Enthalpy and Hess’s Enthalpy and Hess’s LawLaw

From the homework, you may have From the homework, you may have realized that realized that H can have a negative H can have a negative

number.number. It relates to the fact that energy as It relates to the fact that energy as heatheat

has either has either enteredentered or or leftleft the system. the system. If it was If it was positivepositive, it meant that the , it meant that the

heating of the sample heating of the sample requiredrequired energy. energy. Energy Energy inin means means endothermicendothermic So a So a positivepositive H means that the process H means that the process

was was endothermicendothermic.. The The oppositeopposite is true if is true if H was H was negativenegative.. A A negativenegative H means the process was H means the process was

exothermicexothermic (energy (energy leftleft the system) the system)

ThermodynamicsThermodynamics

The branch of science that is The branch of science that is concerned with the concerned with the energyenergy changes changes that accompany that accompany chemicalchemical and and physicalphysical changes changes

To standardize the enthalpies of To standardize the enthalpies of reactions, scientists have agreed reactions, scientists have agreed upon a standard upon a standard thermodynamicthermodynamic temperature of temperature of 25.00 25.00 CC ( (298.15 K298.15 K).).

Equations and EnthalpyEquations and Enthalpy As a result, we can incorporate more As a result, we can incorporate more

information, like information, like H valuesH values and and temperaturestemperatures, into chemical equations., into chemical equations.

For example:For example: Fe (Fe (s, 300 Ks, 300 K) ) Fe ( Fe (s, 1100Ks, 1100K) ) H = H = 20.120.1

kJ/molkJ/mol

HH22 ( (g, 298 Kg, 298 K) + Br) + Br22 ( (l, 298 Kl, 298 K) ) 2 HBr ( 2 HBr (g, g, 298K298K) ) H = H = -72.8-72.8 kJ/mol kJ/mol

How do they get How do they get H values?H values? To get To get H values, scientists use H values, scientists use

calorimetrycalorimetry.. It is defined as the It is defined as the measurementmeasurement of heat- of heat-

related constants (related constants (specificspecific heat, heat, latentlatent heat)heat)

It is an It is an experimentalexperimental measurement of an measurement of an enthalpyenthalpy change. change.

A A calorimetercalorimeter is utilized – a is utilized – a devicedevice used to used to measure the heat measure the heat absorbedabsorbed or or releasedreleased in in a chemical or physical change.a chemical or physical change.

Hess’s LawHess’s Law It is stated that the It is stated that the amountamount of heat of heat

released or absorbed in a chemical released or absorbed in a chemical reaction does reaction does notnot depend on the number depend on the number of of stepssteps in the reaction. in the reaction.

Example: Phosphorous pentachloride Example: Phosphorous pentachloride can be made two ways.can be made two ways.

Way #1 – Way #1 – 11 Step Step PP4 4 (s) + 10 Cl(s) + 10 Cl22 4 PCl 4 PCl55 (g) (g) H = H = -1596 kJ-1596 kJ

Hess’s LawHess’s Law

Way #2 takes Way #2 takes twotwo steps steps PP44 (s) + 6 Cl (s) + 6 Cl22 (g) (g) 4 PCl 4 PCl33 (g) (g) H H =-1224=-1224 kJ kJ PClPCl33 (g) + Cl (g) + Cl22 (g) (g) PCl PCl55 (g) (g) H = H = -93-93 kJ kJ Notice that we need to do the second step Notice that we need to do the second step

44 times to use all the PCl times to use all the PCl3 3 from step 1.from step 1. So, So, -1224 +4(-93) = -1596 kJ-1224 +4(-93) = -1596 kJ Comparing the 1 step way from the last Comparing the 1 step way from the last

slide to the 2 step way from this slide:slide to the 2 step way from this slide: -1596 kJ-1596 kJ = = -1596 kJ-1596 kJ, Hess’s Law , Hess’s Law worksworks

So, what does Hess’s Law So, what does Hess’s Law mean?mean?

H H reactionreaction = = H H productsproducts - - H H reactantsreactants

So we need to use So we need to use StandardStandard enthalpy of enthalpy of formationformation – the enthalpy change in forming – the enthalpy change in forming oneone mole of a substance from mole of a substance from elementselements in in their their standardstandard states. These are listed on states. These are listed on Pg 355 and 833.Pg 355 and 833.

What these tables mean is when a What these tables mean is when a compoundcompound is formed from the is formed from the elementselements that make it up, that make it up, H of H of formationformation will be the will be the value listed.value listed.

That means that we can find each That means that we can find each compound on these lists and use that compound on these lists and use that number in the equation above.number in the equation above.

Things to rememberThings to remember

When compounds are When compounds are multipliedmultiplied by a by a constantconstant (coefficient), the enthalpy (coefficient), the enthalpy change must be change must be multipliedmultiplied by the by the samesame constant. So equations will constant. So equations will have to be have to be balanced.balanced.

H of H of elementselements is is zerozero, so those , so those equations may not be in the book.equations may not be in the book.

Remember the Remember the diatomicsdiatomics and that S and that S by itself is by itself is SS88. . This may help you find This may help you find their values.their values.

Now we can look at an Now we can look at an exampleexample We want to look at the enthalpy change We want to look at the enthalpy change

when carbon and carbon dioxide are when carbon and carbon dioxide are reacted together to make carbon reacted together to make carbon monoxide. (monoxide. (C + COC + CO22 2 CO 2 CO))

So, we need to look at the formation of So, we need to look at the formation of COCO and and COCO22 to complete this problem.to complete this problem.

We see that We see that CO, CO, H = H = -110.5-110.5 kJ/mol and kJ/mol and

COCO22, , H = H = -393.5-393.5 kJ/mol kJ/mol(C, (C, H = H = 00))

H H reactionreaction = = H H productsproducts - - H H reactantsreactants

= = 2(-110.5)-[(0)+(-393.5)]2(-110.5)-[(0)+(-393.5)] = = 172.5 kJ/mol172.5 kJ/mol

ExamplesExamples

Calculate the enthalpy change for the Calculate the enthalpy change for the reaction:reaction:

2H2H22 (g) + 2CO (g) + 2CO22 (g) (g) 2H 2H22O (g) + 2CO (g) O (g) + 2CO (g) What enthalpy change accompanies What enthalpy change accompanies

the reaction:the reaction:

2Al(s) + 3H2Al(s) + 3H22O (l) O (l) Al Al22OO33 (s) + 3H (s) + 3H22 (g) (g) Calculate Calculate H for the decomposition of H for the decomposition of

calcium carbonate into calcium oxide calcium carbonate into calcium oxide and carbon dioxide.and carbon dioxide.

HomeworkHomework

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