20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #5
page 1
Thermochemistry
Much of thermochemistry is based on finding “easy” paths to
calculate changes in enthalpy, i.e. understanding how to work
with thermodynamic cycles.
•
Goal:
To predict ∆H for every reaction, even if it
cannot be carried out in the laboratory
The heat of reaction
∆Hrx
is the ∆H for the isothermal
reaction at constant pressure (the complete transfer from reactants
to products, not to some equilibrium state).
Fe2O3(s,T,p) + 3H2(g,T,p) = 2Fe(s,T,p) + 3H2O(l,T,p)
e.g.
∆Hrx (T , p ) = 2HFe (T , p ) + 3HH O (T , p ) − 3HH (T , p ) − HFe O (T , p )
2
2
2 3
[∆Hrx = H ( products ) − H ( reactants )]
We cannot know H values because enthalpy, like energy, is not an
absolute scale. We can only measure differences in enthalpy.
•
Define a reference scale for enthalpy
H (298.15K, 1 bar) ≡ 0
e.g.
For every element in its most stable
form at 1 bar and 298.15K
HH° (g) (298.15K ) = 0
2
HC°(graphite) (298.15K ) = 0
The “°” means 1 bar
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
•
∆ Hf° (298.15K ) :
Lecture #5
page 2
The heat of formation is the heat of reaction
to create 1 mole of that compound from its constituent elements in
their most stable forms.
Example (T = 298.15 K)
½ H2 (g,T,1 bar) + ½ Br2 (l,T,1 bar) = HBr (g,T,1 bar)
°
∆Hf°,HBr (T ) = ∆Hrx (T ,1bar ) = HHBr
( g,T ) −
1 °
1
HH2 ( g,T ) − HBr°2 ( l,T )
2
2
0 - elements in most stable forms
We can tabulate ∆ Hf° (298.15K ) values for all known compounds.
We can calculate ∆ Hrx° (298.15K ) for any reaction.
e.g. (T=298.15K)
CH4 (g,T,1 bar) + 2O2 (g,T,1 bar) = CO2 (g,T,1 bar) + 2H2O(l,T, 1 bar)
•
First decompose reactants into elements
•
Second put elements together to form products
•
Use Hess’s law [An example of a thermodynamic cycle applied
to thermochemistry]
Reactants
∆ H rx
−∑ν i ∆Hf°,i ( reactants )
i
Products
∑i νi ∆Hf°i ( products )
,
Elements
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #5
page 3
CH4 (g,T,1 bar) = Cgraphite (s,T,1 bar) + 2H2(g,T,p)
∆HI
2O2 (g,T,1 bar) = 2O2 (g,T,1 bar)
∆HII
Cgraphite (s,T,1 bar) + O2 (g,T,1 bar) = CO2 (g,T,1 bar) ∆HIII
2H2(g,T,p) + O2 (g,T,1 bar) = 2H2O(l,T, 1 bar)
∆HIV
____________________________________________
CH4 (g,T,1 bar) + 2O2 (g,T,1 bar) = CO2 (g,T,1 bar) + 2H2O(l,T, 1 bar)
∆Hrx = ∆HI + ∆HII + ∆HIII + ∆HIV
∆HI = HC + 2HH − HCH = −∆Hf°,CH
2
∆HII = HO − HO = 0
2
4
4
2
∆HIII = HCO − HC − HO = ∆Hf°,CO
2
2
2
∆HIV = 2HH O − 2HH − HO = 2∆Hf°,H O
2
∴
2
2
2
∆Hrx = 2∆Hf°,H O + ∆Hf°,CO − ∆Hf°,CH
2
2
4
In general,
∆Hrx = ∑ν i ∆Hf°,i (products ) − ∑ν i ∆Hf°,i (reactants )
i
i
ν ≡ stoichiometric coefficient
•
∆H at constant p and for reversible pV process is ∆H = q p
⇒
The heat of reaction is the heat flowing into the
reaction from the surroundings
If
∆Hrx < 0,
qp < 0
heat flows from the reaction to
the surroundings (exothermic)
If
∆Hrx > 0,
qp > 0
heat flows into the reaction from
the surroundings (endothermic)
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #5
page 4
Calorimetry
The objective is to measure
Reactants (T1)
∆Hrx (T1 )
•
isothermal
=
constant p
Constant pressure
Products (T1)
(for solutions)
adiabatic
constant p
reaction
calorimeter
React. (T1) + Cal. (T1)
∆Hrx
Prod. (T1) + Cal. (T1)
→
constant p
∆HI
adiabatic
∆HII
constant p
constant p
Prod. (T2) + Cal. (T2)
I)
∆HI
React. (T1) + Cal. (T1)
II)
∆HII
Prod. (T2) + Cal. (T2)
adiabatic
=
constant p
=
constant p
Prod. (T2) + Cal. (T2)
Prod. (T1) + Cal. (T1)
______________________________________________________
∆Hrx (T1 )
React. (T1) + Cal. (T1)
=
constant p
∆Hrx (T1 ) = ∆HI + ∆HII
Prod. (T1) + Cal. (T1)
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #5
page 5
(I) Purpose is to measure (T2 - T1)
Adiabatic, const. p
⇒
qp = 0
⇒
∆H I = 0
(II) Purpose is to measure heat qp needed to take prod. + cal.
from T2 back to T1.
T1
qp = ∫ C p ( Prod . + Cal .) dT = ∆HII
T2
∴
•
T2
∆Hrx (T1 ) = − ∫T C p ( Prod . + Cal . )dT
1
Constant volume
(when gases involved)
adiabatic
constant V
reaction
calorimeter
React. (T1) + Cal. (T1)
∆Urx
Prod. (T1) + Cal. (T1)
→
constant V
∆UI
adiabatic
∆UII
constant V
constant V
Prod. (T2) + Cal. (T2)
I)
∆U I
React. (T1) + Cal. (T1)
II)
∆UII
Prod. (T2) + Cal. (T2)
adiabatic
=
constant V
=
constant V
Prod. (T2) + Cal. (T2)
Prod. (T1) + Cal. (T1)
______________________________________________________
∆Urx (T1 )
React. (T1) + Cal. (T1)
=
constant V
Prod. (T1) + Cal. (T1)
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #5
page 6
∆Urx (T1 ) = ∆UI + ∆UII
(I) Purpose is to measure (T2 - T1)
Adiabatic, const. V
qV = 0
⇒
⇒
∆UI = 0
(II) Purpose is to measure heat qV needed to take prod. + cal.
from T2 back to T1.
T1
qV = ∫ CV ( Prod . + Cal . )dT = ∆UII
T2
∴
T2
∆Urx (T1 ) = − ∫T CV ( Prod . + Cal . )dT
1
Now use
H = U + pV
or
∆H = ∆U + ∆ ( pV
Assume only significant contribution to ∆ ( pV
∆ ( pV ) = R ∆ (nT
Ideal gas
⇒
Isothermal
T =T1 ⇒
∴
)
)
is from gases.
)
∆ ( pV ) = RT1 ∆ngas
∆Hrx (T1 ) = ∆Urx (T1 ) + RT1 ∆ngas
T2
∆Hrx (T1 ) = − ∫ CV ( Pr od . + Cal . ) dT + RT1 ∆ngas
T1
e.g.
4 HCl(g) + O2(g) = 2 H2O(l) + 2 Cl2(g)
T1 = 298.15 K
∆Urx (T1 ) = -195.0 kJ
∆ngas = -3 moles
∆Hrx (T1 ) = -195.0 kJ + (-3 mol)( 298.15 K)(8.314x10-3 kJ/K-mol)
= -202.43 kJ
20.110J / 2.772J / 5.601J
Thermodynamics of Biomolecular Systems
Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
20.110/5.60 Fall 2005
Lecture #5
page 7
Temperature dependence of ∆Hrx
Suppose know ∆Hrx at some temperature T1 (e.g. at 298.15 K) and we
want to know it at some other temperature T2.
Generally the difference is small… often we assume that there is no
temperature dependence if the difference between T1 and T2 is
“small”. If the difference between T1 and T2 is large enough, we can
calculate ∆Hrx(T2) from the heat capacities of the reactants and
products (assuming no phase change in any component).
∆Hrx (T2 )
Reactants (T2)
→
constant p
Products (T2)
∆HI
∆HII
constant p
constant p
∆Hrx (T1 )
Reactants (T1)
→
constant p
Products (T1)
∆Hrx (T2 ) = ∆HI + ∆Hrx (T1 ) + ∆HII
T1
T2
T2
T1
∆Hrx (T2 ) = ∆Hrx (T1 ) + ∫ C p (react .)dT + ∫ C p ( prod .)dT
T2
∆Hrx (T2 ) = ∆Hrx (T1 ) + ∫ ⎡⎣C p ( prod .) − C p (react .) ⎤⎦ dT
T1
T2
∆Hrx (T2 ) = ∆Hrx (T1 ) + ∫ ∆C pdT
T1