energy and chemical energy & chemistry reactions ...mhchem.org/221/pdfppt/ch06bweb.pdfmar energy...

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Energy and Chemical Reactions: Thermochemistry

orThermodynamics

Chapter Five

Chemistry 221 Professor Michael Russell

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Energy & Chemistry

Burning peanuts supplies sufficient energy to boil a cup of water

Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)

These reactions are PRODUCT FAVORED They proceed almost completely from

reactants to products, perhaps with some outside assistance.

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Energy and ChemistryENERGY is the capacity to do

work or transfer heat. HEAT is the form of energy that

flows between 2 samples because of their difference in temperature.

Other forms of energy - • light – next chapter• electrical - CH 223• kinetic and potential

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Potential energy - energy a motionless body has by virtue of its position

In chemistry, positive and negative particles (ions) attract one another.

As the particles attract to make a bond, they have a lower potential energy

Potential Energy

NaCl - composed of Na+ and Cl- ions.

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Kinetic Energy

Three forms of kinetic energy:

Translational - from physics, KE = 1/2 mv2 = 0.5(mass)(velocity)2

Vibrational Rotational

translate

rotate

vibrate

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Internal Energy (E)PE + KE = Internal

Energy (E or U) The higher the T

the higher the internal energy

So, use changes in T (∆T) to monitor changes in E (∆E).

Page III-5-1 / Chapter Five Lecture Notes


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Thermodynamics

Thermodynamics is the science of heat (energy) transfer.

Heat energy is associated with molecular motions (Kinetic Molecular Theory)

Heat transfers until thermal equilibrium is established.MAR

Directionality of Heat TransferHeat always transfers from the hot

object to the cooler object. EXOthermic: heat transfers from SYSTEM to SURROUNDINGS.

T(system) goes down T(surr) goes up - this is what we usually measure

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Directionality of Heat TransferHeat always transfers from the hot

object to the cooler object. ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.

T(system) goes up T(surr) goes down - this is what we usually measure

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Energy & ChemistryAll of thermodynamics depends on

the law of CONSERVATION OF ENERGY.

• The total energy is unchanged in a chemical reaction.

• If PE of products is less than reactants, the difference must be released as KE.

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UNITS OF ENERGY1 calorie = heat required to

raise temp. of 1.00 g of H2O by 1.0 oC.

1000 cal = 1 kilocalorie = 1 kcal

1 kcal = 1 Calorie (a food "calorie")

But we use the unit called the JOULE

1 cal = 4.184 joules James Joule 1818-1889

Memorize 4.184! MAR

Specific Heat Capacity

Thermochemistry is the science of heat (energy) flow. A difference in temperature leads to energy transfer.

The heat, q, "lost" or "gained" is related to a) sample massb) change in T andc) specific heat capacity



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Specific Heat Capacity:

Substance Spec. Heat (J/g•K) H2O(l) 4.184

Ethylene glycol(l) 2.39 Al(s) 0.902 glass(s) 0.84

Aluminum

Water

The heat required to raise an object's temperature by 1 ˚C.

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q = mC TImportant!

q = heat "lost" or "gained" m = sample massC = specific heat capacity

T = change in temperatureT = final T - initial T

q is positive when heat flows in (cold surroundings) - endothermic

q is negative when heat flows out (hot surroundings) - exothermic

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Specific Heat CapacityIf 25.0 g of Al cool from

310. oC to 37 oC, how many joules of heat energy are lost by the Al?

q = mC∆T CAl = 0.902 J g-1 K-1 MAR

Specific Heat CapacityIf 25.0 g of Al cool from 310. oC to 37 oC, how

many joules of heat energy are lost by the Al? CAl = 0.902 J g-1 K-1

heat gain/lost = q = (mass)(sp. ht.)(∆T)

Notice that the negative sign on q signals heat "lost by" or transferred OUT of Al.

where ∆T = Tfinal - Tinitial

q = (25.0 g)(0.902 J/g•˚C)(37 - 310.)˚C q = - 6160 J

Notice also that ∆T can be either K or ˚C - the difference in temperatures is the same!

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Specific Heat CapacityA piece of iron (88.5 g) at 77.8 oC is placed in 244 g

of water at 18.8 oC. What is the final temperature of the mixture?

By the law of conservation of energy:

qhot + qcold = 0, or

mFeCFe∆TFe + mwaterCwater∆Twater = 0

CFe = 0.449 J g-1 K-1

Cwater = 4.184 J g-1 K-1 Memorize

Final Temperature (warm) same for Fe & H2OMAR

Specific Heat Capacity

qhot + qcold = 0

mFeCFe∆TFe + mwaterCwater∆Twater = 0

88.5 * 0.449 * (Tf - 77.8) + 244 * 4.184 * (Tf - 18.8) = 0

39.7Tf - 3090 + 1020Tf - 19200 = 0

1060Tf = 22300

Tf = 21.0 oC

Efficient method to determine approximate final temperature of mixture

Fe (88.5 g, 0.449 J/g•K, 77.8 oC) and water (244 g, 4.184 J/g•K, 18.8 oC); final temperature?



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Heat TransferNo Change in State

q transferred = (mass)(sp. ht.)(∆T)

Heat Transfer with Change of State

As matter absorbs heat, its temperature rises until it undergoes a phase change (change of state)

Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water

q = (heat of "something")("mass")

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Heat of fusion of ice = 333 J/gSpecific heat of water = 4.184 J/g•K

Heat of vaporization = 2260 J/g

What quantity of heat is required to melt 500. g of ice at 0.0 oC and heat the water to steam at 100.0 oC?

+333 J/g

Heat & Changes of State

solid (ice) liquid (0 – 100 °C) gas (steam)

+2260 J/g

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What quantity of heat is required to melt 500. g of ice at 0.0 oC and heat the water to steam at 100.0 oC?

1. To melt ice q = (500. g)(333 J/g) = 1.67 x 105 J2. To raise water from 0.0 oC to 100.0 oC q = (500. g)(4.184 J/g•K)(100. - 0.)K = 2.09 x 105 J3. To evaporate water at 100. oC q = (500. g)(2260 J/g) = 1.13 x 106 J4. Total heat energy = 1.51 x 106 J = 1510 kJ

Heat & Changes of State

Heat Transfer covered more in Chapter 12 of CH 222

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CHEMICAL REACTIVITYWhat drives chemical

reactions? How do they occur, and how fast?

The first question is answered by

THERMODYNAMICS, and the second question is answered by KINETICS.

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CHEMICAL REACTIVITYTHERMODYNAMICS dictates if the reaction will

occur or not.

Sand will not decompose into silicon and oxygen

Paper will combine with oxygen to burn



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CHEMICAL REACTIVITYKINETICS dictates how fast the reaction will

occur.Example: diamond into graphite is

thermodynamically favored, but the kinetics of the reaction is too slow to be of useful

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CHEMICAL REACTIVITYWe have already seen a number of

"driving forces" for reactions that are PRODUCT-FAVORED.

• formation of a precipitate• gas formation• H2O formation (acid-base reaction)• electron transfer in a battery

In general, reactions that transfer energy to their

surroundings are product-favored.

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FIRST LAW OF THERMODYNAMICS

∆E = q + w

heat energy transferred

energy change

work done by the system

Energy is conserved! Chemists focus on q more than w

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ENTHALPYMost chemical reactions occur at constant P, so

∆E = q + w and ∆E = ∆H + w (P const)

w is usually small, so ∆H ≈ ∆E

Heat transferred at constant P = qp

qp = ∆H where H = enthalpy

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If Hfinal < Hinitial then ∆H is negativeProcess is EXOTHERMIC

∆H is negative

If Hfinal > Hinitial then ∆H is positiveProcess is ENDOTHERMIC

∆H is positive

ENTHALPY∆H = Hfinal - Hinitial

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Consider the formation of waterH2(g) + 1/2 O2(g) --> H2O(g) + 241.8 kJ

Exothermic reaction - heat is a "product" and ∆H = – 241.8 kJ

USING ENTHALPY



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Making liquid H2O from H2 + O2 involves two exothermic steps.

USING ENTHALPY

H2 + O2 gas

Liquid H2OH2O vapor

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Making H2O from H2 involves two steps.H2(g) + 1/2 O2(g) ---> H2O(g) + 242 kJH2O(g) ---> H2O(liq) + 44 kJ ----------------------------------------------

H2(g) + 1/2 O2(g) --> H2O(liq) + 286 kJExample of HESS'S LAW-If a rxn. is the sum of 2 or more others, the net ∆H is the sum of the ∆H's of the other rxns.

USING ENTHALPY

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Hess's Law & Energy Level Diagrams

Forming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

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Hess's Law & Energy Level Diagrams

Forming CO2 can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

∆H is a STATE FUNCTION - depends only on the state of the system and not how it got there.

Other state functions: V, T, P, and energy

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Standard Enthalpy ValuesMost ∆H values are labeled ∆Ho Measured under standard conditions

P = 1 bar (approx. 1 atm) Concentration = 1 mol/LT = usually 25 oC

with all species in standard statesi.e., C = graphite and O2 = gas

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Enthalpy Values

H2(g) + 1/2 O2(g) --> H2O(g)∆H˚ = -242 kJ

2 H2(g) + O2(g) --> 2 H2O(g)

∆H˚ = -484 kJ H2O(g) ---> H2(g) + 1/2 O2(g)

∆H˚ = +242 kJ H2(g) + 1/2 O2(g) --> H2O(liquid)

∆H˚ = -286 kJ

Depend on how the reaction is written and on phases of reactants and products



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Standard Enthalpy ValuesNIST (Nat'l Institute for Standards and

Technology) gives values of ∆Hf

o = standard molar enthalpy of formation

- the enthalpy change when 1 mol of compound is formed from elements under standard conditions.

These values are available in your text

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∆Hfo, standard molar

enthalpy of formation

H2(g) + 1/2 O2(g) --> H2O(g)

∆Hfo (H2O, g)= -241.8 kJ/mol

By definition,

∆Hfo = 0 for elements in their standard states.

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Using Standard Enthalpy Values

Use ∆H˚ values to calculate the enthalpy change for:

H2O(g) + C(graphite) --> H2(g) + CO(g)

(product is called "water gas")

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Find enthalpy change for this reaction:

H2O(g) + C(graphite) --> H2(g) + CO(g)

From reference books we find:

H2(g) + 1/2 O2(g) --> H2O(g)

∆Hf˚ of H2O vapor = - 242 kJ/mol

C(s) + 1/2 O2(g) --> CO(g)

∆Hf˚ of CO = - 111 kJ/mol

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H2O(g) --> H2(g) + 1/2 O2(g) ∆Ho = +242 kJ

C(s) + 1/2 O2(g) --> CO(g) ∆Ho = -111 kJ

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To convert 1 mol of water to 1 mol each of H2 and CO requires 131 kJ of energy. The "water gas" reaction is ENDOthermic.

H2O(g) + C(graphite) --> H2(g) + CO(g) ∆Honet = +131 kJ

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Using Standard Enthalpy ValuesFind ∆Ho

rxn for CaCO3(s) --> CaO(s) + CO2(g)

using Hess's Law and:

Ca(s) + 3/2 O2(g) + C(s) --> CaCO3(s)

∆Hof = -1206.2 kJ as well as

Ca(s) + 3/2 O2(g) + C(s) --> CaO(s) + CO2(g)

∆Horxn = -1028.6 kJ

CaCO3

CaO



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Using Standard Enthalpy ValuesFind ∆Ho

rxn for CaCO3(s) --> CaO(s) + CO2(g)

CaCO3(s) --> Ca(s) + 3/2 O2(g) + C(s)

∆Ho = -(-1206.2 kJ) = +1206.2 kJ

Ca(s) + 3/2 O2(g) + C(s) --> CaO(s) + CO2(g)

∆Horxn = -1028.6 kJ

----------------------------------------------------------- CaCO3(s) --> CaO(s) + CO2(g), and

∆Horxn = +1206.2 kJ + (-1028.6 kJ) = +177.6 kJ

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In general, when ALL enthalpies of formation are known,

Calculate ∆H of reaction?

∆Horxn =

Σ ∆Hfo (products)

- Σ ∆Hfo (reactants)

Remember that ∆ = final – initial

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Answer: ∆Ho

rxn = {∆Hfo

(CaO) + ∆Hfo

(CO2)} - {∆Hfo

(CaCO3)} ∆Ho

rxn = {-635.1 + -393.5} - {-1206.2}

∆Horxn = +177.6 kJ

Same answer as before with Hess's Law, but simpler

Example: Find ∆Horxn for

CaCO3(s) --> CaO(s) + CO2(g) using:

∆Horxn = Σ ∆Hf

o (prod) - Σ ∆Hf

o (react)

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Using Standard Enthalpy ValuesCalculate the heat of combustion of

methanol, i.e., ∆Horxn for

CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)

∆Horxn = Σ ∆Hf

o (prod) - Σ ∆Hf

o (react)

As before, look up ∆Hfo values for reactants

and products in your text Elements in standard states have ∆Hf

o = 0

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∆Horxn = {∆Hf

o (CO2) + 2 ∆Hf

o (H2O)}

- {3/2 ∆Hfo

(O2) + ∆Hfo

(CH3OH)} = {(-393.5 kJ) + 2 (-241.8 kJ)} - {0 + (-201.5 kJ)} ∆Ho

rxn = -675.6 kJ per mol of methanol

CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g) ∆Ho

rxn = Σ ∆Hfo (prod) - Σ ∆Hf

o (react)



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Measuring Heats of ReactionCALORIMETRY

Constant Volume "Bomb" Calorimeter

• Burn combustible sample.

• Measure heat evolved in a reaction.

• Derive ∆E for reaction.

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Calorimetry

Some heat from reaction warms water qwater = (sp. ht.)(water mass)(∆T)

Some heat from reaction warms "bomb" qbomb = (heat capacity, J/K)(∆T)

Total heat evolved = qtotal = qwater + qbomb

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Calculate heat of combustion of octane: C8H18 + 25/2 O2 -->

8 CO2 + 9 H2O

• Burn 1.00 g of octane • Temp rises from 25.00 to 33.20 oC • Calorimeter contains 1200. g water • Heat capacity of bomb = 837 J/K


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Step 1 Calc. heat transferred from reaction to water.

q = (4.184 J/g•K)(1200. g)(8.20 K) = 41,200 JStep 2 Calc. heat transferred from

reaction to bomb.q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 JStep 3 Total heat evolved 41,200 J + 6860 J = 48,100 JHeat of combustion of 1.00 g of octane =

- 48.1 kJ (-5.48*103 kJ/mol)


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End of Chapter Five

See also: • Chapter Five Study Guide • Chapter Five Concept Guide



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