phase changes and heat calculations
DESCRIPTION
Phase Changes and Heat Calculations. Obj. 1…Vapor Pressure. Vapor pressure (VP) is the P exerted at the surface of a. liquid by particles trying to escape the liquid. Obj. 2…VP and Temperature. As T , KE will. (direct relationship). - PowerPoint PPT PresentationTRANSCRIPT
Phase Changes Phase Changes and Heat and Heat
CalculationsCalculations
Obj. 1…Vapor Obj. 1…Vapor PressurePressure
o Vapor pressure (VP)Vapor pressure (VP) is the P exerted at the is the P exerted at the surface of asurface of aliquid by particles trying to escape the liquid.liquid by particles trying to escape the liquid.
Obj. 2…VP and Obj. 2…VP and TemperatureTemperature
o As T , KE As T , KE will . will .
(direct (direct relationship)relationship)
o If liquid molecules gain enough KE, they will If liquid molecules gain enough KE, they will overcomeovercomethe intermolecular bonds that hold them together.the intermolecular bonds that hold them together.
become a become a gasgas
Obj. Obj. 3…Boiling/Melting 3…Boiling/Melting PointsPoints
o Boiling pointBoiling point (BP) (BP) = the temp. at which the VP = the temp. at which the VP of a liquidof a liquidis equal to the external pressure.is equal to the external pressure.
o BP is directly related to atmospheric BP is directly related to atmospheric pressure.pressure. a pot of water in Denver (mountains…low pressure) will a pot of water in Denver (mountains…low pressure) will
boil at aboil at alower temp. than a pot of water in Houston (sea level).lower temp. than a pot of water in Houston (sea level).
o normal BP normal BP is always measured at sea is always measured at sea level.level.o Melting PointMelting Point (MP) (MP) = the temp. at which a solid = the temp. at which a solid turns into turns into a liquid.a liquid.
KE increases pressure enough to break intermolecular KE increases pressure enough to break intermolecular bonds.bonds.
as KE of solid increases, molecules begin to vibrate as KE of solid increases, molecules begin to vibrate
if vibrations are strong enough, molecules will break if vibrations are strong enough, molecules will break away fromaway fromtheir fixed positionstheir fixed positions liquidliquid
Obj. Obj. 3…Boiling/Melting 3…Boiling/Melting PointsPoints
Obj. 4… Freezing/Melting Obj. 4… Freezing/Melting and Boiling/Condensation and Boiling/Condensation PointsPoints
o The freezing point (FP) and melting point (MP) of The freezing point (FP) and melting point (MP) of aa substance occur at the same temp.substance occur at the same temp.
FP FP (liquid solid)(liquid solid) is used as a substance is used as a substance losesloses KE KE (heat)(heat) molecules get slower and lock into place. molecules get slower and lock into place. MP MP (solid liquid)(solid liquid) is used as a substance is used as a substance gainsgains KE (heat)KE (heat) molecules break away from solid bonds.molecules break away from solid bonds.
o The boiling point (BP) and condensation point The boiling point (BP) and condensation point (CP) of a(CP) of asubstance occur at the same temp.substance occur at the same temp.
BP BP (liquid gas)(liquid gas) is used as a substance is used as a substance gainsgains KE KE (heat)(heat) molecules break away from liquid bonds.molecules break away from liquid bonds. CP CP (gas liquid)(gas liquid) is used as a substance is used as a substance losesloses KE KE (heat)(heat) molecules get slower and more attracted to molecules get slower and more attracted to
each other.each other.
Obj. 5…SublimationObj. 5…Sublimationo sublimationsublimation = a solid changing directly into a = a solid changing directly into a vapor (gas)vapor (gas)w/out passing through the liquid stage.w/out passing through the liquid stage.
o only occurs in certain solids with only occurs in certain solids with high VP. high VP. o Ex…naphthalene (moth balls), COEx…naphthalene (moth balls), CO22 (dry ice) etc… (dry ice) etc…
Obj. 6…Boiling vs. Obj. 6…Boiling vs. EvaporationEvaporation
o for a liquid to boil, the VP of the liquid for a liquid to boil, the VP of the liquid MUSTMUST = = thetheatmospheric pressure.atmospheric pressure.
o to accomplish this, we to accomplish this, we can…can… increase temp. of liquid ( KE = VP) increase temp. of liquid ( KE = VP)
reduce atmospheric pressurereduce atmospheric pressure
** entire pot of water boils at the ** entire pot of water boils at the same time!!!same time!!!o evaporation occurs evaporation occurs w/outw/out changing temp. or changing temp. or
pressure. pressure. surface molecules exposed to more KE surface molecules exposed to more KE (sun/atmosphere)(sun/atmosphere)than particles below surface.than particles below surface. this is a cooling process (high KE molecules leave, this is a cooling process (high KE molecules leave, low KElow KEmolecules stay).molecules stay).
** only occurs at the SURFACE of a ** only occurs at the SURFACE of a liquid!!!liquid!!!
Obj. 7…Volatile vs. Obj. 7…Volatile vs. Non-VolatileNon-Volatile
o volatile substancesvolatile substances evaporate very evaporate very easily and boil easily and boil at lowat lowtemps.temps.
o vapors are typically very strong and distinct.vapors are typically very strong and distinct.
o Ex…ammonia, gasoline, rubbing alcohol, acetoneEx…ammonia, gasoline, rubbing alcohol, acetone
o non-volatile substancesnon-volatile substances contain stronger bonds contain stronger bonds and do and do not evaporate easily.not evaporate easily.
o Ex…molasses, glue, paintEx…molasses, glue, paint
Obj. 8…KE and Obj. 8…KE and Intermolecular BondsIntermolecular Bondso As KE , the strength of intermolecular bonds As KE , the strength of intermolecular bonds will . will . (inverse (inverse relationship)relationship) heat causes KE toheat causes KE to
enough movement eventually breaks enough movement eventually breaks intermolecular bonds. intermolecular bonds.
heat causes KE toheat causes KE to molecules get slower, move less. molecules get slower, move less. eventually lock into place. eventually lock into place. bond strength increases. bond strength increases.
Obj. 9…Heating/Cooling Obj. 9…Heating/Cooling CurvesCurves
TimeTime
Temperature Temperature (KE) (KE)
SolidSolid
MeltingMelting
BoilingBoilingMPMP
BPBP
Heati
ng
H
eati
ng
C
urv
e:
Cu
rve:
• KE is
KE is
(melt
ing
an
d b
oilin
g)
(melt
ing
an
d b
oilin
g)
LiquidLiquid
Gas (vapor)Gas (vapor)• P
late
au
s =
ph
ase c
han
ges!
Pla
teau
s =
ph
ase c
han
ges!
temp. remains constant until EVERY molecule temp. remains constant until EVERY molecule changeschangesphase.phase.
Obj. 9 cont…Obj. 9 cont…C
oolin
g
Coolin
g
Cu
rve:
Cu
rve:
• KE is
KE is
(con
den
sati
on
an
d f
reezi
ng
)(c
on
den
sati
on
an
d f
reezi
ng
)
TimeTime
Temperature Temperature (KE) (KE)
SolidSolid
Gas (vapor)Gas (vapor)
LiquidLiquid
CondensationCondensation
FreezingFreezingCPCP
FPFP
Obj. 10…VocabularyObj. 10…Vocabulary
Obj. 11…Heat CalculationsObj. 11…Heat Calculations
o as a
su
bsta
nce c
han
ges p
hases,
tem
p.
rem
ain
s
as a
su
bsta
nce c
han
ges p
hases,
tem
p.
rem
ain
s
con
sta
nt
con
sta
nt
plateaus on heating/cooling curves. plateaus on heating/cooling curves.
until all molecules have completed the change!until all molecules have completed the change!
o to
calc
ula
te h
eat
gain
ed
/lost
du
rin
g a
to
calc
ula
te h
eat
gain
ed
/lost
du
rin
g a
ph
ase
ph
ase
ch
an
ge
ch
an
ge……
tota
l h
eat
(q)
= m
ass x
Hto
tal h
eat
(q)
= m
ass x
H(f
or
v)
(f o
r v)
heat
of
fusio
n…
h
eat
of
fusio
n…
u
se w
hen
u
se w
hen
m
elt
ing
!m
elt
ing
!
heat
of
heat
of
vap
ori
zati
on
… u
se
vap
ori
zati
on
… u
se
wh
en
boilin
g!
wh
en
boilin
g!
** B
oth
H**
Both
Hff an
d H
an
d H
vv
will b
e g
iven
to
will b
e g
iven
to
you
! you
!
Obj. 11 cont…Obj. 11 cont…
o Ex…
Ex…
How
man
y k
ilojo
ule
s (
kJ)
of
heat
are
req
uir
ed
to
How
man
y k
ilojo
ule
s (
kJ)
of
heat
are
req
uir
ed
to
melt
am
elt
a
10.0 gram ice cube at 0°C and 101.3 kPa? 10.0 gram ice cube at 0°C and 101.3 kPa? (Hf° = 0.334 kJ/g)(Hf° = 0.334 kJ/g)
tota
l h
eat
(q)
= m
ass x
Hto
tal h
eat
(q)
= m
ass x
Hff
tota
l h
eat
(q)
= 1
0.0
xto
tal h
eat
(q)
= 1
0.0
x0.334 0.334 kJ/gkJ/g
==
3.3
4
3.3
4
kJ
kJ
o Th
is c
an
be u
sed
for
an
y p
hase c
han
ge,
as lon
g
Th
is c
an
be u
sed
for
an
y p
hase c
han
ge,
as lon
g
as t
em
p.
as t
em
p.
remains constant remains constant (plateaus).(plateaus).
Obj. 12 and 14…Temp. Obj. 12 and 14…Temp. ChangesChanges
o To c
alc
ula
te a
tem
p.
ch
an
ge (
slo
pe)…
To c
alc
ula
te a
tem
p.
ch
an
ge (
slo
pe)…
heat
(q)
= m
x C
heat
(q)
= m
x C
pp x
x
ΔΔ T T
mass
mass
sp
ecifi
c h
eat
sp
ecifi
c h
eat
cap
acit
y
cap
acit
y
**
giv
en
…ch
an
ges w
/ **
giv
en
…ch
an
ges w
/ p
hases!*
*p
hases!*
*
ch
an
ge in
ch
an
ge in
te
mp
…(T
tem
p…
(Tff –
–
TT
ii))
o Ex…
Ex…
Th
e t
em
p.
of
a 6
4.0
g s
am
ple
of
HTh
e t
em
p.
of
a 6
4.0
g s
am
ple
of
H22O
is r
ais
ed
fro
m
O is r
ais
ed
fro
m
20.0
°C20.0
°Cto 40.0°C. How much heat is required? to 40.0°C. How much heat is required? (C(Cpp water = 4.184 J/g°C) water = 4.184 J/g°C)h
eat
(q)
= m
x C
heat
(q)
= m
x C
pp x
x
ΔΔTT
40-2
0 =
40-2
0 =
20
20
q =
64
q =
64
xx
4.1
84 x
4.1
84 x
20°
20°
==
5360
5360
jou
les
jou
les
Obj. 12 and 14 cont…Obj. 12 and 14 cont…
o W
e c
an
com
bin
e t
he p
hase c
han
ge e
q.
an
d t
he
We c
an
com
bin
e t
he p
hase c
han
ge e
q.
an
d t
he ΔΔ
te
mp
. te
mp
.
eq. to find the total heat absorbed on a heating curve.eq. to find the total heat absorbed on a heating curve.
o Ex…
Ex…
How
mu
ch
heat
is n
eed
ed
to c
han
ge 3
2.0
gra
ms o
f H
ow
mu
ch
heat
is n
eed
ed
to c
han
ge 3
2.0
gra
ms o
f HH
22O
at
O a
t
-30.0°C to 45.0°C? -30.0°C to 45.0°C?
time time
tem
p.
tem
p.
-30°C-30°C
45°C45°C
**
mu
st
do 3
sep
ara
te
** m
ust
do 3
sep
ara
te
eq
uati
on
s…
eq
uati
on
s…
1) 1) ΔΔ temp. from -30°C to temp. from -30°C to 0°C0°C2) Phase change (melting)2) Phase change (melting)
1) 32.0 x1) 32.0 x
3) 3) ΔΔ temp. from 0°C to temp. from 0°C to 45°C45°C
(m x C(m x Cpp x x ΔΔT)T)
(m x C(m x Cpp x x ΔΔT)T)
(m x H(m x Hff))
2016 J2016 J
2) 32.0 x2) 32.0 x 10684.8 J10684.8 J
3) 32.0 x3) 32.0 x 6024.96 J6024.96 J
add together = add together =
18700 J18700 J
113322 2.1 x2.1 x30 =30 =
333.9 =333.9 =
4.184 x4.184 x45 =45 =
(H(Hff water = 333.9 J/g, C water = 333.9 J/g, Cpp ice = 2.1 J/g°C, C ice = 2.1 J/g°C, Cpp water = 4.184 J/g°C) water = 4.184 J/g°C)
o +
q (
heat)
= e
nd
oth
erm
ic
+q
(h
eat)
= e
nd
oth
erm
ic
reacti
on
reacti
on
Obj. 12 and 14 cont…Obj. 12 and 14 cont…
o -q
(h
eat)
= e
xoth
erm
ic
-q (
heat)
= e
xoth
erm
ic
reacti
on
re
acti
on
(heat absorbed)(heat absorbed)
(heat released)(heat released)
o g
rap
hic
ally…
gra
ph
ically…
RR
PP
Enthalpy Enthalpy ((ΔΔH) H)
TimeTime
Enthalpy Enthalpy ((ΔΔH) H)
TimeTime
RR
PP
** R
have
R h
ave less
less
heat
heat
than
P =
than
P =endothermic!endothermic!
** R
have
R h
ave m
ore
more
heat
than
h
eat
than
P
=P
=
lost
heat
lost
heat
=
= exothermic!exothermic!
gain
ed
heat
gain
ed
heat
=
=
Obj. 12 and 14 cont…Obj. 12 and 14 cont…
o to
calc
ula
te h
eat
of
reacti
on
(H
to c
alc
ula
te h
eat
of
reacti
on
(H
rr) f
rom
a g
rap
h…
) fr
om
a g
rap
h…
HHrr =
=
ΔΔHH
pro
du
cts
pro
du
cts –
–
ΔΔHH
reacta
nts
reacta
nts
((ΔΔHH))
TimTimee
RR
PP((ΔΔHH))
TimTimee
RR
PP154 kJ154 kJ
561 kJ561 kJ
HHrr =
154 –
561 =
=
154 –
561 =
--407kJ
407kJ
-q
= e
xoth
erm
ic!
-q =
exoth
erm
ic! 45.2 45.2
kJkJ
113.5 kJ113.5 kJ
HHrr =
113.5
– 4
5.2
=
= 1
13.5
– 4
5.2
=
68.3
68.3
kJ
kJ
+q
=
+q
=
en
doth
erm
ic!
en
doth
erm
ic!
Obj. 13 and 16…Phase Obj. 13 and 16…Phase DiagramsDiagrams
o a
a p
hase d
iag
ram
ph
ase d
iag
ram
rep
resen
ts t
he P
-T r
ela
tion
sh
ips
rep
resen
ts t
he P
-T r
ela
tion
sh
ips
b/n
th
eb
/n t
he
different phases of the same substance.different phases of the same substance.
o each
poin
t on
th
e c
urv
es s
how
s t
he T
an
d P
at
each
poin
t on
th
e c
urv
es s
how
s t
he T
an
d P
at
wh
ich
tw
ow
hic
h t
wo
phases are in equilibrium. phases are in equilibrium. (conditions for phase changes (conditions for phase changes to occur!)to occur!)
Obj. 13 and 16 cont…Obj. 13 and 16 cont…
o th
e p
oin
t at
wh
ich
all 3
cu
rves in
ters
ect
=
the p
oin
t at
wh
ich
all 3
cu
rves in
ters
ect
= t
rip
le
trip
le
poin
tp
oin
t ..
represents T and P at which all 3 phases exist represents T and P at which all 3 phases exist simultaneously! simultaneously! triple point for water is 0.016°C and 0.61 kPa triple point for water is 0.016°C and 0.61 kPa every substance has its own triple point. every substance has its own triple point.
Obj. 15…Liquefying Obj. 15…Liquefying GasesGases
o tw
o w
ays t
o liq
uefy
(con
den
se)
a g
as…
two w
ays t
o liq
uefy
(con
den
se)
a g
as…
atmospheric pressure.atmospheric pressure. VP of gas would be lower than atmospheric VP of gas would be lower than atmospheric pressurepressure
temperature of gas.temperature of gas.
KE of gas decreases causing bond strength to KE of gas decreases causing bond strength to increaseincrease