chemistry 30 – unit 1b thermochemical changes to accompany inquiry into chemistry powerpoint...
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Chemistry 30 – Unit 1B Thermochemical Changes
To accompany Inquiry into Chemistry
PowerPoint Presentation prepared by Robert [email protected]
Section 10.1
• Hess’s Law: The enthalpy change of a chemical or physical process depends only on the initial and final systems
• It is the sum of the of the enthalpy changes of the individual processes
Chapter 10, Section 10.1• The picture illustrates this – the
gravitational Ep change for the two riders is the same
• It doesn’t depend on the pathway
Chapter 10, Section 10.1• This principle, Hess’s Law, can be
used to determine ∆H for reactions when ∆H can’t be measured by calorimetry
• When using Hess’s Law, don’t worry about pure elements – get compounds with their appropriate coefficients on the proper side of the equation and cancel out any new compounds you add
Chapter 10, Section 10.1• Practice Problem 1, page 374 – find
∆H for the reaction:C2H4(g) + H2O(l) C2H5OH(l) ,
using the following equations:
(1) C2H5OH(l) + 3 O2(g) 3 H2O(l) + 2 CO2(g) ∆H=-1366.8 kJ
(2) C2H4(g) + 3 O2(g) 2 H2O(l) + 2 CO2(g) ∆H=-1411.0 kJ
• The desired equation needs C2H5OH(l) on the right, so flip equation 1
• It needs C2H4(g) on the left, so leave equation 2 as is
Strategy:
Chapter 10, Section 10.1
• 3 H2O(l) + 2 CO2(g) C2H5OH(l) + 3 O2(g) ∆H=+1366.8 kJ
• C2H4(g) + 3 O2(g) 2 H2O(l) + 2 CO2(g) ∆H=-1411.0 kJ
C2H4(g) + H2O(l) C2H5OH(l) ∆H= -44.2 kJ
+
Add the equations and their ∆H’sNote the sign change
Chapter 10, Section 10.1• Potential Energy Diagram for
previous example:
Reaction Coordinate
Ep (
kJ)
C2H4(g) + 3 O2(g)
2 H2O(l) + 2 CO2(g) + H2O(l)
C2H5OH(l) + 3 O2(g)
∆H =-1411.0 kJ ∆H =+1366.8 kJ
∆H=-44.2 kJ
Chapter 10, Section 10.1Example: Practice Problem 2, page 374
Find ∆H for:
Pb(s) + PbO2(s) + 2 H2SO4(l) 2 PbSO4(aq) + 2 H2O(l)
using the equations:
(1) Pb(s) + PbO2(s) + 2 SO3(g) 2 PbSO4(aq) ∆Ho = -775 kJ
(2) SO3(g) + H2O(l) H2SO4(l) ∆Ho = -133 kJ
Strategy: Use equation (1) as is, double equation (2) and flip it
Chapter 10, Section 10.1 Pb(s) + PbO2(s) + 2 SO3(g) 2 PbSO4(aq) ∆Ho = -775 kJ
2 H2SO4(l) 2 SO3(g) + 2 H2O(l) ∆Ho =
+266 kJ
Pb(s) + PbO2(s) + 2 H2SO4(l) 2 PbSO4(aq) + 2 H2O(l) ∆Ho = -509 kJ
+
Handout BLM 10.1.3
Do Nelson Chemistry WS 48
Lab 10.A page 375
Chapter 10, Section 10.1• With the right set of reaction
enthalpy data and reaction equations, using Hess’s Law, you could calculate the enthalpy (change) for any chemical reaction
• Chemists have tabulated data and prepared charts of standard molar enthalpies of formation
Chapter 10, Section 10.1• A formation reaction is a reaction that
forms one mole of product from its elements in their standard states at SATP
• Examples:
• S8(s) + O2(g) SO2(g)
• 2 C(s) + 3 H2(g) + ½ O2(g) C2H5OH(l)
18
Chapter 10, Section 10.1•Your Chemistry 30 Data Booklet, on pages 4 and 5, has a chart of Standard Molar Enthalpies of Formation
•Standard molar enthalpies of formation can be used to:• predict thermal stability• write formation equations including energy terms• predict ∆H for other reactions
Chapter 10, Section 10.1• Substances with very large
negative enthalpies of formation are very thermally stable since it takes a lot of heat to decompose them to their elements
• Which compound on your chart is most stable? Least stable?
• Example: question 4a, page 379
• C(s) + 2 H2(g) CH4(g) + 74.6 kJ
Chapter 10, Section 10.1
• Using enthalpies of formation, Hess’s Law becomes a shorter process
• ∆H’s can be found using the following formula:
products reactantso offH n H n H
Σ (sigma) is a Greek letter meaning “sum of”
stands for standard molar enthalpy of formation
n is moles of substance from the balanced equation
of H
Chapter 10, Section 10.1• Why does this work?
Because standard molar enthalpies of formation refer to reactions with only elements on the left side, and you don’t need to worry about elements when using Hess’s Law
Chapter 10, Section 10.1• Calculate the enthalpy change for:2 ZnS(s) + 3 O2(g) 2 ZnO(s) + 2
SO2(g)
Data: p. 4-5 of Data Booklet:
Now use these numbers directly in the formula. No need to change signs.Why?!
206.0o kJmolf ZnSH
350.5o kJmolf ZnOH
2
296.8o kJmolf SOH
Chapter 10, Section 10.1
2 350.5 2 296.8
2 206.0 3 0
882.6
kJ kJmol mol
kJmol
H mol mol
mol mol
H kJ
The subtraction changes the sign of the reactants
Products are unchanged since they are already on the proper side
Discuss significant digits in these calcs
Chapter 10, Section 10.1• Try Practice Problem 10, page 383
10b is an extension of the question, but you have already done this part in Chapter 9
Chapter 10, Section 10.1• Practice Problem 10a, page 383
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g)
products reactants
1 393.5 2 241.8
o off
kJ kJmol mol
H n H n H
mol mol
1 239.2 0
637.9
kJmolmol
kJ
Chapter 10, Section 10.1
3637.9r CH OH
kJH mol
• 10b Since the equation is balanced with 1 mol of CH3OH(l),
• therefore,
and,
3125637.9 2.49 10 or 2.49
32.05
r
kJmolg
mol
H n H
gkJ MJ
Chapter 10, Section 10.1• Example: Practice Problem 12, page 383C7H16(l) + 11 O2(g) 7 CO2(g) + 8 H2O(l)
Good vocabulary question!
7 16
products reactants
4816.7 * 7 393.5 8 285.8
1
o off
kJ
f C H
kJmol mol
H n H n H
kJ mol mol
l Hmo
7 16
7 393.5 8 285.8 4816.7
1
224.2
kJ kJmol mo
f C H
l
kJmol
mol mol kJ
mH
ol
Chapter 10, Section 10.1• Do Nelson Chemistry Worksheet 49
• Do Nelson Chemistry Worksheet 50
• Do Worksheet 50A and Diploma Exam written response questions
• Discuss Review Questions 2, 4, and 5 page 383
Chapter 10, Section 10.2• Canada, as your text says, has a
very large per capita energy requirement (2nd only to the USA)
• Why?
• cold winter
• large physical size leading to high transport costs
• others?
Chapter 10, Section 10.2• In Canada, most of our energy is
obtained from the exothermic combustion of fuels
• A significant percentage comes from nuclear fission reactors
Chapter 10, Section 10.2• In Alberta, even our electricity
comes mostly from combustion of fuels:
• In 2008, 48.8 % came from combustion of coal
• 38.4% came from combustion of natural gas
• 7.1% came from hydroelectricity
Chapter 10, Section 10.2• How do the relative amounts of
energy from phase, chemical, and nuclear changes compare?
• phase changes: 100 to 102 kJ/molsource: intermolecular forces (bonds)
• chemical changes: 102 to 105 kJ/molsource: chemical bonds (also called intramolecular bonds)
• nuclear changes: 106 to 1010 kJ/molsource: nuclear bonds or intranuclear bonds, the bonds that hold particles of the nucleus together
Chapter 10, Section 10.2• Efficiency – the ratio of useful energy
produced (energy output) to energy used in its production (energy input), in % form
100%energy output
efficiencyenergy input
Chapter 10, Section 10.2• Example: Practice Problem 13,
page 387• Note that the molar enthalpy of
combustion chart referred to in the question is not present in your Data Booklet, so we will do this question without the aid of the table
Chapter 10, Section 10.2
products reactants
4 393.5 5 241.8
1 125.7
2657.3
ff
kJ kJmol mol
kJmol
H n H n H
mol mol
mol
kJ
Since equation is balanced with 1 mol of C4H10 , 4 10
2657.3kJmolr C HH
• C4H10(g) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g)
Chapter 10, Section 10.2• If the butane lighter was 100%
efficient in heating the spoon, 0.70 g of butane would release:
0.702657.3 32
58.14
r
kJmolg
mol
H n H
gkJ
• Actual amount of useful energy absorbed:
0.250 0.503 45 5.7
5.7100% 100% 18%
32
kJkg C
Q mc t
kg C kJ
useful output kJefficiency
energy input kJ
Chapter 10, Section 10.2
• Try Practice Problem 14, page 387
• Aluminium’s specific heat capacity, which you’ll need for the question is given on page 3 of the Data Booklet with a few others:
0.897 Jg C
Chapter 10, Section 10.2• Practice Problem 14, page 387
• Energy input:
• Useful energy output:
2.50 50 125kJgg kJ
0.500 3.77 45 0.050 0.897 45
87
soup pot
kJ kJkg C kg C
Q mc t mc t
kg C kg C
kJ
87100% 69%
125kJ
efficiencykJ
Chapter 10, Section 10.2• The previous problem, Practice
Problem 14, page 387 is a more difficult question than you could expect to see on the Diploma Exam or on my test
Chapter 10, Section 10.2 – 10.3
• Read pages 388 to 398
• Why does efficiency decrease dramatically as the number of stages in a process increases?
• Do question 8 page 392 and questions 3, 4, 5, and 6 page 398
Chapter 10, Section 10.3