Some Thermodynamic Terms

© 2006 Brooks/Cole - Thomson

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

ENTHALPYENTHALPY

Most 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 negative

Process is EXOTHERMIC

If Hfinal > Hinitial then ∆H is positive

Process is ENDOTHERMIC

ENTHALPY

∆H = Hfinal - Hinitial

Consider the formation of water

H2(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 kJ

Example 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 kJ

H2(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 Diagrams

Forming 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.

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 Hfo.

• 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/mol

By definition, ∆Hfo = 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)


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 25.23oC

50.0mL of 0.600 M NaOH at 23.35 oCDensity final solution = 1.02 g/mL CH2O = 4.184 J/goC



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

Some Thermodynamic Terms

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