slide 1 of 57 7-7 indirect determination of h: hess’s law h is an extensive property. –...

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7-7 Indirect Determination of H:Hess’s Law

H is an extensive property.–Enthalpy change is directly proportional to the amount of substance in a system.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 7

N2(g) + O2(g) → 2 NO(g) H° = +180.50 kJ

½N2(g) + ½O2(g) → NO(g) H° = +90.25 kJ

• H changes sign when a process is reversed

NO(g) → ½N2(g) + ½O2(g) H° = -90.25 kJ

Hess’s Law and Enthalpies• The ideas on the previous slide are familiar.

∆H is an intensive quantity. → The bigger the steak we want to barbecue the more moles (or grams) of propane we’ll have to burn.

• H changes sign when a process is reversed. An analogy here would be climbing up or down a ladder. Climbing up our gravitational potential energy increases – climbing down our potential energy decreases (mgh rule!).

All ∆H’s Need Not Be Measured• Elemental nitrogen and oxygen react to form

a variety of oxides. We need not measure every ∆H experimentally – some can be calculated using data for a subset of all possible reactions. Reactions forming NXOY molecules can be exothermic or endothermic.

• N2(g) + O2(g) → 2 NO(g) Endothermic

• In the atmosphere lightning supplies the energy needed to form NO(g).

Nature’s Synthesis of NO(g)!

• The reaction of N2(g) and O2(g) to give NO(g) is an endothermic process. In nature, lightning storms supply the needed energy!

All ∆H’s Need Not Be Measured

• The reaction of NO(g) with additional oxygen is exothermic.

• NO(g) + ½ O2(g) → NO2(g) Exothermic

• Knowing the ∆H for this rxn and the one on the previous slide (by experiment) we can calculate ∆H for the rxn

• ½ N2(g) + O2(g) → NO2(g) (red brown, LAX)

Smog in Southern California

• The formation of NO2(g) is one factor in smog formation.

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Hess’s Law Schematically


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• Hess’s Law of Constant Heat Summation


If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

½N2(g) + O2(g) → NO2(g) H° = +33.18 kJ

½N2(g) + O2(g) → NO(g) + ½ O2 (g) H° = +90.25 kJ

NO(g) + ½O2(g) → NO2(g) H°= -57.07 kJ

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7-8 Standard Enthalpies of Formation,

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


Hof

The standard enthalpy of formation of a pure element in its reference state is 0.

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Liquid bromine vaporizing


Br2(l) Br2(g) Hf° = 30.91 kJ

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Diamond and graphite


Element 79!

• Gold (Au), like most metals, is a shiny metal at room temperature (standard state!).

You can Mix a Metal and a Nonmetal

• If the “chemistry” is right! (One standard state?)

Enthalpies of Formation• Heats of combustion of hydrocarbons are

always –ve (exothermic reactions). Heats of formation of compounds can be +ve or –ve (endothermic and exothermic reactions).

• ½ H2(g) + ½ Cl2(g) → HCl(g) ∆Hof = -92.3 kJ

• ½ H2(g) + ½ I2(g) → HI(g) ∆Hof = +26.5 kJ

• If we knew the ∆Hof for every substance we

could calculate the ∆Ho value for every reaction.

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Some standard enthalpies of formation at 298.15 KFIGURE 7-18


number of carbons

Combining Heats of Formation

• The process for calculating a general enthalpy change from heat of formation relies on the fact that enthalpy, H, is a state function. Thus, we can imagine proceeding from the reactants to the products by a one step (direct process) or by a two step process:

• Step 1: Reactants → Constituent Elements• Step 2: Constituent Elements → Products

REACTANTS PRODUCTS

CONSTITUENTELEMENTS

Heats of Formation

• The previous slide outlines a simple process for calculating Heats of Reactions (ΔHo’s) if the relevant Heats of Formation for all reactants and all products are known.

• ΔHoRxn = ΣΔHo

f(Products) – ΣΔHof(Reactants)

• In practice a particular temperature is specified for the Heats of Formation (usually 298K). Care must be taken to ensure that the correct phase for reactants and products has been specified.

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Standard Enthalpies of Reaction


FIGURE 7-20

•Computing heats of reaction from standard enthalpies of formation

Hoverall = -2Hf°NaHCO3

+ Hf°Na2CO3

+ Hf

°CO2

+ Hf°H2O

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Diagramatic representation of equation (7.21)FIGURE 7-21


∆H° = ∑p∆Hf°(products) - ∑r∆Hf°(reactants) (7.21)

Examples: Use of Heats of Formation

• Class examples: Show how to calculate the standard enthalpy change for each of the following reactions using Heats of Formation data.

• (a) CaCO3(s) → CaO(s) + CO2(g)

• (b) 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)

• (c) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

Combustion Reactions – History• Combustion reactions for hydrocarbons are

important sources of energy for many processes. One could easily put together a spreadsheet to calculate the enthalpies of combustion for all hydrocarbons – if the heats of formation of the hydrocarbons were known. Historically this is backwards as enthalpies of combustion are used in most cases to determine heats of formation for hydrocarbons. n-octane will be used as an example.

8CO2(g) + 9H2O(l)

H(Enthalpy) (kJ mol∙ -1)

“Elements”

Products

Combustion of Liquid n-Octane – C8H18(l)

Reactants

Combustion of n-octane

• Class example: The standard enthalpy of combustion of n-octane was determined by experiment to be -5470.3 kJ mol∙ -1 at 298K. Use heat of formation data for the products of combustion to determine the heat of formation of n-octane.

Combining Thermochemical Equations• Unknown ΔH (and ΔU) values can be

calculated by combining known thermochemical equations which are often more complex than those written for ΔHo

f’s.

• It is often necessary to combine several thermochemical equations to obtain a ΔHo value not known from experiments. This requires that we compare the coefficients of substances appearing only once in the given and desired thermochemical equations.

Combining Themochemical Eqtns - Hydrazine

• Example: Determine ∆Ho for the reaction• N2H4(l) + 2 H2O2(l) → N2(g) + 4 H2O(l)

• Given:• N2H4(l) + O2(g) → N2(g) + 2 H2O(l) ∆Ho = - 622.2 kJ

• H2(g) + ½ O2(g) → H2O(l) ∆Ho = - 285.8 kJ

• H2(g) + ½ O2(g) → H2O2(l) ∆Ho = - 187.8 kJ

Combining Thermochemical Equations

Results of Combining EquationsEquation New Balanced Thermochemical Equations

-1 x Eq (1) N2H4(l) + O2(g) → N2(g) +2 H2O(l) ΔHo = -622.2 kJ

-2 x Eq (3) 2 H2O2(l) → 2 H2(g) + 2 O2(g) ) ΔHo = +375.6 kJ

+2 x Eq (2) 2 H2(g) + O2(g) ) → 2 H2O(l) ΔHo = - 571.6 kJ

SUM N2H4(l) + 2 H2O2(l) → N2(g) +4 H2O(l) ΔHo = - 818.2 kJ

slide 1 of 57 7-7 indirect determination of h: hess’s law h is an extensive property. –...

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