Chemistry 5

Chapter-7

Thermochemistry

Part-3

23 October 2002

A thermochemical equation combines••

Reaction enthalpy

Things to pay attention to:

• ∆H is extensive

• ∆H depends on direction

• State of reactants/products

Enthalpies of Reaction

balanced chemical equationreaction enthalpy, ∆H

∆H of reaction– heat change at constant P– ignores work done by/on the system in moving from initial to final state.

The reaction enthalpy is the change in enthalpy for stoichiometric number of moles of reactants in the chemical equation:CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H = -890 kJ

Enthalpy change is directly proportional to amounts of substances in system.

∆H changes sign when a process is reversed; that is, ∆Η is a state function, and reverse intial/final states.

Enthalpy change depends on the state– gas, liquid or solid– of substances in chemical equation.∆H is state function and changing state of reactant/product changes intial/final state.

Thermochemical Equation:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H = -890 kJ

Change Number of Moles

Change Direction of Reaction

Change State of Reactants and Products

Enthalpies of Reaction-- Example

2CH4(g) + 4O2(g) 2CO2(g) + 4H2O(l) ∆H = -1780 kJ(∆H is extensive property)

CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ∆H = +890 kJ(∆H is state function)

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ∆H = -802 kJIt takes heat (∆Hvap) to change water from liquid to gas state, and thus heat of reaction is reduced.

How do we generalize this idea?

Thermochemical Equations:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ∆H = -890 kJ

2H2O(l) 2H2O(g) ∆H = +88 kJ

Hess’s Law

Combining Thermochemical EqnsE

NT

HA

LPY

CH4(g) + 2O2(g) initial state

CO2(g) + 2H2O(l) final state

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ∆H = -802 kJ

The overall reaction enthalpy is the sum of reaction enthalpies of the steps into which the reaction is divided.If a process occurs in steps– even if these are hypothetical– then the enthalpy change for the overall process is the sum of the enthalpy changes of the individual steps!

S(s) + 3/2O2(g) SO3(g) ∆H = ?

Hess’s Law

1. S(s) + O2(g) ∆H1 =

2. SO2(g) + 1/2O2(g) ∆H2 =

∆Hrxn = ∆H1 + ∆H2

Hess’s Law– Example-2

S solid

SO3 gas

+ 3/2 O2

SO2 gas

+O2

+ 1/2O2

SO2(g) -320.5 kJ

SO3(g) -75.2 kJS(s) + 3/2O2(g) SO3(g)

= -320.5 + -75.2= -395.7 kJ

Standard Enthalpy of Formation, ∆Hfo

Pure Elements in Reference Form?

• pure elements, reference form:

Standard Enthalpies

of a substance is the enthalpy change that occurs in the formation of one mole of the substance in the standard state from reference form of the elements in their standard states.

Generally, most stable form of elements in standard state; for example: H is H2(g), O is O2(g), Na is Na(s)

The standard enthalpy of formation of a pure element in its reference form is zero, O.

Formation Reactions:

∆Hfo: Examples

CO2(g)

CH4(g)

NH3(g)

NO(g)

NO2(g)

NaCl(s)

C(s, graphite) + O2(g)

C(s, graphite) + 2H2(g)

1/2N2(g) + 3/2H2(g)

1/2N2(g) + 1/2O2(g)1/2N2(g) + O2(g)

Na(s) + 1/2Cl2(g)

∆Hfo & ∆Ho

rxn

•

•

∆H = Σvp∆Hfo(products) – Σvr∆Hf

o(reactants)

Example: NO(g) + (1/2) O2(g) NO2(g)

• ∆Horxn =

• determine enthalpies of formation

Standard Enthalpies of RXN

NO2(g)

NO(g) + (1/2) O2(g)

(1/2) N2(g) + O2(g)

EN

TH

AL

PYStandard enthalpies of formation tabulated since they can be used to calculate enthalpy of reaction.How? Use state function property of H– does not matter what pathway we take– and design path through elements in their standard forms.

∆Hf(NO2) – ∆Hf(NO) – 1/2∆Hf(O2)

Overall Chemical Reaction

Al + Fe2O3

Is reaction exothermic or endothermic?

Enthalpy of Reaction:

∆H = Σvp∆Hfo(products) – Σvr∆Hf

o(reactants)

Demonstration– Observations?••

RXN of Al & Iron Oxide

Fe2O3(s)

Fe(s)

Al2O3(s)

Al(s)

∆Hfo (kJ/mol)

Al2O3 + Fe2 2

= ∆Hf(Al2O3) + 2∆Hf(Fe) – 2∆Hf(Al) – ∆Hf(Fe2O3)

=

=

-824

-1676

0

0

After starting, the reaction proceeds vigorousA lot of heat is released!!

-1676 + 2(0) – 2(0) - -824

-852 kJ/mol

Standard State

• conditions:

• reference state:

Ionic Reactions in Water?

Many chemical reactions proceed in aqueous solution (e.g., biochemical reactions) and involve ionic species, such as protons and metal cations.Standard State: 1M concentration

To determine enthalpies of formation for ions there is new issue– cannot create ion of single type in chemical reaction.Therefore, define a reference state to which compare enthalpies of formation of other ions:

∆Hfo (H+) = 0

When 50.0ml of 0.100M Na2SO4(aq) is mixed with 25.0ml of 0.200M AgNO3(aq), a white ppt forms.

• Net ionic reaction?

• Given ∆Hfo(Ag2SO4(s)) = -715 kJ/mol, calculate

∆Hrxn; is reaction endo or exothermic?

• If the solution starts at 25.0oC, will the final temperature be high or lower, and by how much?

Enthalpies of Ionic Reactions

2Ag+(aq) + SO42-(aq) Ag2SO4(s)

∆Hrxn = ∆Hf(Ag2SO4) – 2∆Hf(Ag+) – ∆Hf(SO42-)

= -715 – 2(106) – (-909)= -18 kJ/mol

Slightly Exothermictotal enthalpy for experiment = 2.5x10-3mol x -18kJ/mol

= -4.5x10-2 kJ

Slightly HigherAssume: (i) specific heat = that of water, 4.18 J/g-K

(ii) density = that of water 1g/ml(iii) solve using q = m x specific heat x ∆T

Overall Thermochemical Reactions:• Carbohydrate– sucrose:

C12H22O11 + 12O2 12CO2(g) + 11H2O(l)

• FatC12H24 + 36O2 12CO2(g) + 12H2O(l)

+/- of different types of fuels?

∆H(carbohydrate) ~ 17 kJ/g∆H(fat) ~ 38 kJ/gThere is approximate 2x more energy store per gram in

fat than carbohydrates…….hence, if you are a bear sleeping for the winter it is much better to store needed fuel as fat!

Food & Enthalpies of Reactions