Download - Thermochemistry
Thermochemistry
Chapter 8
Energy is the capacity to do work
• Thermal energy is the energy associated with the random motion of atoms and molecules
• Chemical energy is the energy stored within the bonds of chemical substances
• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom
• Electrical energy is the energy associated with the flow of electrons
• Potential energy is the energy available by virtue of an object’s position or composition.
Heat is the transfer of thermal energy between two bodies that are at different temperatures.
Energy Changes in Chemical Reactions
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C400C
greater thermal energy
Thermochemistry is the study of heat change in chemical reactions.
The system is the
specific part of the
universe that is of interest in the study.
open
mass & energyExchange:
closed
energy
isolated
nothing
SYSTEM
SURROUNDINGS
Thermodynamics
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
Potential energy of hiker 1 and hiker 2 is the same even though they took
different paths.
energy, pressure, volume, temperature
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = ms
Heat (q) absorbed or released:
q = mst
q = Ct
t = tfinal - tinitial
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = 0.444 J/g • 0C
t = tfinal – tinitial = 50C – 940C = -890C
q = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
Constant-Pressure Calorimetry
No heat enters or leaves!
qsys = qwater + qcal + qrxn
qsys = 0
qrxn = - (qwater + qcal)
qwater = mst
qcal = Ccalt
Reaction at Constant PH = qrxn
Constant-Volume Calorimetry
No heat enters or leaves!
qsys = qwater + qbomb + qrxn
qsys = 0
qrxn = - (qwater + qbomb)
qwater = mst
qbomb = Cbombt
Reaction at Constant V
H ~ qrxn
H = qrxn
TWO Trends in Nature
• Order Disorder
• High energy Low energy
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
2H2 (g) + O2 (g) 2H2O (l) + energy
H2O (g) H2O (l) + energy
energy + 2HgO (s) 2Hg (l) + O2 (g)
energy + H2O (s) H2O (l)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0Hproducts > Hreactants
H > 0
Thermochemical Equations
H2O (s) H2O (l) H = 6.01 kJ
Is H negative or positive?
System absorbs heat
Endothermic
H > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJ
Is H negative or positive?
System gives off heat
Exothermic
H < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
H2O (s) H2O (l) H = 6.01 kJ/mol ΔH = 6.01 kJ
• The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
• If you reverse a reaction, the sign of H changes
H2O (l) H2O (s) H = -6.01 kJ
• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.
2H2O (s) 2H2O (l) H = 2 mol x 6.01 kJ/mol = 12.0 kJ
H2O (s) H2O (l) H = 6.01 kJ
• The physical states of all reactants and products must be specified in thermochemical equations.
Thermochemical Equations
H2O (l) H2O (g) H = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) Hreaction = -3013 kJ
266 g P4
1 mol P4
123.9 g P4
x3013 kJ1 mol P4
x = 6470 kJ
Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.
f
The standard enthalpy of formation of any element in its most stable form is zero.
H0 (O2) = 0f
H0 (O3) = 142 kJ/molf
H0 (C, graphite) = 0f
H0 (C, diamond) = 1.90 kJ/molf
The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]
H0rxn H0 (products)f= H0 (reactants)f-
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.)
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)
H0rxn H0 (products)f= H0 (reactants)f-
H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]
H0rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ
-6535 kJ2 mol
= - 3267 kJ/mol C6H6
Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn
S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn
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.
rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ
2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)
H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn6.6
Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) H = -2801 kJ/mol
1 cal = 4.184 J
1 Cal = 1 kcal = 1000 cal = 4184 J
The enthalpy of solution (Hsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
Hsoln = Hsoln - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
The Solution Process for NaCl
Hsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
Energy Diagrams
Exothermic Endothermic
(a) Activation energy (Ea) for the forward reaction
(b) Activation energy (Ea) for the reverse reaction
(c) Delta H
50 kJ/mol 300 kJ/mol
150 kJ/mol 100 kJ/mol
-100 kJ/mol +200 kJ/mol
Sub
limat
ion
Dep
ositi
on
H2O (s) H2O (g)
Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid.
Hsub = Hfus + Hvap
( Hess’s Law)
Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance.
Sample Problem• How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 1: Heat the ice Q=mcΔT
Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ
Step 2: Convert the solid to liquid ΔH fusion
Q = 2.0 mol x 6.01 kJ/mol = 12 kJ
Step 3: Heat the liquid Q=mcΔT
Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem• How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C?
Step 4: Convert the liquid to gas ΔH vaporization
Q = 2.0 mol x 44.01 kJ/mol = 88 kJ
Step 5: Heat the gas Q=mcΔT
Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ
Now, add all the steps together
0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ
Bond Enthalpies (Bond Energy)
Strength of a chemical bond is measured by the bond enthalpy, HB
Bond enthalpies are positive, because heat must be supplied to break a bond.
Bond breaking is endothermic (H is positive).
Bond formation is exothermic (H is negative).
H2(g) --> 2 H Ho = +436 kJ
HB = 436 kJ/mol
Mean bond enthalpy: average molar enthalpy change accompanying the dissociation of a given type of bond.
Estimate the enthalpy change of the reaction between gaseous iodoethane and water vapor.
CH3CH2I(g) + H2O(g) --> CH3CH2OH(g) + HI(g)
Reactant: break a C-I bond and an O-H bond
Ho = 238 kJ + 463 kJ = 701 kJ
Product: to form a C-O bond and an H-I bond
Ho = -360 kJ + -299 kJ = -659 kJ
Overall enthalpy change = 701 kJ - 659 kJ = 42 kJ