Sample Problem

Sample Problem
Find ΔH for H2O (s) 0 C → H2O (g) 100 C sublimation:
H
Happens
all
ll around:
d snow, ice
i in
i f
fridge,
id
d
dry iice
This is clearly not a constant T process. Since H is a state function, define any path
to get to product. Make it simple to use. Calculate ΔH for that path. Here is one:
H2O ((g))
H2O (s)
fusion
+40 6 kJ/mol
+40.6
kJ/ l
+ 6 kJ/mol
H2O ()
0 ºC
vaporization
Hvap
Hfusion
Cp,l =33.6 J/mol K
heating
H2O ()
100 ºC
100C
H  H fusion 

C p , dT  H vap
0C


CpT for H 2 O
ConcepTest
p
#0
Find ∆H for the reaction:
CH4(g) + H2O(l)  H2CO(g) + 2H2(g)
using the following standard heats of formation:
CH4(g)
H2CO(g)
H2O(g)
-74
74.8
8 kJ/mol
-108.6 kJ/mol
-241.8 kJ/mol
a.
a
b.
c.
d.
Less than 208.0
208 0 kJ/mol
Equal to 208.0 kJ/mol
G
Greater
t th
than 208.0
208 0 kJ/mol
kJ/ l
Not possible to tell
ConcepTest #0 Solution
Find ∆H for the reaction:
CH4(g) + H2O(l)  H2CO(g) + 2H2(g)
using
us
ng the follow
following
ng standard heats of formation:
format on:
CH4(g)
H2CO(g)
H2O(g)
O( )
-74.8 kJ/mol
-108.6 kJ/mol
-241.8
241 8 kJ/mol
We need this
ΔH = [ΔHf(H2CO(g))+ ΔHf(2H2 (g)) ]
–[ΔH
[ΔHf(CH4(g)) + ΔHf(H2O (l)) ]
Ho(H2O(l) ) = Ho(H2O(g) ) + Hfus(H2O(g) )
= -241.8
-241 8
+ F (a negative number)
ΔH =[-108.6 +2x0] –[-74.8 +(-241.8 +F)] = 208 –F
Since F is negative, ΔH must be > 208 kJ/mol
Thermochemistry Revisited
Consider some chemical reaction is taking place at constant Pex, and the
temperature changes from Ti to Tf as the reaction proceeds. We can write
Hrxn  Hrxn Ti   C p  p
products  T
where Hrxn is the enthalpy change for the stated conditions, and ΔT = (Tf – Ti).
STANDARD ENTHALPY CHANGE
R f s to
Refers
t enthalpy
th l change
h
under
d set
s t of
f standard
st d d conditions.
diti s The
Th standard
st d d
enthalpy change ΔHrxn for a process A → B is the change in enthalpy HB – HA when
A and B are in their standard states. The components of A and B are also
p
from each other.
separated
That’s double talk! What is a standard state??
STANDARD STATE
“The standard state of a substance at a specified temperature is its
pure form at that temperature.
temperature ”
Moreover, for the very special case that T=298.14 K and P = 1 atm, we call Hrxn
the standard reaction enthalpy and designate this condition with a superscript o.
H process 298  Ho process at P = 1atm
(or sometimes  or even )
However, the standard enthalpy change of a Rxn can be defined at any temperature
Rx Enthalpy as f(T)
Consider the reaction 2A + 3B  C + 2D
at two temperatures T1 and T2. What is the difference
in the standard enthalpies of formation at these
temperatures?
For constant Cp,
r H2  r H1   r CP dT  r CP T2 T1 
o
o
T2
o
o
T1
r CP 
o

 CP,m 
products
o

 CP,m
o
reactants
The  are the coefficients in the reaction above. m is molar.
Ammonia Heat of Formation
Thus, suppose we want to determine the heat of formation of NH3 at 400 K.
½ N2 + 3/2 H2  NH3
Hf  400  
Hf
 CP

T



? 400  298
kJ
46.11
mol
How to obtain ΔCP?
Cp(NH3) = 35.06 J/mol K
Cp(N2) = 29.1 J/mol K
Cp(H2) = 28.8 J/mol K
all from tabulated values

1
 J
3
J
So CP  35.06   29.1  28.8 
 22.7
2
mol K
2
  mol K

and Hf  400   46.11
kJ

mol
22.7
22 7
kJ
kJ
400  298 
 48.43

1000
mol
mo l

get J to kJ
This approach assumes that the CP are all constant over this temperature range.
This assumption is certainly valid for ideal gases; it is certainly false if there is
a phase change, and it is not exact for real gases!
Propene Hydrogenation/Combustion
What’s propene? H2C=CH-CH3
Does it matter where the double bond goes? What’s the hydrogenation Rxn?
H
H
C
H
C
C
H
+ H2
H
H
H
H
C
C
C
H
H
H
H
H
H
We know (from table) that ΔH for this reaction is -124 kJ/mol at standard
conditions. All components are gases.
Now we use Hess’ Law to find the standard heat of reaction for the combustion of
propene,
9
C3 H 6  g   O 2  g   3CO 2  g   3H 2 O   
2
We need to find a series of reactions that have known ΔH and add up
p to this one.
Propene Hydrogenation/Combustion
C3H6 
9
O2  3CO2  3H2O   
2
C3H6  H2  g   C3H8
C3H8  5O2  3CO2  4H2O   
-124
-2220  3(-393.5)+
3(-393 5)+ 4  285.8
285 8    103
103.8
8
1
O g 
286  -Hf H2O   
2 2
_____________________________________________
9
kJ
C3H6  O2  3CO2  3H2O   
-2058
2
mol
H2O     H2  g  


P
Pressure
ConcepTest
p
#1
1
4
3
2
An id
A
ideall gass goess
through the four step
process shown. In which
p
stages is work done?
Vl
Volume
a. In 1 and 2 only
d. In none of the stages
b. In 1 and 3 only
e. In all of the stages
c. In 2 and 4 only
ConcepTest
p
#2
Pressurre
The curve shown at left
r pr s nts th
represents
the expansion
xp nsi n
of an ideal gas. If this
process is isothermal then
we can also say:
Volume
a. There is no heat involved in this process.
b There is no work involved in this process.
b.
process
c. There is a transfer of thermal energy, or heat, into the gas.
d. There is a transfer of thermal energy, or heat, out of the gas.
ConcepTest
p
#3
Two 10 kg weights sit on a piston, compressing the air
underneath. One of the weights
g
is removed, and
the air underneath expands from 18.3 to 20.0L.
Then the second weight is removed and the air
expands from 20.0 to 22.0L. How does the
amount of work done compare if instead both
weights were removed at once? Assume the same
total change in volume.
a. More work is done removing one weight at a time than removing
both weights at once.
b Less
b.
L
work
k is
i done
d
removing
i one weight
i h at a time
i
than
h in
i removing
i
both weights at once.
c. The same amount of work is done removing one weight at a time,
or if both are removed at once.
once
Adiabatic Expansion/Compression
Consider the three paths, A, B and C
P1, V1, T1
U A  U B  U C
B
P
A
P3, V3, T3
C
P2, V2, T2
V
This is why we call U or H state functions.
and
H A  H B  H C
but
 q A   qB   qC
and
 wA   wB   wC
Suppose that path A in the 2 D slice in PV above characterizes the
expansion, with P1, V1 and T1 being the initial condition. Further, let path
A be a reversible adiabatic path. What can we say experimentally?
P1, V1, T1 → P2, V2, T2
Is V1 < V2 or is V1 > V2 ?
Is T1 < T2 , is T1 =T2,
or is T1 > T2 ?
Is P1 < P2 or is P1 > P2 ?
Now we calculate the cooling in a the reversible adiabatic expansion of an ideal gas.