SCH4U Equations with
Energy Terms
Molar Enthalpies, Potential Energy Diagrams,
Calorimeters
Page 308 # 1-3
310 # 4,5
311 # 6-10
312 # 1-5
319 # 1 - 5
320 # 1 - 4
Exothermic process is any process that gives off heat –
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 + H 2O (s)
energy + 2HgO
(s)
H2O (l)
2Hg (l) + O2 (g)
Change of Enthalpy, D H
DH= Hfinal - Hinitial
Exothermic
If the products have less energy than the reactants,
i.e. Hfinal < Hinitial , then DH is negative. Heat has been
released. Exothermic processes have a negative DH.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Enthalpy diagram for exothermic process.
Change of Enthalpy,
D H
D H= H final - H initial
Endothermic
If the products have more energy than the reactants,
i.e. Hfinal > Hinitial , then DH is positive. Heat has been
absorbed. Endothermic processes have a positive DH.
H
O(
s
)
2
H
2
O( l )
Enthalpydiagramforendothermicprocess.
Drawing Enthalpy Diagrams and Determining the Sign of Ä H
PROBLEM:
In each of the following cases, determine the sign of Ä H, state
whether the reaction is exothermic or endothermic, and draw
an enthalpy diagram.
(a)
H2(g) + 1/2 O2(g)
H2O(l)
O(l)++2
285.8kJ
PLAN:
Determine whether heat is a reactant or a product.
The reactants are at a higher energy than the products.
The reaction is an exothermic reaction when heat is a product.
SOLUTION:
(a)
H2(g) +
The reaction is exothermic.
1/2
O2(g)
EXOTHERMIC
H2O(l)
(reactants)
DH = -285.8kJ
(products)
(b)
40.7kJ ++HH 2 O( l)
The reaction is
H 2O(g)
ENDOTHERMIC
H 2O(l)
H
2 O( g)
endothermic
(products)
DH = +40.7kJ
(reactants)
.
Examples
kJ as reactant
= endo
+ = endo
H2O(s) + 6.01kJ
H2O(l)
Or…
H2O(s)
H2O(l)
DH = + 6.01kJ
kJ as product
= exo
2C4H10(g) + 13O2(g)
Or…
8CO2(g) + 10H2O(g) + 5314kJ
2C4H10(g) + 13O2(g)
8CO2(g) + 10H2O(g) DH = -5314kJ
Calculate the molar heat of combustion (H CO) for butane
2C 4 H 10(g) + 13O
2(g)
8CO
2(g)
+ 10H
2
O (g) + 5314kJ
Solution – whenweareattemptingtheseÄH problems,firstseeifthe
reaction'senergyisprovidedinthequestionI.fitis,procee d to solve. If it
isnot,thenwehaveamulti-stepproblem.
Theenergy given may be used for every substance in the reaction
I.e. It applies to the butane, oxygen, carbon dioxide and steam
2C 4 H 10(g) + 13O
2(g)
8CO
2(g)
+ 10H
2
O (g) + 5314kJ
The energy given may be used for every substance in the reaction
So,accordingtothereaction,2molesofbutanecorrespondto5314kJ.
We need the amount of energy for 1 mol of butane;
\H CO = 5314kJ / 2mol butane gives us H CO =
- 2657kJ/mol
PE diagram of ice melting
H 2O(l)
PE diagram of combustion of butane
2C 4H 10(g) + 13O 2(g)
DH = -5738kJ
DH = +6.01kJ
8CO 2(g) + 10H 2O (g)
H 2O(s)
Reaction Progress (min)
Reaction Progress (min)
**Anytime you need to answer a theory or calculation question, i t is always a
good idea to sketch a graph, if applicable.
*Do products/reactants have more potential energy?
* Are the products or reactants more stable?
Some Important Types of Enthalpy Change
heat of formation (
K( s ) +
1/2
Br 2( l)
13/2
O2(g)
H
f)
KBr( s )
D
heat of combustion (
C4H10(l) +
D
H
comb )
4CO2(g) + 5H2O(g)
Some Important Types of Enthalpy Change
D
heat of fusion (
NaCl(
s)
fus
NaCl(
D
heat of vaporization (
C 6H 6( l)
H
C
H
)
l)
vap)
H
(
g
)
6
6
Writing Formation Equations
PROBLEM:
Writb
ealancee
dquationfsotrhfeormatioo
nm
1f ootlh
f feollowing
compoundfr
fs
srom
tthheee
eirirl
leement
mentitn
hsitn
sheseir
irtstans
andar
dta
sar
dta
tdete
saand
snin
in
dcclude
lude
DH0f.
(a)
Silvecrhloride,AgCla,solidasttandardconditions.
Usethetableofheatsoformationforvalues.
PLAN:
SOLUTION:
(a)
Ag(s) + 1/2Cl2(g)
AgCl(s)
DH0f = -127.0kJ
(b)
Calciucm
arbonatCea,CO 3a,solidasttandardconditions.
Ca(s) + C(graphite+
)3/2O2(g) CaCO3(s)
(c)
DH0f =-1206.9kJ
HydrogencyanideH
, CNa,gasasttandardconditions.
1/2H2(g)+C(graphite)+1/2N2(g)
HCN(g)
DH0f1=35kJ
Specific Heat Capacity
Wheanonbjecatbsorbhseagitt,ethsotter.
Thm
e orh
eeaa
itbsorbtshh
eotteg
irtets,
i.e.q ~DT or q =constant xDT or q /cD=o
Tnstant
This capacity of an object to absorb heat is called its:
D (units of J/K)
The amount of heat to change the temperature by 1 K
heatcapacityq
=/T
SpeciH
fie
cCaatpac(icty)
A related property is
This is the amount of heat to change the temperature of 1 gram o f
substance by 1 K
c q
D (units of J/g-K)
= /massx T
Another property is MolH
aerCaatpac(itCy)
This is the amount of heat to change the temperature of 1 mole o f
substance by 1 K
C q
D (units of J/mol-K)
= m
/ olesx T
Sample Problem 6.3
PROBLEM:
Calculating the Quantity of Heat from the
Specific Heat Capacity
lA
ayce
oorfppe
wreldeth
o
dbeottosa
om
kfillw
eteigh1s25g.
Hom
wuchheniaset0ederto
a
d0isth
ete
emperatutrohefe
copplaeyrfreo2rm
5 tC
3o00.TC
hs?
epecifhicecaatpacity
o
(c
C)f0
iu
s.387J/g-K.
Givtm
e
hn
e
asspse
, chife
cicap
t acacin
th
ydatn
ein
g
meperature,
w
e
canuse
q=cxmassxDtTfointdhaenswer.
DTin0Cisthesameasfor
K.
SOLUTION:
.87J x125g x3
0C 3x10 4J
1.3=
q= 03
(
0
0-2
5
)
gK
*
PLAN:
Calorimetry
Calorimetry is a way to measure heat from a process by
directing the heat into “surroundings” where the increase of
heat can be measured. The surroundings will be a container
called a calorimeter.
Theheatlostbythesystemequalstheheatgainedby
thcealorimeter.
- q sample = qcalorimeter (minu
sisgfhonerla
otst)
Cofca
e-clo
urpimeter
- q sample = qcalorimeter
- (mass x heat capacity x
(mass x heat capacity x
Dsamp
Dca.l
T)
T)
=
Sample Problem 6.4
Determining the Specific Heat Capacity of a
Solid
2A5.64sgampso
laeoflw
idahseateta
ien
dtsutbt1e
o00.00 oCin
boiliw
ngataecnradrefualldycdoatefofdee-ccuaplorimeter
containi
5
n0
g.w
T
0
o
w
0
h
a
fa
g
e
t
e
t
t
e
e
r
.
m
r
peratu
in
re
c
r
easfe
r
o
d
m
o
oCh.a
251
.02
tC
o8.49W
tishtsepecifh
iceactapacitotyhfseolid?
(Assuma
thehlleegiasatinebth
dw
yeater).
ihIstelpfutuolstae
ablte
soummariztehdeatgaive
Tnh.ewor
n tkhe
problerm
ealizinth
gh
ae
t laotbsthytseystem
mubsee
tquta
that
ol
gstauhirbn
eryo
eu
dndings.
PROBLEM:
PLAN:
Mass(g)
c(J/g-K)
solid
25.64
c
2
50.00
4.184
HO
SOLUTION:
T
inital
n
fial
T
100.00 28.49
25.10
28.49
D
T
-71.51
3.39
25.6
x 4g
cx -715
.1K =- 50.0
x 0g
4.18J/g-Kx 3.K
39
csodil = - 50.0
x 0g
4.18J/g-Kx 3.K
39 = 0.387J/g-K
25.6
x 4g
-715
.1K
boAcmablorimeter
-qsampel =qcaolrm
i eter
qcaolrm
i eter h
=eactapacitx
DyT
Sample Problem 6.5
Calculating the Heat of Combustion
mAanufacturcela
r im
thnistae
sdtw
ietedtie
cssehr“atfsewer
tha1nC
0aloriepsesrervinTg
toe.”sthtcelaim
cah,emisat
th
Deepartme
Coonft
summer Affaip
rslaceosnseervinaing
bomb
cao
l rm
i ater anbdurniiO
n
st 2(thheea
ca
t pacio
t0h
yfe
calorim
8.e
1=tT
5teekhrm
Je/p
Ke).ratiunrcereas
4e.9s37 C.
m
thaeInsufacturcelcra'o
sim
rrect?
-qsampel =qcaolrm
i eter
PROBLEM:
PLAN:
SOLUTION:
qcaolrm
=eactapacitDxT
y
i eter h
8=.1514kx.9
J/3K7K
40.=
24kJ
402
.4kJ kcal 9.=
62k
Cocaa
rlolries
41
.8kJ
Tmhaenufacturectrrluai'seim
.