Transcript CHE 106 Chapter 5x

CHE 106
Chapter 5
THERMOCHEMISTRY
Important Terminology:
Force:
Work:
Heat:
Energy:
Important Terminology cont…
Kinetic Energy:
Formula:
Potential Energy:
Kinetic vs. Potential Energy
Energy Transfers
System:
Surroundings:
Example:
Energy and the 1st Law
First Law of Thermodynamics:
* Heat:
* Work
Example:
Internal Energy
Definition:
DE = Efinal – Einitial
DE: Magnitude, Sign, Unit:
Example:
Reaction Profiles
CH4 + O2
H2O + CO2
EXOTHERMIC
DPE
PE
CH4 + O2
N2 + O2
2 NO
ENDOTHERMIC
2 NO
DPE
H2O + CO2
N2 + O2
Energy: Heat and Work
DE can be depends on the energy conversions
and exchanges within a chemical system.
Equation:
Sign Conventions for q and w
Sign Conventions: Two
Cases
HEAT
HEAT
WORK
SYSTEM
q = positive (+)
w = positive (+)
DE = (+) endothermic
WORK
SYSTEM
q = negative (-)
w = negative (-)
DE = (-) exothermic
Internal Energy
Example: Calculate DE for the following
situations:
A. A system does 195 kJ of of PV work and
absorbs 38 J of heat.
B. A chemical reaction in a piston chamber
gives off 500 j of heat to its surroundings.
The expanding gas moves the piston
upward and does 240J of work.
C. The chemical reaction from (B) was rerun with the piston in a locked position
and generates 740 J of heat.
Internal Energy and State
Functions
Internal Energy and State
Functions
Example: Sweat Evaporating – Two Ways
Case 1:
System:
Surroundings:
Case 2:
System
Surroundings:
Internal Energy and State
Functions
State Functions
Definition:
Example:
Variables that are State Functions:
Variables that are NOT State Functions:
State Functions
Internal Energy
Although energy is a state function, it is also
considered an extensive property.
Example: The combustion of octane
Internal Energy and Enthalpy
Making math work for us:
DE = q + w
Internal Energy and Enthalpy
Enthalpy Definition:
Relationship between internal energy and
enthalpy: at constant pressure.
H = E + PV
Equation Manipulation:
Final Equation:
Internal Energy and Enthalpy
ENDOTHERMIC:
EXOTHERMIC:
ACTIVATION ENERGY:
Activation Energy
Activation Energy - minimum
energy to initiate a chemical
reaction - barrier to reaction
Activation E
non-state function
A
Eact
DPE
state function
B
Enthalpy and Reaction
Profiles
Enthalpy and Reaction
Profiles
Enthalpy: Heats of Reaction
Most reactions take place at constant pressure
and the change in volume is nearly zero. In
these cases, DE = DH.
There are 4 things to remember when dealing
with enthalpy.
Enthalpy: Heats of Reactions
1. Enthalpy is a state function.
DH = Hproducts – Hreactants
2. Enthalpy is an extensive property.
3. DH for a reaction is equal in magnitude but
opposite in sign for the reverse reaction.
4. DH for a reaction depends on the state of
products and reactants.
Enthalpy: Heats of Reaction
Consider the reaction:
2 N2 (g) + O2 (g)  2 N2O (g) DH = +163.2 kJ
1. Is the reactions exothermic or endothermic?
2. Calculate the amount of heat transferred
when 12.8 grams of N2O (g) forms at constant
pressure.
3. How many grams of N2(g) must react to
produce a DH of 1.00 kJ
Enthalpy: Heats of Reaction
Sample exercise: Hydrogen peroxide can
decompose to form water and oxygen by the
reaction:
2 H2O2 (l)  2 H2O (l) + O2 (g)
DH = -196 kJ
Calculate the value of q when 5.00 g of
hydrogen peroxide decomposes at constant
pressure.
Enthalpy: Heats of Reaction
Hindenburg:
https://www.youtube.com/watch?v=CgWHbpMVQ1U
2H2 + O2  2 H2O DH = -484 kJ
7,062,000 cubic feet of H2 at STP produces how much
energy, given 1 ft3 contains 28.3 L of H2 gas.
Enthalpy: Heats of Reaction
Example: Oxygen may be generated on small scales in
the laboratory by the thermal decomposition of
potassium chlorate:
Equation:
DH = - 89.4 kJ
For this reaction, calculate DH for the formation of
- 6.45 g of O2 (g)
- 9.22 g of KClO3 (s)
Enthalpy: Heats of Reaction
Example: Which of the following has the highest
enthalpy at a given at a given temperature and
pressure:
H2O(s) vs. H2O (l) vs. H2O (g)
Example: Formation of gaseous water vs liquid water:
Enthalpy: Heats of Reaction
When solutions containing silver ions and
chloride ions are mixed, silver chloride
precipitates:
Ag+(aq) + Cl-(aq)  AgCl (s) DH = -65.5 kJ
1. Calculate the DH for the formation of 0.200 mol
AgCl from this reaction.
2. Calculate the DH for the formation of 2.50 g of
AgCl.
3. Calculate DH when 0.350 mol of AgCl dissolves
in water.
Calorimetry
Calorimetry: Measure of heat flow. Can
calculate DH or DE.
Heat Capacity:
Molar Heat Capacity:
Specific Heat Capacity
Calorimetry
Substance
Specific Heat (J °C-1 g-1)
H2O(l)
H2O(s)
4.18
2.03
Al(s)
C(s)
Fe(s)
Hg(l)
CCl4(l)
CaCO3(s)
0.89
0.71
0.45
0.14
0.86
0.85
Calorimetry
Example: A swimming pool, 10.0 m by 4.0 m,
is filled to a depth of 3.0 m with water at 20.2
°C. How much energy is required to raise the
water temperature to 30.0°C?
Example: How much energy is required to
raise the temperature of an 8.50 x 102 g block
of aluminum from 22.8°C to 94.6°C? What is
the molar heat capacity of aluminum?
Calorimetry
• Example: Ti metal is used as a structural
material in many high tech applications
including jet engines.
• What is the specific heat of Ti if it takes
89.7 kJ to raise a 33.0 Kg block by 5.20 °C?
• What is the molar heat capacity of Ti?
Calorimetry
Example: Large beds of rocks are used in
some solar-heated homes to store heat.
A) Calculate the quantity of heat absorbed by
50.0 kg of rocks if their temperature increases
by 12.0°C. Specific heat of rocks is 0.82 J/g-K.
B) What temperature change would these
rocks undergo if they absorbed 450 kJ of heat?
Calorimetry
Two types of calorimetry: Constant Pressure vs. Constant
Volume
Constant Pressure:
Constant Volume:
Calorimetry
Calorimetry
Calorimetry
Consider mixing 50 mL of 1.0 M HCl solution
with 50 mL of 1.0 M NaOH solution, both at
25oC. Upon mixing, the temperature of the
new solution is 31.9oC.
a) How much energy is released in the
problem?
b) How much energy is released in the
sample problem on a molar basis?
Calorimetry
Calorimetry
Heat of Dilution Example:
Reaction:
H2SO4 (conc) + n H2O  H2SO4 (dil)
Where n = moles H2O / Mole H2SO4
10 mL
H2SO4
(conc)
20 mL
H2SO4
(conc)
100 mL
H2O
30 mL
H2SO4
(conc)
100 mL
H2O
Measure Temperature Changes
100 mL
H2O
Calorimetry
n
Heat of dilution of sulfuric acid (H2SO4)
10 mL
H2SO4
(conc)
20 mL
H2SO4
(conc)
100 mL
H2O
30 mL
H2SO4
(conc)
100 mL
H2O
100 mL
H2O
Initial
Temp
25° C
25° C
25° C
Final
Temp
50° C
73° C
95° C
Calorimetry
• Heat of dilution of sulfuric acid (H2SO4)
Reaction:
H2SO4(conc) + n H2O
H2SO4(dil)
(where n = moles H2O/mole H2SO4)
mL H2SO4
10 mL
20 mL
30 mL
mL H2O n
100 mL 31.0
100 mL 15.6
100 mL 10.5
DT (°C)
25.0
48.0
70.0
DHrxn
-11.0 kJ
-26.0 kJ
-30.0 kJ
DHrxn values are dependent upon the amount of
material present
Calorimetry
Example: A coffee cup calorimeter contains
125 g of water at 24.2oC. After KBr (10.5 g) at
24.2oC is added, the temperature becomes
21.1oC.
What is the heat of solution of KBr? Given that
no heat is transferred to the surroundings
from the calorimeter and the specific heat of
the solution is 4.18 J/goC.
Calorimetry
Example: When 50.0 mL of 0.100M AgNO3
and 50.0 mL of 0.100 HCl are mixed in a
constants pressure calorimeter, the
temperature of the mixture increases from
22.30oC to 23.11oC. The temperature increase
is caused by this reaction:
AgNO3 + HCl  AgCl + HNO3
Calculate DH for this reaction.
Bomb Calorimetry
Typically used to study combustion reactions:
System: Sample (hydrocarbon) and oxygen
Surroundings: Bomb and Water
When the reaction takes place, the heat released is
absorbed first by the reaction chamber (bomb)
and then transferred to the surrounding water.
Bomb Calorimetry
Because we are working with constant volume, our
calculations will lead us to the DE. However, because the
work done is equal to zero, the DE and DH are relatively
close to one another.
Equation:
qrxn = -Ccal x DT
The Ccal must be determined experimentally, using a
reaction with a known DH.
Bomb Calorimetry
Example: The combustion of 0.1624g of benzoic acid
(C7H7O2) raises the temperature of a calorimeter by 2.71oC.
Given: DHcomb benzoic acid = -26.73 kJ/g.
What is the heat capacity of the calorimeter?
ANSWER: Ccal = 1.60 kJ/oC.
If 0.2138g of vanillin (C8H8O3) is burned in the calorimeter
and the temperature increases by 3.28oC, what is the heat
of combustion per gram and per mole?
ANSWER: -24.6 kJ/g, -3740 kJ/mol.
Bomb Calorimetry
A 0.1964g sample of quinone (C8H4O2) is burned in a bomb
calorimeter that has a heat capacity of 1.56 kJ/oC. The temperature of
the calorimeter increases by 3.2oC. Calculate the heat of combustion
of quinone per gram and per mole.
ANSWER: -25.4 kJ/g , -2740 kJ/ mol
Under constant volume conditions, the heat of combustion of
glucose is -15.57 kJ/g. A 2.500g sample is burned and the
temperature increased from 20.55oC to 23.25oC.
What is the total heat capacity of the calorimeter?
If the calorimeter contains 2.70Kg of water, what is the heat capacity
of the dry calorimeter.
ANSWER: Ctotal = 14.4 kJ/oC, Cdry = 3.11 kJ/oC
Bomb Calorimetry
A 0.5865g sample of lactic acid is burned in a calorimeter
whose heat capacity is 4.812 kJ/oC. The temperature
increases from 23.10oC to 24.95oC. Calculate the heat of
combustion of lactic acid per gram.
ANSWER: -15.2 kJ/g
Hess’s Law
It’s inefficient to use calorimetric calculations to determine
the enthalpy for every possible reaction. Instead, we take
advantage of enthalpy being a state function, and use preexisting databases to calculate the enthalpy for a reaction
that we haven’t necessarily carried out.
DH is independent of the pathway, so whether a reaction
takes place in 2 steps or 5 steps, the net DH will be the
same.
Hess’s Law: If a reaction is carried out in a series of steps,
DH for the reaction will be equal to the sum of DH’s for
the individual steps.
Hess’s Law
For the reaction: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Step 1: CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) DH = -802 kJ
Step 2: 2 H2O(g)  2 H2O (l) DH = -88 kJ
SUM OF REACTION:
CH4(g) + 2 O2(g) + 2 H2O(g)  CO2(g) + 2 H2O(g) + 2 H2O (l)
DH = -802 kJ + -88 kJ = -890 kJ
Hess’s Law
Determine the DH for A+B C if you are given:
A+ B D
DH1
DE
DH2
EC
DH3
A+B  C
DH(A+B=C) = DH1+ DH2 + DH3
Hess’s Law
Example: Haber process for the formation of ammonia:
3 H2(g) + N2(g)  2 NH3(g)
DHo = -92.2kJ
Step 1: 2 H2(g) + N2(g)  N2H4 (g)
DHo = ?
Step 2: N2H4 (g) + H2(g)  2 NH3(g)
DHo = -187.6 kJ
NET: 3 H2(g) + N2(g)  2 NH3(g)
DHo = -92.2kJ
Solve for DHo(rxn 1)
ANSWER: +95.4 kJ
Hess’s Law
What is the enthalpy of combustion of carbon(s) going to
carbon monoxide (CO)?
The following are known:
C(s) + O2(g)  CO2(g)
CO(g) + 0.5 O2(g)  CO2(g)
ANSWER: -110.5 kJ
DH = -393.5 kJ
DH = -283.0 kJ
Hess’s Law
Water gas is a very important mixture of CO and H2
prepared by passing steam over hot charcoal at 1000°C.
Calculate DH° for the water-gas rxn.
Given:
(1) C(s) + H2O(g)  CO(g) + H2(g) (Water Gas Rxn)
(2) C(s) + O2(g)  CO2(g)
DH° = -393.5 KJ
(3) 2 H2(g) + O2(g)  2H2O(g)
DH° = -483.6 kJ
(4) 2 CO(g) + O2(g)  2 CO2(g)
DH = -566.0 kJ
ANSWER: +131.3 kJ
Hess’s Law
Ex: Calculate DH for 2F2(g) + 2H2O(l)  4HF(g) + O2(g)
• Given:
(1) H2(g) + F2(g)  2HF(g)
(2) 2H2(g) + O2(g)  2H2O(l)
DH = -537 kJ
DH = -572 kJ
ANSWER: -502 kJ
Ex: Carbon occurs in two forms, graphite and diamond.
The enthalpy of combustion of graphite is -393.5 kJ/mol,
and that of diamond is -395.4 kJ/mol. Calculate DH for
the conversion of graphite to diamond:
C(graphite) C(diamond)
ANSWER: 1.9 kJ/ mol
Hess’s Law
Example: Calculate DH for the reaction:
NO + O NO2
Given:
NO + O3
NO2 + O2
DH = -198.9
O3 3/2 O2
DH = -142.3
O2 2O
DH = 495.0
ANSWER: -304.1 kJ
Enthalpy of Formation
Hess’s Law can be used with other thermodynamic data to calculate the
enthalpy for many different reaction types:
Heat of Formation (DH compound from elements) labeled DHf
Heat of formation (DHf) is usually given for reactants and products
in standard states (since DH depends on the state of these items).
When in standard state, the denotation is DH°f
Ex: 2C(s) + 3H2(g) + 1/2O2  C2H5OH(l)
Heat of Vaporization (DH for liquid to gas)
Heat of Fusion (DH for melting solids)
Heat of Combustion (DH reaction with O2)
DH°f = -277.7 kJ
Enthalpy of Formation
Things to remember:
DH is as state function, so we are allowed to apply Hess’s
Law and add together appropriate reactions to calculate
the DHof
The standard enthalpy of formation of the most stable
form of any element is zero. (That way we don’t need to
find out how much energy is required to produce each
element involved in a reaction).
DH°rxn = •nDH°f (products) - mDH°f (reactants)
Where m and n are stoichiometric coefficients from
balanced equation.
Enthalpy of Formation
For example, if we have a simple chemical equation with the
variables A, B and C representing different compounds:
A+B⇋C
And we have the standard enthalpy of formation values as such:
ΔHfo[A] = 433 KJ/mol
ΔHfo[B] = -256 KJ/mol
ΔHfo[C] = 523 KJ/mol
The equation would be:
ΔHoreaction = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B])
ANSWER: 346 kJ
Enthalpy of Formation
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
Given:
DHof of C3H8: -103.85 kJ/mol
DHof of H2O: -285.8 kJ/mol
DHof of CO2: -393.4 kJ/mol
DHorxn: -2220 kJ
Enthalpy of Formation
C6H12O6(s)  2 C2H5OH(l) + 2 CO2(g)
Given:
DHof of glucose: -1260 kJ/mol
DHof of ethanol: -277.7 kJ/mol
DHof of carbon dioxide: -393.5 kJ/mol
ANSWER: -82 kJ
Enthalpy of Formation
Oxyacetylene welding torches burn acetylene gas, C2H2.
Calculate DH° (kJ) for the combustion of acetylene.
2C2H2(g) + 5O2(g)  2H2O(g) + 4CO2(g)
DH°rxn = •nDH°f (products) - •mDH°f (reactants)
Given:
DHof of acetylene gas: 226.7 kJ/mol
DHof of H2O(g): -241.8 kJ/mol
DHof of CO2(g): -393.5 kJ/mol
ANSWER: -2511 kJ for every 2 moles of acetylene.
Enthalpy of Formation
Using the standard enthalpies of formation, calculate
the enthalpy change for the combustion of 1 mol of
ethanol:
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l)
ANSWER: -1367 kJ