Transcript Thermochemistry PPT

Thermochemistry
Chapter 5
1
Thermochemistry
• Thermochemistry:
Relationships between chemical reactions
and energy changes
• Thermodynamics:
• Therm: heat
Dynamics: power
2
Energy
Energy: the capacity to do work
•
Radiant energy
•
Thermal energy
•
Chemical energy
•
Nuclear energy
•
Potential energy- relative to position- stored in bonds
electrostatic potential energy (C) = kQ1Q2
d
•
Kinetic energy Ek = ½ mv2
•
Units of Energy: J = 1kg-m2/s2 1 cal= 4.184 J
3
Energy Changes in Chemical Reactions
Which one has more thermal energy?
900C
Coffee cup?
900C
Bathtub?
4
Energy Changes in Chemical Reactions
Which one has more thermal energy?
900C
What additional
information is
needed?
400C
How does it relate to heat?
5
Energy Changes in Chemical Reactions
Heat:
the transfer of _______ between two bodies that are
at ________temperatures.
Temperature is a measure of the ________.
Measure of the ______________
Temperature = Thermal Energy
6
Energy Changes
The system is the specific part of the universe that is of
interest in the study.
SYSTEM
SURROUNDINGS
open
Exchange: mass & energy
closed
isolated
energy
nothing
7
Energy Changes
8
Enthalpy
Exothermic process: is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat is supplied
to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
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Enthalpy
• Enthalpy: a thermodynamic quantity used to
describe heat changes(absorbed/released) at
constant pressure (most reactions occur at constant
pressure).
H = E + PV
Equation:
ΔHreaction = ΔHproducts - ΔHreactants
10
Enthalpy: Enthalpy Diagrams
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
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DH > 0
Enthalpy: Enthalpy Diagrams
ΔHrxn= ΔHproducts – ΔHreactants
12
Enthalpy: Examples
Indicate the sign of enthalpy change in the
following processes carried under atmospheric
pressure, and indicate whether the process is
exothermic or endothermic:
1.
2.
3.
An ice cube melts
1 g of butane (C4H10) is combusted in sufficient
oxygen to give complete combustion to CO2 and
H2O
A bowling ball is dropped from a height of 8 ft into a
bucket of sand.
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Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s) → H2O (l)
DH = 6.01 kJ
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6.3
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
DH = -890.4 kJ
15
6.3
Thermochemical Equations: Rules
Stoichiometric coefficients: equal the number of moles
of a substance
H2O (s)
H2O (l)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l)
H2O (s)
DH =
- 6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 12.0 kJ
2 x 6.01kJ/mol
=12.0 KJ/2 mol
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Thermochemical Equations: Rules
When a reaction
is reversed, the
magnitude of ΔH
remains the same,
but its sign
changes.
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Thermochemical Equations: Rules
The physical states of all reactants and
products must be specified in thermochemical
equations.
H2O(s) → H2O(l)
H2O(l)
→ H2O(g)
ΔH = 6.00 kJ
ΔH = 44.0 kJ
• What will be the ΔH when 1 mole of ice at
0ºC is changed into one mole of steam at the
boiling point?
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Thermochemical Equations: Rules
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
-3013 kJ
1 mol P4
DH = -3013 kJ
[ -6470 kJ]
19
Enthalpy: Examples
4. The complete combustion of liquid octane, C8H18, to
produce carbon dioxide and liquid water at 25 ºC and
at constant pressure gives off 47.9 kJ per gram of
octane. Write a chemical equation to represent this
information.
[2C8H18(l) + 25 O2(g) → 16 CO2(g) + 9H2O(l)
ΔH = -1.09 x 104 kJ]
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Enthalpy: Examples
5.
Given the thermochemical equation:
SO2(g) + ½ O2(g) →SO3(g) ΔH = -99.1 kCal
Calculate the heat evolved when 74.6 g of SO2
(MM = 64.07 g/mol) is converted to SO3.
[ -115 kJ]
6.
Hydrogen peroxide can decompose to water and oxygen
by the following reaction:
2H2O2(l) → 2H2O(l) + O2(g) ΔH = -196 kJ
Calculate the value of q (heat evolved or absorbed) when
5.00 g of H2O(l) decomposes at constant pressure.
[ -14.4 kJ]
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Enthalpy: Examples
7.
Given the equation
H2(g) + I2(s) → 2HI(g)
ΔH = +52.96 KJ
Calculate ΔH for the reaction
HI(g) → ½ H2(g) + ½ I2(s)
[ -26.48 kJ]
8.
Given the equation:
3O2(g) → 2O3(g)
ΔH = +285.4 kJ
Calculate ΔH for the following reaction:
3/2 O2(g) → O3(g)
[ 142.7 kJ]
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Calorimetry and Heat Exchanges in a
Chemical Reaction
• Read Section 6.4 (Pages 213-2200)
• Examples: pp. 216 – 220: 6.1 – 6.3
• Exercises: pp. 240: 6.19, 6.23, 6.26, 6.28
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Calorimetry
Calorimetry is
performed in
devices called
calorimeters
Calorimetry is
based on the
Law of
Conservation of
Energy
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Calorimetry
9. The specific heat of aluminum is 0.895 J/g ºC.
Calculate the heat necessary to raise the
temperature of 40.0 g of aluminum from -20.0
ºC to 32.3 ºC. Specific heat of aluminum is
0.215-cal/g ºC.
[1.87 x 103 Joules]
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Calorimetry: Heat Capacity
•
•
•
Calorimetry:
Experimental measurement of heat produced
in chemical and physical processes
Heat Capacity, C, of a system:
the quantity of heat needed to raise the
temperature of matter 1 K.
Units: J/ºC
q
C
DT
• Molar Heat Capacity:
The heat required to raise 1 mole of a
substance, 1 K.
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Calorimetry: Specific Heat
• Specific heat of a substance
(also represented by cp):
the amount of heat (q) required to raise the
temperature of one gram of the substance by
one degree Celsius ( or 1 K).
heat capacity C
Specific heat 

mass
m
Specific heat :
q
but C 
DT
C
q 1
q
cp 

( ) 
m
DT m
mDT
q  mc p DT
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Calorimetry: Definitions
Heat (q) absorbed or released:
q = mcpDt
q = CDt
Dt = tfinal - tinitial
When q>0 endothermic reaction
When q<0 exothermic reaction
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Calorimetry: Specific Heat
10.
How much heat, in joules and kilojoules, does it
take to raise the temperature of 225 g of water from
25.0 ºC to 100.0 ºC
[ 7.05 x 104 J; 70.5 kJ]
11. Calculate the heat capacity, specific heat and the
molar heat capacity of an aluminum block that has
mass of 5.7 g that must absorb 629 J of heat from
its surroundings for its temperature to rise from 22
ºC to 145 ºC.
[C = 5.11 J/ºC; cp = 0.90 J g-1 ºC-1, 24 J mol-1 ºC-1 ]
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Calorimetry: Problems
12. What will be the final temperature if a 5.00-g
silver ring at 37.0 ºC gives off 25.0 J of heat to
the surroundings? Specific heat of silver = 0.235
J g-1 ºC-1
[15.7 ºC]
13. A 15.5 g sample of a metallic alloy is heated to
98.9 ºC and then dropped into 25.0 g of water in
a calorimeter. The temperature of the water
rises from 22.5 ºC to 25.7 ºC. Calculate the
specific heat of the alloy.
[0.29 J g-1 ºC-1]
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Calorimetry: Problems
14. Suppose a piece of gold with a mass
of 21.5 g at temperature of 95.00 ºC is
dropped into an insulated calorimeter
containing 125.0 g of water at 22.00 ºC.
What will be the final temperature of the
water?
[22.4 ºC]
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Measuring Enthalpy Changes for
Chemical Reactions: ConstantVolume Calorimetry
Bomb Calorimeter
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Constant Volume Calorimeter: Isolated
System
Isolated system:
No exchange of
matter and energy
Uses Cp of water
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Constant-Volume Calorimetry: Isolated system
qsys = qwater + qbomb + qrxn = 0
There is no heat exchange with the surroundings
qrxn = - (qwater + qbomb)
Cbomb= mbombx cbomb
qbomb = CbombDt
qwater = mwcpΔt
qrxn = -(qwater + CbombΔt) = -(mwcpΔt + CbombΔt)
qrxn = - (mwcp + Cbomb)Δt
Measured heat is not enthalpy:
constant volume
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Constant Volume Calorimetry: Problems
15. In a preliminary experiment, the heat
capacity of a bomb calorimeter assembly is
found to be 5.15 kJ/°C. In a second
experiment. A 0.480 g sample of graphite
(carbon) is placed in the bomb with an excess
of oxygen. The water, bomb, and other
contents of the calorimeter are in thermal
equilibrium at 25.00 °C. The graphite is
ignited and burned, and the water
temperature rises to 28.05 °C. Calculate ΔH
for the reaction.
C(graphite) + O2(g) → CO2(g)
[ΔH = -393 kJ]
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Measuring Enthalpy Changes for
Chemical Reactions
Coffee-Cup Calorimeter:
Constant pressure
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Constant-Pressure Calorimetry
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcalor)
qwater = mcpDt
qcal = CcalDt
Reaction at Constant P
ΔH = qrxn = - (qwater + qcalor)
No heat enters or leaves!
Insulated system!
Pressure is constant.
Measured heat is the
enthalpy
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Calorimetry
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Constant Pressure Calorimetry: Problems
16. (a) A 50.0 mL sample of 0.250 M HCl at 19.50 °C is
added to 50.0 mL of 0.250 M NaOH, also at 19.50 °C, in
a calorimeter. After mixing, the solution temperature
rises to 21.21 °C. Calculate the heat of this reaction.
Assume te volumes of the solutions are additive. Ignore
the heat absorbed by the calorimeter.
[-715 J]
(b) Determine the value of ΔH that should be written in
the equation for the neutralization reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔH = ?
[-57.2 kJ]
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Calorimetry: Problem
17. A quantity of 1.00 x 102 mL of 0.500M HCl is
mixed with 1.00 x 102 mL of 0.500 M NaOH in a
constant pressure calorimeter having a heat
capacity of 335 J/ ºC. The initial temperature of the
HCl and NaOH solutions is the same, 22.50 ºC, and
the final temperature of the mixed solution is 24.90
ºC. Calculate the heat change for the neutralization
reaction and the molar heat of the neutralization
reaction. Assume s = 4.18 J/gºC for the solution.
[-2.81 kJ; -56.2 kJ/mol]
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Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol
1 cal = 4.184 J
One food Calorie:
1 Cal = 1000 cal = 4184 J
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Standard Enthalpy of Formation
Absolute value of the enthalpy of a substance cannot be
measured. We need a frame of reference.
Standard states for elements and compounds has to be
established.
Standard state for a solid or liquid substance is the pure
element or compound at 1 atm at a given temperature.
Standard state for gas: pure gas behaving as an ideal
gas at 1 atm and temperature of interest.
Standard enthalpy of formation: the enthalpy change for
a reaction in which the reactants in their standard state
yield products in their standard states. Standard state
denoted with a superscript (º)
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Standard Enthalpy of Formation
Standard enthalpy of formation ΔHºf (also called heat of
formation
enthalpy change that occurs in the formation of 1 mole of
the substance from its elements when both products and
reactants are in their standard states (f stands for
formation).
Reference forms:
standard enthalpy of formation of a pure element in its
reference form is 0.
standard enthalpy of formation (ΔH0) as a reference
point for all enthalpy expressions is the formation of H2(g)
= 0 kJ/mol
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Standard Enthalpy of Formation
Standard enthalpy of formation (DH0f) is the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm and 25 ºC .
The standard enthalpy of formation of any element in
its most stable form is zero.
DH0f (O2) = 0
DH0
f
(O3) = 142 kJ/mol
DH0f (C, graphite) = 0
DH0f (C, diamond) = 1.90 kJ/mol
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standard
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Determination of Heat of Formation
1. Direct method: From elements when it can be easily
measured.
C(graphite) + O2(g) →CO2(g) ΔHrxn = -393.5 kJ
ΔHrxn= ΔHoCO2 – ΔH0O2 – ΔHgraphite = -393.5 kJ
ΔH0CO2 = -393.5 kJ
2. Indirect method: Using the heat of reaction and
heats of formations
Given CH4(g) +2O2(g) → CO2(g) + 2H2O(g) ΔHrxn=-890.3 kJ
Calculate the heat of formation of CH4
[-74.8kJ/mol]
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Standard Enthalpy of Reaction
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm and 25ºC.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
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Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
= - 2973 kJ/mol C6H6
2 mol
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Standard Enthalpy of Reaction
18. Synthesis gas is a mixture of carbon monoxide and
hydrogen that is used to synthesize a variety of
organic compounds, such as methanol. One reaction
for producing synthesis gas is shown here:
3CH4(g) + 2H2O(l) + CO2(g) → 4CO(g) + 8H2(g)
ΔHº = ?
[747.5 kJ]
19. The combustion of isopropyl alcohol, common
rubbing alcohol, is represented by this equation:
2(CH3)2CHOH(l) + 9O2(g) → 6CO2(g) + 8H2O(l)
ΔHº = -4011 kJ
use this equation and data from Thermodynamic
tables to establish the standard enthalpy of formation
for isopropyl alcohol.
[ -318 kJ/mol]
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Heat of Formation: Examples
20. Pentaborane-9, B5H9, is a colorless liquid. It is highly
reactive substance that will burst into flame or even
explode when exposed to oxygen:
2B5H9(l) + 12O2 → 5B2O3(s) + 9H2O(l)
It was considered a potential rocket fuel as it release large
amount of heat per gram. Calculate the kJ of heat
released per gram of the compound reacted with oxygen.
The standard enthalpy of formation of B5H9 is 73.2 kJ/mol
[-71.58 kJ/gB5H9]
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Heat of Reactions (Formations):
Hess’s Law of Constant heat of Summations
Hess’s Law: When reactants are converted to
products, the change in enthalpy is the same
whether the reaction takes place in one step
or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
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Hess’ Law: Example
Given the following reactions:
CO(g) + ½ O2(g) → CO2(g)
C(graphite) + O2(g) → CO2(g)
ΔH = -283.0 kJ
ΔH = -393.5 kJ
Calculate the enthalpy for the formation of CO
from its elements
C(graphite) + ½ O2(g) → CO(g)
ΔH = ?
52
Hess’s Law: Example
53
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
C(graphite) + 2S(rhombic)
6.6
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
CS2 (l)
DH0rxn= -393.5 + 2(-296.1) + 1072 = 86.3 kJ
54
Hess’ Law: Examples
21. Given
C(graphite) + O2 (g)→ CO2(g)
ΔHºrxn= -393.5 kJ
H2(g) + ½ O2(g) → H2O(l)
2C2H2(g) + 5O2(g) → CO2(g) + 2H2O(l)
ΔHºrxn = -285.8 kJ
ΔHºrxn = -2598.8 kJ
Calculate the heat of formation of C2H2(g)
2C(graphite) + H2(g) → C2H2(g)
[52.3 kJ]
55
Hess’ Law: Conceptual Example
Without performing a
calculation, determine
which of the two
substances should
yield the greater
quantity of heat upon
Complete combustion,
on a per mole basis:
ethane, C2H6(g), or
ethanol, CH3CH2OH(l)
56
The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788kJ – 784kJ = 4 kJ/mol
57
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hfinal soln - Hcomponents
Which substance(s) could be
used for melting ice?
Which substance(s) could be
used for a cold pack?
Which substances could be
used for hot pack?
58
Heat of Solution and Dilution
The process: NaCl(s) → Na+(aq) + Cl-(aq)
The steps:
1. Separate ions in the crystal into ions in
gaseous state
2. Hydration of the gaseous ions
Energy involved:
Lattice Energy, U
Hydration energy, ΔH hydration
ΔH solution= U + ΔH hydration
59
Heat of Solution and Dilution
Lattice energy: energy required to completely
separate one mole of solid ionic compound into
gaseous ions (endothermic process)
U + NaCl(s) →Na+(g) + Cl-(g)
Hydration of the gaseous ions
Heat of hydration: energy released when
ions interact with water molecules.
Na+(g) + Cl-(g) → Na+(aq) + Cl-(aq) + Energy
60
Heat of Dilution
The heat change associated with the
dilution of a solution.
Can be exothermic or endothermic
61
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