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Thermo Chemistry Lecture: Calculating Heat of Reactions and Phase Transformations, Study notes of Materials Physics

Illustrations and exercises on how to calculate heat of reactions and phase transformations using given data. It covers the calculation of heat of reaction for various components at different temperatures, as well as the determination of heat content in a substance during phase transformations. The document also includes the calculation of enthalpy changes for reactions and the enthalpy increment for heating a substance.

Typology: Study notes

2012/2013

Uploaded on 04/26/2013

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Lecture7Thermochemistry:illustration
Content:
Illustrationi
Doyourselfi
Illustrationii
Doyourselfii
Illustrationiii
Doyourselfiii
Illustrationiv
Illustration‐i
Calculateheatofthefollowingreaction/moleat550K.
ZnO󰇛S󰇜C
󰇛S󰇜Zn
󰇛g󰇜CO󰇛g󰇜
Given:
Component H1500‐H298
(kcal/kgmol)
∆H
°
(kcal/kgmol)
ZnO(s)
C(s)
Zn(g)
CO(g)
14800
9291
37150
5480
83310
‐
‐
‐29160
Forthereaction1
∆H
° H
°CO H
°ZO
Substitutingthevaluesweget
∆H
° 54150 k cal/kg mol.
󰇛∆H󰇜 H
° ∑󰇛H H
󰇜 ∑󰇛H H
󰇜
Substitutingthevalueweget
󰇛∆H󰇜 72689 kcal/kg mol
Doyouself–i
Calculateheatofthefollowingreactionat1200K.
FeO3CO2Fe3CO

pf3
pf4

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Lecture 7 Thermo chemistry: illustration

Content:

Illustration‐i Do yourself‐i Illustration‐ii Do yourself‐ii Illustration‐iii Do yourself‐iii Illustration‐iv

Illustration‐ i

Calculate heat of the following reaction/ mole at 550K.

ZnOሺSሻ ൅ CሺSሻ ൌ Znሺgሻ ൅ COሺgሻ

Given:

Component H 1500 ‐H 298 (kcal/kg mol)

∆H ୤°

(kcal/kg mol) ZnO(s) C(s) Zn(g) CO(g)

For the reaction 1

൫∆H (^) ୤°^ ൯ଶଽ଼ ൌ ൫∆H (^) ୤°^ ൯CO െ ൫∆H (^) ୤°^ ൯Z୬O

Substituting the values we get

൫∆H (^) ୤°^ ൯ଶଽ଼ ൌ ൅54150 k cal/kg mol.

ሺ∆H (^) ୤ ሻଵହ଴଴ ൌ ൫∆H (^) ୤°^ ൯ଶଽ଼ ൅ ∑ሺHଵହ଴଴ െ H (^) ଶଽ଼ ሻ (^) ୮୰୭ୢ୳ୡ୲ୱ െ ∑ሺHଵହ଴଴ െ H (^) ଶଽ଼ ሻ (^) ୰ୣୟୡ୲ୟ୬୲

Substituting the value we get

ሺ∆H (^) ୤ ሻଵହ଴଴ ൌ 72689 kcal/kg mol

Do you self –i

Calculate heat of the following reaction at 1200K.

Feଶ O (^) ଷ ൅ 3CO ൌ 2Fe ൅ 3CO (^) ଶ

Component H 1200 ‐H 298 (kcal/kg mol)

∆H ୤°

(kcal/kg mol) Fe 2 O 3 CO CO 2 Fe

Illustration –ii

Calculate heat content in 1 kg mol nickel at 3500Ԩ.

At 3500Ԩ ሺNiሻ^ ୱ ൌ NiሺVሻ.

Nickel undergoes phase transformation from α to β at360Ԩ.

Melting point of Ni ൌ 1455Ԩ, boiling point is 2730Ԩ

Latent heat of melting ൌ 4248 (^) ୩୥ ୫୭୪୩ୡୟ୪

Latent heat of vaporization ൌ 87910 kcal/kg mol

ሺC (^) P ሻ (^) N୧ α ൌ 12.535 ൅ ሺ35.815 ൈ 10ି ଷ^ Tሻ െ ሺ6.95x10ହ^ Tି ଶ^ ሻ (^) ୩୥ ୫୭୪ୣ KKJ

ሺCP ሻ (^) N୧ β ൌ 25.104 ൅ ሺ7.531 ൈ 10ି ଷ^ Tሻ െ (^) ୩୥ ୫୭୪ୣ KKJ

Latent heat of transformation from α to β ൌ 0

Solution

Heat content in 1 kg mol nickel vapour

݂ ݅ܰ ݊݅ ݐ݊݁ݐ݊݋ ܿ ݐܽ݁ܪሺ 25 ݉݋ݎԨ 360 ݋ݐԨሻ ൅ ሺlatent heat of transformationሻ ൅ ሺHeat content in Ni from 360Ԩto 1455Ԩሻ ൅ ሺlatent heat of meltingሻ ൅ ሺHeat content in Ni from 1455Ԩto 2730Ԩሻ ൅ ሺlatent heat of vaprizationሻ ൅ ሺHeat content in Ni from 2730Ԩto 3500Ԩሻ

One has to calculate all the values to give heat content ൌ 119217.5 (^) ୩୥ ୫୭୪୩ୡୟ୪ in Ni vapour at 3500Ԩ

which is equal to (^2020) ୩୥ N୧୩ ୡୟ୪

Do yourself – ii

For a solid silver, the molar heat capacity at constant pressure

C (^) P ൌ 21.3 ൅ ሺ8.535 ൈ 10ି ଷ^ Tሻ ൅ ሺ1.506 ൈ 10ହ^ Tି ଷ^ ሻJ/g mol k

Illustration iv

Calculate heat of reaction at 1100Ԩ

Cuଶ Sሺsሻ ൅ ሺO (^) ଶ ൅ 3.76 N (^) ଶ ሻ (^) ୥ ൌ 2Cuሺlሻ ൅ SO (^) ଶ ሺgሻ ൅ 3.76N (^) ଶ ሺgሻ

Reactants and products are at 1100Ԩ. Air is supplied at 25Ԩ. Melting point of copper is 1084Ԩ.

C (^) P ሺCuଶ sሻ ൌ 81.588 (^) ୫୭୪ ୩.J

CP ሺCuୱ୭୪୧ୢ ሻ ൌ 22.635 ൅ ሺ6.27 ൈ 10ି ଷ^ Tሻ^ ୫୭୪ ୩J.

CP ൫Cu୪୧୯୳୧ୢ ൯ ൌ 31.380 (^) ୫୭୪ ୩୨

CP ሺSO (^) ଶ ሻ ൌ 43.430 ൅ ሺ10.62 ൈ 10ି ଷ^ Tሻ െ ሺ3.766 ൈ 10ହ^ Tି ଶ^ ሻ

CP ሺN (^) ଶ ሻ ൌ 27.865 ൅ ሺ4.268 ൈ 10ି ଷ^ Tሻ

Latent heat of copper ൌ 11193 (^) ୫୭୪୨

൫∆H C୳°^ మ ୱ൯ଶଽ଼ ൌ െ79496 ୫୭୪୨ , ∆H SO°^ మൌ െ296810 ୫୭୪୨

∆H R ሺ1373ሻ ൌ ∆H R°^ ൅ ∑ሺHଵଷ଻ଷ െ H ଶଽ଼ ሻ P െ ∑ሺHଵଷ଻ଷ െ H ଶଽ଼ ሻ R

On solving ∆H (^) R ሺ1373ሻ ൌ െ40405 joule