Two Last Lectures
First Law of Thermodynamics
Heat Capacities allow calculations of
energy transfer as heat
Work is a mode of energy transfer
Enthalpy is a state function that relates
to the potential for heat transfer at
constant pressure.
Example: work depends on path
We compare the work energy by expanding 1 mol
of Ne from 1 to 2 atm in:
a) One big expansion (2 to 1 atm)
b) Two expansions (2 to 1.5 atm, then
1.5 to 1 atm)
Solution, step 1
First-the single-step expansion from 2 atm to 1 atm pressure.
V=
nRt
,
P
1 mol 298 K 0.0821
Vinitial =
2 atm
1 mol 298 K 0.0821
Vfinal =
1 atm
L-atm
mol K =12.2329 L;
L-atm
mol K =24.4658 L;
work = -PV= 1 atm (24.4658 12.2329 L) = 12.2329 L-atm;
or about -1235.5 Joules (1 L-atm 101 Joules)
Solution, step 2
Now, the two-step expansion from 2 atm to 1.5 atm, then 1.5 to 1 atm pressure.
V2 atm = 12.2329 L; V1atm =24.4658 L; (from the previous calculation)
1 mol 298 K 0.0821
V1.5 atm =
1.5 atm
L-atm
mol K =16.3105 L;
work1 = -PV= -1.5 atm (16.3105 L- 12.2329) = -6.1164 L-atm
work 2 = -1.0 atm (24.4658 16.3105 L) = -8.1553 L-atm;
total work = -6.1164-8.1553 = -14.2717 L-atm;
-1441.4 Joules
We can create a state function from q and w
because the internal energy, U, does not depend on
path.
U = q - PV
H = U + PV = q
at constant pressure
H is a state function, meaning that it doesn't depend
on path and can be tabulated.
Enthalpy
Heat released or absorbed at constant P
Standard enthalpy
5 Pa on pure
measured at 1 atm = 1.01325 x 10
substances
symbol: H
Coffee Cup Calorimeter
A simple calorimeter.
Well insulated and therefore isolated.
Measure temperature change.
qrxn = qcal
= C (Tfinal Tinitial )
i.e. The heat of the reaction is simply the
negative of the heat producing the T change in
the calorimeter!
Bomb Calorimeter
qrxn = -qcal
qcal = qbomb + qwater + qwires +
qcal = CT
How do we know C?
Combustion of a standard:
e.g. benzoic acid C6H5COOH
Heat combustion = -3227 kJ/mol
Rigid insulating walls
fixed volume and isolated
Measure T change
heat of the chemical reaction
Example, Heat of Reaction
0.1g of liquid benzene is burned in a closed calorimeter.
The temperature of the calorimeter increases by 2.609
C.
The calorimeter has a heat capacity of Cp=1.602 kJ/C)
Find the molar heat of combustion of benzene.
C6 H 6 (l ) +
15
2
O 2 (g ) . 6CO 2 (g ) + 3H 2 O(l )
Solution
Heat Released:
qrxn = qcal
qcal = C T = 1.602 103 J / C 2.609C = 4179.618 J
qrxn = 4179.618 J
Please note that it is exothermic!
Molar heat of combustion:
# moles C6 H 6 burned, n =
qrxn
qrxn 4179.618 J
=
=
0.00128mol
n
0.1g
0.1g
=
= 0.00128mo
(6 12.011+6 1.0079) g / mol 78.1134 g / mol
= - 3.265106 J / mol
= 3265 kJ / mol
Phase changes are a simple reaction
Energy is needed to break bonds:
solid
U
liquid
U
gas
Total q can be calculated from Cp, Csp,
Hvap and Hfus
temperatures are constant during
phase changes
Example
Calculate the total heat needed to change
18 g (1 mol) of ice at -25 C to vapor at
+125 C.
Csp(Ice) = 2.09 J/gC
Hfus = 6.01 kJ/mol
Csp(Water) = 4.184 J/gC
Csp(Vapor) = 1.84 J/gC Hvap = 40.7 kJ/mol
o
Step 1--heating of ice from -25 to 0 C
J = 2.09 J/gC x (0- (-25))oC 18 x g of Ice
= 940 J
o
Step 2-- melting ice at 0 C
J = Hfus = 6.01 kJ/mol x 1 mol H2O
= 6010 J
Step 3- heating water from 0 C to 100oC
o
J = 4.184 J/gC x 18 g x (100-0) C
= 7531.2 Joules
o
Step 4 - boiling water at 0 C
= 40.7 kJ/mol x 1 mol
= 40,700 Joules
Step 5- heating water vapor from 100 C to 125oC
o
J = 1.84 J/gC x 18 g x (125-100)oC
= 828 Joules
Add the heats for each step:
total heat = 940 J + 6010 J
+ 7531 J + 40,700 J + 828 J
Each element has a reference form that is
defined to have enthalpy of formation equal to
zero if at standard conditions.
Hf (O2(g)) 0 kJ/mol
Hf (H2(g)) 0 kJ/mol
This definition allows chemists to build a table
of enthalpies from calorimetrically measured
heats and from heat capacities.
Standard Enthalpy of Formation,
Hf
The enthalpy change that occurs in the formation of one
mole of a substance in the standard state from the
reference forms of the elements in their standard states.
Standard state = 1 bar pressure
H2(g) + O2(g)
H2O(l)
Hf (H2O(l)) = -285.83 kJ/mol
Reference Forms of the Elements
O:
O2(g) - a diatomic gas at room temperature and 1 bar
3O2
Br:
Br2(l) - a diatomic liquid at room temperature and 1 bar
Br2(l)
C:
Br2(g),
H=+30.9 kJ
C(graphite)
C(graphite)
S:
H=+285 kJ
2O3,
C(diamond),
H=+1.9 kJ
S8(s) - many other allotropes, e.g. S7
Usually the reference state is the most stable form. Exception:
P:
P4(s,white)
P4(s,white)
P(s,red),
H=-17.6 kJ
Hess Law
(this is powerful)
Reaction enthalpies are additive:
Hf(reverse) = -Hf(forward)
Hcondensation = Hliquid - Hvapor
You can use them in a balanced
chemical reaction.
Hesss Law and balanced chemical
reactions
N2(g) + O2(g) NO(g)
H = +90.25 kJ
NO(g) + O2(g) NO2(g)
H = -57.07 kJ
N2(g) + O2(g) NO2(g)
H = +33.18 kJ
Tabulated enthalpies can be used among different reactions.
This tabulation lets you predict things, like heats of
reactions and even the temperature dependencies of equilibria.
Hess Law
If a process occurs in stages or steps (even hypothetically),
the enthalpy change for the overall process is the sum of
the enthalpy changes for the individual steps.
A
?
AC
B
H (AC)
H (DB)
C
D
H (CD)
H (AC)
CD
H (CD)
DB
H (DB)
AB
H (AB)
Recall: H is a state function: path independent!
Reaction Enthalpy
Reactants Products
Hf =nHf (products) - nHf (reactants)
Hf positive Endothermic
Hf negative Exothermic
Hf Standard enthalpy of formation
H is extensive: H = U + PV
PCl5(l) PCl3(l) + Cl2(g);
Ho = +123.8 kJ
Can multiply:
4PCl5(l) 4PCl3(l) + 4Cl2(g);
Ho = 4(+123.8 kJ)
= +495.2 kJ
Can reverse:
PCl3 (l) + Cl2 (g) PCl5 (l);
H = -123.8 kJ
Example - Combustion
Is the combustion of propane C3H8(g)
endothermic or exothermic?
Hf (O2) = 0 kJ/mol
Hf (H2O) = -285.83 kJ/mol
Hf (CO2) = -393.51 kJ/mol
Hf (C3H8) = -2220 kJ/mol
Solution
Reaction: Complete combustion all propane is converted to carbon
dioxide (gas) and water (liquid at room temperature)
C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2O (l )
Enthalpy :
H combustion = 3H f (CO2( g ) ) + 4H f ( H 2O( l ) ) H f (C3 H 8( g ) ) 5H f (O2( g ) )
= (-1180.53)
+ (-1143.32)
(-103.8)
= -2220 kJ/mol, EXOTHERMIC (of course)
(0) kJ/mol