Energy & Chemistry

Energy & Chemistry2H2H2(g)2(g) + O + O2(g)2(g) → 2H → 2H22OO(g)(g) + heat and + heat and

lightlightThis can be set up to provide This can be set up to provide

ELECTRIC ENERGY in a ELECTRIC ENERGY in a fuel cellfuel cell..Oxidation:Oxidation:

2 H2 H22 → 4 H → 4 H++ + 4 e + 4 e--

Reduction: Reduction: 4 e4 e-- + O + O22 + 2 H + 2 H22O → 4 OHO → 4 OH--

H2/O2 Fuel CellEnergy, page 288

Energy & ChemistryENERGYENERGY is the capacity to do work or transfer is the capacity to do work or transfer

heat.heat.HEATHEAT is the form of energy that flows between is the form of energy that flows between

2 objects because of their difference in 2 objects because of their difference in temperature.temperature.

Other forms of energy —Other forms of energy —• lightlight• electricalelectrical• kinetic and potentialkinetic and potential• Positive and negative

particles (ions) attract one another.

• Two atoms can bond • As the particles attract

they have a lower potential energy

NaCl — composed NaCl — composed of Naof Na++ and Cl and Cl-- ions.ions.

Potential & Kinetic Energy

Kinetic energy Kinetic energy — energy of — energy of motion.motion.

translate

rotate

vibrate

© 2006 Brooks/Cole - Thomson

Internal Energy (E)Internal Energy (E)

• PE + KE = Internal energy (E or U)• Internal Energy of a chemical system

depends on•number of particles•type of particles•temperature• The higher the T the higher the

internal energy• So, use changes in T (∆T) to monitor

changes in E (∆E).

Thermodynamics• Thermodynamics is the science of heat (energy)

transfer.Heat transfers until thermal equilibrium is

established.∆T measures energy transferred.

• SYSTEM– The object under study

• SURROUNDINGS– Everything outside the

system

Directionality of Heat TransferDirectionality of Heat Transfer• Heat always transfer from hotter object to cooler

one.• EXOthermic: heat transfers from SYSTEM to

SURROUNDINGS.

T(system) goes downT(system) goes downT(surr) goes upT(surr) goes up

Directionality of Heat TransferDirectionality of Heat Transfer• Heat always transfers from hotter object to cooler

one.• ENDOthermic: heat transfers from

SURROUNDINGS to the SYSTEM.

T(system) goes upT(system) goes upT (surr) goes downT (surr) goes down

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.Energy Change in Chemical

ProcessesReactants

Products

Kinetic Energy

PE

Potential Energy of Potential Energy of system dropped. Kinetic system dropped. Kinetic energy increased. energy increased. Therefore, you often Therefore, you often feel a Temperature feel a Temperature increase.increase.

HEAT CAPACITYThe heat required to raise an object’s T

by 1 ˚C.

Which has the larger heat capacity?

Specific Heat Capacity

How much energy is transferred due to Temperature difference?The heat (q) “lost” or “gained” is related to a) sample massb) change in T andc) specific heat capacity

Specific heat capacity = heat lost or gained by substance (J)(mass, g) (T change, K)


Substance Phase cp J g-1 K-1 Cp J mol-1 K-1

Air (typical room conditionsA) gas 1.012 29.19

Aluminium solid 0.897 24.2

Argon gas 0.5203 20.7862

Copper solid 0.385 24.47

Diamond solid 0.5091 6.115

Ethanol liquid 2.44 112

Gold solid 0.1291 25.42

Graphite solid 0.710 8.53

Helium gas 5.1932 20.7862

Hydrogen gas 14.30 28.82

Iron solid 0.450 25.1

Lithium solid 3.58 24.8

Mercury liquid 0.1395 27.98

Nitrogen gas 1.040 29.12

Neon gas 1.0301 20.7862

Oxygen gas 0.918 29.38

Uranium solid 0.116 27.7

Water

gas (100 °C) 2.080 37.47

liquid (25 °C) 4.1813 75.327

solid (0 °C) 2.114 38.09

All measurements are at 25 °C unless noted. Notable minimums and maximums are shown in maroon text.

Aluminum

http://en.wikipedia.org/wiki/Phase_(matter)

http://en.wikipedia.org/wiki/Aluminium

http://en.wikipedia.org/wiki/Argon

Specific Heat Capacity

If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?

Heat/Energy TransferHeat/Energy TransferNo Change in StateNo Change in State

q transferred = (sp. ht.)(mass)(∆T)

Heat Transfer• Use heat transfer as a way to find specific heat capacity, Cp

• 55.0 g Fe at 99.8 ˚C• Drop into 225 g water at 21.0 ˚C• Water and metal come to 23.1 ˚C• What is the specific heat capacity of the metal?

Heating/Cooling Curve for Water

Note that T is Note that T is constant as ice constant as ice melts or water melts or water

boilsboils

Chemical ReactivityBut energy transfer also allows us to predict

reactivity.In general, reactions that transfer energy

to their surroundings are product-favored.

So, let us consider heat transfer in chemical So, let us consider heat transfer in chemical processes.processes.

FIRST LAW OF FIRST LAW OF THERMODYNAMICSTHERMODYNAMICS

∆∆E = q + wE = q + w

heat energy transferredheat energy transferred

energyenergychangechange

work donework doneby the by the systemsystem

Energy is conserved!Energy is conserved!


The First Law of Thermodynamics

• Exothermic reactions generate specific amounts of heat.

• This is because the potential energies of the products are lower than the potential energies of the reactants.


The First Law of Thermodynamics• There are two basic ideas of importance for

thermodynamic systems.1. Chemical systems tend toward a state of minimum

potential energy.

2. Chemical systems tend toward a state of maximum disorder.

• The first law is also known as the Law of Conservation of Energy.

– Energy is neither created nor destroyed in chemical reactions and physical changes.

heat transfer outheat transfer out(exothermic), -q(exothermic), -q

heat transfer inheat transfer in(endothermic), +q(endothermic), +q

SYSTEMSYSTEM

∆E = q + w

w transfer inw transfer in(+w)(+w)

w transfer outw transfer out(-w)(-w)


Some Thermodynamic Terms• Notice that the energy change in moving from

the top to the bottom is independent of pathway but the work required may not be!

• Some examples of state functions are:– T (temperature), P (pressure), V (volume),

E (change in energy), H (change in enthalpy – the transfer of heat), and S (entropy)

• Examples of non-state functions are:– n (moles), q (heat), w (work)∆H along one path = ∆H along

another path• This equation is valid because ∆H is a STATE FUNCTION• These depend only on the state of the system and not how

it got there.• V, T, P, energy — and your bank account!• Unlike V, T, and P, one cannot measure absolute H. Can

only measure ∆H.


Some Thermodynamic Terms• The properties of a system that depend only on the state of the

system are called state functions.– State functions are always written using capital letters.

• The value of a state function is independent of pathway.• An analog to a state function is the energy required to climb a

mountain taking two different paths.– E1 = energy at the bottom of the mountain– E1 = mgh1

– E2 = energy at the top of the mountain– E2 = mgh2

E = E2-E1 = mgh2 – mgh1 = mg(h)


Standard States and Standard Enthalpy Changes• Thermochemical standard state conditions

– The thermochemical standard T = 298.15 K.– The thermochemical standard P = 1.0000 atm.

• Be careful not to confuse these values with STP.• Thermochemical standard states of matter

– For pure substances in their liquid or solid phase the standard state is the pure liquid or solid.

– For gases the standard state is the gas at 1.00 atm of pressure.• For gaseous mixtures the partial pressure must be 1.00

atm.– For aqueous solutions the standard state is 1.00 M

concentration.∆Hf

o = standard molar enthalpy of formation• the enthalpy change when 1 mol of compound is formed from elements under standard conditions.

See Table 6.2 and Appendix L


ENTHALPYENTHALPYMost chemical reactions occur at constant P, so

and so ∆E = ∆H + w (and w is usually small)∆H = heat transferred at constant P ≈ ∆E∆H = change in heat content of the system∆H = Hfinal - Hinitial

Heat transferred at constant P = qp

qp = ∆H where H = enthalpy


If Hfinal < Hinitial then ∆H is negativeProcess is EXOTHERMIC

If Hfinal > Hinitial then ∆H is positiveProcess is ENDOTHERMIC

ENTHALPY

∆H = Hfinal - Hinitial


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

USING ENTHALPY

Exothermic reaction — heat is a “product” and ∆H = – 241.8 kJ


Making Making liquidliquid H H22O from HO from H22 + O + O22 involves involves twotwo exoexothermic steps. thermic steps.

USING ENTHALPY

H2 + O2 gas Liquid H2OH2O vapor

Making HMaking H22O from HO from H22 involves two steps. involves two steps.HH2(g)2(g) + 1/2 O + 1/2 O2(g)2(g) → H → H22OO(g)(g) + 242 kJ + 242 kJ HH22OO(g)(g) → H → H22OO(l)(l) + 44 kJ + 44 kJ HH2(g)2(g) + 1/2 O + 1/2 O2(g)2(g) → H → H22OO(l)(l) + 286 kJ + 286 kJExample of Example of HESS’S LAWHESS’S LAW——If a rxn. is the sum of 2 or more others, the net ∆H is 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.the sum of the ∆H’s of the other rxns.


Enthalpy Values

H2(g) + 1/2 O2(g) → H2O(g) ∆H˚ = -242 kJ 2H2(g) + O2(g) → 2H2O(g) ∆H˚ = -484 kJ

H2O(g) → H2(g) + 1/2 O2(g) ∆H˚ = +242 kJH2(g) + 1/2 O2(g) → H2O(l) ∆H˚ = -286 kJ

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


Hess’s Law & Energy Level DiagramsForming H2O can occur in a single step or in a two steps. ∆Htotal is the same no matter which path is followed.

Active Figure 6.18


Thermochemical Equations• Thermochemical equations are a balanced chemical reaction

plus the H value for the reaction.– For example, this is a thermochemical equation.

• The stoichiometric coefficients in thermochemical equations must be interpreted as numbers of moles.

• 1 mol of C5H12 reacts with 8 mol of O2 to produce 5 mol of CO2, 6 mol of H2O, and releasing 3523 kJ is referred to as one mole of reactions.

mole 1 moles 6 moles 5 moles 8 mole 1

kJ 3523 OH 6 CO 5O 8 HC )(22(g)2(g))12(5


Hess’s Law• Hess’s Law of Heat Summation, Hrxn = H1 +H2 +H3 + ..., states

that the enthalpy change for a reaction is the same whether it occurs by one step or by any (hypothetical) series of steps.– Hess’s Law is true because H is a state function.

• If we know the following Ho’s

kJ 1648H OFe 2 O 3 Fe 4 3

kJ 454H FeO 2O Fe 2 2

kJ 560H OFe 2O FeO 4 1

o3(s)22(g)(s)

o(g)2(g)(s)

o(s)322(g)(s)


Hess’s Law• For example, we can calculate the Ho for reaction [1] by properly adding (or

subtracting) the Ho’s for reactions [2] and [3].• Notice that reaction [1] has FeO and O2 as reactants and Fe2O3 as a product.

– Arrange reactions [2] and [3] so that they also have FeO and O2 as reactants and Fe2O3 as a product.• Each reaction can be doubled, tripled, or multiplied by half, etc.• The Ho values are also doubled, tripled, etc.• If a reaction is reversed the sign of the Ho is changed.


Hess’s Law• Given the following equations and Hovalues

calculate Ho for the reaction below.

H kJ

2 N O N O 164.1

[2] N + O NO 180.5

[3] N + 2 O NO 66.4

o

2 g 2 g 2 g

2 g 2 g g

2 g 2 g 2 g

[ ]1 2

2

2


Hess’s Law• Use a little algebra and Hess’s Law to get the appropriate Hovalues


Thermochemical Equations• This is an equivalent method of writing thermochemical

equations.

H < 0 designates an exothermic reaction. H > 0 designates an endothermic reaction

kJ 3523 - H OH 6 CO 5O 8 HC orxn)(22(g)2(g))12(5


Standard Molar Enthalpies of Formation, Hf

o

• The standard molar enthalpy of formation is defined as the enthalpy for the reaction in which one mole of a substance is formed from its constituent elements.– The symbol for standard molar enthalpy of formation is Hf

o.• The standard molar enthalpy of formation for MgCl2 is:



o

• Standard molar enthalpies of formation have been determined for many substances and are tabulated in Table 15-1 and Appendix K in the text.

• Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero.

• Example 15-4: The standard molar enthalpy of formation for phosphoric acid is -1281 kJ/mol. Write the equation for the reaction for whichHo

rxn = -1281 kJ.P in standard state is P4

Phosphoric acid in standard state is H3PO4(s)


Hess’s Law• Hess’s Law in a more useful form.

– For any chemical reaction at standard conditions, the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products (each multiplied by its coefficient in the balanced chemical equation) minus the corresponding sum for the reactants.

tscoefficien tricstoichiomen

Hn Hn H 0reactants f

0products f

0rxn

nn


Hess’s Law


∆Hfo, standard molar enthalpy of

formationH2(g) + ½½ O2(g) → H2O(g) ∆Hf˚̊ (H2O, g)= -241.8 kJ/mol

CC(s)(s) + ½ O + ½ O2(g)2(g) → CO → CO(g)(g) ∆H∆Hff˚ of CO = - 111 kJ/mol˚ of CO = - 111 kJ/molBy definition, ∆Hf

o = 0 for elements in their

standard states.Use ∆H˚’s to calculate enthalpy change for Use ∆H˚’s to calculate enthalpy change for HH22O(g) + C(graphite) → HO(g) + C(graphite) → H22(g) + CO(g)(g) + CO(g)


Using Standard Enthalpy Values

Calculate the heat of combustion of methanol, Calculate the heat of combustion of methanol, i.e., ∆Hi.e., ∆Hoo

rxnrxn for for

CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) → CO(g) → CO22(g) + 2 H(g) + 2 H22O(g)O(g)

∆∆HHoorxnrxn = = ∆H ∆Hff

oo (prod) - (prod) - ∆H ∆Hff

oo (react)(react)



o• Calculate the enthalpy change for the reaction of one

mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.



o• Calculate the enthalpy change for the reaction in

which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.


Hess’s Law• Calculate the H o

298 for the following reaction from data in Appendix K.

)(22(g)2(g)8(g)3 OH 4 CO 3 O 5 HC


Hess’s Law• Application of Hess’s Law and more algebra allows us to

calculate the Hfofor a substance participating in a reaction

for which we know Hrxno , if we also know Hf

ofor all other substances in the reaction.

• Given the following information, calculate Hfo for H2S(g).

2 H S + 3 O 2 SO + 2 H O H = -1124 kJ

H ? 0 - 296.8 - 285.8(kJ / mol)

2 g 2 g 2 g 2 298o

fo

l


Thermochemical Equations• Write the thermochemical equation for the reaction

for CuSO4(aq) + 2NaOH(aq) Cu(OH)2(s) + Na2SO4(aq)

50.0mL of 0.400 M CuSO4 at 23.35 oC Tfinal 25.23oC50.0mL of 0.600 M NaOH at 23.35 oCDensity final solution = 1.02 g/mL CH2O = 4.184 J/goC

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