Thermochemistry:
- Energy
- 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
closed system can exchange energy (but not matter) with
its surroundings
- surroundings -- container & all
outside environment
- enthalpy change for forward rxn is
equal in magnitude but opposite in sign for the reverse rxn
CH4(g) ®
C(s) + 2H2(g)
D
H = +74.8 kJ/mol
C(s) + 2H2(g)
® CH4(g) D
H = - 74.8 kJ/mol
- enthalpy change for a reaction depends
n the state of the reactants and products
C(g) + 2H2(g)
® CH4(g)
D
H = -793.2 kJ/mol
2H2(g) + O2(g)
® 2H2O(g)
D
H = -486.6 kJ/mol
2H2(g)+ O2(g)
® 2H2O(l)
D
H = -571.7 kJ/mol
- Calorimetry
experimental determination of D
H using heat flow
- heat 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
- 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 D 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 J
g K
- Constant-Pressure Calorimetry
- D H = qp
at constant pressure as in coffee cup calorimeter
heat gained by solution = qsoln
\ qsoln
= (S.H.soln)(gsoln)(DT)
heat gained by solution must that which is given off
by reaction
\ qrxn
= - qsoln
= - (S.H.soln)(gsoln)(DT)
- 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 22.30°C
to 23.11°C. Calculate D 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 0.81°C = 3.39 x 102
J
g oC
qrxn = - qsoln
= - 3.39 x 102 J
= - 68,000 J or - 68 kJ/mol
0.00500
mol
A 0.5865 g sample of lactic acid, HC3H5O3,
is burned in a calorimeter with C = 4.812 kJ/°C.
Temp. increases from 23.10°C to 24.95°C.
Calculate heat of combustion per gram and per mole. D
T = +1.85°C
qrxn = - (4.812
kJ/°C) (1.85°C)
= - 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)
D H = - 802 kJ
CH4(g) + 2O2(g)
® CO2(g)
+ 2H2O(l)
D H = - 890 kJ
- Practice Ex. 5.6:
- Calculate D
H for the conversion of graphite to diamond:
Cgraphite ®
Cdiamond
Cgraphite +
O2(g) ®
CO2(g) D
H = -393.5 kJ
Cdiamond +
O2(g) ®
CO2(g) D
H = -395.4 kJ
Cgraphite +
O2(g) ®
CO2(g) D
H = -393.5 kJ
CO2(g) ®
Cdiamond + O2(g)
D
H = 395.4 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,
DHf
- standard enthalpy of formation, DHfo,
are values for a rxn that forms 1 mol of the compound from its elements
under standard conditions, 298 K, 1 atm
- elemental forms
eg. C(s)
graphite, Ag(s)
, H2(g) , O2(g)
, etc.
DHfo,
for any element is = 0
- used for calculation of enthalpies
of reaction, DHrxn
- DHrxn
= S DH prod
- S DH react
Given this standard enthalpy of reaction, use the standard
enthalpies of formation to calculate the standard enthalpy of formation
of CuO(s)
CuO(s) + H2(g) ®
Cu(s) + H2O(l) DHo
= -130.6 kJ
DHrxn
= S DH f o prod - S
DH f o react
-130.6 kJ = [(0) + (-285.8)] - [(CuO) + (0)]
DHfo
CuO = -155.2 kJ/mol