CALCULATING HEATS OF RXNSCALCULATING HEATS OF RXNS There are three ways to calculate There are three ways to calculate

the energy of a reaction.the energy of a reaction.– H=mCH=mCT T (takes a temperature (takes a temperature

change, mass, and specific heat change, mass, and specific heat constant to calculate a constant to calculate a ΔΔHHrxnrxn; uses ; uses conservation of energy)conservation of energy)

– Enthalpy of Formation Enthalpy of Formation (takes data (takes data from a table and uses it to from a table and uses it to calculate the energy of a reaction)calculate the energy of a reaction)

– Hess’s Law Hess’s Law (allows us to take two or (allows us to take two or more chem rxns with known more chem rxns with known ΔΔHHrxnrxn and combine them in such a way to and combine them in such a way to calculate the enthalpy of a target calculate the enthalpy of a target reaction)reaction)

HEATS OF FORMATIONHEATS OF FORMATION Another method of calculating the Another method of calculating the

enthalpy of a reaction is by using enthalpy of a reaction is by using heats of formation.heats of formation.– There are tables of There are tables of HHformform that we can that we can

gather information fromgather information fromo Elements are always 0Elements are always 0o HHformform is dependent on the number is dependent on the number

of moles of moles– We also need to use the equation We also need to use the equation

presented earlier:presented earlier:

HHrxnrxn = = ∑H∑Hproductsproducts - ∑H - ∑Hreactantsreactants

HEATS OF FORMATIONHEATS OF FORMATIONCalculate Calculate H for the following reaction: H for the following reaction:

8 Al(s) + 3 Fe8 Al(s) + 3 Fe33OO44(s) (s) 4 Al 4 Al22OO33(s) + 9 (s) + 9 Fe(s)Fe(s)8(0)8(0) 3(-1118.4)3(-1118.4) 4(-1675.7)4(-1675.7) 9(0)9(0)

HHrxnrxn = = ∑H∑Hproductsproducts - ∑H - ∑Hreactantsreactants

HHrxnrxn= = {4(-1675.7)+9(0)} – {8(0)+3(-{4(-1675.7)+9(0)} – {8(0)+3(-

1118.4)}1118.4)}HHrxnrxn= (= (-6,702.8) – (-3355.2)-6,702.8) – (-3355.2)

HHrxnrxn= = -3,347.6 kJ-3,347.6 kJ

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CHCH44 + 2O + 2O22 CO CO22 + 2H + 2H22OO2C2C22HH66 + 7O + 7O22 4CO 4CO22 + 6H + 6H22OOCC33HH88 + 5O + 5O22 3CO 3CO22 + 4H + 4H22OO

2C2C44HH1010 + 13O + 13O22 8CO 8CO22 + 10H + 10H22OO

Use heats of formations calculations to Use heats of formations calculations to determine the combustion of which determine the combustion of which hydro-carbon will produce the most hydro-carbon will produce the most energy per mole…energy per mole…

(CH(CH44= -74.81 kJ/mol; C= -74.81 kJ/mol; C22HH66= -84.68 = -84.68 kJ/mol; CkJ/mol; C33HH88= -104.5; C= -104.5; C44HH1010= -126.5 = -126.5

kJ/mol)kJ/mol)

HESS’ LAWHESS’ LAW The change in energy of a process The change in energy of a process

or reaction is a state function, or reaction is a state function, meaning that regardless of the path meaning that regardless of the path to reach your goal, the energy to get to reach your goal, the energy to get there is constant. there is constant.

For instance if you want to vaporize For instance if you want to vaporize a solid, you have a solid, you have two pathwaystwo pathways..– You can melt it into a liquid and then You can melt it into a liquid and then

vaporize it into a gas.vaporize it into a gas.– Or you can sublime the solid directly Or you can sublime the solid directly

into a gas.into a gas. Either path gets the desired results Either path gets the desired results

and either path requires the same and either path requires the same amount of heat energy, this is amount of heat energy, this is Hess’s LawHess’s Law. .

The idea that we can calculate The idea that we can calculate HHsublimationsublimation by combining the by combining the HHfusfus with the with the HHvapvap is an illustration of is an illustration of Hess’ LawHess’ Law. .

During any Hess’s Law calculation, During any Hess’s Law calculation, there are two things that we are there are two things that we are allowed to do to the given reactions allowed to do to the given reactions in order to manipulate the.in order to manipulate the.− We can reverse the reaction in order We can reverse the reaction in order

to make the products reactants, as to make the products reactants, as long as we change the sign of the long as we change the sign of the enthalpyenthalpy

− We can also increase or decrease We can also increase or decrease the amounts of reactants or the amounts of reactants or products by multiplying by a factor, products by multiplying by a factor, as long as we multiply the enthalpy as long as we multiply the enthalpy by the same factorby the same factor

The key is to keep our eye on the The key is to keep our eye on the prize, the goal reactionprize, the goal reaction

2 N2 N2(g)2(g) + 5 O + 5 O2(g)2(g) 2 N 2 N22OO5(g) 5(g) HHff = ? = ?

For example, use Hess’s Law to For example, use Hess’s Law to calculate the enthalpy of formation calculate the enthalpy of formation for the following reaction equation: for the following reaction equation:

2NO2NO(g)(g) + O + O2(g)2(g) 2NO 2NO2(g) 2(g) HH°°rxnrxn= = -114kJ/mol-114kJ/mol4NO4NO2(g)2(g) + O + O2(g)2(g) 2N 2N22OO5(g) 5(g) HH°°rxnrxn= = -110kJ/mol-110kJ/mol

NN2(g)2(g) + O + O2(g)2(g) 2NO 2NO(g) (g) HH°°rxnrxn= +181kJ/mol= +181kJ/mol

Given the following reaction Given the following reaction equations: equations:

22

22

2(-114 kJ)+(-110 kJ)+2(181 kJ) = 2(-114 kJ)+(-110 kJ)+2(181 kJ) = 24 kJ24 kJ

2H2H22OO2H2H22+O+O2 2 484kJ/mol484kJ/mol

Example 2:Example 2:Given the following information:Given the following information:

2C2C22HH66 + 7O + 7O22 4CO 4CO22 + 6H + 6H22OO

CC22HH66CC22HH44 + H + H2 2 137kJ/mol137kJ/mol

2H2H22O+2COO+2CO22CC22HH44+3O+3O2 2 1323kJ/mol1323kJ/mol

Find the value of Find the value of H° for the H° for the reaction:reaction:

2222

22 22

2H2H22OO2H2H22+O+O2 2 484kJ/mol484kJ/mol

Example 2:Example 2:Rearranging and multiplying:Rearranging and multiplying:

2C2C22HH66 + 7O + 7O22 4CO 4CO22 + 6H + 6H22OO

2 C2 C22HH6 6 2 C 2 C22HH44 + 2 H + 2 H2 2 274kJ/mol274kJ/mol

2C2C22HH44+6O+6O22 4H4H22O+4COO+4CO2 2

-2646kJ/mol-2646kJ/molFind the value of Find the value of H° for the H° for the

reaction:reaction:

2H2H22 + O + O222H2H22OO - 484kJ/mol- 484kJ/mol

Example 2:Example 2:Rearranging and multiplying:Rearranging and multiplying:

2C2C22HH66 + 7O + 7O22 4CO 4CO22 + 6H + 6H22OO

2 C2 C22HH6 6 2 C 2 C22HH44 + 2 H + 2 H2 2 274kJ/mol274kJ/mol

2C2C22HH44+6O+6O22 4H4H22O+4COO+4CO2 2 - - 2646kJ/mol2646kJ/mol

Find the value of Find the value of H° for the H° for the reaction:reaction:

(274kJ)+(-484kJ)+(-2646kJ) = (274kJ)+(-484kJ)+(-2646kJ) = HHrxnrxn

-2856 kJ = -2856 kJ = HHrxnrxn

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CC(s)(s) + ½O + ½O2(g)2(g) CO CO(g)(g) HH = -110.53 = -110.53 kJ kJ

C(s) + HC(s) + H22O(g) O(g) CO(g) + H CO(g) + H22(g)(g)

Before pipelines were built to deliver Before pipelines were built to deliver natural gas, individual towns and cities natural gas, individual towns and cities contained plants that produced a fuel contained plants that produced a fuel known as town gas by passing steam known as town gas by passing steam over red-hot charcoal.over red-hot charcoal.

COCO(g)(g) + ½O + ½O2(g)2(g) CO CO2(g)2(g) HH = -28 = -2822..98 98 kJkJCC((ss)) + O + O2(g)2(g) CCOO2 (2 (gg)) HH = -393.51 = -393.51 kJkJ

HH22(s)(s) + + ½O½O2(g)2(g) HH22OO(g)(g) HH = -241.82 = -241.82 kJkJ

Calculate Calculate HH for this reaction from the for this reaction from the following information..following information..

PHASE CHANGES & HEATPHASE CHANGES & HEAT Energy is required to Energy is required to change the change the

phasephase of a substanceof a substance– The amount of heat necessary to The amount of heat necessary to

melt 1 mole of substancemelt 1 mole of substance o Heat of fusion (Heat of fusion (HHfusfus))o It takes It takes 6.00 kJ6.00 kJ of energy to melt 18 of energy to melt 18

grams of ice into liquid water.grams of ice into liquid water.– The amount of heat necessary to boil The amount of heat necessary to boil

1 mole of substance 1 mole of substance o Heat of vaporization (Heat of vaporization (HHvapvap))o It takes It takes 40.6 kJ40.6 kJ of energy to boil of energy to boil

away 18 grams of water.away 18 grams of water.

MOLAR HEAT DATA FOR SOME COMMON MOLAR HEAT DATA FOR SOME COMMON SUBSTANCESSUBSTANCES

SUBSTANCESUBSTANCE HHfusfus HHvapvap

Mercury, HgMercury, Hg 2.29kJ/mol2.29kJ/mol 59.1kJ/mol59.1kJ/molEthanol, CEthanol, C22HH55OHOH 5.02kJ/mol5.02kJ/mol 38.6kJ/mol38.6kJ/mol

Water, HWater, H22OO 6.00kJ/mol6.00kJ/mol 40.6kJ/mol40.6kJ/mol

Ammonia, NHAmmonia, NH33 5.65kJ/mol5.65kJ/mol 23.4kJ/mol23.4kJ/molHelium, HeHelium, He 0.02kJ/mol0.02kJ/mol 0.08kJ/mol0.08kJ/molAcetoneAcetone 5.72kJ/mol5.72kJ/mol 29.1kJ/mol29.1kJ/molMethanolMethanol, , CHCH33OHOH 3.16kJ/mol3.16kJ/mol 35.3kJ/mol35.3kJ/mol

There are 5 distinct sections we can There are 5 distinct sections we can divide the curve intodivide the curve into

–Ice (solid only)Ice (solid only)–Water & ice (solid & liquid)Water & ice (solid & liquid)–Water only (liquid only)Water only (liquid only)–Water & steam (liquid & gas)Water & steam (liquid & gas)–Steam only (gas only)Steam only (gas only)

We can calculate the We can calculate the amount of amount of energy energy involved in each stageinvolved in each stage

There are two types of calculationsThere are two types of calculations–Temperature changes use Temperature changes use H=mCH=mCTT

–Phase changes use Phase changes use (#mols)(#mols)HHfusfus or or (#mols)(#mols)HHvapvap

If we journey through all of the 5 If we journey through all of the 5 stages of the heating we have 2 stages of the heating we have 2 phase changes and 3 increases in phase changes and 3 increases in temperaturestemperatures

– Each stage has its own amnt of Each stage has its own amnt of energy to absorb or release to energy to absorb or release to make the change necessarymake the change necessary

– The total energy of the entire The total energy of the entire process can be calculated by process can be calculated by combining the energies of each combining the energies of each stagestage

HHtotaltotal = =mCmCsolidsolidT+n(T+n(HHfusfus))

++mCmCliquidliquidT+T+HHvapvap++HHgasgas

HHtotaltotal = =HHsolidsolid HHmeltingmelting HHliquidliquid HHvaporizingvaporizing HHgasgas

HHtotaltotal = =HHsolidsolid HHmeltingmelting HHliquidliquid HHvaporizingvaporizing HHgasgas

mCmCsolidsolid TT+n(+n(HHfusfus))+mC+mCliquidliquidTT+n(+n(HHvapvap))+mC+mCgasgasTT

Let’s say we have 180.0g of ice at –Let’s say we have 180.0g of ice at –10°C, & we begin heating it on a 10°C, & we begin heating it on a hot plate with sustained continuous hot plate with sustained continuous heat.heat.

Heat energy absorbs into the ice Heat energy absorbs into the ice increas-ing theincreas-ing the vibrational or kinetic vibrational or kinetic energy energy of the ice moleculesof the ice molecules– The temp will increase & will The temp will increase & will

continue to increase until just continue to increase until just before the ice has enough energy to before the ice has enough energy to change change from solid to liquidfrom solid to liquid (to (to the melting point)the melting point)

SOLID ICESOLID ICE

HHiceice =(180g)(2.09J/g°C)(0°C-(-=(180g)(2.09J/g°C)(0°C-(-10°C))10°C))HHiceice= 3762J= 3762J

HHiceice=mC=mCiceice TT

We can calc the energy absorbed by We can calc the energy absorbed by the the iceice to this pointto this point– Use CUse Ciceice=2.09J/g°C=2.09J/g°C

Any additional heat absorbed by the Any additional heat absorbed by the ice goes into ice goes into partially breaking the partially breaking the connectionsconnections between the ice between the ice molecules.molecules.

There is no change in the KE of the There is no change in the KE of the molecules (molecules (graph flattens outgraph flattens out))– No change in tempNo change in temp– All of the energy goes into breaking All of the energy goes into breaking

the connectionsthe connections

WATER & ICE (MELTING)WATER & ICE (MELTING)

As long there is solid ice present, the As long there is solid ice present, the temp cannot increase.temp cannot increase.– The solid & liquid are in The solid & liquid are in equilibriumequilibrium

if they are both present if they are both present

The energy required to change from The energy required to change from the solid to a liquid is called the the solid to a liquid is called the heat heat of fusion & depends on the molsof fusion & depends on the mols of of the substance (the substance (HHfusfus of of HH22O=6000J/mol or 6kJ/molO=6000J/mol or 6kJ/mol)) Using the formula: Using the formula: HHmeltingmelting == (mol) (mol) HHfusfus

180g H180g H22OO1 mol H1 mol H22OO

6000J6000J

18g H18g H22OO

1 mol H1 mol H22OO = 60,000J= 60,000J

Now all of the particles are Now all of the particles are free to free to flowflow, , – The heat energy gained now goes The heat energy gained now goes

into the into the vibrational energy vibrational energy of the of the molecules.molecules.

– The temp of the water increasesThe temp of the water increases The rate of temp increase now The rate of temp increase now

depends on the depends on the heat capacity of heat capacity of liquid waterliquid water– CCwaterwater=4.18 J/g°C=4.18 J/g°C

ALL WATERALL WATER

HHwaterwater=(180g)(4.18J/g°C)(100°C-0°C)=(180g)(4.18J/g°C)(100°C-0°C)

HHwaterwater= 75,240 J= 75,240 J

The temp continues to increase until The temp continues to increase until it just reaches the it just reaches the boiling boiling point point (for water = (for water = 100˚C)100˚C)– again, again, HHwaterwater=mC=mCwaterwaterTT

Any additional heat absorbed by the Any additional heat absorbed by the water goes into completely breaking water goes into completely breaking the connections between the water the connections between the water molecules.molecules.

Again the heat does not increase the Again the heat does not increase the KE of the molecules so theKE of the molecules so the temp does temp does not change,not change, – the energy is used to vaporize the the energy is used to vaporize the

waterwater

STEAM & WATER (STEAM & WATER (VAPORIZINGVAPORIZING))

If there are still connections to break or If there are still connections to break or there is liquid present, the temp there is liquid present, the temp cannot increase.cannot increase.

The energy required to change from The energy required to change from the liquid to the vapor phase is called the liquid to the vapor phase is called the the heatheat of vaporization; using of vaporization; using HHboilingboiling=(mol)=(mol)HHvapvap

– HHvapvap of H of H22O=40,600J/molO=40,600J/mol

180g H2O18g H2O

1 mol H2O

1mol H2O

40,600J= 406,000 J

STEAM ONLY (STEAM ONLY (VAPOR PHASEVAPOR PHASE)) Again the heat energy goes into the Again the heat energy goes into the

vibrational energy of the molecule. vibrational energy of the molecule. – Rate of temp increase depends on Rate of temp increase depends on

C CHH22O vaporO vapor=1.84 J/g°C=1.84 J/g°C The temp can increase indefinitely, The temp can increase indefinitely,

or until the substance decomposes or until the substance decomposes ((plasmaplasma))– We’ll stop at 125°C.We’ll stop at 125°C.

HHsteamsteam=(180g)(1.84J/g°C)(125°-100°)=(180g)(1.84J/g°C)(125°-100°)

HHsteamsteam= 8280 J= 8280 J

HHsteamsteam=(m)(C=(m)(Csteamsteam)()(T)T)

To figure out how much energy we To figure out how much energy we need would need all together to heat need would need all together to heat up the water this much, we just need up the water this much, we just need to add up the energy of each step.to add up the energy of each step.

Notice, the majority of the energy is Notice, the majority of the energy is needed for the vaporization step. needed for the vaporization step. – The connections between molecules The connections between molecules

of Hof H22O must be broken completely to O must be broken completely to vaporizevaporize

HHtotaltotal=(3760 J+60,000J+75,240 J+ =(3760 J+60,000J+75,240 J+

406,000 J+8280 J) 406,000 J+8280 J) HHtotal total = 553,280 J= 553,280 J

mCmCliquid waxliquid waxTTHHtotaltotal== (50g)(2.31J/g˚C)(62˚C-85˚C) + (50g)(2.31J/g˚C)(62˚C-85˚C) +

(50g/352.7g/mol)(-70,500J/mol)+ (50g)(50g/352.7g/mol)(-70,500J/mol)+ (50g)(2.18J/g(2.18J/g˚̊C)(25C)(25˚̊C-62C-62˚̊C)C)

How much energy must be lost for How much energy must be lost for 50.0 g of liquid wax at 85.0˚C to cool 50.0 g of liquid wax at 85.0˚C to cool to room temperature at 25.0˚C? to room temperature at 25.0˚C? (C(Csolid solid

waxwax= 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, C= 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, Cliquid liquid

waxwax=2.31 J/g˚C; MM = 352.7 g/mol, =2.31 J/g˚C; MM = 352.7 g/mol, DHDHfusionfusion=70,500 J/mol)=70,500 J/mol)

Example:Example:

HHliquid waxliquid wax

HHsolidificationsolidification

HHsolid waxsolid wax

HHtotal= total= HHliquid wax liquid wax + + HHsolidification solidification + + HHsolid solid

waxwax

n(n(HHfusionfusion))

mCmCsolid waxsolid waxTT

HHtotal= total= mCmCliquid waxliquid waxT+n(T+n(HHfusionfusion)+ )+ mCmCsolid waxsolid waxTT

HHtotaltotal== (-2656.5 J) + (-9994.3 J)+ (-(-2656.5 J) + (-9994.3 J)+ (-4033 J)4033 J)HHtotaltotal== -16,683.8 J-16,683.8 J

We have a collection of steam at We have a collection of steam at 173°C that occupies a volume of 173°C that occupies a volume of

30.65 L and a pressure of 2.53 atm. 30.65 L and a pressure of 2.53 atm. How much energy would it need to How much energy would it need to lose to end up as a block of ice at lose to end up as a block of ice at

0.00°C?0.00°C?

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