Thermochemistry
First Law of Thermodynamics
The First Law of Thermodynamics explains the exchange of energy
(∆E) between a system and its surroundings.
∆E= q + w
q= heat (+ endothermic, - exothermic)
w= work (+ when done on system, - when by system)
W= -P x ∆V
W= (-) because work flows out of the system
P= external pressure
∆V= a positive volume, except for when gas is
compressed where it is (-) which makes w (+)
Heat
Heat flows from hot to cold and is represented by q.
Endothermic: Absorbs heat from surroundings
Example: melting ice, cooking food
Exothermic: Releases heat to surroundings
Example: Burning wood, freezing water
Heat capacity is measured in either calories (c) or joules (J)
The greater the mass of the object, the greater its heat capacity
Specific heat capacity is represented by C.
Change in enthalpy: q= mc∆T
m= mass in grams (g)
c= specific heat
∆T= change in temperature (Tfinal - Tinitial)
Q= heat (J)
Specific heat of water is 4.18 J/g x °C or 1.00cal/g x °C
One of the highest specific heats known; water can easily
absorb
or give up large amounts of heat
Calorimetry
A calorimeter measures the amount of heat absorbed or released
during a chemical reaction.
∆H is negative in exothermic reactions
∆H is positive in endothermic reactions
For a problem including water:
-(Qlost)= Qgained
-(Heat lost by _____)= Heat gained by water
Phase Changes
Phase changes can be either kinetic or potential.
Phase change diagram.
In the flattened areas, PE is increasing and breaks the bonds
between molecules and change the sate; energy is used to
melt or vaporize the liquid while the temp stays constant until
the substance has completely changed to the next phase. Once
over, the heat is used to change the KE which changes the temp.
Words you need to know:
Thermochemistry: The study of heat changes that occur during chemical reactions
Heat (q): Transferred from one object to another because of temp difference.
Energy: Capacity to do work or supply heat; can be kinetic (KE, due to motion) or
potential (PE, stored within structural units)
First Law of Thermodynamics: The energy of the universe is constant
Heat Capacity: Amount of heat to change an object's temperature by 1°c.
Joule: SI unit of heat and energy
Specific Heat Capacity: Amount of heat needed to raise temp of 1g substance to 1°c.
Calorimetry: Measurement of heat changes for chemical and physical processes
Enthalpy (∆H): Amount of heat a substance has at a given temperature and pressure;
change of heat; Q
Phase Change: Change of heat (energy); constant temperature; energy used to melt,
vaporize, freeze, or solidify the substance
Molar heat of fusion: Heat absorbed by 1 mole of a substance in melting (+∆H)
Molar heat of solidification: Heat lost by 1 mole substance in freezing (-∆H)
Molar heat of vaporization: Heat absorbed by 1 mole of a substance in vaporization
(+∆H)
Molar heat of condensation: Heat lost by 1 mole of a substance in condensation (-∆H)
Thermochemical Equation: An equation that includes ∆H (enthalpy)
Hess's Law: If you add 2 or more thermochemical equations to give a final equation,
then you can add their enthalpies to give the final heat of reaction
Standard Enthalpy of Formation: ∆H°f ; change in enthalpy that accompanies the
formation of one mole of a compound from its elements with all substances in their
standard states
Subsequently, the amount of energy that it takes to melt the ice is the same as the
amount to freeze the same quantity of water.
Molar heat is the amount of energy needed per 1 mole and is a change in energy,
not temperature
Molar heat of fusion: Heat absorbed by 1 mole of a substance in melting (+∆H)
Molar heat of solidification: Heat lost by 1 mole substance in freezing (-∆H)
Molar heat of vaporization: Heat absorbed by 1 mole of a substance in
vaporization (+∆H)
Molar heat of condensation: Heat lost by 1 mole of a substance in
condensation (-∆H)
Molar heat of solution: Heat lost by one mole of a substance when it is
dissolved
Exothermic: temp of the solution increases
Endothermic: the substance absorbs heat
Helpful Constants
Phase
J/g°C
J/kgK
Gas
2.02
2.03 x103
Liquid 4.18
4.18 x 103
Solid
2.03
2.03 x 103
Phase change equations:
Latent heat of fusion: Q=m∆Hfus (m-∆Hsolid)
Latent heat of vaporization: Q=m∆Hvap (m-∆Hcond)
Phase change constants:
∆Hfus(H2O)= 6.02 kJ/mol= 334 J/g
∆Hvap(H2O)= 40.7 kJ/mol= 2260 J/g
Endothermic phase changes: Fusion and vaporization (absorb energy)
Exothermic phase changes: Condensation and solidification (release energy)
Hess's Law
If a reaction is reversed, the sign of ∆H is reversed too; if the coefficients in a
balanced reaction are multiplied by an integer, the value of ∆H are multiplied by
the same integer; ∆H1 + ∆H2 = ∆H3
Standard Heat of Formation
Standard heat of a free element is set at 0 (diatomics and graphite form of carbon)
Formula: ∆H°= ∆H°f (products) - ∆H°f (reactants)
Review Questions
#1. Karen decides to go on a hot air ballooning trip for two on Valentines Day. To impress her
chemist boyfriend NAtHeN, she decides to find out the amount of energy exchanged between the
balloon and its surroundings. If the volume of the balloon increases from 9.0 x 1017 L to 3.43 x 1020
L at a constant pressure of 1.0 atm, calculate ∆E for the process if there is an addition of 4.9 x 1023
J of energy as heat. (Tip: ∆V= Vfinal- Vinitial) (1L x atm= 101.3 J)
#2. A car's engine is powered by the combustion of propane with air. When 1 gallon of 1.00 M
propane solution at 25°C combusts with 3 L 1.00 M O2 at 25°C, the temperature inside the engine
rises to 29 °C. If the specific heat capacity is .918 J/°C x g and the density of the final solution is
1.0 g/mL, solve the enthalpy change per mole of CO2. (3.78 L= 1 gallon)
#3. At a renaissance fair, a blacksmith is creating iron armor for a joust happening later. If he
takes 200 g of iron and heats it until it becomes malleable at 700 °C, shapes it, and transports it to
a slack tub containing 600g of water at 25 °C and both come to the same temperature, what will
the final temperature be if the specific heat of iron is 0.444 J/g°C?
#4. Substance X has a molar mass of 55.0 g/mol and has the following constants:
Solid= 4.3 J/gC° , liquid= 3.5, gas= 2.4
∆Hfus= 49 kJ/mol
∆Hvap= 25 kJ/mol
melting point= -5 °C boiling point=149°C
a. Draw the phase diagram.
b. Calculate the energy required to convert 430.0 g of the substance from a solid at -70°C to a
gas at 180 °C.
Review Answers
#1. To solve, use the three equations ∆E= q + w, W= -P x ∆V, and ∆V= Vfinal- Vinitial. You already have q,
equaling 4.9 x 1023J. Because the two given volumes are not exponentially equal, you must convert 9.0 x 1017 to .009 x
1020 so that you can subtract it from 3.43 x 1020, which gives you
∆V= 3.43-.0090= 3.421 x 1020.
Then you can find W by using W= -P x ∆V, which gives you
W= -1 x 3.421 x 1020= -3.421 x 1020 L x atm.
You then sum q and w but must convert w to joules
w= -3.421 x 1020 L x atm x 101.3 J/L x atm = -3.465 x 1022 J
Finally, you can do
∆E= q + w= (4.9 x 1023 J) + (-3.465 x 1022 J)= 4.5 x 1023 J.
#2. This problem requires a series of multiple steps. The first step is to create a net ionic equation for the combustion
of propane, which you must know as C3H8. You then get the net ionic equation
C3H8(g) + 5O2(g) p 3CO2(g)+ 4H2O(g) + heat
Now you must find the total solution volume. Because you are given liters and gallons, you must convert from gallons
to liters first to get
1 gallon x 3.78 L/1 gallon = 3.78 L
You then add 3.78 L to 3 L oxygen to get 6.78 L, and to convert to get the final mass of the solution you do
6.78 L x 1000 mL/1L x 1.0g/mL = 6.78 x 103 g
To calculate temperature increase, you do ∆T= Tfinal - Tinitial to get
29-25= 4 °C.
Finally, you can solve mc∆T, which comes out to
(6.78 x 103 g)(.918 J/°C x g)(4°C) = 2.49 x 102 J.
We aren't done yet, because we must originally determine whether the reaction is endothermic or exothermic and it
must be in kJ. Because heat is being released in the combustion of propane, q must be negative, therefore giving the
final answer of q= -2.49 x 10-1 kJ/mol.
#3. You will use the equation -(Heat lost by iron)= Heat gained by water, which will give you
-(200)(x-700)(.444)=(600)(4.18)(x-25)
where the x stands for the wanted temperature. Using algebra, do
-88.8(x-700)=2508(x-25)
-88.8x + 62160= 2508x - 62700
124860=2596.8x
x=48.1°C
#4. a.
180149T
-5-
E
D
C
B
-70-A
Heat energy
b. A= m(65)(c) = (4.3)(65)(430) = 120.2 kJ
430 g x 1mole/55g= 7.8 moles
C= (430)(3.5)(154)= 231.8 kJ
E= (430)(31)(2.4)= 32 kJ
B= (49)(7.8) = 382.2 kJ
D=(25)(7.8)= 195 kJ
Total= 961.2 kJ