conditions of constant volume. Let thermal effects for a constant volume process be labeled q v , and
∆E = q v
note the important consequence that under this condition,
(because the work term will be zero)
Under more common laboratory conditions where volume can change but pressure is (relatively) constant at 1 atm, then
∆E = q p + w p
applies.
Continue the consideration with this set of conditions and this expression.
q
P∆V
As noted previously, heat effects for a constant pressure process,
as noted above
work( system ) = − P∆V( system ) .
Introducing these into the expression
Clearly,
∆E = q p + w p
p , lead to enthalpy (
H
) on a per mole basis, and
units (i.e., Liters-atm) must be ENERGY units.
leads to the following important relation:
∆E = ∆H − P∆V
.
This relation finds very practical application in making conversions from ∆E values to ∆H values. Heat effects are
commonly and conveniently measured in "bomb calorimeters". However, bomb calorimeters are constant volume
devices so they lead to ∆E values. But chemists need ∆H information because most laboratories operate at a
relatively constant pressure of 1 atm. How are ∆E values converted to ∆H values? The simple expression,
∆H = ∆E + P∆V is used to perform such conversions.
Consider one additional point in this regard. For cases of chemical reactions, it will become apparent that the P∆V
term is in general small relative to ∆E and ∆H . Furthermore, P∆V (as the work term) only contributes to ∆H
when there is a difference between volumes of gas products and gas reactants.
Recall the ideal gas law, PV = nRT . For a process at constant pressure and temperature, any change in volume is
accompanied by a change in moles of gas products compared to reactants. This is Avogadro's Hypothesis, i.e.,
P∆V = ∆nRT
Also recall that
∆E
and
∆H
where
∆n = ∆n(gas ) = [ n( gas products ) − n( gas reactants) ]
∆n( gas) can be read directly in
∆H = ∆E + P∆V becomes
are energy quantities on a per mole basis, and note that
moles from coefficients of gas substances in balanced chemical equations. So
∆H = ∆E + ∆n( gas) RT
The value already used for the gas constant
8.31
Joules
K − mole
R , ( 0.082
L − atm
) is not applicable here.
K − mole
Rather, it is replaced by
whenever Joule units are required. Consider the following problem:
ex. 13 When 0.9437 g of n-octane ( C8 H18 ) is burned in a bomb calorimeter, a temperature rise of 2.93deg.C is noted
and recorded. In a separate experiment the Heat Capacity of the calorimeter was determined to be
15.38 kJ/deg.C. Determine values of
a.
calorimeter
surroundings
b.
system
q
v,
∆E , ∆n( gas) , P∆V , work( system ) , and ∆H
for this process.
15.38 kJ
q(calorimeter) = C ∆ T = 
(for 0.9437 g cmpd)
(2.93 deg) = + 45.06 kJ
 deg 
the calorimeter experienced an increase in temperature so heat was added to it.
The process (w/r to the calorimeter) was endothermic and q(calorimeter) will
have a positive sign.
the reaction is the system, and it generated heat gained by the calorimeter. This makes
the magnitude of heat change for the system equal to that experienced by the
calorimeter. However, they are opposite in sign; i.e., endothermic (positive) for
surroundings means exothermic (negative) for the system…
q(sys) = − 45.06 kJ
q(sys) = - q(calorimeter)
(for 0.9437 g)
This reaction took place in a bomb calorimeter, so volume was constant. Accordingly,
q
c.
∆E
v = q(sys) = − 45.06 kJ
In a constant volume process
∆E
is directly related to
q
v ( b/c no work can be done),
and only differs in that the energy is expressed on a per mole basis…
?
∆E
 − 45.06 kJ   114 g 
= 

 = − 5444 kJ/mole
 0.9437 g  1 mole 
d.
∆n( gas)
C 8 H 18 (liquid) + 25/2 O 2 (gas) à 8 CO 2 (gas) + 9 H 2 O (liquid)
Note that a total of 8 moles of gas product ( 8 CO 2 ) are formed, and a total of 25 / 2
moles of gas reactant ( 25 / 2 O 2 ) are used. So …
∆n = ∆n(gas ) = [ n( gas products ) − n( gas reactants) ] = [ 8 - 25/2 ] = − 9 / 2 moles gas
Note, this is a compression because n( gas products) is less than n(gas reactants ) .
e.
P∆V
relate
P∆V
∆n( gas) RT
to
Note, the decrease in moles of gases (or equivalently, volume of gases) in this reaction,
so the system experienced a COMPRESSION. Thus
∆n( gas)
is negative. so
P∆V
must also be negative.
P∆V = ∆n( gas) RT
= ( − 9 / 2 mole)(8.31
J
)(298 deg) = − 11,144 J
mole − deg
Remember - units of P∆V (Liter-atmospheres) are energy units. So, it is not
surprising that gas constant values of R = 0.082 L-atm / K-mol , and
R = 8.31 Joules / K-mol, are equivalent because both specify units of
ENERGY / K-mol. The gas constant
f.
work( system )
recall that
R
can be cast in a variety of energy units.
work( system ) = − P∆V( system )
In order to compress system gas, the surroundings had to perform work on the
system, so the system gains energy.
work( system ) = - ( - 11,144 J)
g.
∆H
obtain
∆H
from
∆E
Thus,
work( system )
should be positive.
= + 11,144 J
by using the expression
∆H = ∆E + ∆n( gas) RT
All parts needed here have been already determined, simply substitute…
∆H = - 5444 kJ / mole
A reminder about the Practical Application of
+ ( - 11 kJ) = - 5455 kJ / mole
∆H = ∆E + ∆n( gas) RT
Bomb calorimeters offer straightforward measurement of thermal energies for chemical processes.
However, as a
constant volume device its measurements lead to total energy values ( E ), as opposed to enthalpies ( H ). Is there
some way to work with readily obtainable E values, and convert them to the H values that chemists' find so useful? Of
course there is. The expression
∆H = ∆E + ∆n( gas) RT
is used, as illustrated by problem 13 shown above.
most enthalpies values ( H ) started out as total energy values ( E )
∆H = ∆E + ∆n( gas) RT .
So
which were subsequently converted using
Ordinarily, chemists work under conditions of relatively constant temperature and pressure (about 25 degrees C and
1 atm. in most laboratories) rather than conditions of constant volume and markedly changing pressures (as met inside
bomb calorimeters), so thermochemical information tied to the former condition is of greatest practical use. Consequently,
enthalpy ( H ) is the energy term of importance to working scientists.
Enthalpies of formation, ∆H 0f , and reaction,
∆H 0Rxn , for hundreds of thousands of reactions at 25 deg.C and 1 atm. have been tabulated, and new enthalpy values
are constantly being added to this body of knowledge.
calorimeters.
Most probably began as
∆E values measured in bomb