Download Chapter Six Energy Relationships in Chemical Reactions

Chapter Six
Energy Relationships in
Chemical Reactions
1
Energy (U): Capacity to Do Work
Radiant energy
Energy from the sun
Nuclear energy
Energy stored in the nucleus of an atom
Thermal energy
Energy associated with temperature
Kinetic energy: molecular movement
Chemical energy
Energy stored in chemical bonds
Potential energy: Object postion
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Thermochemistry
Study of heat change in chemical reactions
System: The part of the universe being studied
Open: Energy & matter exchange with surrounding
Closed: Only energy exchange with surroundings
Isolated: No energy or matter exchange- rare
Surroundings: Part of the universe not being studied
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Heat Transfer in a System
Heat, q: Transfer of energy due to temperature difference
Exothermic Reaction: System gives off heat (Exiting)
Ex: Methane Burning: CH4 + O2 → CO2 + H2O
Bonds stronger in CO2 and H2O molecules than in CH4 + O2
Heat goes from system to surroundings
Endothermic Reaction: System gains heat (Entering)
Ex: Ice Melting: H2O (s) → H2O (l)
Energy needed to break attractions between H2O molecules
Heat goes from surroundings into the system
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Thermodynamics
Study of the conversions between
heat & energy
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Using Thermodynamics
State of a system:
Describes specific conditions defined by state functions
Measurement of these properties
Potential energy of ball at the top of the mountain all
State functions:
Properties defined by final - initial values only
Composition, temperature, pressure, energy, volume
Height of mountain
Path functions:
final - initial values vary by process
Heat, work: interdependent variables
# steps taken, Time to top
energy per step, etc.
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First Law of Thermodynamics
Energy can be converted from 1 form to another but
cannot be created or destroyed
Usystem= Ufinal-Uinitial
Internal Energy of System:
Usystem = Ukinetic+ Upotential
Kinetic energy: Amount of molecular motion
Associated with temperature
Potential energy: Energy stored in bonds
Strength of chemical bonds
Cannot separate two types of energy
For chemical reactions: Ureaction = Uproducts- Ureactants
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Work and Heat
Energy transfer is caused by changes in heat & work in a system
Usystem = q + w
Work (w) = force x distance = -PV
Negative work (-w): System loses energy to surroundings
Positive work (+w): System gains energy from surroundings
Work (w) = -PV
V = Vf -Vi
For this situation:
+V
-w
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Work and Heat as Path Functions
A gas expands in volume from 2L to 4L at STP. 350J of energy
are needed to maintain the new conditions.
Usys = Ufinal- Uinitial =350J = q + w
V = 2L
P = 1atm
T= 273K
a. Calculate work done against a vacuum & a pressure of 1atm.
w = -PV = -(0atm x 2L) = 0atm L = 0J
w = -PV = -(1atm x 2L) = -2atm L = -202.6J
(1Latm= 101.32J)
b. Calculate the heat required for each system.
q = Usys - w
Under a vacuum: q = 350J-0J = 350J (All energy is heat)
Under 2.0atm:
q = 350J-(-202.6J) = 552.6J (need extra heat!)
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Enthalpy
of
Chemical Reactions
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Pressure/Volume Relationships with U
For systems at constant pressure (usually ~1atm):
U = q + w = qp - PV
Some work used to expand system against surroundings.
Enthalpy: (H) Experimentally measured and tabulated
At constant pressure:
Assume negligible volume change:
H= U + PV
H= U = qp
There will be slight differences between H and U
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Real Differences between U and H
H measured experimentally as heat
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
H=-367.5kJ/mol
Volume change of H2 gas
V=nRT/P
T= 298K, P=1.0atm, n=1.0mol H2
V= 24.5L
U = H- PV
H = -367.5kJ
PV = 1.0atm x 24.5L = 24.5Latm x 0.101kJ/Latm = 2.5kJ
U = -367.5kJ -2.5kJ= 370kJ
Difference between H and U is only 2.5kJ
~0.7% difference considered negligible
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Enthalpy of Reaction (H = qrxn)
Amount of heat exchanged between system and
surroundings during a chemical reaction
Hreaction = Hproducts- Hreactants
Exothermic Reactions
Temperature increases in an isolated system.
Heat released to surroundings is treated as a product
2SO2 (g) + O2(g) 2SO3 (g) + 197.8kJ
H = -197.8kJ
Endothermic Reactions
Temperature decreases in an isolated system.
Heat absorbed by system is treated as a reactant
197.8kJ + 2SO32SO2 (g) + O2(g)
State function: Independent of path
H = +197.8kJ
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Rules for Working with Heat and Enthalpy
1. Use same multiplier as used in chemical reaction
2SO2 (g) + O2(g) 2SO3 (g)
H = -197.8kJ
Decrease or increase reactants and products, do the same for H
1SO2 (g) + 1/2O2(g) 1SO3 (g)
H = -197.8kJ/2= -98.9kJ
Note: Write H with units that match mole ratio in equation
H = -98.9 kJ /mol SO3(g)
2. Reverse the reaction, reverse sign of H
2SO3
 2SO2 (g) + O2(g)
H = +197.8kJ
Use rules to calculate the H for any amount of product or reactant
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Calculate how much heat is required to
decompose15.0g NO2(g) according to this reaction:
2NO(g) + O2(g)  2NO2(g) H = -114 kJ
Heat required for following reaction:
H = +114 kJ
2NO2(g)  2NO(g) + O2(g)
H = +114 kJ is needed to decompose of 2 moles NO2
Calculate molar mass of NO2 to convert mass to moles of NO2.
moles NO2 
15.0 g NO2
1
x
1mol NO2
46.0 g NO2
 0.326mol NO2
Find the amount of heat for the amount of NO2 actually present.
H rxn
 114kJ 0.326mol NO2

x
 18.6kJ
2mol NO2
1
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Calorimetry
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Calorimetry
Calorimetry: Measurement of heat changes
Calorimeter:
Device to measure heat (T) produced by a chemical reaction
T= Tfinal –Tinitial
Units: C
Specific Heat (s): Heat needed to raise T of 1 gram by 1C.
s = q/(m x T)
Units: J/gC
swater = 4.184 J/gC
Heat capacity (C): Quantity of heat needed to raise T by 1C
C=mxs
Units: J/C or J/K
Heat of Reaction (qrxn) = m x s x T
Units: J
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Commonly Used Specific Heats
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Calorimetry Calculations
Reaction (system) heat is transferred to water (surroundings)
qsys = -qH2O
1. Measure T of water
qH2O = mH2O x sH2O x TH2O
Use q H2O to find system variables
Crossover: qH2O = -qsys
qrxn= msys x ssys x Tsys
Always have 2 sets of variables!
1 for water (surroundings)
1 for reaction (system)
Make 2 columns of variables
System: s, m, Tinitial, Tfinal
Water: s, m, Tinitial, Tfinal
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A 30.0g sample of metal is heated to 100.0C by placing it in a
boiling water bath. The metal is removed from the boiling water,
and placed in 25.0g of water (4.18 J/gC) at 22.0C. The final
temp was 27.6C, what is the specific heat of the metal?
What is your equation?
Smetal = qmetal/mT
qmetal = - qwater
qwater = masswater x Swater x Twater
Twater = 27.6C – 22.0C = 5.6C
qwater = 25.0g x 4.18 J/gC x 5.6C = 585 J
qmetal = - qwater = -585 J
What is the metal mass? 30.0g
What is qmetal?
What is Tmetal ?
What is S. H. metal?
(27.6C -100.0C) = -72.4C
SH metal 
 585 J
1
1
0.27 J

x
x
g C
1
30.0 g  72.4 C
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Heat Evolved During a Chemical Reaction
Reaction conducted in a Styrofoam cup calorimeter
The calorimeter does not absorb or give off heat.
Heat absorbed by the solution in the calorimeter, qcal is the
heat that has been given off by the chemical reaction, -qrxn
Two solutions are mixed in the calorimeter
40.0 mL of 1.00 M KOH(aq)
40.0mL of 0.500M H2S04(aq)
Ti of both solutions = 21.00C
Data on final solution
Density: 1.02 g/mL Volume:
80.0 mL
S.H.: 4.00 J/gC Temperature: 27.85C
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Calculate the enthalpy change, H, of this reaction
per mole of H2O formed.
What is the chemical reaction?
What is the equation for H?
Calculate Heat of Reaction (q).
2 KOH + H2SO4  K2SO4(aq) + 2H2O
Hrxn= qrxn = -qH2O
q = S x m x T
s = 4.00 J/gC (assume water)
m = 81.6g (mass of both solns)
T = 6.85C (+ temp went up)
m = 1.02 g/mLx 80.0 mL = 81.6g
T = 27.85C - 21.00C = 6.85C
qH2O= 4.00 J/gC x 81.6g x 6.85C = 2236 J
qrxn = -qH2O= -2236 J
Note: this is heat that was transferred into the water IN CALORIMETER, not
water in chemical reaction!!!
Amount of heat generated for 81.6g of reactants, not per mole H2O!
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Calculating the H per mole H2O
How many moles of water are produced in the reaction?
2 KOH + H2SO4  K2SO4(aq) + 2H2O
40.0 mL of 1.00 M KOH (aq) = 0.040 mol KOH =
40.0 mL of 0.500 M H2S04(aq) = 0.020 mol H2S04 =
0.040 mol H2O
0.040 mol H2O
Calculate H per 1 mole of water produced.
Hrxn
H per mole water
H per mole water
= qrxn = -2236 J
= -2236 J / 0.040 mol H2O = -55900 J/mol
= -55.9kJ/mol
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Heats of Reactions
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Standard Enthalpy of
Formation and Reaction
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The Direct Method
Standard Enthalpies of Formation, Hf
The standard state of an element: Hf=0
Pure element in its most stable form at a pressure of 1 atm and 20C
For solvents in aqueous solution, they are at a concentration of 1 M
H2(g) N2(g) O2(g) Cl2(g) Br2(l) Hg(l) Na(s)
The standard molar enthalpy of formation, Hf
Hf for 1 mol of substance produced from elements in standard states
Tabulated for compounds at 1atm and 20C
C(s) + O2(g) CO2(g) Hf = -393.5kJ/mol
Calculating H of a chemical reaction
Subtract Hf values of reactants from Hf values of products
Multiply by the stoichiometric coefficient for that species.
C(s) and O2(g)
Hf = 1 x 0kJ/mol = 0kJ/mol each
CO2(g)
Hf = 1 x -393.5kJ/mol = -393.5kJ/mol
Hf = H products - H reactants= -393.5 - (0+0) = -393.5kJ/mol
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Calculate H for the combustion of C2H5OH(l)
Hfvalues are given below
C2H5OH(l)+3 O2(g)2CO2(g) + 3H2O(l)
Hf Reactants
C2H5OH(l) = -277.7 kJ/mol x 1 mol
O2(g) = 0 kJ/mol
x 3 mol
Total Reactants:
= -277.7kJ
= 0.0kJ
= -277.7kJ
Hf Products
CO2(g) = -393.5 kJ/mol x 2 mol
H2O (l) = -285.8 kJ/mol x 3 mol
Total Products:
= -787.0kJ
= -857.4kJ
=-1644.4kJ
H=products - reactants = (-1644.4kJ)-(-277.7kJ) = -1366.7kJ
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The Indirect Method:
Hess’s Law Of Constant Heat Summation
The heat of a reaction, H, is constant, whether the reaction is
carried out directly in one step or through a number of steps.
The equation for the overall reaction is the sum of two or more
other equations (reaction steps).
Hrxn = H1 + H2 + H3 +...
Reverse a chemical equation: Reverse the sign of H.
Hright = - Hleft
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An Enthalpy Diagram
1. Find each reactant in 1 equation only
Change equation if necessary
2. Find each product in 1 equation only
Change equation if necessary
3. Add reactions
Cross out product/reactant pairs
Add multiples
4. Check final equation
Find: C(graphite) + O2(g)  CO(g) + 1/2O2(g)
a. CO(g) + 1/2O2(g)  CO2
b. C(graphite) + O2(g)  CO2
H= ?
H= -283.0kJ
H= -393.5kJ
1. C(graphite) + O2(g)  CO2
 CO(g) + 1/2O2(g)
2. CO2
3. C(graphite) + O2(g)  CO(g) + 1/2O2(g)
H= -393.5kJ
H= +283.0kJ
H= -110.5kJ
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A Styrofoam cup calorimeter was used to determine Hsol, of KOH(s).
Hsol = -58.4 kJ for the reaction: KOH(s)KOH(aq)
What is Hsol,when KOH(s) reacts with H2S04 (aq)?
Use the H from the earlier calorimetry example
2 KOH(aq) + H2SO4(aq)K2SO4(aq) + 2H2O(l)
H = 2 moles x –55.9 kJ/mol = -111.8 kJ
Look up H of KOH(s) to KOH(aq) and convert for 2 moles KOH
KOH(s)KOH(aq)
H = -58.4 kJ/mol
H = 2 moles x -58.4 kJ/mol = -116.8 kJ
Add the Hs for all steps of the reactions
2 KOH(s)
2 KOH(aq)
2 KOH(aq) + H2SO4(aq)K2SO4(aq) + 2H20(l)
2 KOH (s) + H2SO4(aq)K2SO4(aq) + 2H20(l)
H = -116.8 kJ
H = -111.8 kJ
H = -228.6 kJ
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