chapter 17 energy and chemical change
DESCRIPTION
Chapter 17 Energy and Chemical Change. Thermochemistry. The study of heat changes in chemical reactions. Law of Conservation of Energy. Defn – energy can be converted from one to another, but neither created nor destroyed Ex: potential to kinetic solar to chemical. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 17 Energy and Chemical Change
Thermochemistry
• The study of heat changes in chemical reactions
Law of Conservation of Energy
• Defn – energy can be converted from one to another, but neither created nor
destroyed
• Ex: potential to kinetic
solar to chemical
Energy and Chemical Change
• Energy (defn) – ability to do work or produce heat
• Exists in two formsa) potential energy – stored energy
b) kinetic energy – energy in motion
Chemical Potential Energy
• Defn – energy stored in a substance based on composition
• Ex: methane (CH4) vs. propane (C3H8)
C HH
H
H
C HH
H
H
C
H
H
C
H
H
Propane has more energy b/c it has more bonds
Heat (q)
• Defn – energy that goes from warmer object to colder object
• Energy transfer moves from an area of high energy to an area of low energy
Units of Heat
• calorie (cal) – amount of heat required to raise the temp of 1 gram of water one degree Celsius– Calorie vs calorie
1 Calorie = 1000 calories = 1 kilocalorie (kcal)• Joule – SI unit of heat and energy
1 calorie = 4.18 J
Comparing Specific Heatwater (liquid)
aluminum
iron
water (ice)
4.18
2.03
0.897
0.449
ethanol 2.44
Specific Heat
• Defn – amount of heat required to raise temp of one gram of any substance by one degree Celsius
q = m c ΔT
heat mass specificheat
changeIn temp
Unit:J
g ̊ C
Specific Heat
• If heat is released from the reaction: -q
• If heat is absorbed into the reaction: +q
• Change in temperature = tfinal - tinitial
Sample problem (a)
• The temperature of a 10.0 g sample of iron is changed from 50.4°C to 25.0°C with the release of 114 J of heat. What is the specific heat of iron?
q = mcΔT
114 J = (10.0 g) c (25.4°C)
c = 0.449 J/g·°C
Sample problem (b)
• If the temperature of 34.4 g ethanol increases from 25°C to 78.8°C, how much heat is absorbed by ethanol? (specific heat of ethanol = 2.44 J/g°C)
q = mcΔT
= (34.4 g) ( 2.44 J/g°C) (53.8°C)
= 4515 J
Heat In Chemical Reactions
• Calorimeter –
insulated device used to measure amount of heat absorbed or released during a chemical or physical process
• Thermochemistry – study of heat changes in chemical reactions
3 parts we look at
• 1) system – specific part of universe that contains the reaction
• 2) surroundings – everything in universe other than the system
• 3) universe – system + surroundings
Enthalpy (H)
• Defn – heat content of a system
• Enthalpy (heat) of rxn (ΔHrxn)
a) defn – change in enthalpy for a reaction
b) formula
ΔHrxn = Hproducts – Hreactants
A + B C + D
Enthalpy (H)
• c) endothermic vs exothermic rxns
if +ΔHrxn = endothermic rxn
if –ΔHrxn = exothermic rxn
Reaction Energy Diagrams
• This is an exothermic reaction – the reactants have more energy than the products, so energy has been released
Reaction Energy Diagrams
• This is an endothermic reaction – the reactants have less energy than the products, so energy has been absorbed
Reaction Energy Diagrams
• A: energy held by the activated complex
• B: energy of the reactants
• C: energy of the products
• F: heat of reaction (ΔH)
• I: activation energy
H H
products
reactants products
reactants
EXOTHERMIC ENDOTHERMIC
ΔH ΔH
ΔH < 0 ΔH > 0
Example reactions
• 4 Fe + 3 O2 2 Fe2O3 + 1625 kJ
i) exo- or endo-?
ii) what is ΔHrxn?
Exothermic (heat written on right side of equation)
ΔHrxn = -1625 kJ
Another way to write equation:4 Fe + 3 O2 2 Fe2O3 ΔHrxn = -1625 kJ
2 Fe2O3
4 Fe + 3 O2
ΔH = -1625 kJH
Example reactions
• 27 kJ + NH4NO3 NH4+ + NO3
1-
i) exo- or endo-?
ii) what is ΔHrxn?
Endothermic (heat written on left side of equation)
ΔHrxn = +27 kJ
Another way to write equation:NH4NO3 NH4
+ + NO31- ΔHrxn = +27 kJ
H
NH4+ + NO3
-
NH4NO3
ΔH = +27 kJ
Standard Enthalpy of Formation (ΔHf
°)• Defn – change in enthalpy when one mole
of a compound is formed from its elements in their standard states
• Standard State – normal physical state of substance at room conditions (25°C and 1 atm)
ex: standard state of Hg is liquid
N2 is gas
Standard Enthalpy of Formation (ΔHf
°)• Examples
H2 (g) + S (s) H2S (g) ΔHf° = -21 kJ
S (s) + O2 (g) SO2 (g) ΔHf° = -297 kJ
Hess’s Law
• Defn – overall enthalpy change of reaction is equal to the sum of the enthalpy
changes of individual steps
A D ΔH = ?
A + B C ΔH = x
C D + B ΔH = y
Overall: A D ΔH = x + y
Example problem #1
• Calculate the enthalpy of reaction, ΔH, for the reaction:
2 H2O2 2 H2O + O2
(a) H2 + O2 H2O2 ΔH = -188 kJ
(b) 2 H2 + O2 2 H2O ΔH = -572 kJ
Example problem #1
2 H2O2 2 H2 + 2 O2 ΔH = +376 kJ
2 H2 + O2 2 H2O ΔH = -572 kJ
2 H2O2 + 2 H2 + O2 2 H2O + 2 H2 + 2 O2
2 H2O2 2 H2O + O2
ΔH = -196 kJ
1
Example problem #2
• Calculate the enthalpy of reaction, ΔH, for the reaction:
2 S + 3 O2 2 SO3
(a) S + O2 SO2 ΔH = -297 kJ
(b) 2 SO3 2 SO2 + O2 ΔH = 198 kJ
Example problem #2
2 S + 2 O2 2 SO2 ΔH = -594 kJ
2 SO2 + O2 2 SO3 ΔH = -198 kJ
2 S + 3 O2 + 2 SO2 2 SO2 + 2 SO3
2 S + 3 O2 2 SO3 ΔH = -792 kJ
Heat of Reaction (ΔHrxn)
• Defn – amount of heat lost or gained in a reaction
• Formula
ΔHrxn = Σ ΔHf (products) – Σ ΔHf (reactants)
Example problem
• Find ΔHrxn for the following reaction. Is reaction endo- or exothermic?
H2S + 4 F2 2 HF + SF6
ΔHrxn =
= -1745 kJ
[(2)(-273) + (1)(-1220)] - [(1)(-21) + (0)(4)]
exothermic
-273 kJ -1220 kJ-21 kJ 0 kJ
Entropy (S)
• Defn – measure of disorder or randomness in a system
• Formula
ΔS = Sproducts – Sreactants
• Reaction Tendency– Nature favors a disordered state– The more entropy/disorder, the greater ΔS
Unit: J/K
Entropy (S)
• Predicting ΔS
- go to more disorder + ΔS
- go to less disorder - ΔS• Ex problem: predict ΔSsystem for these
changes
a) H2O (s) H2O (l)+ ΔS; Solid to liquid is more disorder
Entropy (S)
b) 2 SO3 (g) 2 SO2 (g) + O2 (g)
• Keep in mind: the reverse reactions have opposite signs
There are more product particles (3) than reactant particles (2)
+ ΔS
Entropy example problem
• Calculate entropy change for this reaction:
2 PbO2 (s) 2 PbO (s) + O2 (g)
ΔSrxn =
= +209.2 J/K
[(2)(68.7) + 205] - [(2)(66.6)]
68.7 20566.6
Entropy, the Universe, and Free Energy
• The universe entropy
ΔSuniverse > 0
• Two natural universe processes
a) things tend to go towards lower energy (-ΔH, exothermic)
b) things tend to go towards higher disorder (+ΔS)
Spontaneous Process
• Defn – physical or chemical change that occurs with no outside intervention
Free Energy (G)
• Defn – energy available to do work• Gibbs Free Energy Equation
ΔG = ΔH – TΔS
T is in Kelvin• What does ΔG tell you?
- ΔG spontaneous rxn
+ ΔG not spontaneous rxn
Free Energy Problem
• calculate the change in free energy, ΔG, for the reaction at 25°C. Is the reaction spontaneous or nonspontaneous?
N2(g) + 3 H2 (g) 2 NH3 (g)
ΔHrxn = -91800 J, ΔSrxn = -197 J/K
Free Energy Problem
ΔGrxn = ΔHrxn – TΔSrxn
= -91800 J – (298 K)(-197 J/K)
= -33100 J Spontaneous
How does ΔH and ΔS affect spontaneity?
+ΔS
-ΔS
-ΔH +ΔH
Alwaysspontaneous
neverspontaneous
Spontaneity depends on
temp
Spontaneity dependson temp