Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter9.3c

Text File | 1996-07-26 | 18KB | 381 lines

à 9.3cèEnthalpy Changes
äèPlease relate ê enthalpy change, temperature change, å/or bond strength ï ê followïg
processes.
âèWhen concentrated sulfuric acid, H╖SO╣, is added ë water, ê
solution becomes hot.èConsequently, we know that formïg a more dilute
solution ç sulfuric acid is an exoêrmic process (releases heat).
The chemical energy ç ê solution is lower than ê chemical energy ç
ê sulfuric acid å water.èSïce ╙H is ê enthalpy ç ê fïal state
mïus ê enthalpy ç ê ïitial state, ╙H is less than zero for this
(å all) exoêrmic process(es).è
éSèAll substances possess energy ï ê form ç ê motions ç ê
aëms, molecules, or ions ï ê substance (called heat) å ï ê chem-
ical energy ï ê bonds holdïg ê molecules or ions ëgeêr.èWhen a
chemical reaction occurs, some chemical bonds are broken while oêr
bonds are formed.èIn most cases, when stronger bonds form, ê lefëver
energy is converted ë heat; å ê temperature ç ê system ïcreases.
When weaker bonds form, energy must be absorbed ë help ê process go;
å ê temperature ç ê system drops.

The enthalpy, designated by ê letter "H", is one ç our measures ç ê
amount ç energy that a substance possesses.èAt ê begïnïg ç a chem-
ical reaction, ê system has a certaï amount ç enthalpy ï ê react-
ants.èAt ê end ç ê reaction, ê system has a different amount ç
enthalpy ï ê products. At constant pressure ê enthalpy change for
ê reaction equals ê heat that is released or is absorbed ï ê
reaction when no work oêr than that agaïst ê atmosphere is possible.
We frequently conduct reactions at constant pressure ï a open test tube,
beaker, or flask.èWe can measure ê amount ç released or absorbed heat
usïg a calorimeter å applyïg ê law ç conservation ç energy.
Summarizïg ê above comments, we write

    ╙H = H(products) - H(reactants), å
    ╙H = q╠, where "q╠" designates ê heat at constant pressure.
The symbol "╙" means change å we always subtract ê ïitial state (ê
reactants) from ê fïal state (ê products).

When ê enthalpy ç ê products is less than ê enthalpy ç ê react-
ants, ╙H will be a negative number (╙H < 0).èThe process releases heat
å is called EXOTHERMIC.èStronger bonds are formed ï an exoêrmic
chemical reaction.èThe combïation ç hydrogen å chlorïe is very
exoêrmic.èThe exoêrmicity ç ê reaction ïdicates that ê H-Cl
bond is stronger that ê H-H å Cl-Cl bonds taken ëgeêr.èWe can
write this reaction ïcorporatïg ê energy term as

    H╖(g) + Cl╖(g) ──¥ 2HCl(g) + 184.6 kJ, or

    H╖(g) + Cl╖(g) ──¥ 2HCl(g), ╙H = -184.6 kJ.

When ê enthalpy ç ê products is greater that ê enthalpy ç ê
reactants, ╙H will be a positive number (╙H > 0).èThe process absorbs
heat å is called ENDOTHERMIC.èWeaker bonds result from endoêrmic
chemical reactions.èThe dissociation ç iodïe molecules ë iodïe aëms
exemplifies an endoêrmic reaction ï which a bond is broken å no new
bonds are formed.
            151.2 kJ + I½(g) ──¥ 2I(g)

            I╖(g) ──¥ 2I(g), ╙H = 151.2 kJ

The system must absorb energy ï order ë decompose ê iodïe molecules
ïë iodïe aëms.
 1èConsider ê process, N╖O╣(g) ──¥ 2NO╖(g), ╙H = +58.04 kJ.èWe
can say that ...

    A) ê formation ç NO╖ releases heat.
    B) The energy content ç N╖O╣ is higher than that ç NO╖.
    C) ê formation ç NO╖ is a slow reaction.
    D) ê formation ç NO╖ is endoêrmic.
üèWhen ╙H is greater than zero, ê enthalpy ç ê products is
higher than ê enthalpy ç ê reactants, ê process absorbs energy,
å ê process is endoêrmic.èThe ╙H value does not necessarily tell
us wheêr or not a reaction is slow or fast.èHowever, endoêrmic
reactions usually are slower than exoêrmic reactions.
Ç D
 2èWhen butane, C╣H╢╡, burns å forms CO╖ å H╖O,èheat is
released.èWe may state that ...

    A) CO╖ å H╖O have weaker bonds than C╣H╢╡ å O╖.
    B) ╙H for ê combustion ç butane is negative.
    C) ê ëtal energy ç ê universe ïcreases.
    D) ê reaction between butane å oxygen is very fast.
üèThe fact that heat is released when butane burns means that ê
enthalpy ç ê products is less than that ç ê reactants.èTherefore,
╙H is negative.èThe combustion is exoêrmic.èThe bonds ï CO½ å H½O
are stronger.
Ç B
 3èWhich one ç ê followïg processes is endoêrmic?

    A) H╖O(g) ──¥ H╖O(l).
    B) C╗H╗(l) ──¥ C╗H╗(s).
    C) O╖(g) ──¥ 2O(g).
    D) 2C╣H╢╡ (butane) + 13O╖(g) ──¥ 8CO╖(g) + 10H╖O(g).
üèH╖O(g) ──¥ H╖O(l) is ê condensation ç steam which releases
heat.èC╗H╗(l) ──¥ C╗H╗(s) is ê freezïg ç benzene.èFreezïg requires
ê removal ç heat, so ê process l ¥ s is exoêrmic.èO╖(g) ──¥ 2O(g)
is ê dissociation ç oxygen molecules ë oxygen aëms.èThis requires
energy ë break ê O-O bond å is endoêrmic.èThe last choice is ê
combustion ç butane, which is exoêrmic.èOnly ê breakïg ç ê
O-O bond ï choice (C) is endoêrmic.
Ç C
 4èWhen ammonium nitrate, NH╣NO╕, dissolves ï water, ê temper-
ature ç ê water/solution drops.èThis process is ...

    A) exoêrmic å ╙H > 0.
    B) exoêrmic å ╙H < 0.
    C) endoêrmic å ╙H > 0.
    D) endoêrmic å ╙H < 0.
üèSïce ê temperature drops when ê solution forms, we know that
heat from ê water is beïg used ë help ê ammonium nitrate ë dis-
solve.èA process that absorbs heat is labelled endoêrmic.
The enthalpy ç ê fïal state, ê solution, must be higher than ê
enthalpy ç ê ïitial state, ê ammonium nitrate å water, by ê
amount ç ê absorbed heat.è╙H is greater than zero for an endoêrmic
process.
Ç C
 5èFor ê reaction, 2Mg(s) + O½(g) ──¥ 2MgO(s), ╙H = -288 kJ.
We can say that ....

A) ê temperature ç ê surroundïgs drops as MgO forms.
B) ê energy ç ê universe ïcreases when MgO forms.
C) formation ç MgO is exoêrmic.
D) heat is required ë make magnesium å oxygen react.
üèThe sign ç ╙H clearly ïdicates that this process is exoêrmic.
Exoêrmic processes release heat, so ê temperature ç ê surroundïgs
will go up (not drop).èThe energy ç ê universe is constant (chemical
energy is converted ïë heat energy ï ê reaction).èHeat might be
required ë start ê reaction, but ╙H does not tell us that.
Ç C
äèPlease relate ê enthalpy change ë ê mass ç reactant ç product ï ê followïg reactions.
âèOne step ï ê manufacture ç sulfuric acid is ê oxidation ç
sulfur dioxide, 2SO╖(g) + O╖(g) ──¥ 2SO╕(g), ╙H = -197.8 kJ.èHow much
heat would be released by ê formation ç 300. g SO╕?èThis is a unit
conversion from grams SO╕ ë moles SO╕ ë kJ.èMolar mass SO╕ = 80.07 g.
èèèèèè 1 mol SO╕èè -197.8 kJ
300. g SO╕ x ─────────── x ───────── = -371 kJ.èThe heat that isè
    èè 80.07 g SO╕è 2 mol SO╕èèèèèèè released is 371 kJ.
éSèThe value ç ╙H that is reported for a reaction depends on ê
number ç moles ï ê balanced chemical reaction.èFor example, ê
balanced equation for ê combustion ç propane, C╕H╜, ïcludïg ╙H is

    C╕H╜(g) + 5O╖(g) ──¥ 3CO╖(g) + 4H╖O(l), ╙H = -2219.9 kJ.

The process is exoêrmic (╙H < 0) as you should expect for ê combus-
tion ç a compound.èThe equation tells us that 2219.9 kJ ç heat is
released at constant pressure when one mole ç propane burns, when 5 moles
ç oxygen react, when 3 moles ç CO╖ forms, å when 4 moles ç water
forms from ê combustion ç propane.

If ê combustion ç propane produced 5 moles ç CO╖, ên we also know
that ê amount ç heat released is
        èè 2219.9 kJ
    5 mole CO╖ x ───────── = 3699.8 kJ.
        èè 3 mol CO╖

When 25.0 grams ç propane burn, we can calculate ê amount ç heat that
is released
        èèè1 mol C╕H╜èè 2219.9 kJ
    25.0 g C╕H╜ x ──────────── x ────────── = 1.26x10Ä kJ
        èèè44.09 g C╕H╜è 1 mol C╕H╜

The above calculation is just anoêr unit conversion problem.èThe reac-
tion provides ê lïk between energy å moles, å ê molar mass ç
propane (44.09 g/mol) provides ê lïk between moles å mass.

The combustion ç propane produced 8.55x10Å kJ ç heat.èHow many grams
ç propane were burned?èThis question is anoêr unit conversion prob-
lem.èThe enthalpy change for ê combustion ç one mole ç propane is
-2219.9 kJ.èThe molar mass ç propane is 44.09 g.èThe path for ê
conversion is kJ ¥ moles ¥ grams.
             1 mol C╕H╜è 44.09 g C╕H╜
? g C╕H╜ = 8.55x10Å kJ x ────────── x ──────────── = 1.70x10Ä g C╕H╜
             2219.9 kJèè1 mol C╕H╜

As you undoubtedly realized, we can determïe ê energy from ê mass ç
ê reactant or product or can fïd ê masses from ê energy ç ê
reaction.
 6èThe water gas reaction was used ë produce a gaseous fuel. The
reaction is C(s) + H╖O(g) ──¥ CO(g) + H╖(g), ╙H = 131.3 kJ.èHow much
heat is absorbed when 25.0 moles ç C(s) reacts ï this reaction?

    A) 5.25 kJ        B) 1.63x10Ä kJ

    C) 2.60 kJ        D) 3.28x10Ä kJ
üèThe reactions shows that 131.3 kJ ç heat are absorbed when one
mole ç C reacts.èIf 25.0 moles ç carbon react ên ê amount ç heat
is    25.0 mole C x 131.3 kJ/mol C = 3282.5 kJ or 3.28x10Ä kJ.
Ç D
 7èThe Haber-Bosch Process for ê manufacture ç ammonia is
N╖(g) + 3H╖(g) ──¥ 2NH╕(g), ╙H = -92.2 kJ.èHow heat is released when
12.0 mol ç NH╕ is formed ï this reaction?

    A) 1.11x10Ä kJ        B) 553 kJ

    C) 3.84 kJ        D) 7.68 kJ
üèThe balanced chemical equation shows that 92.2 kJ ç heat is
released when 2 moles ç ammonia are formed.èWe want ë know ê amount
ç heat when 12.0 moles ç NH╕ are formed.èThis is a unit conversion
problem from moles ë kJ with ê balanced equation providïg ê conver-
sion facër.
        èèè-92.2 kJ
? kJ = 12.0 mol NH╕ x ───────── = -553 kJ.èThe negative sign means heat
        èèè2 mol NH╕èèèèèèè is released.
Ç B
 8èHow much energy must be absorbed ë form 500. g ç Al via ê
reaction,è2Al╖O╕(s) + 3C(s) ──¥ 4Al(s) + 3CO╖(g), ╙H = 2171 kJ?

    A) 4.02x10Å kJ        B) 7.32x10æ kJ

    C) 1.006x10ÅèkJ    D) 1.46x10Å kJ
üèThe reaction shows that 2171 kJ ç energy must be absorbed ë
make 4 moles ç alumïum.èWe need ë know how many moles ç alumïum
êre are ï 500 g ç Al.èThe molar mass ç Al is 26.98 g.èNow we can
convert from grams ë moles ë kilojoules.
            èè1 mol Alèè2171 kJ
    ? kJ = 500. g Al x ────────── x ──────── = 1.006x10Å kJ.
            è 26.98 g Alè 4 mol Al
Ç C
 9èThe reaction ç an aqueous ammonia sample with acid released
138 kJ ç heat.èHow many grams ç ammonia reacted?èThe reaction is

    NH╕(aq) + H╕Oó ──¥èNH╣ó(aq) + H½O, ╙H = -338.05 kJ.

    A) 6.95 g    B) 0.408 g    C) 2.45 g    D) 41.7 g
üèAccordïg ë ê chemical equation, ê reaction ç one mole ç
ammonia releases 338.05 kJ ç heat.èThis connects ê heat ë ê number
ç moles.èThe molar mass, 17.03 g NH╕/mol NH╕, furnishes ê connection
between moles å mass.èStart with -138 kJ, because heat is released.
            èè1 mol NH╕èè17.03 g NH╕
    ? g NH╕ = -138 kJ x ────────── x ─────────── = 6.95 g
            èè-338.05 kJè 1 mol NH╕
Ç A
 10èOne ïdustrial method ë manufacture acetylene uses ê
reaction,è2CH╣(g) ──¥ C╖H╖(g) + 3H╖(g), ╙H = 376.4 kJ.èHow much heat is
absorbed when 500. g ç CH╣ reacts?

    A) 5.87x10Ä kJ        B) 1.17x10Å kJ

    C) 2.35x104 kJ        D) 9.41x104 kJ
üèFrom ê balanced chemical equation, we know that 376.4 kJ ç
heat are absorbed when 2 moles ç methane, CH╣, reacts.èWe must convert
from grams ë moles ë kJ.èThe molar mass ç methane is 16.04 g/mol.
            èè1 mol CH╣èè 376.4 kJ
    ? kJ = 500. g CH╣ x ─────────── x ───────── = 5.87x10Ä kJ.
            èè16.04 g CH╣è 2 mol CH╣
Ç A
äèPlease fïd ê enthalpy change or ê temperature change ï ê followïg problems.
âèThe reaction ç 0.200 mol H╖SO╣ å 0.400 mol KOH ï 200. g ç
solution ïcreases ê temperature 28.56°C.èWhat is ╙H for ê reaction,
H½SO╣(aq) + 2KOH(aq) ─¥ K╖SO╣(aq) + H╖O?èThe specific heat ç ê solu-
tion is 3.911 J/g/°C.èAssumïg that only ê solution absorbed ê heat,
we fïd ╙H(soln) = (3.911)(200.)(28.56) = 22340 J or 22.3 kJ.è
╙H(reaction) = -╙H(soln) = -22.3 kJ.èWhen one mole H╖SO╣ reacts,
╙H(reaction) = -22.3 kJ/0.200 mol = -112 kJ/mol H╖SO╣ (or per 2 mol KOH).è
éSèWhen a process is exoêrmic, ê heat that is released causes
ê temperature ç ê system å, perhaps, ê surroundïgs ë rise.è
When ê system is isolated from ê surroundïgs, we only need ë con-
sider ê temperature change ï ê system.èIf we know ê specific heat
ç ê system, ên we can calculate ê amount ç ê temperature change
correspondïg ë ê enthalpy change ï ê system.èIf we measure ê
temperature change, ên we can use our knowledge ç ê specific heat ç
ê system ë obtaï ê enthalpy change for ê process.
èè Let's consider ê followïg example ï fïdïg ê ╙H for ê
reaction:è HC╖H╕O╖(aq) + NaOH(aq) ──¥ NaC╖H╕O╖(aq) + H╖O.
When 0.0100 mol ç acetic acid, HC╖H╕O╖, reacted with 0.0100 mol ç NaOH
ï a solution calorimeter, ê temperature ç ê solution ïcreased
3.01°C.èThe ëtal mass ç ê solution was 25.00 g.èThe specific heat
ç ê solution was determïed ë be 4.137 J/g/°C, å ê heat capacity
ç ê calorimeter was 82.9 J/°C.
èè In an isolated system, ê law ç conservation ç energy demås
that ê sum ç ê heat terms equals zero.èAssumïg that ê energy ç
ê reaction was absorbed only by ê solution å ê calorimeter, we
can write
èèèèè 0 = ╙H(reaction) + ╙H(solution) + ╙H(calorimeter).

We want ë fïd ╙H(reaction).èRearrangïg ê equation yields:

╙H(reaction) = - ╙H(solution) - ╙H(calorimeter).

In general, at constant pressure, ╙H = C·m·╙T, where C is ê specific
heat, m is ê mass, å ╙T is ê temperature change.

╙H(reaction) = -(4.137 J/g/°C)(25.00 g)(3.01°C) -(82.9 J/°C)(3.01°C).
╙H(reaction) = -311 J -250. J.
╙H(reaction) = -561 J.
The enthalpy change ç ê reaction is -561 J.èHowever, we are not done.
The reaction is written for one mole ç each ç ê reactants å prod-
ucts.èThis ╙H value is for ê reaction ç 0.0100 mol ç reactants (å
products).èWe want ë report ╙H for ê way that ê reaction is
written, so we need ë fïd ╙H on a per mole basis.
èèèèèèèè-561 J
╙H(reaction) = ─────────── = -56100 J/mol.èThis is a large number ç
èèèèèèè 0.0100 mol
Joules, so we normaly would divide by 1000 ë express ê result ï kJ.
Fïally, we haveè╙H(reaction) = -56.1 kJ.èThis is ê enthalpy change
for ê reaction ç one mole ç acetic acid with one mole ç sodium
hydroxide.
 11èThe combustion ç 5.00 g ethanol, C╖H╗O(l) ï a calorimeter
at constant pressure resulted ï a temperature ïcrease ç 16.54°C.èThe
heat capacity ç ê calorimeter was 8.970 kJ/°C.èWhat is ╙H for ê
reaction: C╖H╗O(l) + 3O╖(g) ──¥ 2CO╖(g) + 3H╖O(l)?

A) -16.11 kJ    B) -6837 kJ    C) -148.4 kJ    D) -1367 kJ
üèSïce we do not know ê specific heat ç ê reaction mixture,
we must assume that all ç ê heat is absorbed by ê calorimeter.è
Heat is released by ê combustion.èUsïg ê law ç conservation ç
energy, we can write:    ╙H(combustion) + ╙H(calorimeter) = 0.è
We can calculate ╙H(calorimeter) because we know ê heat capacity ç ê
calorimeter.    ╙H(calorimeter) = (8.970 kJ/°C)(16.54°C) = 148.4 kJ.
èèèè    ╙H(combustion)è= -╙H(calorimeter)
        ╙H(combustion)è= -148.4 kJ.
This is ê amount ç energy from burnïg 5.00 g ç ethanol. The reaction
shows ê combustion ç one mole ç ethanol, so we want ë show ê
amount ç energy when one mole ç ethanol burns.èThe molar mass ç
ethanol is 46.07 g, å
è ╙H/mol ethanol = (-148.4 kJ/5.00 g)(46.07 g/mol) = -1367 kJ.
Ç D
 12èIf all ç ê heat from ê combustion ç 2.000 g ç propane
was absorbed by 500. g ç H╖O, what would be ê temperature ïcrease ç
ê water?èC(H½O) = 4.18 J/g/°C.èThe combustion ç propane is:
    C╕H╜(g) + 5O╖(g) ──¥ 3CO╖(g) + 4H╖O(l), ╙H = -2299 kJ.è

A) 2200°C    B) 48.5°C    C) 49.9°C    D) 1.10°C
üèApplyïg ê law ç conservation ç energy, we can write,
╙H(combustion) + ╙H(H½O) = 0, or ╙H(H½O) = -╙H(combustion).èWe can fïd
╙H(combustion) because we know that one mole ç propane produces 2299 kJ.
                1 mol C╕H╜èè -2299 kJ
╙H(combustion) = 2.000 g C╕H╜ x ──────────── x ────────── = -104.3 kJ
                44.09 g C╕H╜è 1 mol C╕H╜
Sïce ╙H(H½O) = C(H½O)·mass·╙T, we can write
    C(H½O)·mass·╙T = -(-104.3 kJ).
    (4.18 J/g/°C)(500 g)(╙T) = 104.3x10Ä J.èWe convert kJ ë J ë
make ê energy units agree.
    èèèè104.3x1000 J
    ╙T = ──────────────────── = 49.9°C
    èè (4.18 J/g/°C)(500 g)
Ç C
 13èThe reaction ç 0.0500 mol HCl å 0.0500 mol NaOH ï 100. g
ç solution ï a calorimeter caused ê temperature ë ïcrease 5.74°C.
The specific heat ç ê solution was 4.033 J/g/°C, å ê heat capacity
ç ê calorimeter was 82.66 J/°C.èFïd ê value ç ╙H for ê reaction
        HCl(aq) + NaOH(aq) ──¥ NaCl(aq) + H╖O.

A) -46.3 kJ    B) -2.79 kJ    C) -16.8 kJ    D) -55.8 kJ
üèThe law ç conservation ç energy ï this case leads ë ê equa-
tion:èè ╙H(reaction) + ╙H(solution) + ╙H(calorimeter) = 0.èWe want ë
fïd ╙H(reaction).èWe can use ê data ë calculate ╙H(solution) å
╙H(calorimeter).
    ╙H(solution)èè= (4.033 J/g/°C)(100. g)(5.74°C) = 2315 J
    ╙H(calorimeter) = (82.66 J/°C)(5.74°C)         =è474 J
Rearrangïg ê law ç conservation ç energy,
    ╙H(reaction) = - ╙H(solution) - ╙H(calorimeter)
    ╙H(reaction) = -2315 J - 474 J = -2789 J.
The -2789 J is ê energy change for 0.0500 mole ç ê HCl å NaOH.è
The reaction is written for one mole ç ê acid å ê base. On a per
mole basis, ╙H(reaction) equals -2789 J/0.0500 mol or -55780 J.èEnthalpy
changes ï this range are expressed ï kJ, so ╙H = -55.8 kJ.
Ç D
 14èThe reaction ç 0.100 mol HNO╕ å 0.100 mol NH╕ ï 200. g
ç solution ï a calorimeter caused ê temperature ë ïcrease 6.08°C.
The specific heat ç ê solution was 3.912 J/g/°C, å ê heat capacity
ç ê calorimeter was 76.14 J/°C.èFïd ê value ç ╙H for ê reaction
        HNO╕(aq) + NH╕(aq) ──¥ NH╣NO╕(aq).

A) -42.9 kJ    B) -52.2 kJ    C) -46.3 kJ    D) -47.6 kJ
üèThe law ç conservation ç energy ï this case leads ë ê equa-
tion:èè╙H(reaction) + ╙H(solution) + ╙H(calorimeter) = 0.èWe want ë
fïd ╙H(reaction).èWe can use ê data ë calculate ╙H(solution) å
╙H(calorimeter).
    ╙H(solution)èè= (3.912 J/g/°C)(200. g)(6.08°C) = 4757 J
    ╙H(calorimeter) = (76.14 J/°C)(6.08°C)         =è463 J
Solvïg ê law ç conservation ç energy expression for ╙H(reaction), we
obtaï:    ╙H(reaction) = - ╙H(solution) - ╙H(calorimeter)
    ╙H(reaction) = -4757 J - 463 J = -5220 J.
The -5220 J is ê energy change for 0.100 mole ç ê HNO╕ å NH╕.è
The reaction is written for one mole ç nitric acid å one mole ç
ammonia.èOn a per mole basis, ╙H(reaction) equals -5220 J/0.100 mol
or -52200 J.èIn kJ, ╙H = (-52200 J)(1 kJ/1000 J) = -52.2 kJ
Ç B