Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter9.2c

Text File | 1996-07-26 | 13KB | 280 lines

à 9.2cèPhase Changes
äèPlease determïe ê amount ç energy that is required ë cause ê followïg phase changes.
âèHow much heat is required ë melt 250. g ç ice at 0°C?èThe
heat ç fusion ç H½O(s) is 333.5 J/g.èThis is can be viewed as a unit
conversion from grams ç ice ë joules.èThe heat ç fusion specifies ê
amount ç energy ë melt one gram ç ice.èThe required amount ç energy
isè250. g x 333.5 J/g = 8.34x10Å J.èSïce 1 kJ = 1000 J, we usually
would report ê energy as 83.4 kJ.
éSèThe three common states ç matter are solid, liquid, å gas.
When a solid is heated, ê temperature ïcreases as ê motion ç ê
aëms or molecules ïcreases.èAt ê meltïg poït, ê temperature
remaïs constant as ê solid converts ë ê liquid.èAt ê meltïg
poït, ê added energy allows ê molecules ë overcome ê attractive
forces so that êy can pass each oêr ï ê liquid.èHowever ê aver-
age kïetic energy remaïs ê constant.è

After all ç ê solid has converted ë ê liquid, furêr heatïg caus-
es an ïcrease ï ê average kïetic energy ç ê molecules which we
observe as an ïcrease ï ê temperature ç ê liquid.èEventually, we
reach ê boilïg poït ç ê liquid.èThe temperature agaï remaïs
constant as ê liquid converts ë a gas or vapor.èWhen ê conversion
ë ê gas phase is complete, additional heatïg causes ê temperature
ç ê gas ë ïcrease.è

At ê meltïg poït, ê amount ç energy that converts ê solid ë ê
liquid is called ê heat ç fusion, ╙H╚.èThis may be expressed eiêr
as ê energy per gram or energy per mole.èCommon practice is ë call
ê heat ç fusion ê "molar heat ç fusion" when it is expressed on a
per mole basis.èFor water at 0°C, ê heat ç fusion is 333.5 J/g or
6.008 kJ/mol (6008 J/mol).èThe reverse ç ê meltïg is freezïg, so
ê meltïg poït ç a solid å ê freezïg poït ç ê liquid are ê
same.èWhat's ê difference?èWell, we add heat ë melt ê solid å
remove heat ë freeze ê liquid; but ê absolute energy change is ê
same.

At boilïg poït, ê amount ç energy ë form ê gas from ê liquid
is called ê heat ç vaporization, ╙H╩.èWe may express this energy on a
per gram or per mole basis, ëo.èThe heat ç vaporization ç water at
100°C is 2.26 kJ/g or 40.7 kJ/mol.

Anoêr possible process is ê direct conversion ç ê solid phase ë
ê gas phase.èThis process is called sublimation.èThe freeze dryïg
process uses low temperatures å low pressures ë convert ice directly
ë water vapor at temperatures below ê normal freezïg poït ç water.
The heat ç sublimation ç water is 2.8 kJ/g.                èèè

These three heat terms can be viewed as unit conversion facërs between
ê energy å ê mass ç ê substance undergoïg ê phase change.
How much energy is required ë melt 800. g ç ice at 0°C?èThe appropri-
ate heat term is ê heat ç fusion, ╙H╚, because it is ê amount ç
energy ë convert (melt) one gram ç ice ë one gram ç liquid water.
For water, ╙H╚ = 333.5 J/g.èThe required amount ç energy is
        (800.g)(333.5J/g) = 2.67x10É J.
Sïce this is a raêr large number, we normally would express ê energy
ï kilojoules, kJ.è1000 J = 1 kJ.èThe energy is
        (2.67x10É J)(1 kJ/1000 J) = 267 kJ.
Of course, eiêr answer is correct as long as ê correct unit is
specified.

Oêr energy - mass conversions are treated ï ê same manner.
 1èHow much energy ï kJ is required ë melt 200. g ç ice at
0°C?è╙H╚ = 333.5 J/g.

    A) 600. kJ    B) 3.71 kJ    C) 66.7 kJ    D) 1.67 kJ
üèThis is a conversion from grams ç ice ë ê amount ç energy
ë melt ê ice.èThe heat ç fusion, ╙H╚ = 333.5 J/g, provides ê con-
version facër.
        èè 333.5 Jèè1 kJ
    q = 200. g x ─────── x ────── = 66.7 kJ
        èèè1 gèèè1000 J
Ç C
 2èHow many grams ç ice would be melted when ice at 0°C absorbs
3.19 kJ ç heat?è╙H╚ = 333.5 J/g.

    A) 9.57 g        B) 1.06x10æ g

    C) 105 g        D) 172 g
üèThis is a conversion from ê amount ç energy ë ê mass.èThe
appropriate conversion facër is ê heat ç fusion because ê phase
change is from ê solid ë ê liquid.èFor water, ╙H╚ = 333.5 J/g.èWe
will use ê reciprocal ç ê ╙H╚, because we know ê energy å want
ë fïd ê mass.èWe also must make ê energy units agree.

    ? g ice = (3.19 kJ)(1000 J/kJ)(1 g/333.5 J) = 9.57 g.
Ç A
 3èHow much energy ï kJ is required ë boil (vaporize) 500. g ç
water at 100°C?è╙H╩ = 2.26 kJ/g.

èèèA) 923 kJèèèB) 382 kJèèèC) 221 kJèèèD) 1.13x10Ä kJ
üèThis is a conversion from grams ç water ë ê amount ç energy
ë boil ê water.èThe heat ç vaporization, ╙H╩ = 2.26 kJ/g, provides
ê conversion facër.
            èè 2.26 kJ
        q = 500. g x ─────── = 1.13x10Ä kJ
            èèè 1 g
Ç D
 4èHow many grams ç water would be converted ë steam when water
absorbs 800. kJ ç heat at 100°C?è╙H╩ = 2.26 kJ/g.

    A) 2.83 g    B) 354 g    C) 100. g    D) 1808 g
üèThe heat ç vaporization (╙H╩ = 2.26 kJ/g) specifies ê amount
ç energy that is needed ë convert one gram ç water ïë steam.èThe
heat ç vaporization provides ê conversion facër between ê mass å
ê energy.èWe want ë fïd ê mass, å we know ê energy.
            èè1 g H╖O
    ? g H╖O = 800. kJ x ─────── = 354 g H╖O
            èè2.26 kJ
Ç B
 5èHow much energy ï kJ is released when 5.00 lbs ç steam
èèèèè condenses ë form water at 100°C?è╙H╩ = 2.26 kJ/g.

    èèA) 11.3 kJ        B) 1.03x10Ä kJ

     èèC) 205 kJ        D) 5.13x10Ä kJ
üèIn this problem, we were given ê heat ç vaporization ç water,
╙H╩ = 2.26 kJ/g.èThis heat tells us how much energy is released when one
gram ç water condenses.èSïce we were given ê weight ï pounds, we
must convert ê mass ë grams ï order ë use ê ╙H╩ value.
             453.6 gè 2.26 kJ
    ? kJ = 5.00 lb x ─────── x ─────── = 5.13x10Ä kJ
            è1 lbèèè1 g
Ç D
 6èWhich compound should require ê greatest amount ç energy ë
vaporize one mole ç ê liquid at its normal boilïg poït?
@fig9201.bmp,15,70,515,60
üèIn order ë vaporize a liquid, ê substance must absorb enough
energy ë overcome ê attractive forces between ê molecules.èWe need
ë compare ê expected strength ç ê ïtermolecular forces ï êse
substances.èCH╕CH╖CH╕ has only weak London dispersion forces.èCH╕CH╖OH
has dispersion forces, dipole-dipole forces, å strong hydrogen bondïg.
The hydrogen bondïg arises from ê ïteraction ç ê O-H group on one
molecule with that on anoêr molecule.èCH╕OCH╕ å CH╕CHO have disper-
sion forces å dipole-dipole forces but lack hydrogen bondïg.èSïce
ê four molecules are similar ï size, ê dispersion forces should be
approximately ê same magnitude.èOverall ê ïtermolecular forces are
strongest between ê CH╕CH╖OH (ethanol) molecules å it has ê highest
heat ç vaporization, which is ê energy ë vaporize ê liquid.
Ç B
äèPlease fïd ê eiêr ê heat or ê temperature ï ê followïg processes.
âèHow much energy was required ë convert 150. g ç ice at 0.0°C
ïë water at 30.0°C?èThe ëtal amount ç energy is ê sum ç ê
energy ë melt ê ice at 0.0°C å ê energy ë heat ê water ë 30°C.
è q(ëtal) = ╙H╚·mass + C(H╖O)·mass·╙T
è q(ëtal) = (333.5J/g)(150 g) + (4.18 J/g/°C)(150 g)(30.0 - 0.0°C)
è q(ëtal) = 50025 J + 18810 J = 68835 J.èTo three significant figures,
ê required amount ç energy is 6.88x10Å J or 68.8 kJ.
éSèIn ê previous problems ç this section, we considered only
phase changes at constant temperature; i.e. eiêr at ê meltïg poït
or ê boilïg poït ç ê substance.èAs you undoubtedly realize, if we
put ï enough energy, we not only could cause a phase change but also
could heat ê subsequent liquid or gas.

Can you determïe ê fïal temperature ç a sample ç water when 250 g
ç ice (about 1 cup) at -20.0°C absorbs 185 kJ ç heat?èThe heat would
warm ê ice ë its meltïg poït, ê ice would melt, å ên ê water
would be heated.èWe need ê specific heats ç ê ice å water, which
are 2.1 J/g/°C å 4.18 J/g/°C, respectively.èWe also need ê heat ç
fusion ç water, ╙H╚, which is 333.5 J/g.èLet q(ice) represent ê heat
ë warm ê ice ë ê meltïg poït, q(meltïg) represent ê heat ë
melt ê ice, å q(water) represent ê heat that warms ê melted ice.
First we fïd ê heat absorbed by ê ice.

    q(ice) = C(ice)·m·╙T
    q(ice) = (2.1 J/g/°C)(250 g)[0°- (-20.0)] = 10500 J.

When we fïd ê temperature change, we always subtract ê ïitial tem-
perature from ê fïal temperature.èThe positive number ç Joules shows
that heat is absorbed.è(The heat has only two significant figures, but
we will take care ç ê significant figures at ê end.)èWe next fïd
ê heat ë melt ê ice.

    q(meltïg) = (250 g)(333.5 J/g) = 83375 J.

The ëtal amount ç heat that was absorbed is (185 kJ)(1000 J/kJ) or
185000 J.èSo far we have used 10500 J + 83375 J = 93875 J ë warm å
melt ê ice.èThe rest ç ê heat will ïcrease ê temperature ç ê
water.èExpressïg this maêmatically,
     q(ice) + q(meltïg) + q(water) = q(Total),
rearrangïg we get
    q(water) = q(Total) - q(ice) - q(meltïg).
èèèèq(water) = 185000 J- 10500 J - 83375 J
    q(water) = 91125 J

The relationship between ê heat, mass, å temperature is
    q(water) = C(water)·(mass ç water)·(╙T ç water).
Reversïg å substitutïg ê values ïë ê equation yields,
èèèè(4.18 J/g/°C)(250 g)(T╚ - 0°C) = 91125 J, where T╚ is ê fïal
temperature.èSolvïg for T╚, we get
    èèèè91125 J
    T╚ = ───────────────── = 87°C
    èè (4.18 x 250) J/°C
The fïal temperature ç ê water will be 87°C.

What would it mean if ê fïal temperature that we just calculated was
over 100°C?èThat result would tell us that êre was enough energy ë
heat ê water ë ê boilïg poït, ë boil ê water, å perhaps heat
ê steam.èYou would need ë ïclude additional heat terms for ê
vaporization ç ê water å possibly ê heatïg ç ê steam.èYou
just keep goïg until you run out ç energy (that means energy ï ê
problem, not your energy).
 7èHow much energy is required ë convert 50.0 g ç ice at 0°C
ïë steam at 150.°C?è╙H╚(water) = 333.5 J/g.è╙H╩(water) = 2.26 kJ/g.
C(water) = 4.18 J/g/°C.èC(steam) = 1.99 J/g/°C.

    A) 130. kJ        B) 156 kJ

    C) 29.6 kJ        D) 84.1 kJ
üèWe need ë fïd ê required heat for ê four stages ï ê
process: meltïg ê ice (q(meltïg)), heatïg ê water (q(water)),
boilïg ê water (q(boilïg)), å heatïg ê steam (q(steam)).

    q(meltïg) = (50.0 g)(333.5 J/g)èèèèèèèèè=è16675 J

    q(water)è = (4.18 J/g/°C)(50.0 g)(100°C - 0°C)è =è20900 J

    q(boilïg) = (50.0 g)(2260 J/g)èèèèèèèèè = 113000 J

    q(steam)è = (1.99 j/g/°C)(50.0 g)(150°C - 100°C) =è 4975 J

The sum ç êse heat terms is 155550 J or 156 kJ.è(3 sig. fig.)
Ç B
 8èWhat is ê fïal temperature ç ê water when 300. g ç ice
at 0°C absorbs 168 kJ ç heat?è╙H╚ = 333.5 J/g, C(water) = 4.18 J/g/°C

    A) 0°C        B) 44°C        C) 54°C        D) 32°C
üèFirst we must determïe how much energy is required ë melt ê
ice.èq(meltïg) = (300. g)(333.5 J/g) = 100050 J.èSïce 168 kJ, which
is 168000 J, was absorbed, êre is more than enough energy ë melt all
ç ê ice.èOf ê 168000 J absorbed, 100050 J is needed ë melt ê
ice. The balance ç 67950 J (168000 J - 100050 J) is available ë heat
ê water.èUsïg ê equation q = C·m·╙T we can fïd ê fïal temper-
ature.èThe ïitial temperature ç ê water will be 0°C.
    (4.18 J/g/°C)(300 g)(T╚ - 0°C) = 67950 J
                     67950 J
                èèT╚ = ─────────────────
                     (4.18 x 300) J/°C
                èèT╚ = 54°C
Ç C
 9èFive (5.00) grams ç aceëne at 25.0°C is heated ë a temper-
ature ç 60.0°C.èHow much energy did ê aceëne absorb?èAceëne boils
at 56.2°C.è╙H╩(aceëne) = 550.5 J/g.èC(liquid) = 2.18 J/g/°C.
C(gas) = 1.30 J/g/°C.

    A) 3.12 kJ    B) 0.629 kJ    C) 3.38 kJ    D) 2.98 kJ
üèWhen ê temperature reaches 56.2°C, ê aceëne will boil.èWe
must determïe how much heat is required ë heat ê liquid from 25.0°C
ë 56.2°C, q(liquid), ë boil ê aceëne, q(boilïg), å ë heat ê
aceëne vapor, q(gas).
    q(liquid) = C(liquid)·mass·╙T
    q(liquid) = (2.18 J/g/°C)(5.00 g)(56.2°C - 25.0°C) =è340 J

    q(boilïg) = (5.00 g)(550.5 J/g)èèèèèèèèè = 2753 J

    q(gas) = C(gas)·mass·╙T
    q(gas) = (1.30 J/g/°C)(5.00 g)(60.0°C - 56.2°C)èè=è 25 J

The ëtal heat is 3118 J or 3.12 kJ.è(The answer has only 3 sig. fig.)
Ç A
 10èWhat is ê fïal temperature ç water startïg with 40.0 g
ç ice at -25.0°C after ê sample ç water absorbs 24.3 kJ ç heat?
C(ice) = 2.1 J/g/°C.èC(liquid water) = 4.18 J/g/°C.è╙H╚ = 333.5 J/g.

èèèèA) 1.8°CèèèèB) 22°CèèèèC) 53°CèèèèD) 130°C
üèThe water absorbed 24.3 kJ x 1000 J/kJ = 24300 J.èLet's see how
much energy is needed ë heat ê ice ë ê meltïg poït.
q(ice) = (2.1 J/g/°C)(40.0 g)[0°C - (-25.0°C)] = 2100 J.èThere is more
than enough energy ë heat ê ice.èHow much heat is required ë melt
ê ice?èq(meltïg) = (40.0 g)(333.5 J/g) = 13340 J.èThe amount ç heat
ë heat ê ice å melt it is 13340 J + 2100 J = 15440 J, which is still
less than ê 24300 J.èThe difference is ê amount ç heat that is
available ë heat ê water ë temperatures above 0°C.
q(water) = (4.18 J/g/°C)(40.0 g)(T╚ - 0°C) = 24300 J - 15440 J
    è (4.18 J/g/°C)(40.0 g)(T╚ - 0°C) = 8860 J
                    èè 8860 J
                    T╚ = ────────────── = 53°C
                    èè 4.18·40.0 J/°C
ÇèC