1. Chapter 5
Thermochemistry ( )

2. Energy Every chemical change has an accompanying change of energy
Combustion of fossil fuels
Metabolism of foods
Discharging of a battery
Chemical Cold Pack

3. Kinetic and potential energy
Physics Context
Kinetic Energy is the energy of motion
KE = ½ mv2
Potential Energy is stored energy; (GPE = gravitational PE)
Chemistry Context
There is chemical potential (internal) energy in the bonds between atoms in a compound. When the bonds are broken, energy is absorbed. When bonds are formed, energy is released.
Temperature is the measure of relative chemical energy

4. Thermochemistry
The study of energy and its transformations is THERMODYNAMICS
Study of the relationship between heat, work and energy in a system
In this class we will study the relationships between chemical reactions and energy changes involving heat – THERMOCHEMISTRY

5. SI unit for energy is the Joule Another unit of energy is the calorie
Units of energy
SI unit for energy is the Joule
1kg moving at 1 m/s = 1 J of energy
Another unit of energy is the calorie
1 g water raised 1ºC
1 cal = J

6. System and Surroundings
In a chemical reaction:
System – the objects or chemicals we are studying
Surroundings – container and everything else beyond the chemicals
Universe – total of the system and the surroundings
Transferring Energy: Work and Heat
Work (w) is energy used to push or pull an object against a force
Heat (q) is energy transferred from a hotter object to a cooler one

7. 5.2 First Law of Thermodynamics
Definition
The First Law of says that in any process, the total change in energy of the system, is equal to the sum of the heat absorbed, q, and the work, w, done on the system
This law is often referred to as the Law of Conservation of Energy
Internal Energy (E)
The sum of all the kinetic and potential energy in the system
We measure the change in internal energy (∆E)
∆E = Efinal –Einitial
(+) ∆E = system gained energy from surroundings = endothermic
(-) ∆E = system lost energy to surroundings = exothermic

8. Relating ∆E to heat and work
Two ways to calculate work
Work done against force of gravity: w = F x d
Pressure-volume work (compression & expansion of gases) w=-P∆V
Equation for the First Law: ∆E = q + w
Sign Convention:
Heat (q) is positive if energy enters the system; it is negative when energy leaves the system
Work (w) is positive if work is done on the system; it is negative if work is done by the system.
∆E is positive when there is a net gain of energy by the system; it is negative when there is a net loss.

9. Sign conventions for q, w, and ∆E+ - q
Gains heat
Loses heat W
Work done on system
Work done by system ∆E
Net gain of energy by the system
Net loss of energy by the system

10. Examples Calculate the ∆E for a gas in which the gas:
Absorbs 35 J of heat and does 10 J of work
Q is absorbed = +; w is released = -
∆E = w + q = -10 J + 35 J = 25 J
Produces 15 J of heat and has 12 J of work done on it as it contracts
q is released = -; w is absorbed = +
∆E = +12 J – 15 J = -3 J

11. Energy diagrams Terminology
Exothermic – a reaction that releases energy, ∆E is negative
Endothermic – a reaction that absorbs energy, ∆E is positive
Activation energy, Ea – the minimum energy that must be possessed by a pair of molecules if collision is to result in reaction
Activated complex – a species, formed by collision of energetic particles, that can react to form products
Energy of reaction, ∆E – the difference between the energies of the products & reactants.

14. State Properties
The state of a system refers to the temperature, pressure, and composition of that system
25.00 g of hydrogen gas at 1.00 atm at 25C
State properties are quantites that depend only upon the “state” of a system, not upon the way it reached that state
Ex: volume, temperature
Heat and work are not state properties – they do depend on the path

15. Enthalpy
Chem II

16. Enthalpy
The chemical and physical changes that occur around us essentially occur at constant atmospheric pressure.
Most commonly, the only kind of work produced by chemical and physical changes open to the atmosphere is the mechanical work associated with change in the volume of the system.

17. Enthalpy, H From a Greek word meaning “to warm” ∆H
Commonly called “heat of reaction.”
In most cases ∆E and ∆H are equal, except with gases. In these cases the pressure or volume of a system is changing: In these cases ∆H = ∆E + P∆V
For most reactions the difference in ∆H and ∆E are so small because P ∆ V is small.

18. Enthalpy
1. Enthalpy is an extensive property→∆H is directly proportional to the amount of reactant consumed in the process.
2. The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to the ∆H for the reverse reaction.
3. The enthalpy change for a reaction depends on the state of the reactants and products. Important to note the states in your equations.

19. Enthalpy of Reaction ∆Hrxn = ∆Hproducts– ∆Hreactants
Enthalpy of reaction–the enthalpy change that accompanies a reaction
Sometimes called = Heat of reaction

20. Thermochemical equations
Balanced chemical equations that show the associated enthalpy change
2 H2(g) + O2(g) → 2 H20 (g) ∆H = kJ
Would this be exothermic or endothermic?
exothermic

21. Example NH4NO3(s) → NH4+(aq) + NO3-(aq) ∆H= +28.1 kJ
Is this reaction exothermic or endothermic?
endothermic
If 2.00g of solid ammonium nitrate react, how much energy is needed?
2.00g NH4NO mole kJ = kJ
80.06 g mol

22. You try it… H2(g) + Cl2(g) → 2 HCl(g) ∆H = -185kJ
Is this reaction exothermic or endothermic?
exothermic
How much energy is created when 3.50 g of HCl are formed?
3.50g HCl mol HCl kJ = kJ
36.46 g mol HCl

23. And again…
Hydrogen peroxide can decompose to water and oxygen by the following reaction:
2H2O2 (l) →2H2O (l) + O2 (g) ∆H= -196 kJ
How much energy is created when 5.00 grams of hydrogen peroxide decomposes at constant pressure?
5.00g H2O mole kJ = mole H2O2
34.02 g mol H2O2

24. Calorimetry
Chem II
5.5

25. Calorimetry
Calorimeters are used to measure heat flow or the temperature change experienced by an object when it absorbs a certain amount of heat.
Heat Capacity –the amount of heat required to raise its temperature by 1 K.
The greater the heat capacity, the greater the heat required to produce a given increase in temperature.
It is a unique value for all substances.

26. Molar Heat Capacity The heat capacity of one mole of a substance.
Cmolar
Specific heat capacity (specific heat) –the heat capacity of one gram of a substance q
c = or q = c x m x ∆T
m x ∆T

27. Specific Heat Values Examples: N2(g) = 1.04 J/g-K Al(s) = 0.90 J/g-K
Fe(s) = 0.45 J/g-K
H2O (l) = 4.18 J/g-K
What do you notice about the value of water compared to others?
Why would this be important to us?

28. You Try It…
How much heat is needed to warm 250 grams of water from 22 C to near its boiling point, 98 C?
What is the molar heat capacity of water?
q = cm∆T
q = (4.18 J/g-K)(250 g)(371 – 295)
= J = 79 kJ
250 g H2O 1 mole = = 14 moles H20
18.02 g
79.4 kJ/13.87 mole = = 5.7 kJ/mole H2O

29. Constant Pressure Calorimetry
Uses a “coffee-cup calorimeter”
Not sealed so reaction occurs at
constant pressure –atmospheric
In a simple calorimeter in which
essentially all of the heat given off by
a reaction is absorbed by the known
amount of water, qsolution = -qreaction

30. NH4NO3(s) → NH4+(aq) + NO3-(aq) ∆H= + 29 kJ
What is the qreaction when a 1.00g sample of ammonium nitrate is dissolved in water if the temperature of 40.0 g of water changed from 25.0⁰C to 22.9⁰C?
Q = cm∆T
Qreaction = -(4.184J/g⁰C) (41.0 g) (-2.1⁰C) = 360 J
How much heat would be evolved if one mole of ammonium nitrate dissolved? (This is called the ∆H for the reaction.) Write the thermochemical equation for this process.
1 mol NH4NO g J = 29,000J
1 mol g
NH4NO3(s) → NH4+(aq) + NO3-(aq) ∆H= + 29 kJ

31. If 25. 0 g of NH4NO3 are placed in 125 grams of water at 22
If 25.0 g of NH4NO3 are placed in 125 grams of water at 22.0⁰C, what is the final temperature of the water?
25g mol ,000J = 9000J = 9.0 x 103 J
80.06g mol
∆T = qsolution = J = -14⁰C
cm (4.184J/gC)(150g)
so Tfinal = 22.0 – 14 = 8⁰C

32. What is the heat of reaction (ΔH) for the neutralization of HCl if 100
What is the heat of reaction (ΔH) for the neutralization of HCl if mL of 1.0 M NaOH is added to mL of 1.0 M HCl with an initial temperature of 24.6⁰C and after the reaction is complete, the temperature is 31.3⁰C? Assume the densities of the solutions are 1.00 g/mL & the specific heat is J/g ⁰C.
q = c x m x ΔT
q = J/g ⁰C x g x (31.3 ⁰C ⁰C)
q = 5600 J
HCl + NaOH --> H2O + NaCl
L x 1.0 M = moles of HCl & NaOH
ΔH = J / moles = -56,000 J/mol OR
-56 kJ/mol of reaction

33. 1. Theory - a device used to measure heat flow in which a reaction is
Bomb Calorimeter
1. Theory - a device used to measure heat flow in which a reaction is
carried out with a sealed metal container
2. Calculations: qreaction = - Ccalorimeter x ΔT
a. The reaction between 1.00 g octane & oxygen during
combustion causes the temp. of the calorimeter to rise from
25.00⁰C to 33.20⁰C.
If Ccalorimeter = 837 J/ ⁰C, how much heat is produced in the reaction?
-(837J/g⁰C)( ) = = J (per 1 g of octane)
* The negative sign means that heat is released from the reaction.

34. b. How much energy is produced if one mole of octane was burned?
Write the thermochemical equation for this process.
1.00 mol g J kJ = kJ
1 mol g J
C8H18(l) O2(g) → 8 CO2(g) + 9 H2O(g) ΔH = -784kJ

35. Hess’s Law
Chem II
5.6

36. Hess’s Law
If a reaction is carried out in a series of steps, ∆H for the reaction will equal the summ of the enthalpy changes for the individual steps.
Enthalpy is a state function.
It is independent of the path.
We can add equations to come up with the desired final product, and add the ∆H
Two rules
If the reaction is reversed the sign of ∆H is changed
If the reaction is multiplied or divided, so is ∆H

37. Using Hess’s Law to Calculate ΔH
The following information is known:
C(s) + O2(g) →CO2(g) ΔH1 = kJ
CO(g) + ½ O2(g) → CO2(g) ΔH2 = kJ
Using these data, calculate the enthalpy for:
C(s) + ½ O2(g) → CO(g)

38. Answer The following information is known:
C(s) + O2(g) →CO2(g) ΔH1 = kJ
CO(g) + ½ O2(g) → CO2(g) ΔH2 = kJ
Using these data, calculate the enthalpy for:
C(s) + ½ O2(g) → CO(g)
(Reverse Rxn2)
ΔHrxn = (+283.0) = kJ

39. Practice with Hess’s Law
Calculate ∆H for the reaction
2C(s) + H2(g) →C2H2(g)
Given the following chemical equations and their respective ∆H.
C2H2(g)+ 5/2O2 → 2CO2(g) + H2O(l) ∆H = kJ
C(s) + O2(g) → CO2(g) ∆ H = kJ
H2(g) + ½ O2(g) → H2O(l) ∆H = kJ

40. Answer Reverse Rxn1 Double Rxn 2 ΔH1 = + 1299.6
ΔHrxn =

41. You Try It… Calculate ΔH for the reaction
NO(g) + O(g) →NO2(g)
Given the following information:
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = kJ
O3(g) → 3/2 O2(g) ΔH = kJ
O2(g) → 2O(g) ΔH = kJ

42. Answer Reverse Rxn 2 Reverse Rxn 3 Cut Rxn 3 in half ΔH1 = -198.9 kJ

43. Remember…
H is a state function, so for a particular set of reactants and products, ΔH is the same whether the reaction takes place in one step or in a series of steps.

44. Standard Enthalpies of Formation
∆Hf is the enthalpy change associated with the formation of a given compound
∆Hffor a number of compounds is found in Appendix C starting on pg. 1041
Standard Enthalpies of formation is the change of enthalpy for the reaction that forms 1 mol e of the compound from its elements in standard states.
Ex: 2C(s) + 3H2(g) + ½ O2(g)→ C2H5OH(l)
∆ 𝐻𝑓 ⁰ = kJ

45. Standard Enthalpy of Formation
The standard enthalpy of formation of the most stable form (elemental) of any element is zero.
Thus, the ∆Hf⁰ for most individual elements is 0. The exception would be the diatomic gases of “Dr. HOFBrINCl. H2, O2, F2, Br2, I2, N2, and Cl2 are zero and the individual elements of hydrogen, oxygen, fluorine, bromine, iodine, nitrogen and chlorine are not zero.