Lecture 4: 1st Law: Chemical Reactions
Review:

State variables during p-V work (mechanical work)
and heat

State Variable during a phase transitions

Application of 1st Law for chemical reactions
Today:

Application to biochemical reactions
o Example: Biochemical oxidation of Sucrose
o Example: Biochemical oxidation of Glycine

Begin discussion of 2nd Law of thermodynamics
o Carnot cycle
Changes involved in chemical reactions
Consider a general reaction
n A A  n B B  nC C  n D D
H  H Pr oducts  H Re ac tan ts
H  nC H  n D H  n A H  n B H
o Heat effects depend on whether the reaction takes
place under conditions of PV work or not. If only
work involved is PV type, then ΔE=qV, ΔH=qP=
ΔE+Δ(PV). In general for gaseous reactant we
assume ideal gas law so: ΔH=ΔE+Δ(PV) becomes
ΔH= ΔE+Δ(nRT)
o If the reaction takes place with accompanying rise
in temperature or evolution of heat (q negative)
then the reaction is called as exothermic. Similarly,
if the reaction results in flow of heat from
surrounding into system (q positive) then the
reaction is called as endothermic.
o Ħ is the standard heat of formation at 25ºC. Tables
of heat of formation are in appendix A5-A7.
Generally these heats are determined using Bomb
calorimetry.
Temperature dependence of ΔH of a reaction
H  nC H C  nD H D  n A H A  nB H B
so
d H C 
d H D 
d H A 
d H B 
d H 
 nC
 nD
 nA
 nB
dT
dT
dT
dT
dT
 C P ,C  C P , D  C P , A  C P , B
int egrating
H (T2 )  H (T1 )  C P (T2  T1 )
Note this form is exactly same as we found for ΔH for the
phase change. However the meaning of specific heat change is
different.
H (T2 )  H (T1 )  C P (T2  T1 )
 H (T1 )  (C P ,C  C P , D  C P , A  C P , B )(T2  T1 )
 H (T1 )  (nC C P ,C  nD C P , D  n AC P , A  nB C P , B )(T2  T1 )
 H (T1 ) 

Pr oducts

ni C P ,i  Re ac tan t n j C P , j (T2  T1 )
Using these equation we can calculate the heat of reaction at
different temperature. Next we consider few examples of how
we can use these equations.
Oxidation of Sucrose.
Consider oxidation of sucrose by oxygen given by following:
C12 H 22O11  12O2  12CO2  11H 2O
f
H Sucrose
( s )  2222.1kJ mol 1
f
H CO
(aq)  413.8kJ mol 1
2
H Hf 2O (l )  285.83kJ mol 1
H Of2  0 kJ mol 1
H Oxidation  (12  413.8  11  285.83)  (2222.1  12  0) kJ mol 1
 5887.63 kJ mol 1  17.2 kJ gm 1
 4.11 kcal gm 1  4.11Calorie gm 1
Average person needs about 8000kJ-10000kJ per day for
sustenance. This corresponds about 450-600 gms of sugar if
all the energy is derived from sugar. Note the magnitude of
nutritional Calorie is different from thermal calories!
What is important to note is that the energy released, as
calculated, is the same whether it involves chemical or
biochemical oxidation mediated by series of enzymatic
processes. This is because H is a state variable depending only
on the initial and final states.
In above case, we used the heat of formation of solid
sugar. In biochemical reactions we are interested in the heat
of formation for the solvated sugar molecule. To realize the
corresponding H, one needs to measure the heat of solution.
Biochemical Oxidation of Glycine.
Chemical oxidation
3O2 ( g )  2 NH 2 CH 2 COOH ( s )  4CO2 ( g )  2 H 2 O(l )  2 NH 3 ( g )
H 1  1163.5 kJ mol 1
However, we know from biochemistry the reaction product
is Urea. So let us calculate the heat of hydrolysis of urea to
yield CO2 and NH3:
H 2 O(l )  H 2 NCONH 2 ( s )  CO2 ( g )  2 NH 3 ( g )
H 2  133.3kJ mol 1
Subtracting the two equations and the corresponding ΔH’s
and rearranging we obtain:
3O2 ( g )  2 glycine ( s)  urea ( s)  H 2 O(l )  4CO2 ( g )  2 H 2O(l )
 2 NH 3 ( g )  CO2 ( g )  2 NH 3
H 1  H 2
3O2 ( g )  2 glycine ( s)  Urea( s)  3CO2 ( g )  3H 2 O
Thus, using a combination of two chemical reactions we have
estimated a heat of biochemical reaction. However, we can
still fine-tune the above reaction by taking into account the
heat of solution for both urea and glycine as follows.
glycine ( s)  H 2 O(l )  Glycine(aq) H 3  15.69 kj mol 1
Urea( s)  H 2 O(l )  Urea(aq)
H 4  13.43 kj mol 1
Now subtracting twice the hydration of glycine reaction and
adding solvation reaction of urea we can show:
3O2 ( g )  2 glycine ( s)  2 glycine ( s)  H 2 O  Urea( s)  H 2 O
 Urea( s)  3CO2 ( g )  3H 2 O  2 glycine (aq)  Urea(aq)
3O2 ( g )  2 glycine (aq)  Urea(aq)  3CO2 ( g )  3H 2 O
H Total  H 1  H 2  2H 3  H 4  1314kJ mol 1
Second Law of Thermodynamics
Introduces a concept of entropy, a new state variable. It is
perhaps the most widely term common language. It indicates
an extent of disorder. It was developed during the studies of
steam engines. Heated steam (“ideal gas”) does mechanical
work of moving piston but it also looses some of the heat to
surrounding environment wasting “useful” energy. Issue is
what is the theoretically maximum efficiency of the process.
To understand Carnot’s construction, we have to
understand adiabatic paths. Adiabatic path is the one where
the state of the system changes without loss of heat to
surrounding. According to 1st law ΔE=q+W, for adiabatic
process q=0. For small change in volume there will be work
and the change in the temperature of the system is given by:
CV dT   p.dV   RT
CV
dV
V
dT
dV
V
T 2 
 R
 CV ln     R ln 2 but
T
V
V1
 T1 
T2 P2V2

T1 P1V1
 V 
 P 
V
CV ln  2   ln  2     R ln 2
V1
 P1  
  V1 
P 
V
V
CV ln  2   (CV  R) ln 2  (CV  R) ln 1
V1
V2
 P1 
 P2   V1 
 P   V 
 1  2
CV  R
CV
 PV   const  
CV  R
CV
Note for adiabatic path pVγ is the equation of state.
Carnot Cycle
Carnot cycle considers following cyclic path for a reversible
engine.
Gas at
P2,V2,T1
Isothermal expansion
Ideal Gas
T1,P1,V1
Adiabatic
Compression
Adiabatic
expansion
Isothermal Compression
Gas at
P4,V4,T2
Gas
P3,V3,T2
Adiabatic path
I
P
2
4
3
V
Isothermal Path
Calculation of q and W for the Carnot Cycle.
o Path I. Isothermal reversible expansion of gas at
T1. Since this increases the final volume of the gas
the work is negative. In isothermal reversible
expansion the net change in the internal energy is
zero.
V 
E  0  q  w1  q1   P.dV  q1  RT1 ln  2 
 V1 
o Adiabatic reversible Expansion
E  w2  CV T2  T1 
o Isothermal reversible compression.
V 
E  0  q  w3  q3   P.dV  q3  RT2 ln  4 
V3 
o Adiabatic reversible compression
E  w4  CV T1  T2 
Thus total heat absorbed and work done is:
qTotal
V4 
V2 
 q1  q 2  q3  q 4  RT1 ln    RT 2 ln  
 V1 
V3 
V 
V 
wTotal  w1  w2  w3  w4   RT1 ln  2   RT 2 ln  4 
 V1 
V3 
qTotal   wTotal or ETotal  0