Download Chemistry 12 – Unit 3 – Chapter 5 – Thermochemistry

Chemistry 12
Notes on Thermochemistry
Chemistry 12 – Unit 3 – Chapter 5 – Thermochemistry
5.1 – Changes in Matter and Energy
Heat Changes and Energy Changes
thermal energy: form of kinetic energy due to particle motion
surroundings: all matter around system capable of absorbing or releasing thermal energy
heat:…
temperature:…
Exothermic vs. Endothermic
exothermic: releasing thermal energy as heat flows out of system
Diagram
endothermic: absorbing thermal energy as heat flows into system
Diagram
Unit 3 – Chapter – 5 – Thermochemistry
Page 1
Chemistry 12
Notes on Thermochemistry
System and Surroundings:
When energy changes are measured in a chemical reaction, the system is the reaction mixture being studied,
and the surroundings are the flask, the room, and the rest of the universe. The energy change is the difference
between final and initial states (E = Efinal - Einitial). Any energy that flows from the system to the surroundings
has a negative sign because Efinal is smaller than Einitial, and any energy that flows into the system from the
surroundings has a positive sign because Efinal is larger than Einitial.
Types of systems
open system: system in which both matter and energy can move in or out
isolated system: ideal system in which neither matter nor energy can move in or out
closed system: system in which energy can move in or out, but not matter
Changes in in enrgy
Phase change:
Chemical Change
Nuclear Change:
Read Chapter 5.1 Page 291 2-5, 7-9
Unit 3 – Chapter – 5 – Thermochemistry
Page 2
Chemistry 12
Notes on Thermochemistry
5.2 – Molar Enthalpies
Calorimetry: Measuring Energy Changes
calorimetry: process of measuring energy changes in a chemical system
Diagram of a Calorimeter A calorimeter for measuring the heat flow in a
reaction at constant pressure (H). The reaction takes place inside an
insulated vessel outfitted with a loose-fitting top, a thermometer, and a
stirrer. Measuring the temperature change that accompanies the reaction
makes it possible to calculate H.
quantity of heat, q:
q = H = mcT
where q = heat absorbed by the water in joules (J)
m = mass of substance (usu. water)
T = Tfinal - Tinitial
c = specific heat capacity of water
c water = 4.184 J/g . 0C
specific heat capacity:…
problem of heat transfer:
A chemical reaction in a calorimeter raises the temperature of the 2.0 L mixture from 25.00C to 28.10C.
Assuming that the reaction mixture is a dilute solution, that the calorimeter absorbs a negligible
quantity of heat, and that the density of the mixture is 1.00 g/mL, calculate the heat transfer into the
mixture.
Solution:…
p. 297 #1-3 p. 306 #2-7
Unit 3 – Chapter – 5 – Thermochemistry
Page 3
Chemistry 12
Notes on Thermochemistry
Molar Enthalpies
Thermochemical Equation: Specify H in kilojoules/mole (kJ/mol).
Example of exothermic chemical reaction…:
The preceding is based upon the law of conservation of energy (James Joule, 1818-1889, Joule also developed
the First Law of Thermodynamics): energy is neither created nor destroyed in ordinary chemical or physical
changes.
This law can be adapted to calorimetry: The total energy change of the chemical system is equal to the total
energy change of the surroundings.
Hsystem =…
molar enthalpy, Hx : enthalpy change associated with a physical, chemical, or nuclear change involving
one mole of a substance
Various molar enthalpies of reaction, Hx
physical changes …
solution,
hydration,
vaporization,
freezing,
melting (fusion)
Hsol
Hhyd
Hvap
Hfr
Hfus
dissolving solute in liquid
dissolving gas in water
change of state
change of state
change of state
chemical changes …
neutralization,
formation,
combustion,
Hneut
Hf
Hcomb
neutralization reaction
formation reaction
combustion reaction
Unit 3 – Chapter – 5 – Thermochemistry
Page 4
Chemistry 12
Notes on Thermochemistry
Properties of Enthalpy Change
1. The magnitude of H is directly proportional to the amount of reactant or product.
-Thus H can be used as a conversion factor in a balanced equation to obtain amounts of
reactant/product or H itself. (mole to mole ratios).
2. H for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction.
Problems with molar enthalpies:
H Calculation using formula … ___________________
Problems with molar enthalpies:
H Calculation using dimensional analysis
1) When 1 mol of methane is burned at constant pressure, 890.3 kJ of energy is released as heat. Calculate
H for a process in which a 5.8 gram sample of methane is burned at constant pressure.
Solution:…
Because there is an energy release by the chemical system, H = ____________
2) For the above reaction of methane with oxygen, calculate the H in kJ if 5.8 grams of oxygen are
consumed in the process.
Solution:…
Because there is an energy release by the chemical system, H = ____________
Unit 3 – Chapter – 5 – Thermochemistry
Page 5
Chemistry 12
Notes on Thermochemistry
3) Ammonium nitrate, NH4NO3, is commonly used as an explosive. It dissolves in water by the following
reaction:
NH4NO3 (S) → NH4NO3 (aq)
If 2.7 grams of NH4NO3 raises the temperature of 150 mL of water from 23.1 0 C to 21.1 0 C, what is the
molar enthalpy of solution? (answer: Hsol = 37 kJ/mol)
Solution:…
Because there is an energy release by the chemical system, H = ____________
5.2 Representing Enthalpy Changes
Method 1: Thermochemical Equations with Energy Terms
Method 2: Thermochemical Equations with ∆H Values
Method 3: Molar Enthalpies of Reaction
Method : Graphically
5.2 (p. 301) 1, 2, 3, 4, (p.306) 2-7
Dry Lab Quantity of enthalpy change (Due in two days)
Unit 3 – Chapter – 5 – Thermochemistry
Page 6
Chemistry 12
Notes on Thermochemistry
5.3 Bond Enthalpy (Bond Energy)
The Bond Enthalpy is the energy required to break a chemical bond. It is usually expressed in units
of kJ mol-1, measured at 298 K.
This basically means that you add up all the energies of the broken bonds; add up all the energies of the bonds
that are reformed and subtract one from the other, you can determine the change in enthalpy
∆H = ∑ ∆H(bonds broken) - ∑ ∆H(bonds formed)
Example 1: Find
H for the following reaction given the following bond energies:
Method 1:
Bond (kJ/mol)
H-H
O=O
O-H
Unit 3 – Chapter – 5 – Thermochemistry
436
499
463
Page 7
Chemistry 12
Notes on Thermochemistry
Method 2:
∆H = ∑ ∆H(bonds broken) - ∑ ∆H(bonds formed)
EXAMPLE #2
The complete combustion of propane can be represented by the following equation:
or we could redraw it to represent the bonds present:
Method 1:
Method 2:
Bond
kJ/mol
C-H
C-C
O=O
C=O
H-O
413
347
498
799
467
∆H = ∑ ∆H(bonds broken) - ∑ ∆H(bonds formed)
∆H= [(8x413)+(2x347)+(5x498)] - [(6x799)+(8x467)]
∆H= - 2042 kJ mol-1
5.1 (p 312) 1-4 (p 313) 1-8
Dry Lab Quantity of enthalpy change
Unit 3 – Chapter – 5 – Thermochemistry
Page 8
Chemistry 12
SCH 4U1- ___
Notes on Thermochemistry
Name: ________________________________________
Date: ___________
Lab: Calorimetry of Physical Change … Enthalpy of Solution
Purpose: To calculate the molar enthalpy of dissolving for two ionic solids.
Materials:
•
Reagent grade calcium chloride
•
200 mL coffee cup as lid
•
300 ml coffee cup
•
graduated cylinder
•
water (distilled is
preferred but not
absolutely
necessary)
•
thermometer
•
stirring rod.
Procedure:
1.
Measure out 100 mL of distilled water in a graduated cylinder Add the water to
the coffee cup. and record it's temperature.
2.
Measure out 5 g of calcium chloride,
3.
Add this amount to the water in the coffee cup and place the lid on top. Stir the
solid until it dissolves.
4.
Record the lowest/greatest temperature reached .
5.
Empty out the coffee cup , and rinse and collect the data for a second and third sample.
Table 2.
Calorimetry of calcium chloride
Calcium Chloride
Trial #1
m1 Mass of water in calorimeter (1.0 mL has a
mass of 1.0g)
m2 Mass of solid used
n2
Moles of solid used
t1
Initial temperature
t2
Final temperature
∆T Temperature change
Unit 3 – Chapter – 5 – Thermochemistry
Page 9
Chemistry 12
Notes on Thermochemistry
1.
Calculate the molar enthalpy of the solid
2.
Internet Research for calcium chloride:
3.
Percent Error Calculations:
∆H0sol =
% Error = Accepted value – Avgerage. Experimental value x 100%
Accepted value
4.
Research the Internet for the uses and chemistry of hot and cold packs.
Unit 3 – Chapter – 5 – Thermochemistry
Page 2
Chemistry 12
Notes on Thermochemistry
5.4 – Hess’s Law of Additivity of Reaction Enthalpies
P.E. Diagram for Multistep Reaction
½ N2 + ½ O2 → NO
NO + ½ O2 → NO2
H= 90 kJ
H= -56 KJ
N2 + O2 → NO2
H0total = H1 + H2 = 34 KJ
NO(g)
intermediate(s)
∆H2 -56 kJ
EP
∆H1 = +90 kJ
(kJ)
NO2(g)
product(s)
∆Htotal +34 kJ
N2(g) + O2(g)
reactant(s)
Reaction Progress
Enthalpy Characteristics
Some characteristics of enthalpy are...
1. If the reaction is reversed then the sign of ∆H is also reversed.
For example, for the reaction
CH4(g) + 2 O2(g) --> CO2(g) + 2 H2O(g)
∆H0 is -802 kJ,
In contrast, the reverse reaction
CO2(g) + 2H2O(g) --> CH4(g) + 2H2O(g)
∆H0 of +802 kJ,
Iit is now endothermic.
Unit 3 – Chapter – 5 – Thermochemistry
Page 3
Chemistry 12
Notes on Thermochemistry
2. ∆H0 is extensive. If the coefficients in a balanced reaction are multiplied by an integer, then the value
of ∆H0 is multiplied by the same integer.
Xe(g) + 2F2(g) --> XeF4(s) ∆H0 = -251 kJ
2 Xe(g) + 4F2(g) --> 2XeF4(s) ∆H0 = -502 kJ
but...
Predicting ∆H Using Hess’ Law
Since enthalpy is a state function, the change in enthalpy in going from some initial state to some final state is
independent of the pathway. This leads us to Hess' Law...
Hess’ Law…
Another way to state Hess' Law - In going from a particular set of reactants to a particular set of products,
the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
Question 1.
Given the following equations and ∆H0 values, determine the ∆H0 (kJ) at 298 K for the overall reaction:
CS2(l) + 2H2O(l)
H2S(g) + 3/2O2(g)
CS2(l) + 3O2(g)
CO2(g) + 2H2S(g)
H2O(l) + SO2(g)
CO2(g) + 2SO2(g)
∆H01 = -562.6 kJ
∆H02 = -1075.2 kJ
Solution:
Reverse equation 1 and multiply by 2:…
Use the original equation 2:…
Now add equations 1 and 2 to produce the overall equation:…
Unit 3 – Chapter – 5 – Thermochemistry
Page 4
Chemistry 12
Notes on Thermochemistry
Question 2.
Given the following equations and ∆H0 values, determine the ∆H0 (kJ) at 298 K for the overall reaction:
2N2(g) +5O2(g)
N2(g) + 3O2(g) + H2(g)
N2O5(s) + H2O(l)
2H2(g) + O2(g)
2HNO3(aq)
2HNO3(aq)
2N2O5(s)
∆H01 = -414.0 kJ
∆H02 = -86.0 kJ
∆H03 = -571.6 kJ
2H2O(l)
Solution:
Read 5.4 and answer Practice (p.317): 1-4
and Questions (p.318): 1,-6
Unit 3 – Chapter – 5 – Thermochemistry
Page 5
Chemistry 12
Notes on Thermochemistry
5.5 – Standard Enthalpies of Formation
Writing Formation Equations
formation reaction: forming a compound from stable elements at SATP (250C and
one mole of compound in equation
100 kPa) in kJ/mol for
O.K.
1
/2N2(g) + O2(g)
NO2(g)
∆H0f = 34 kJ/mol
stable forms of elements
but
not O.K.
N(g) + 2O(g)
NO2(g)
∆H0f = 34 kJ/mol
unstable forms of elements
summary for writing a formation equation:
a) Write one mole of product in the specified state.
b) Write the reactant elements in their standard states.
c) Choose equation coefficients for the reactants to give a balanced equation yielding one mole of
product.
Practice:
Write the formation equations for:
CH4
NH4Cl
Unit 3 – Chapter – 5 – Thermochemistry
Page 6
Chemistry 12
Notes on Thermochemistry
Using Standard Enthalpies of Formation
Enthalpies (or Heats) of Formation:
Molar enthalpy of formation (∆H0f) is equal to the enthalpy change, ∆H when one mole of the compound
is formed from the elements in their stable forms at SATP (25oC and 1 atm). It is ∆H0 (pronounced 'delta h
naught'). ∆H0 of a solution is for a 1.0 M solution, at 1 atm and 25 oC.
Enthalpies of formation are usually negative quantities.
The standard enthalpy change for a reaction can be calculated from the standard enthalpy changes of each
of the substance in the reaction using this formula.
∆H0reaction =…
SAMPLE PROBLEM #1:
Calculate ∆H0rxn for the following reaction in kJ per mole of cyclopropane, C3H6(g), burned.
2C3H6(g) + 9O2(g)
6CO2(g) + 6H2O(l)
*See Appendix C6*
∆H0reaction = Σ n∆H0f products - Σ n∆H0f reactants
∆H0rxn = [6 mol x ∆H0f of H2O(l) + 6 mol x ∆H0f of CO2(g)] – [2 mol x ∆H0f of C3H6(g) + 9 mol x ∆H0f of O2(g)]
∆H0rxn = [ 6 mol x(-285.8 kJ/mol) + 6 mol x (-393.5 kJ/mol) ] – [ 2 mol x (17.8 kJ/mol) + 9 mol x (0 kJ/mol)]
∆H0rxn = [ -1714.8 kJ + -2361 kJ ] - [35.6 kJ ]
∆H0rxn = -4111.4 kJ/2 mol
= -2055.7 kJ/mol
H0rxn for the reaction in kJ per mole of cyclopropane, C3H6(g), is – 2.0557 x 103 kJ/mol.
Unit 3 – Chapter – 5 – Thermochemistry
Page 7
Chemistry 12
Notes on Thermochemistry
SAMPLE PROBLEM #2:
Calculate ∆H0rxn for the following reaction in kJ per mole of aluminum, Al(s), involved in this single
displacement reaction.
2Al(s) + Cr2O3(s)
Al2O3(s) + 2Cr(s)
*See Appendix C6*
∆H0reaction = Σ n∆H0f products - Σ n∆H0f reactants …
Read 5.5 and answer Practice (p.332): 1, 2
5.5 Questions (p.324): 1 -6
Problem Set #1: Communicating Enthalpies (Due in 2 days)
5.6 – The Energy Debate
Compare, with examples, the energy produced from fossil fuels, syngas, hydrogen fuel and bio fuels. It is
suggested that you read chapter 5.6 and review the questions on page 330 to help guide you.
Problem Set #1: Communicating Enthalpies
Unit 3 – Chapter – 5 – Thermochemistry
Page 8
Chemistry 12
SCH 4U1- ___
Notes on Thermochemistry
Name: ________________________________________
Date: ___________
Dry Lab: Quantity of enthalpy change
Examine the following chart of information about the energy released (enthalpy change) by the burning of
propane in oxygen to produce carbon dioxide and water. The reaction is written below.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
Mass of C3H8(g) burned
* Enthalpy change (∆H)
10 g
227 J
20 g
455 J
40 g
905 J
80 g
1816 J
160 g
3600 J
320 g
7300 J
(* measured using calorimeter)
Answer the following questions on the back of the graphing paper
1) Graph the data that you see by placing enthalpy change on the y-axis and mass of C3H8(g) on the x-axis. Do
you get a straight line? If you did what is the slope of the line?
Unit 3 – Chapter – 5 – Thermochemistry
Page 9
Chemistry 12
Notes on Thermochemistry
2) What does the graph indicate about the relationship between the number of grams of C3H8(g) burned and
enthalpy change?
3) Calculate the number of moles of reactant (C3H8(g)) there are in ten grams. What is the enthalpy change per
mole of C3H8(g) ?
4) Now let's write it into the equation. Since heat is given off chemists add the enthalpy change to the
products of the reaction. What would be the balanced equation with the enthalpy change included?
5) If 2 moles of C3H8(g) is burned what will be the energy change?
6) What then is the relationship between the coefficients of the balanced equation and the energy change in the
reaction?
7) Referring to the reaction below, how much is the change in enthalpy if 120 g of potassium metal reacts?
2 K(s) + 2 H2O(l)
8) The
160 kJ
+ H2(g) + 2 KOH(aq)
reaction below, releases 22 KJ of energy when 3.6 g of calcium is used.
Ca + 2HCl
CaCl2 + H2
Rewrite the chemical equation, incorporating the energy change into the equation.
Unit 3 – Chapter – 5 – Thermochemistry
Page 10
Chemistry 12
Notes on Thermochemistry
SCH 4U1- ___
Name: ________________________________________
Date: ___________
Problem Set #1: Communicating Enthalpies
1) Rewrite the equation in " H " notation for one mole of the underlined substance.
Fe2O3 (s)
+
3 CO(g)
3 CO2(g)
+
2 Fe(s)
+
25 kJ
Answer:
1/3 Fe2O3 (s)
1. 4 NH3 (g)
+
CO(g)
+
CO2(g)
5 O2 (g)
+
2/3 Fe(s)
4 NO (g)
+
2.
2 HCl (g)
+
96 kJ
H2(g)
3.
N2 (g)
+
3 H2 (g)
2 NH3(g)
4.
2 CO2 (g)
5.
4 Al (s)
+
+
566 kJ
3 O2 (g)
H = - 8.3 kJ
6 H2O (l)
+
+
1170 kJ
Cl2 (g)
+
92 kJ
2 CO (g)
2 Al2O3 (s)
+
+
O2 (g)
3360 kJ
2) Draw a scaled potential energy diagram of the energy change that occurs in each of the above
balanced equations. (Place potential energy and enthalpy change on the y-axis and course of the reaction
on the y-axis. Scale the y-axis only.)
3) Rewrite the equation using whole number balanced equations and energy as a term.
3 FeCl3 (s)
3 FeCl2(s)
+
3/2 Cl2 (g)
H
=
+ 173 kJ
=
- 36 kJ
Answer;
6 FeCl3 (s)
+
346 kJ
+
1/2 Br2 (g)
6 FeCl2(s)
+
3 Cl2 (g)
Questions;
1.
1/2 H2 (g)
2.
H2O(g)
3.
3/4 C (s)
4.
1/2 C2H5OH (l)
5. ½ N2H4 (g)
+
SO2 (g)
+
1/2 Fe2O3 (s)
+
+
3/2 O2 (g)
1/2 O2 (g)
Unit 3 – Chapter – 5 – Thermochemistry
HBr (g)
H2S (g)
Fe (s)
H
+ 3/2 O2 (g)
+
CO2 (g)
1/2 N2 (g)
H
3/4 CO2 (g)
+
3/2 H2O (l)
+ H2O (l)
=
+ 520 kJ
H=
+ 117 kJ
H=
- 684 kJ
H = - 1530 kJ
Page 11