Heat EffectThermodynamics Course Chapter 4

Agung Ari Wibowo S.T., M.Sc

State Polytechnic of Malang

Malang

2017

Learning ObjectiveChapter 4

Sensible Heat

Heat Capacity &

Integral Evaluation of Sensible

Heat

Laten Heat of Pure

Subtance

Standard Heat of

Reaction and

Formation

Standard Heat of

Combustion

Standard Enthalpy as function of

Temperature

Heat Effect in Industrial Process

Literature UsedJ M Smith, H C Van Ness, M M Abbott, “Introduction to Chemical Engineering Thermodynamics 7th”

What happen when you add heat to the water ?

Sensible HeatChapter 4

Temperature

1 • Temperature rise

2 •Water Start to boil

3• Continous Evaporation

(phase change to vapor)

Boil Water

Sensible HeatChapter 4

What is Sensible Heat ??

Heat which causing the rise in temperature to the certain degree with no change in phase and composition

𝑑𝐻 =𝜕𝐻

𝜕𝑇𝑃

𝑑𝑇 +𝜕𝐻

𝜕𝑃𝑇

𝑑𝑃

At constan Pressure Process

𝑄 = ∆𝐻 = න𝑇1

𝑇2

𝐶𝑝 𝑑𝑇

Heat CapacityChapter 4

∆𝐻 = න𝑇1

𝑇2

𝐶𝑝 𝑑𝑇

𝐶𝑃𝑅= 𝐴 + 𝐵𝑇 + 𝐶𝑇2 + 𝐷𝑇−2

Each substance have their own A, B, C and D

constant. These constant are listed in Appendiks C

Heat capacity of ideal gas can be written as 𝐶𝑣𝑖𝑔

and 𝐶𝑝𝑖𝑔

𝐶𝑝𝑖𝑔

𝑅= 𝐴 + 𝐵𝑇 + 𝐶𝑇2 + 𝐷𝑇−2

𝐶𝑣𝑖𝑔

𝑅=𝐶𝑝𝑖𝑔

𝑅− 1

Heat CapacityChapter 4

Mixture of gasesWhat is the heat capacity of mixture?

A B C

Mixed ABC

𝑪𝒑 𝒄𝒂𝒎𝒑𝒊𝒈

= 𝒚𝑨𝑪𝒑𝑨𝒊𝒈

+ 𝒚𝑩𝑪𝒑𝑩𝒊𝒈

+ 𝒚𝑪𝑪𝒑𝑪𝒊𝒈

Where “y” is mol fraction of the gas in mixture

Do you get it??

Integral Evaluation of Sensible HeatChapter 4

∆𝐻 = න𝑇1

𝑇2

𝐶𝑝 𝑑𝑇

What we know??

How we get the exact number??

න𝑇0

𝑇 𝐶𝑝

𝑅𝑑𝑇 = න

𝑇0

𝑇

𝐴 + 𝐵𝑇 + 𝐶𝑇2 + 𝐷𝑇−2 𝑑𝑇

It is still in Integral Equation

Can we get simpler equation??

න𝑇0

𝑇 𝐶𝑝

𝑅𝑑𝑇 = න

𝑇0

𝑇

𝐴 + 𝐵𝑇 + 𝐶𝑇2 + 𝐷𝑇−2 𝑑𝑇

= 𝐴 +𝐵

2𝑇0 𝜏 + 1 +

𝐶

3𝑇02 𝜏2 + 𝜏 + 1 +

𝐷

𝜏𝑇02 𝑇 − 𝑇0

𝐶𝑝 𝐻

𝑅= 𝐴 +

𝐵

2𝑇0 𝜏 + 1 +

𝐶

3𝑇02 𝜏2 + 𝜏 + 1 +

𝐷

𝜏𝑇02

ሻ∆𝑯 = 𝑪𝑷 𝑯 𝒙 (𝑻 − 𝑻0

OK,ok.. Here much simpler form for you…

𝜏 − 1 =𝑇−𝑇0

𝑇0

Lest back to our first case

Latent Heat of Pure SubtanceChapter 4

Can you see the bubles??

Lest back to our first case

Latent Heat of Pure SubtanceChapter 4

Amount of energi needed to change the phase

Latent Heat of Pure SubtanceChapter 4

Here are the equationsHow can we

know the

number??1. Clayperon Eq: ∆𝐇 = 𝐓∆𝐕

𝐏𝐬𝐚𝐭

𝐝𝐓

∆𝐻 = 𝐿𝑎𝑡𝑒𝑛𝑡 𝐻𝑒𝑎𝑡;∆𝑉 = Volume change occupied phase change;𝑃𝑠𝑎𝑡 = saturated vapor pressure (evaluated by Antoine Eq)

2. Troutons Eq :∆𝐇

𝐑𝐓𝐧~ 10

Tn is absolut boiling point.

3. Riedel Eq :

∆𝑯

𝑹𝑻𝒏=

𝟏,𝟎𝟗𝟐 (𝒍𝒏 𝑷𝒄−𝟏,𝟎𝟏𝟑ሻ

𝟎,𝟗𝟑𝟎−𝑻𝒓𝒏

Pc is critical pressure and Trn is reduced temperature.(Tr = T/Tc)

4. Watson Eq:

∆H2

∆H1=

1 − Tr21 − Tr1

0.38

5. Using Steam tables (only for water)

∆𝐻𝑒𝑣𝑎𝑝 = 𝐻𝑣 − 𝐻𝑙

Heat of ReactionChapter 4

1

2𝑁2 +

3

2𝐻2 ⟶𝑁𝐻3 ∆𝐻 = −46,110 𝐽

Consider this reaction

∆𝐻 = Δ𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡 − Δ𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡

∆𝐻𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡 𝑜𝑟 ∆𝐻𝑝𝑟𝑜𝑑𝑢𝑐𝑡 = න𝑇1

𝑇2

𝐶𝑝 𝑑𝑇

T reference, usually 25 C

Actual T of materials

Standard Heat of ReactionChapter 4

1

2𝑁2 +

3

2𝐻2 ⟶𝑁𝐻3 ∆𝐻298

0 = −46,110 𝐽

Consider this reactionExothermic reaction

What is the meaning of this

symbol?

Standard heat of reaction is written as ∆H2980 , 298

means this heat were calculated at 298,15 K

And What is standard

condition?

Standard Condition :

- Gas : Pure subtance in ideal gas condition at Pressure 1 bar

- Liquid dan solid : Pure subtance in real condition at

Pressure 1 bar.

Standard Heat of FormationChapter 4

Consider this reaction

Definition :

Formation reaction was defined as reaction to produce

1 compound from its molecular element, in which the

product is 1 mole in amount as basis

𝐶 +1

2𝑂2 + 2𝐻2 → 𝐶𝐻3𝑂𝐻

𝐻2𝑂 + 𝑆𝑂3 → 𝐻2𝑆𝑂4

Which one is formation reaction?

𝐶 𝑠 + 𝑂2 𝑔 → 𝐶𝑂2 𝑔 ∆𝐻𝑓2980 = −339,509 𝐽

Generally formation reaction is written in form of :

• 𝑓 stand for “formation”

• The chosen condition is T 298.15 K or

25℃

Standard Heat of FormationChapter 4

𝐶𝑂2 𝑔 + 𝐻2 𝑔 → 𝐶𝑂 𝑔 + 𝐻2𝑂 𝑔

Calculate the heat of reaction from reaction below

𝐶𝑂2 𝑔 → 𝐶 𝑠 + 𝑂2 𝑔 ∆𝐻𝑓2980 = 339,509 𝐽

𝐶 𝑠 +1

2𝑂2 𝑔 → 𝐶𝑂 𝑔 ∆𝐻𝑓298

0 = −110,525 𝐽

𝐻2 𝑠 +1

2𝑂2 𝑔 → 𝐻2𝑂 𝑙 ∆𝐻𝑓298

0 = −285,830 𝐽

𝐻2𝑂 𝑙 → 𝐻2𝑂 𝑔 ∆𝐻𝑓2980 = 44,012𝐽

Break down the reaction to get heat of formation of each compound

Where is H2??

Answer :

H2 is only constructed from 1 kind of

element, so its ∆Hf2980 = 0 𝐶𝑂2 𝑔 + 𝐻2 𝑔 → 𝐶𝑂 𝑔 + 𝐻2𝑂 𝑔 ∆𝐻298

0 = 41,66 𝐽

Wait, waittt

Standard Heat of CombustionChapter 4

Definiton : Heat of Combustion (enthalpy of combustion) of a substance is the heat liberated

when 1 mole of the substance undergoes complete combustion with oxygen at constant pressure.

Have you ever heard methanol gel?

Wedding party food heater

Standard Heat of CombustionChapter 4

The use of heat of combustion

4𝐶 𝑠 + 5𝐻2 𝑔 → 𝐶4𝐻10 𝑔

Wait sir, C and H just only consist of 1 element… How can I know its

∆𝐻𝑓2980 ?

• 4𝐶 𝑠 + 4𝑂2 𝑔 → 4𝐶𝑂2 𝑔 ∆𝐻𝑓2980 = 4 𝑥(−339,509ሻ𝐽

• 5𝐻2 𝑔 + 2,5 𝑂2 𝑔 → 5𝐻2𝑂 𝑙 ∆𝐻𝑓2980 = 5 𝑥(−285,830ሻ 𝐽

• 4𝐶𝑂2 𝑔 + 5𝐻2𝑂 𝑙 → 𝐶4𝐻10 𝑔 + 6,5 𝑂2 𝑔 ∆𝐻2980 = 2877,396 𝐽

4𝐶 𝑠 + 5𝐻2 𝑔 → 𝐶4𝐻10 𝑔 ∆𝐻𝑓2980 = −125,790 𝐽

Temperature Dependent of ∆𝐻0

Chapter 4

What about standard reaction at different temperature or not at

298,15 K?

Heat of reaction at certain temperature can be calculated if the heats of reaction at

standard condition were determined first

Reference : T = 298,15 K

What about at T = 800 ℃

𝐶𝑂 𝑔 + 2𝐻2 → 𝐶𝐻3𝑂𝐻 𝑔

Chemical Reaction :𝑣1𝐴1 + 𝑣2𝐴2 +⋯ → 𝑣3𝐴3 + 𝑣4𝐴4 +⋯

∆𝐻0 =

𝑖

𝑣𝑖𝐻𝑓𝑖0

∆𝐻0° = ∆𝐻298

° = −200,660 − −110,525= −90,135 𝐽

∆H° = ∆H0° + RT0

T ∆CP°

RdT

Calculating Cp

∆𝑪𝑷°

𝑯

𝑹= ∆𝑨 +

∆𝑩

𝟐𝑻𝟎 𝝉 + 𝟏 +

∆𝑪

𝟑𝑻𝟎𝟐 𝝉𝟐 + 𝝉 + 𝟏 +

∆𝑫

𝝉𝑻𝟎𝟐

൯∆𝑯° = ∆𝑯0° + ∆𝑪𝑷

°𝑯(𝑻 − 𝑻0

i vi A 103B 106C 10-5D

CH3OH 1 2,211 12,216 -3,450 0

CO -1 3,376 0,557 0 -0,031

H2 -2 3,249 0,249 0 0,083

∆𝐴=(1)(2,211)+(-1)(3,376)+(-2)(3,249) = -7,663

∆𝐵=10,815 x 10-3 ; ∆𝐶=-3,450x10-6 ; ∆𝐷=-0,135x105

∆H° = ∆H0° + RT0

T ∆CP°

RdT

= -90,135 + 8,314 (-1615,5)

= -103,556 JT = 800 ℃

Heat Effect of Industrial ReactionsChapter 4

Reactor Temperature 377 K

Process Flow Diagram of Cumene Production

Hmmm,,, I smell problem here…

Heat Effect of Industrial ReactionsChapter 4

4𝐻𝐶𝑙 𝑔 + 𝑂2 𝑔 → 2𝐻2𝑂 𝑔 + 2𝐶𝑙2 𝑔

Material entering reactor ( R ) : HCl, O2

Material Out ( P ) : HCl, O2 Cl2, H2O

1 2

∆𝐻 = σ𝑖𝑙 𝑛𝑖 𝐶𝑝𝑖

° (T – 298,15)

𝐶𝑝𝑖° = 𝑅 𝑥

∆𝑪𝑷°

𝑯

𝑹

∆𝐻 = ∆𝐻𝑅 + ∆𝐻298° + ∆𝐻𝑝

Total enthalpy change :

∆𝐻𝑅= perubahan entalpi reaktan pada suhu T1 ke suhu referens

298,15 K

∆𝐻𝑅

∆𝐻298°

∆𝐻𝑝

∆𝐻

T reference = 298,15 K

∆𝐻298° = entalpi reaksi standar pada suhu 298,15 K

∆𝐻𝑝= perubahan entalpi produk dari suhu referens 298,15 K ke

suhu T

https://www.google.com.tr/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjZgJ-U4L_TAhXLvo8KHYBhCkgQjRwIBw&url=https://www.planwallpaper.com/thank-you&psig=AFQjCNE8mfdxv0oktKUhmv1Rm39mzDcgcA&ust=1493214593242467