STANDARD STATES & STANDARD ENTHALPY CHANGES:
a) For
a pure liquid or solid, the standard state IS the pure
liquid or solid.
b) For a
gas, the standard state IS the gas
at 1 atm; in gas
mixtures, its partial pressure must be 1 atm.
c) For
solutions, the standard state is 1
M concentration.
DHo means at standard
pressure and T (STP) =
1 atm & 25oC (298K)
DHfo is standard molar enthalpy of formation: 1 mole of a
substance is
formed from its elements in their standard
states. The DHfo for an element in its standard state = 0
-DH = means giving OFF heat =
EXOthermic
+DH = means REQUIRING heat =
ENDOthermic
Calculate how much heat is
absorbed or evolved in the
following reaction:
2 NH3(g) + CO2(g) ----> NH2CONH2(aq) + H2O(l)
The DHo at 25oC
(in kJ/mol): NH3(g) -45.9
CO2(g) -393.5
NH2CONH2(aq) -319.2
H2O(l) -285.8
DHo = [319.2 - 285.8] - [2(-45.9) - 393.5] = -119.7 kJ.
Because DHo is
negative, we conclude that heat is
evolved and this reaction is exothermic.
Hess'
Law of Heat Summation:
Enthalpy change
of a reaction is the same whether it
occurs by one step or by many different steps.
Ex.: C(s) + 2H2(g) ---> CH4(g) DHo = -4.28 kJ
C(s) +
O2(g) ---> CO2(g) DHo = -22.48 kJ
2H2(g) + O2(g) ---> 2H2O(l) DHo = -32.66 kJ
CO2(g) + 2H2O(l) ---> CH4(g) + 2O2(g) DHo =+50.86kJ ________________________________
C(s) +
O2(g) + 2H2(g) + O2(g) + CO2(g) + 2H2O(l) --->
CO2(g) + 2H2O(l) + CH4(g) + 2O2(g)
Therefore,
C(s) + 2H2(g) ---> CH4(g)
DHo = -4.28 kJ/mol reaction
Therefore:DHo (reaction) = DHo (1) + DHo (2) + DHo (3) + ...
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Calculate DHo (reaction)
for
C2H4(g)
+ H2O(l) ---> C2H5OH(l) from:
1)
C2H5OH(l)
+ 3O2(g) ---> 3H2O(l) + 2CO2(g)
DHo = -1367 kJ
2) C2H4(g) + 3O2(g) ---> 2CO2(g) + 2H2O(l)
DHo = -1411 kJ
Take
equation 2 as it is, and add to it equation 1
(backwards)!! to get:
C2H4(g) + + 3O2(g) + 3H2O(l) + 2CO2(g) --->
2CO2(g) + 2H2O(l) + C2H5OH(l) + 3O2(g)
Therefore:DHo (reaction)
= DHo(2) - DHo(1) =
= -44 kJ/mol reaction
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DH°(reaction) = S nDH°(products) - S mDH°(reactants)
For CaO(s) + CO2(g) ---> CaCO3(s)
DH°(reaction) = S n[DH°CaCO3(s)] -
S m[DH°CaO(s)+DH°CO2(g)]
DH°(reaction) = -1207 kJ - [-635.5kJ + (-393.5kJ)]
DH°(reaction) = -1207 kJ + 1029 kJ
= -178 kJ/mol reaction
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Calculate DH°(reaction) for:
SiH4(g) + 2O2(g) ---> SiO2(g) + 2H2O(l)
DH°(reaction) = S [DH°SiO2(g) + 2DH°H2O(l)] -
S [DH°SiH4(g)+2DH°O2(g)]
= [-217.72 kJ + 2(-68.315 kJ)] - [+8.2 kJ + 2(0) kJ]
= -217.72 - 136.63 - 8.2 = -362.6 kJ/mol reaction
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CalculateDHf°[PbO(s)] from:
PbO(s) + CO(g) ---> Pb(s) + CO2(g)
Given: DH°(reaction) = -65.69
kJ/mol reaction
DHf° CO(g) = -110.5
kJ/mol
DHf° CO2 = -393.5 kJ/mol
DH°(reaction) = S [DH°
Pb(s) + DH°
CO2(g)] -
S [DH° PbO(s) + DH° CO(g)] =
-65.69 kJ = [0 kJ + (-393.5 kJ)]
- [DHf° PbO(s) + -110.5 kJ]
-65.69 + 393.5 - 110.5 = - DHf° PbO(s)
+217.3 kJ/mol = - DHf° PbO(s)
-217.3 kJ/mol = DHf° PbO(s)
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Calculate DHf°[NH3(g)] from:
4NH3(g) + 7O2(g) ---> 4NO2(g) + 6H2O(l)
Given: DH°(reaction) = -1396 kJ/mol reaction
DHf° NO2(g) = 34 kJ/mol
DHf° H2O(l) = -286 kJ/mol
DH°(reaction) = S [4DH° NO2(g) + 6DH° H2O(l)] -
S [4DH° NH3(g) + 7DH° O2(g)]
-1396 kJ/mol = [4 (34 kJ) + 6(-286 kJ)] -
[4DHf° NH3(g) + 7(0) kJ]
-1396 kJ/mol = + 136 - 1716 - [4DHf° NH3(g) + 0]
+184 kJ/mol = - 4DHf° NH3(g)
-46 kJ/mol = DHf° NH3(g)
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Bond Energies:
Amount of energy required to break 1
mole of bonds in a gaseous covalent substance to
form one mole of gaseous products at constant T & P.
H2(g) --- 2H(g)
In order to break the H-H
bond, we must supply
436 kJ/mol.
DH°(reaction)
= DH° (H-H) = +436 kJ mol = endothermic
Therefore,
we can get at the bond energies of simple
molecules by measuring the heat required when bonds
are ruptured.
Ex.:
CH4(g) + 2Cl2(g) + 2F2(g) --->
CF2Cl2(g) + 2HF(g) + 2HCl(g)
CalculateDH°(reaction).
Reactants ------------> Atoms -----------------> Products
(energy required) (energy
released)
Reactant
bonds broken:
forCH4(g): 4 mol C-H = 4 mol (413 kJ/mol) = 1652 kJ
for 2Cl2(g): 2 mol Cl-Cl = 2 mol (242 kJ/mol) = 484 kJ
for 2F2(g): 2 mol F-F = 2 mol (155 kJ/mol) = 310
kJ
Total Energy required = 2446 kJ
Product
bonds formed:
for CF2Cl2(g): 2 mol C-F
= 2 mol (485 kJ/mol) = 970 kJ
2 mol C-Cl = 2 mol (339 kJ/mol) = 678 kJ
for HF(g): 2 mol H-F = 2 mol (565 kJ/mol) = 1130 kJ
for HCl(g): 2 mol H-Cl = 2 mol (432 kJ/mol) = 864 kJ
Total Energy released = 3642 kJ
DH°(reaction) = Energyrequired
to break bonds -
Energy released in forming bonds
= 2446 kJ - 3642 kJ = -1196 kJ/mol CF2Cl2(g) formed
Therefore, since DH° is - , this reaction is EXOthermic.
Changes in Internal Energy DE:
DE = E(final) - E(initial) = q + w ,
where q = heat
and w = work
work = fd (force x distance)
DE = (am't of heat absorbed by system) +
(am't of work done on the system)
When
q = + : heat is absorbed by system from the
surroundings
When
q = - : heat is released by system to the
surroundings
When
w = + : work is done on the system by the
surroundings
When
w = - : work is done by system on the
surroundings
The ONLY
TYPE OF WORK in physical & chemical changes
is PRESSURE-VOLUME work.
w = - PDV DE = E(final) - E(initial) = q + w = q - PDV
1) compression
of gas: work done BY surroundings ON
the system (to compress a gas, you must SUPPLY
energy)
w = -P(Vf-Vi): since Vf is
smaller than Vi , w = +
*decrease in the number of moles of gas;
therefore: Dn = -1
Ex.: 2H2(g) + O2(g) ---> 2H2O(g); Dn = 2 - 3 = -1
2) expansion
of gas: Work done BY the system ON the
surroundings (if you let a gas expand, you will get
energy released)
w = -P(Vf-Vi): since Vi is
smaller than Vf , w = -
*increase in
the number of moles of gas;
therefore: Dn = +1
Ex.: 2H2O(g) ----> 2H2(g) + O2(g); Dn = 3 - 2 = 1
When
the VOLUME is constant, NO PDV work done.
Therefore, DE = qv
When
the PRESSURE is constant, -PDV = -DnRT =
work done.
Therefore, the
only time there is work done is when Dn is
NOT zero; i.e.,
when there is a change in the number of
moles of gaseous products and reactants, so that the
volume of the
system changes.
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Bomb Calorimeter: A
closed volume system where a
sample is
burned, usually in the presence of oxygen.
Of course,
the pressure increases, but so also does the T.
The heat
liberated here is
qv = DE (constant
volume) =
INTERNAL ENERGY
Ex.: C6H6(l) + 15/2[O2(g)] ----> 6CO2(g) + 3H2O(l)
When 0.187 g of benzene, C6H6, is burned in a bomb
calorimeter,
the surrounding water bath rises in
temperature
by 7.48oC. Assuming that the
bath contains
250.0 g of
water and that the calorimeter itself absorbs
2.56 kJ,
calculate the combustion enthalpy for benzene
in kJ/mol.
The amount of heat
responsible for the increase in T of the
water is:
heat required = (7.48oC)(4.184 J/g-oC)(250.0 g)
= 7824 J
The amount of heat responsible
for the heating the
calorimeter = 2.56 kJ = 2560 J
Total
amount of heat absorbed by calorimeter and water =
10384 J
Therefore, the combustion of
0.187 g of benzene liberates
10.384 kJ of energy.
DE (internal energy) = -10.384 kJ/0.187 g
benzene
Therefore, DE =
(-10.384 kJ/0.187 g benzene)X
(78.0g benzene/mol) =
DE = -4331 kJ/mol
benzene OR /mol reaction.
Remember: Constant V !!!!
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Relationship between DH and DE:
DH = DE + PDV (at constant T & P)
But we know that DE = q + w
Therefore, DH = q + w + PDV (at constant T & P)
At constant P, w = -PDV
Therefore, DH =
q -PDV + PDV = qp
The difference between DE and DH is the amount of
expansion work (PDV work): DH = DE + (Dn)RT
(at constant T & P)
Ex.: CH4(g) + 2O2(g) -----> CO2(g) + 2H2O(g)
DHo (at 25oC and 1 atm)= -890
kJ. Calculate DE.
Since Dn = (2-2) = 0,
there is NO
PDV work; DE = DHo =
-890 kJ
Ex.: N2(g) + 3H2(g) -----> 2NH3(g); DHo
= -91.8 kJ.
Calculate DE at
25oC.
DH = DE + DnRT; DE = DH - (Dn)RT
= -91.8 kJ -[(2-4 moles)(8.314J/mol-K)(298K)
= -91.8 kJ + 4955 J = -91.8 kJ + 5.0 kJ
= -86.8 kJ/mol reaction
Ex.: C2H5OH(l) + 3O2(g) ----> 2CO2(g) + 3H2O(l)
Dn = 2 - 3 =
-1 REMEMBER!!! ONLY MOLES OF GAS!!!
BOTTOM
LINE: DH = qp at constant P (usually
atmospheric P)
DE = qv at
constant V (bomb calorimeter experiment)
DH = DE when there is NO WORK
done; no PDV, no (Dn)RT
DH = DE + (Dn)RT when there IS work done
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Spontaneous Processes:
Occur without outside
intervention, e.g., drop two eggs...
they break!
The reverse
process is non-spontaneous.
The direction of a spontaneous
process can depend on
temperature.
Ice
turning to water is spontaneous at T > 0°C.
Water
turning to ice is spontaneous at T < 0°C.
Allowing 1 mol of ice to warm
up is an irreversible process.
To get the reverse process to
occur, the water temperature
must be lowered to 0°C (heat must be removed).
In any spontaneous process, the
path between reactants
and products
is irreversible, unless outside intervention
occurs.
E.g., water can be
frozen...then thawed by adding heat.
Thermodynamics gives us the direction
of a process, but
not the speed
at which it will occur (this latter is
KINETICS).
Why
are endothermic reactions spontaneous?
Entropy
and the Second Law of Thermodynamics:
Consider an initial state: two
flasks connected by a closed
stopcock. One flask is evacuated and the other contains
1 atm. of a
gas. Then open the stopcock....The final
state: two
flasks connected by an open stopcock.
Each
flask contains the gas at 0.5 atm. The expansion
of the gas
is isothermal (i.e., constant temperature).
Therefore,
the gas does NO work and heat is not
transferred.
Why does a gas expand?
ENTROPY!
S: is a measure of the disorder of a system.
The bigger
the disorder.....bigger entropy.
More order.....lower entropy (ice is more ordered
than water, which is more ordered than steam).
Spontaneous
reactions proceed to lower energy or
higher
entropy.
In the
connected flasks, is it more likely that all the gas
molecules
stay in one flask or that they spread out
randomly
to fill both flasks? The molecules are more
likely to
fill both flasks, and the system has moved to a
state of
higher entropy.
What about thawing ice?
In ice, the
molecules are very well ordered because of
the H-bonds
throughout the system (and the molecules
of water are
in fixed positions); therefore, ice has a low
entropy. As ice melts, the intermolecular forces are
broken
(which requires energy), but the order is
interrupted,
so entropy increases. Water is more random
than ice, so
ice spontaneously melts at room
temperature. There is always a balance between energy
and entropy
considerations.
Generally, an increase in entropy in one process is
associated
with a decrease in entropy in another; the
increase in
entropy usually dominates.
Entropy is a state function.
For a system, DS = Sfinal - Sinitial
If DS > 0, the randomness increases (less order);
if DS < 0, the order
increases.