Transcript Introduction to Thermochemistry and Specific Heat

Solid
Liquid
Gas
Heat
Heat = AMOUNT of internal energy
Temperature = a MEASURE of the average molecular
kinetic energy
10 g Pb
T = 40 °C
100 g Pb
T = 40 °C
Both blocks are at the same temperature.
Do they both contain the same amount of heat?
Which substance requires more heat to increase
the temperature by 5 °C?
Pb
100 g
Specific heat capacity (Cp): amount of heat(q) required
to raise 1 g of substance by 1 °C
Cp(paraffin) = 2.1 J/g°C
Cp(Pb) = 0.126 J/g°C
How much heat is required by the 100 g candle to increase
the temperature by 5 °C?
Cp(paraffin) = 2.1 J/g°C
q = Cp(mass)(DT)
q = (2.1 J/g°C)(100 g)(5 °C)
q = 1050 J
If the same amount of heat was used to heat 100 g of
water [Cp(liquid water) = 4.184 J/g°C], what would be the
DT of the water?
q = Cp(mass)(DT)
1050 J = (4.184 J/g°C)(100g)(DT)
DT = 2.5 °C
For the same amount of heat and
mass, DT decreases as the specific
heat of the substance increases
100 kg Pb
If the temperature of the lead is
327°C before it hits the water,
what is the final temperature of
the lead after hitting the water?
q = mCpDT
200 ft
DT = Tf - Ti
-qPb = qH2O
-(1x105 g)(0.13 J/g°C)(Tf – 327°C) =
(1x104 g)(4.18 J/g°C)(Tf – 20°C)
10 kg H2O
Ti = 20 °C
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Tf = 93 °C
Cp(Pb) = 0.13 J/g°C
Cp (H2O) = 4.18 J/g°C
Melting one 14-gram Al soda can requires
5.55 kJ of energy. What is its molar heat
of fusion?
14 g Al
 0.519 mol Al
26.982 g/mol
5.55 kJ
1 can
ΔHfus(Al) 
x
 10.7 kJ/mol
1 can 0.519 mol Al
105,000 cans are recycled in the US every minute.
How many kJ/s are being used in recycling Al cans?
1.05x105 cans 0.519 mol Al 10.7 kJ
x
x
 9.72x10 3 kJ/s
60 s
1 can
1 mole Al
That’s equivalent to burning 2300 food Calories/s!
Experiment: Heat two beakers containing 18 g of water at
the same rate, and monitor their temperatures.
Question: Will their temperatures increase at the same
rate?
- 10
0 °C
18 g H2O = 1 mole H2O
100
90 °C
Experiment: Heat two beakers containing 18 g of ice and
water at the same rate, and monitor their temperatures.
Question: Will their temperatures increase at the same
rate?
Answer: It takes twice as long to increase the
temperature of the liquid water by 10 °C than it
does to increase the temperature of the ice by
the same amount.
Heating curve of water
Temperature (°C)
Gas warming
100
liquid warming
0
liquid + gas present
solid + liquid present
solid warming
Heat (kJ/s)
Temperature (°C)
Heating curve of water
100
0
boiling/condensation point
melting/freezing point
Heat (kJ/s)
Temperature is constant during phase transitions!!
All heat energy goes to changing the state of matter.
Temperature (°C)
Heating curve of water
100
DHvap
(heat of vaporization)
0
DHfus (heat of fusion)
Heat (kJ/s)
DHfus = the amount of heat needed to covert a solid into its
liquid phase
DHvap = the amount of heat needed to convert a liquid into
its gaseous phase
Temperature (°C)
Heating curve of water
100
And vice versa
0
A greater DHfus = more time to melt
Heat (kJ/s)
H2O:
DHfus = 6.01 kJ/mol
DHvap = 40.7 kJ/mol
H2PEw:
DHfus = 20.2 kJ/mol
DHvap = 10.3 kJ/mol
Heating Curve Wrap Up:
• The specific heat capacity (Cp)of a substance determines the
temperature change observed when heat is added or withdrawn from
the substance.
• Temperature is INVARIANT during phase transitions.
• The amount of heat required to convert one mole of the substance
from one phase to another is its molar enthalpy of transition (DHfus,
DHvap, DHsub). What is the sign for all three?
+DH
• The amount of heat given off for one mole of a substance during a
phase transition while cooling is its molar enthalpy of transition
(DHcond, DHsol, DHdep). What is the sign for all three?
-DH
• The shape of a heating curve depends upon the heating rate, specific
heat capacities of the phases involved, and the enthalpies of
transition.
1.0 gram of solid sodium metal is added to 100 g of water. The
reaction produces sodium hydroxide and hydrogen gas. Calculate
the molar heat of reaction if the water’s temperature increased
by 2C.
Step 1: Write out the chemical equation and balance it.
2 Na (s) + 2 H2O (l)
+1
-1
2 Na OH (aq) +
H2 (g)
Step 2: Determine if there’s a limiting reagent.*
1H
1.0 g Na
x 2  1.26 mol H
2
22.98977 g/mol 2 Na
100.0 g H O
1H
2 x
2  2.77 mol H
2
18.015 g/mol 2 H O
2
Na is the limiting
reagent. Only
1.26 mol of H2
will be formed.
*Choose a product that has a coefficient of 1 for best results.
Step 3: Determine the amount of heat involved in the reaction.
q = mCpDT
q=?
m = 100 g
Cp = 4.184 g/JC
DT = 2 C
q  (100 g H O)(4.184 J/gC)(2C)  837 J
2
Step 4: Calculate your molar heat of reaction.
If a reaction that produced 1.26 moles of H2 also released
837 J of heat, then the molar enthalpy (heat) change for
this reaction would be:
ΔHrxn  837 J  664 J/mol
1.26 mol
A simpler problem:
How much heat is given off when 1.6 g of CH4 are burned in
an excess of oxygen if DHcomb = -802 kJ/mol?
Step 1: Write the reaction equation.
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
Step 2: Calculate molar amount involved
1.6 g CH
4  0.100 moles CH combusted
4
16.042 g/mol
Step 3: Calculate amount of heat given off
DHrxn = (-802 kJ/mol)(0.100 mol CH4) = -80.2 kJ
Q: Is this an exothermic or endothermic reaction?
EX 3: What is the molar heat of combustion of propene (C3H6) if
burning 3.2 g releases 156 kJ of heat?
Step 1: Write the reaction equation.
2 C3H6 (g) + 9 O2 (g)  6 CO2 (g) + 6 H2O (g)
C3H6 (g) + 4.5 O2 (g)  3 CO2 (g) + 3 H2O (g)
• This reaction equation involves the combustion of 2 moles
of C3H6 and we want to find out what it is for one mole.
• Save yourself a headache and simplify future calculations by
dividing the reaction equation through by 2.
Step 2: Convert grams of propene to moles.
3.2 g C3H6
 0.076 mol C3H6
42.081 g/mol
Step 3: Divide the heat released by moles of propene.
156 kJ
ΔH

  2053 kJ/mol
comb
0.076 mol C3H6
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What have we learned?
Sometimes heat is given off during a chemical reaction.
This makes it feel hotter.
Sometimes heat is absorbed during a chemical reaction.
This makes it feel colder.
What causes it to be different?
• Chemical bonds contain energy.
• Add the energy of all of the bonds in the
reactants together to find their total energy.
• Add the energy of all of the bonds in the
products together to find their total energy.
• If the two numbers aren’t the same (and they
almost never are), then there will be heat
energy given off or taken in.
2 H2 + O2
2 H2O
2 H2 + O 2
Energy barrier
2 H2O
Energy
• If the products contain less energy, energy must
have been given off during the reaction.
2 H2O
2 H2 + O2
2 H2 + O2
2 H2O
Energy barrier
Energy
• If the products contain more energy, energy must
have been absorbed during the reaction.
If heat energy is given off during a reaction, it is
called an EXOTHERMIC REACTION.
Heat exits = exothermic
Exothermic reactions can be recognized by a
temperature INCREASE.
If heat energy is absorbed during a reaction, it is
called an ENDOTHERMIC REACTION.
Heat enters = endothermic
Endothermic reactions can be recognized by a
temperature DECREASE.
2 AlBr3 + 3 Cl2
2 AlCl3 + 3 Br2
2 AlBr3 + 3 Cl2
But how do we
determine the heat
content in the first
place?
2 AlCl3 + 3 Br2
Energy
DHrxn = Heat content of products – heat content reactants
DHrxn < 0
Reaction is exothermic
Heat of formation, DHf
• The DHf of all elements in their standard state equals zero.
• The DHf of all compounds is the molar heat of reaction for
synthesis of the compound from its elements
DHf (AlBr3): 2 Al + 3 Br2
2 AlBr3
DHrxn = 2DHf(AlBr3)
DHrxn
DHf(AlBr3) =
2
• Since the DHrxn can be used to find DHf, this means that DHf
can be used to find DHrxn WITHOUT having to do all of the
calorimetric measurements ourselves!!
The Law of Conservation of Energy strikes again!!
Hess’s Law:
DHrxn = S DHf(products) – S DHf(reactants)
6 CO2 (g) + 6 H2O (l)
C6H12O6 (s) + 6 O2 (g)
DHrxn = [DHf(C6H12O6) + 6 DHf(O2)] – [6 DHf(CO2) + 6 DHf(H2O)]
From DHf tables:
DHf(C6H12O6) = -1250 kJ/mol
DHf(CO2) = -393.5 kJ/mol
DHf(H2O) = -285.8 kJ/mol
DHrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]
DHrxn = +2825.8 kJ/mol
Using Hess’ Law with DHrxn
What is the DHcomb for ethane?
2 C6H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)
C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)DHrxn = -1323 kJ/mol
C2H4 (g) + H2 (g) → C2H6 (g)
DHrxn = -137 kJ/mol
H2O (l)
H2O (g)
H2O (g)
H2O (l)
DHvap = +40.7 kJ/mol
Energy
• Water will spontaneously evaporate at room temperature
even though this process is endothermic.
• What is providing the uphill driving force?
a measure of the disorder or
randomness of the particles
that make up a system
• Water will spontaneously evaporate at room temperature
because it allows the disorder of the water molecules to
increase.
• The entropy, S, of gases is >> than liquids or solids.
• If Sproducts > Sreactants, DS is > 0
Predict the sign of DS:
ClF (g) + F2 (g)
CH3OH (l)
ClF3 (g)
DS < 0
CH3OH (aq)
DS > 0
Are all +DS reactions spontaneous?
2 H2O (l)
2 H2 (g) + O2 (g)
DS is large and positive…
…but DH is large and positive as well.
• Gibb’s Free Energy, DG, allows us
to predict the spontaneity of a
reaction using DH AND DS.
If –DG  spontaneous reaction
2 H2O (l)
2 H2 (g) + O2 (g)
What is DG for this reaction at 25C?
DHrxn = SHf(products) –SHf(reactants)
DHrxn = [2(0) + 0] - 2(-285.83 kJ/mol) = 571.66 kJ/mol
DSrxn = SSf(products) –SSf(reactants)
DSrxn = [2(130.58 J/molK) + 205.0 J/molK] - 2(69.91 J/molK)
DSrxn = 326.34 J/molK = 0.32634 kJ/molK
DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK)
DGrxn = +474.41 kJ/mol
2 H2O (l)
2 H2 (g) + O2 (g)
What is the minimum temperature needed to make this
reaction spontaneous?
DGrxn = DHrxn – TDSrxn = 571.66 kJ/mol - T(0.32634 kJ/molK)
Set DGrxn = 0 to find minimum temperature
0 = 571.66 kJ/mol - T(0.32634 kJ/molK)
T = (571.66 kJ/mol)/(0.32634 kJ/molK) = 1751 K
T > 1479 C