chapter 5: thermochemistry. thermochemistry: – energy kinetic & potential – first law of...
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Chapter 5:Chapter 5:
ThermochemistryThermochemistry
Thermochemistry: Energy
• Kinetic & Potential First Law of Thermo
• internal energy, heat & work• endothermic & exothermic processes• state functions
Enthalpy Enthalpies of Reaction
Calorimetry• heat capacity and specific heat• constant-pressure calorimetry• bomb calorimetry (constant-volume
calorimetry) Hess’s Law Enthalpies of Formation
• for calculation of enthalpies of reaction Foods and Fuels
Energy• work is a form of energy w = F x d
• energy is the capacity to do work or transfer heat • Kinetic Energy
•energy of motion E = ½ mv2 • potential energy
•energy of position•applies to electrostatic energy•applies to chemical energy (energy of
bonds)
• energy units•one joule = energy of a 2 kg mass moving at 1
m/s• E = ½ mv2 (½)(2 kg) (m/s)2 = kg m2/s2 = 1 J•1 cal = 4.184 J 1 kcal = 1 food calorie (Cal)
Systems & Surroundings• system -- chemicals in the reaction
• surroundings -- container & all outside environment
•closed system can exchange energy (but not matter) with its surroundings
2H2(g) + O2(g)
2H2O(l)+ energy
(system)
energy(as heat or work)
no exchg of matterwith surroundings
Closed System
First Law of Thermo.• Energy is always conserved
•any energy lost by system, must be gained by surroundings
• Internal Energy -- total energy of system•combination of all potential and kinetic
energy of system•incl. motions & interactions of of all
components•we measure the changes in energy
E = Efinal - Einitial
• + E = Efinal > Einitial system has gained E from surroundings
• - E = Efinal < Einitial system has lost E to surroundings
• Relating E to heat and work
E = q + w q is positive if heat goes from surroundings to system w is positive if work is done on system by surroundings
system
surroundings
workhe
at
+q +w
system
surroundingsw
orkhea
t
- q - w
Endothermic•system absorbs heat or heat flows into the
systemExothermic
•system gives off heat or heat flows out of the system
State Function•a property of a system that is determined
by specifying its condition or state (T, P, etc.)
•internal energy is a state function, E depends only on Efinal & Einitial
Enthalpy• for most reactions, most of the energy exchanged
is in the form of heat, that heat transfer is called enthalpy, H
• enthalpy is a state function• like internal energy, we can only measure the
change in enthalpy, H
H = qp when the process occurs under constant pressureH = Hfinal - Hinitial = qp
• - H exothermic process • +H endothermic process
system
surroundings
system
surroundings
H > 0
H < 0
Enthalpies of Reaction Hrxn = Hprod - Hreact
• enthalpy is an extensive property
•magnitude of H depends directly on the amount of reactant
C(s) + 2H2(g) CH4(g) H = -74.8
kJ/mol
2C(s) + 4H2(g) 2CH4(g) H = -149.6
kJ/2mol
• enthalpy change for forward rxn is equal in magnitude but opposite in sign for the reverse rxn
CH4(g) C(s) + 2H2(g) H = +74.8 kJ/mol
C(s) + 2H2(g) CH4(g) H = - 74.8 kJ/mol
• enthalpy change for a reaction depends on the state of the reactants and products
C(g) + 2H2(g) CH4(g) H = -793.2 kJ/mol
2H2(g) + O2(g) 2H2O(g) H = -486.6 kJ/mol
2H2(g)+ O2(g) 2H2O(l) H = -571.7 kJ/mol
H2O(g)
H2O(l) Ent
halp
y 44 kJ
-285.8 kJ
-241.8 kJ
H = Hfinal - Hinitial
+
-
Practice Ex. 5.2:• Hydrogen peroxide can decompose to water and oxygen .
Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.
2H2O2(l) 2H2O(l) + O2(g) H = -196 kJ
5.00 g H2O2(l) x 1 mol = 0.147 mol H2O2(l)
34.0 g H2O2(l)
0.147 mol H2O2(l) x -196 kJ H2O2(l) = -14.4 kJ
2 mol
Calorimetry•experimental determination of H using
heat flowheat capacity
•measures the energy absorbed using temperature change
•the amount of heat required to raise its temp. by 1 K
•molar heat capacity -- heat capacity of 1 mol of substance
specific heat•heat energy required to raise some mass of a
substance to some different temp. specific heat = quantity of heat trans.
(g substance) (temp. change)
= q . m T
S.H. = joule g K
•q = (S.H.) (g substance) ( T)
remember: this is change in temp.
Practice Ex. 5.3:• Calculate the quantity of heat absorbed by 50.0 kg
of rocks if their temp. increases by 12.0 C if the specific heat of the rocks is 0.82 J/gK.
S.H. x g x T = joules
What unit should be in the solution?joules -- quantity of heat
0.82 J x 50.0 x 103 g x 12.0 K = 4.9 x 105 Jg K
Constant-Pressure Calorimetry• H = qp at constant pressure as in coffee cup
calorimeter
•heat gained by solution = qsoln
qsoln = (S.H.soln)(gsoln)(T)•heat gained by solution must that which is
given off by reaction
qrxn = - qsoln = - (S.H.soln)(gsoln)(T)
must be opposite in signmust be opposite in sign
if T is positive then qrxn is exothermic
Practice Ex. 5.4:• When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M
HCl are mixed in a c.p. calorimeter, the temp. of the mixture increases from 22.30C to 23.11C. Calculate H for this reaction, assuming that the combined solution has a mass of 100.0 g and a S.H. = 4.18 J/g C.
AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq)
qsoln = 4.18 J x 100.0 g soln x 0.81C = 3.39 x 102 J g C
qrxn = - qsoln = - 3.39 x 102 J = - 68,000 J or 0.00500 mol - 68 kJ/mol
rxn
soln
q
insulating cup
Bomb Calorimetry (Constant-Volume)
• bomb calorimeter has a pre-determined heat capacity
• sample is combusted in the calorimeter and T is used to determine the heat change of the reaction
• qrxn = - Ccalorimeter x T
because rxn is exothermic
heat capacity of calorimeter
rxn
water
insulation
thermometer
Practice Ex. 5.5:• A 0.5865 g sample of lactic acid, HC3H5O3, is burned in a
calorimeter with C = 4.812 kJ/C. Temp. increases from 23.10C to 24.95C. Calculate heat of combustion per gram and per mole. T = +1.85C
qrxn = - (4.812 kJ/C) (1.85C) = - 8.90 kJ per 0.5865 g lactic acid -8.90 kJ = - 15.2 kJ/g 0.5865 g
- 15.2 kJ x 90 .1 g = - 1370 kJ/mol 1 g 1 mol
Hess’s Law• rxns in one step or multiple steps are additive
because they are state functions
eg.
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = - 802 kJ
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = - 890 kJ
2H2O(g) 2H2O(l) H = - 88 kJ
Practice Ex. 5.6:• Calculate H for the conversion of graphite to
diamond:
Cgraphite Cdiamond
Cgraphite + O2(g) CO2(g) H = -393.5 kJ
Cdiamond + O2(g) CO2(g) H = -395.4 kJ
Cgraphite + O2(g) CO2(g) H = -393.5 kJ
CO2(g) Cdiamond + O2(g) H = 395.4 kJ
Cgraphite Cdiamond H = + 1.9 kJ
Enthalpies of Formation• enthalpies are tabulated for many processes
– vaporization, fusion, formation, etc.
• enthalpy of formation describes the change in heat when a compound is formed from its constituent elements, Hf
• standard enthalpy of formation, Hfo, are values for
a rxn that forms 1 mol of the compound from its elements under standard conditions, 298 K, 1 atm
• For elemental forms:
eg. C(s) graphite, Ag(s) , H2(g) , O2(g) , etc.
Hfo, for any element is = 0
• used for calculation of enthalpies of reaction, Hrxn
• Hrxn = Hfo prod - Hf
o react
Practice Ex. 5.9:• Given this standard enthalpy of reaction, use the
standard enthalpies of formation to calculate the standard enthalpy of formation of CuO(s)(
CuO(a) + H2(g) Cu(s) + H2O(l) Ho = -130.6 kJ
Hrxn = H f o
prod - H f o
react
-130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)]
fo CuO = -155.2 kJ/mol