Transcript Thermochemistry - Kuwait University

Thermochemistry
Chapter 6
Dr. Ali Bumajdad
Chapter 6 Topics
Thermochemistry
Study of heat change in a chemical reaction
• Nature and Type of Energy
•Energy Changes in Chemical Reaction
•Intro. to Thermodynamics: 1st Law, Wark,
Heat, Enthalpy
•Calorimetry, Specific Heat, Heat Capacity
•Sstandard enthalpy of formation and reaction
•Hess Law
•Enthalpy of solution
Dr. Ali Bumajdad
Energy is the capacity to do work
Work = Fd
Unit = J = Kgm2s-2
•
Radiant energy comes from the sun (also called solar
energy)
•
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
•
Potential energy is the energy object position relative to
other objects (e.g. gravity result in pot. E to objects)
•
Kinetic Energy is the energy of motion
Energy convert from one form to another (Law of Conservation of Energy)
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Q) Water cup in the fridge, what is the direction of heat
transfer?
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
900C
400C
greater thermal energy
Thermochemistry is the study of heat change in chemical
reactions.
• System is the specific part of the universe that is of interest in
the study
•Surrounding is every things except the thing we are studying
Type:
Exchange:
Open
system
mass & energy
Closed
system
energy
Isolated
system
nothing
Exothermic process is any process that gives off heat or
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 has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
Thermodynamics
‘The study of the interconversion of heat and other kinds of
energy’
• State functions properties determined by the current state of
the system, regardless of how that condition was achieved
e.g. energy , pressure, volume, temperature
DE = Efinal - Einitial
Work and Heat
Not State Function
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
• First law of thermodynamics:
1) energy can be converted from one form to another, but
cannot be created or destroyed
DEsystem + DEsurroundings = 0 OR DEsystem = -DEsurroundings
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
2) DEsys = q + w
(1)
Work done on (+) or by (-) the system
W=-PDV
(2)
Heat transfer to (+) or from (-) the system
Change in internal energy of a system
Combustion reaction release heat
q=-ve
Vaporization absorbed heat
q=+ve
Condensation release heat
q=-ve
Freezing release heat
q=-ve
Melting absorbed heat
q=+ve
Work Done On the System
w=Fxd
w = -P DV
PxV=
F
x d3 = F x d = w
2
d
DV > 0
-PDV < 0
wsys < 0
pressure vol
Work is not a
state function!
Dw = wfinal - winitial
initial
final
Q) A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L
at constant temperature. What is the work done in joules if the gas
expands (a) against a vacuum and (b) against a constant
pressure of 3.7 atm? 1L atm =101.3 J
w = -P DV
(a)
DV = 5.4 L – 1.6 L = 3.8 L
(2)
P = 0 atm
W = -0 atm x 3.8 L = 0 L•atm = 0 joules
(b)
DV = 5.4 L – 1.6 L = 3.8 L
P = 3.7 atm
w = -3.7 atm x 3.8 L = -14.1 L•atm
101.3 J = -1430 J
1L•atm
 Work is not state Function
w = -14.1 L•atm x
Enthalpy and the First Law of Thermodynamics
•Enthalpy (H) used to quantify heat flow into or out of a system
in a process that occurs at constant pressure.
DE = q + w
At constant volume:
w = 0 and DE = qv
At constant pressure:
DH = qp and w = -PDV
DE = DH - PDV
DH = DE + PDV
State Function
(3)
Enthalpy of Reactions
DH = H (products) – H (reactants)
(4)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
Is DH negative or positive?
H2O (s)
System absorbs heat
H2O (l)
Endothermic
DH > 0, DH =+ve
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/mol
Is DH negative or positive?
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) System gives off heat
Exothermic
DH < 0, DH =+ve
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/mol
Thermochemical Equations
DH = 6.01 kJ/mol
H2O (s)
H2O (l)
1) If you reverse a reaction, the sign of DH changes
H2O (l)
H2O (s)
DH = -6.01 kJ/mol
 DHforward =-DHbackward
2) If you multiply both sides of the equation by a factor n, then
DH must change by the same factor n.
DH extensive property
2H2O (s)
2H2O (l) DH = 2 x 6.01 = 12.0 kJ/mol
3) The physical states of all reactants and products must be
specified in thermochemical equations
DH = 6.01 kJ/mol
H2O (s)
H2O (l)
H2O (l)
H2O (g)
DH = 44.0 kJ/mol
DH H2O(g) > DH H2O(l) > DH H2O(l)
Q) How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
P4O10 (s)
DH = -3013 kJ/mol
= 6470 kJ
A Comparison of DH and DE
2Na (s) + 2H2O (l)
DE = DH - PDV
2NaOH (aq) + H2 (g) DH = -367.5 kJ/mol
At 25 0C, 1 mole H2 = 24.5 L at 1 atm
and DVVgas
PDV = 1 atm x 24.5 L = 2.5 kJ
DE = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
DE > DH
Some of Internal energy released used to do gas expansion work, this is why
For reactions involving gases DE slightly different than DH
For reactions do not involve gases DE  DH (DV is very small)
•Calorimetry, Specific Heat, Heat Capacity
•Calorimetry The measurement of heat flow
•Specific heat (s) is the amount of heat (q) required to raise
the temperature of one gram of the substance by one degree
Celsius.
(5)
q
s=
q = m x s x DT
mDT
J/g.°C
• Heat capacity (C) is the amount of heat (q) required to raise
the temperature of a given quantity (m) by one degree Celsius.
q
DT
C=
J/°C
(6)
q = C x DT
From (5) and (6):
C=mxs
(7)
Q) 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
J/g.°C
s=
q
mDT
(5)
q = m x s x Dt
= -34,000 J
q = m x s x Dt
(1) Constant-Volume Calorimetry
(Good for combustion reaction)
•Heat produced by combustion and absorbed by surrounding
(calorimeter)
qsys = qcal + qrxn
qsys = 0
qrxn = - qcal
qcal = Ccal x DT
qrxn = -Ccal x DT
(8)
For Reaction at Constant V
DH = qrxn
No heat enters or leaves (Adiabatic)
DE = qrxn
•Molar heat of combustion: Heat of combustion of one mole
qmolar
rxn = -Ccal x Dt
n
(8b)
(2) Constant-Pressure Calorimetry
•Heat produced (or absorbed) by reaction and absorbed (or
produced) by the surrounding (solution)
Exoth.
Endo.
qsys = qsolution + qrxn
qsys = 0
qrxn = - qsolution
qcal = Ccal x DT
qrxn = - Ccal x DT
(8)
Reaction at Constant P
DE  qrxn
DH = qrxn
No heat enters or leaves! (Adiabatic)
qmolar
rxn = -Ccal x Dt
n
(8b)
Sa. Ex.5.6a: How much heat is needed to warm 250g of water
from 22°C to 98°C. (water specific heat =4.184 J/g.°C)
J/g.°C
s=
(5)
q
mDT
q = m x s x DT
=79kJ
Sa. Ex.5.6b: What is the molar heat capacity of water?
Cmolar= q
nDT
or
C = m x s (7)
n
qwater = m x s x DT =-qPb
qPb = m x s x DT
• Reaction is exothermic
• qrxn = - qsolution
• when density 1
volume = mass
• mol of NaOH = MNaOH VNaOH
• mol of HCl = MHCl VHCl
Standard Enthalpy of Formation (DH0f)
(It is the heat change when 1 mole of a compound formed from
its elements at 1 atm)
• It is a reference point for all enthalpy expressions (similar to
the sea level for measuring heights)
• The standard enthalpy of formation of any element in its
most stable form is zero
DH0f (O2) = 0
DH0f (C, graphite) = 0
DH0f (O3) = 142 kJ/mol
DH0f (C, diamond) = 1.90 kJ/mol
Standard Enthalpy of Reaction (DH0rxn)
(It is the enthalpy of a reaction carried out at 1 atm)
aA + bB
cC + dD
(9)
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
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.
• If a reaction is carried out in steps, DH will equal the sum of
the enthalpy changes for the individual steps.
(10)
DHrxn= DHstep1 + DHstep2+ DHstep3+……
C (graphite) + 1/2O2 (g)
CO (g) + 1/2O2 (g)
C (graphite) + O2 (g)
CO (g)
CO2 (g)
CO2 (g)
Q) 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)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ
Q) 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, for CO2 is -393.5 and for H2O is -285.8 kJ/mol
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
(9)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6535 kJ
-6535 kJ
= - 3268 kJ/mol C6H6
2 mol
DHrxn= DHstep1 + DHstep2+ DHstep3+……
(10)
DHf°Fe2O3(s)=-822.2kJ/mol,
DHf°Al2O3(s)=-1669.8 kJ/mol
0
DHrxn
= SnDHf0 (products)- SmDHf 0 (reactants)
(9)
Divide by 2 to have the molar value then divide by
the molar mass of Al to get it kJ/g
Enthalpy of solution (DHsoln)
The heat generated or absorbed when a certain amount of
solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol