---
title: "Metallurgical Thermodynamics and Kinetics - MML216"
tags : "SEM4, MME"
---

Dr.Jatin Bhatt
HOD - Metallurgical and Material Engineering
[phone]
[e-mail]

# Metallurgical Thermodynamics and Kinetics - MML216

Making new materials
Metal - Commerical Metal, Pure Metal
Commerical Metal - No control in the production practices
Pure Metal - Not commerically viable - strategic application
Thermodynamic - Imperical - Simulate, Model
Temperature is important to change property of material


New Technology - New Material
- Solid State Electronics
- Nuclear Material
- Aviation Sector
- Space Sector
- Special Purpose Material

Systems - Reactive and Non-Reactive

Thermodynamics - helps in Energy Audit

Solution at high temperature needs
- Experimental data of Enthalpy and Gibbs Free Energy at high temperature
- Understanding of Phases
- Volume change
- Crystal change
- Understandfing of Equilibrium among the pahses
- Integrity of the atoms in the structure
- Chemical basis of phases

Entalphy is a function of specific heat that can be determined using caloriemeter.

## System and Surrounding

- Open
- Exchange matter and energy
- Closed
- Mass is fixed, energy can be transferred
- Isolated
- No Mass exchange or energy transfer

Book(History of thermodynamics with ingo muller)

## Classical Thermodynamics

- Macrosopic size
- Cannot address behavior of atoms and molecules
- Continuum mechanics
- Structure Insesitive (we do not consider internal structure)

Thermodynamics changes with dimension

### State of System

Macrostate

- Defines by state variables (Temperature, Pressure, Volume, Density etc)

- Intensive Variables
- Extensive Variables:
- dependent on mass of system
- Volume, Density

### Equilibrim

- Characterization of system is only possible under this condition
- We make martensite using fast cooling (quenching) i.e. metastable conditions
- Substance which is under thermodynamic equilibrim is know as </B>thermodynamic</B> substance.

#### Types

- Mechanical Equilibrium
- Uniformity in pressure throughout the system
- Thermal Equilibrium
- Uniformity in temperature throughout the system
- Chemical Equilibrium
- Chemical potential is same throughout the system

Contrubution of changes is comming from microscopic change.

Emperical relationship is termed as apparatus

#### Process

- Physical Process
- System is under thermal equilibrium + mechanical equilibrium
- Chemical Process
- Thermal Equilibrium + Chemical Equilibrium
- Physio- Chemical Process
- Thermal equilibrium + Mechanical Equilibrium + Chemical equilibrium

#### Pseudo Equilibirum

Metastable condition or partial chemical equilibrium
System is always under constant change

Example: Metallic liquid

- System
- Open, Close, Isolated
- Homogenous - Ice at subzero temperature
- Hetrogenous - Ice floating in water
- Unary, Multicomponent
- Process
- Physical(Non-Reactive)
- Physico-chemical(reactive)

#### Closed or isolated system

- No transfer of mass or energy.
- Fixed mass-> Fixed composition -> fixed structure.

**Example**: Magnesium has high vapour presure hence cannot be melted in open atmosphere.

## First Law of Thermodynamics

- Deals with Internal Energy and Enthalpy
- Enthalpy
- Greek word which means warm

$\Delta U = q - w$

$\Delta U$ - Change in internal energy

- Based on Law of Conservation of energy
- q amount of heat is supplied to system, w amount of work is done on the surrounding
- q is not equal to w ?
- q-w is not equal to zero. - Hidden energy is introduced to save law of conservation of energy

### Existence of Internal Energy

- Macroscopic Kinetic energy due to motion of system
- Potential Energy - due to position
- Kinetic Energy - Transalation, Rotation, Vibration (Thermal Energy)
- Interaction energy - atoms and molecules
- Columbic Energy - Electrons and nuclei in atoms
- Energy content of electrons and nuclei

Chemical thermodynamics at elevated temperature
- Kinetic energy of atoms due to traslation, rotation and vibration (thermal energy)+interaction among atoms and molecules is taken into consideration.

### Ideal Gas and Ideal Solution

- Gas - obeys Boyle's Law, energy is a function of Temperature
- Solution - activity of each component is equal to its mole fraction, obeys Raoult's Law

### Characterization of Process

Conventional Process
- Take into consideration KE of atoms and interaction among atoms

Non Conventional Process
- Coulombic interaction and energy of electrons

#### For Finite Process

Absolute value of U
- Cannot be determined
- Not a path property - State Property
- If we say free energy for fixed mass and composition then we are indirectly invoking closed system

#### For Cyclic Process

$\sum \Delta U = 0$
$\sum\delta q - \sum\delta W = 0$

- Variable in system are P, V, T, composition and Structure
- For non-reactive

Invoking Property of exact differential

If U = f(V,T)
$dU = \dfrac{\delta U}{\delta V}_T$

### Significance of First Law of thermodynamics

- Based on law of conservation of energy
- Brings concept of Internal Energy
- Separates heat interaction and work interaction
- Treats internal energy as state property unlike q and w which are path properties

### For Cyclic Process

- $\sum \Delta U = 0$
- Variables in system are P, V, T, composition and Structure
- For non-reactive - structure and composition reamin constant
- U = f (V,T)
- U = f (P,T)
- U = f (P,V)

### Invoking property of exact differential

- In case of U = f(V,T)
$\qquad dU = \left(\dfrac{\delta U}{\delta V}\right)_T dV + \left(\dfrac{\delta U}{\delta T}\right)_VdT$
- In case of U = f(P,T)
$\qquad dU = \left(\dfrac{\delta U}{\delta P}\right)_T dP + \left(\dfrac{\delta U}{\delta T}\right)_PdT$
- In case of U = f(P,V)
$\qquad dU = \left(\dfrac{\delta U}{\delta P}\right)_V dP + \left(\dfrac{\delta U}{\delta V}\right)_PdV$

### Reversible Process

- A reversible process may be defines as the hypothetical passage of system through a series of equilibrium stages
- A reversible process is very slow and mostly impractical


## Enthalpy

$H = U + PV$
H - Enthalpy
U - Internal Energy
P - Pressure
V - Volume

- H,U are state properties
- P,V are the state variables
- Change in enthalpy due to chane in temperature

dH = dU + PdV + VdP
dH = dU + PdV (if P is Constant)

dU = δq - δW (According to First law of Thermodynamics)

dU = δq - PdV
dH = δq - PdV + PdV
q = ∆H (for finite process)

- This equation is the basis for Heat Balance
- Heat Balance is important in metallurgical (manufacturing) industry for analysis of process - Energy Audit
- q is path property
- ∆H is state property

### Property of Internal Energy

- In case of U = f(P,T)
$\qquad dU = \left(\dfrac{\delta U}{\delta P}\right)_T dP + \left(\dfrac{\delta U}{\delta T}\right)_PdT$

At const temp dT=0, dU=0
$\left(\dfrac{\delta U}{\delta P}\right)_T = 0$

Now,
$H = U + PV$
$\left(\dfrac{δ H}{δ P}\right)_T= \left(\dfrac{δU}{δP}\right)_T + \left(\dfrac{δ(PV)}{δ P}\right)_T$

For ideal gas at constant temperature
dT=0, dU=0, δ(PV)=0

$\left(\dfrac{\delta H}{\delta P}\right)_T = 0$

- Enthalpy is independent of pressure at constant Temperature and Volume at constant temperature
- Enthalpy is a fucntion of T for fixed Mass i.e. independent of volume

### Molar Heat Capacity

$C_v = \left(\dfrac{\delta q}{\delta T}\right)_V$

$C_p = \left(\dfrac{dH}{dT}\right)_P$

- dH = δq, indicates change in enthalpy due to change in temperature
- This enthalpy is called **sensible heat**

### Specific heat

- Specific heat is the heat required to raise the temperature of the unit mass of a given substance by unit degree celsius

$C_v = \left(\dfrac{\delta q}{\delta T}\right)_V = \left(\dfrac{\delta U}{\delta T}\right)_V$

$C_p = \left(\dfrac{\delta q}{\delta T}\right)_P = \left(\dfrac{\delta H}{\delta T}\right)_P$

$H_1 - H_2 = \overset{T_2}{\underset{T_1}{\int}}C_p\,dT$

- Aggregation of substance must be considered

### Standard State

- Pure element or compound - Stablest state - At temperature under consideration and 1 atm pressure
- Enthalpy Calculation Methods
- Hess' Law
- The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the sequence of steps taken
- Kirchoff's Law
- Enthalpy change that is taking place with change in temperature is known as Kirchhoff’s Law



### Limitations of 1st Law of Thermodynamics

- It is not enough to characterize system
- Only gives idea on various equivalence of energy
- Does not give idea on feasibility of Process

> Tutorial 2 - 24.01.22
>> Q1. The heat formation of FeO at 298K is -267.3kJ/mol. Obtain an expression for $\Delta H°_{298}$ of FeO
>> A. $\Delta H_{T_2} = \Delta H_{T_1} +$
>
>> Q2. Calculate the heat of reaction for the reaction "Al~2~O~3~ $\rightarrow$ 2Al + 3CO
>> A. Kirchoff's Law
>
>> Q3. Calculate the heat of formation of Al~2~O~3~ at 773K. 2Al(s) + 3/2O~2~(g) $\rightarrow$ Al~2~O~3~(s)
>> A. Kirchoff form room temperature to 773K
>
>> Q4. Compare the heat of reaction of the two SiO~2~(cristobalite) formation reactions at 1600°C
>> A. Change in enthalpy, Heat = Enthalpy, at constant pressure
>
>> Q5. Calculate the molar enthalpy change of Mn, $\Delta H_{Mn}$, between 1373K and 1573K

## 2nd Law of Thermodynamics

- Spontaneous process are not thermodynamically reversible
- Now for reversible process temperature is constant
- Entropy - change
- Reversible process
$\quad dS = \displaystyle\frac{(\delta q)_{\text{rev}}}{T}$
$\delta q$ - Heat
- This restiction that entropy change can be calculated only through reversible path

### Consistency in system

- System can be characterized systematically when there is consistency
- If consistency is not present then it is difficult to understand the system
- This is the strength of reversible process when system is considered at constant temperature


When entropy is combined with temperature it gives you nature of bound energy

$dS = \dfrac{\delta q_{rev}}T = C_v\frac{dT}T = C_v d(\text{ln}T)$ - At constant Volume

$dS = \dfrac{\delta q_{rev}}T = C_p\frac{dT}T = C_p d(\text{ln}T)$ - At constant Pressure

$\Delta S_M^° = \dfrac{\Delta H^°_M}T$

$H = U + PV$

$dH = dU + PdV + VdP$

$\delta H = \delta q + VdP$
$\delta H = \delta q$ - if reversible (const pressure usually)


The graph showing sharp change is due to pure form of the material

### Entropy change for **reversible** and **irreversible** process


(dS)~system~=(δq)~rev~/T~sys~
(dS)~surr~=(δq)~surr~/T~surr~

We concider here material to have large heat capacities so that the material maintain constant tempreature.(Important criteria)

(dS)~system~+(dS)~surr~=(δq)~rev~/T~sys~-(δq)~rev~/T~surr~
=(δq)~rev~(1/T~sys~ - 1/T~surr~)
T~sys~ = T~surr~
(dS)~sys~ + (dS)~surr~ =0

Change in energy is dependent on mass, change in temperature and the nature of system

### For irreversible process

- T~surr~ > T~sys~ - Siginificantly higher
- T~surr~ is significantly higher than that of T~sys~
- (dS)~sys~ + (dS)~surr~ > 0

$\Delta H$ - Makes criteria easy
- Easy to quantify $\Delta H$
- It is related to C~p~ and it can be determined using calorimeter

For Finite Process
- Entropy has unit of J/K, if temperature is taken into consideration to normalise entropy then it will become “TS”
- Joule

Why is enthalpy determination making our criteria easier?
- Enthalpy is easy to quantify
- It is related to C~p~
- Can be determined using caloriemeter

### Implication of bound energy (TS)

- More is the increase in entropy of system, less work will be done when a quantity of heat is supplied
- Bound energy - less work will be done

Physical Process - Operation
Chemical Reaction - Process

### Helmholtz Free Energy

A = U - TS

A - Helmholtz Free Energy
U - Internal Energy
T - Temperature
S - Entropy
TS - Bound Energy

#### Combined Equations

**dU = TdS – PdV - δW’**

δW’ - Electrical, magnetic work

H = U + PV
dH = dU + PdV + VdP
dH = TdS – PdV - δW’+ PdV + VdP
**dH = TdS + VdP - δW’**

A= U – TS
dA = dU – TdS –SdT
dA = TdS – PdV - δW’ – TdS –SdT
**dA = - PdV – SdT - δW’**

G = H -TS
dG = dH – TdS – SdT
dG = TdS + VdP - δW’ – TdS – SdT
**dG = VdP – SdT - δW’**

### Free Energy

- TS - Bound Energy
- Can't be utilised to do work
- Gets dissipated as heat
- Free Energy is the energy available for doing work
- H is more Qualified than Q
- H = U + PV

##### Apply Maxwell Equation to thermodynamic equation

• (dT/dV)~s~ = -(dP/dS)~V~
• (dT/dP)~s~ = (dV/dS)~P~
• (dS/dV)~T~ = (dP/dT)~V~
• (dV/dT)~P~ = -(dS/dP)~T~


## How to develop a thermodynamic apparatus

- System
- Apply hydrostatic pressure from all sides
- There will be increase in temp
- This process is considered to be reversible - the pressure is applied slowly
- such that under adiabatic conditions elastic deformation happens
- This means we are intrested in $\ ( \delta P/\delta T)$~q~




*The above proofs are important
*important stuff

### Hydrostatic Compressive to Tensile
- Case of hydrostatic pressure where pressure was applied from all sides of an adiabatic system

$\left(\dfrac{\delta T}{\delta P}\right)_q = \dfrac{VT\alpha}{\left[C_v + \dfrac{VT\alpha^2}{\beta}\right]}= \dfrac{VT\alpha}{C_p}$


#### Uniaxial
- Now concider uniaxial stress is applied then P will get replaces by $\sigma$ and uniaxial stress being opposite to hydrostatic pressure in unidirection

## Iso-Structure(Homogeneous)

- Hyrdostatic prressure is compressive which is opposite to tensile stress

$\left(\dfrac {\delta T}{\delta\sigma}\right)_q = -\dfrac{VTα}{3C_p}$

## Criteria for thermodynamic equilibrim

- No change in system - system remains stable - process is slow
- Reversible process happen through series of equilibrium stages
- Again taking the combined expression from 1st and 2nd Law:

$dU = TdS - PdV$
$dH = TdS + VdP$
$dA = -PdV - SdT$
$dG = VdP - SdT$

No electrical work is concidered
$dW =0$

- Suppose if we keep variable fixed then we arrive to the criteria of equilibrium
$dU = TdS - PdV$
$(dU)_{S,V} = 0$
$dH = TdS + VdP$
$(dH)_{S,P} = 0$
$dA = -PdV - SdT$
$(dA)_{V,T} = 0$
$(dG)_{T,P} = 0$ as $dG = VdP - SdT$

- Rule of Mixtures does not work in phase diagram
- (dG)~T,P~ = 0 - It is relatively easy to maintan constant temperature and pressure
- For irreversible process, (dG)~T,P~<0 - Going from state 1 to state 2 - state 2 is more stable

### Application of G-H equation

To calulate $\Delta H$ and $\Delta S$ from given value of $\Delta G$

$dG = VdP - SdT$
Condition:at constant $T,
dG = VdP$

To calculate $\Delta G$ at any temperature

$dG = RT(\dfrac{dP}{P})$
$dG = RT \text{ ln}\dfrac{P_2}{P_1}$

For Non-Ideal Gas
- We use fugacity
dG = RT(dlnf)
### One component system

- Gibbs Free Energy
Equilibirum
- System
- Reactive - Accomplishes by chemical reation
- Non Reactive

Transformation at constant T

$\Delta G$ = G~A~(II) - G~A~(I) = 0

Change in P results in change in T

- Pressure changes when the system is distrubed
- P~tr~ to P~tr~ + dP
- To maintain equilibrium T~tr~ will change to T~tr~ + dT
- This also causes a change in Gibbs energy
- G~A~(I) changes to G~A~(I) + dG~A~(I)
- G~A~(II) changes to G~A~(II) + dG~A~(II)

$\Delta G$ = G~A~(II) + dG~A~(II) - (G~A~(I) + dG~A~(I))

For equilibirum, $\Delta G$ = 0

dG~A~(I) = dG~A~(II)

$\left(\dfrac{dT}{dP}\right)_{eq} = \dfrac{\Delta V_{tr}}{\Delta S_{tr}}$

$\Delta S_{tr} = \dfrac{\Delta H_{tr}}{T_{tr}}$

$\left(\dfrac{dT}{dP}\right)_{eq} = \dfrac{T_{tr}\times\Delta V_{tr}}{\Delta H_{tr}}$


More stable less perodic

## Physio-chemical process

- Example: Phase transformation,diffusion
- Behvaiour of each component is considered separately
- Note:Re-call Clausius-Claphron equation dealt with single component

### Solid State Phase Transformation Example


### Molar Fraction

$\ X_i = \dfrac{n_i}{n_1+ n_2+ ... + n_i + n_j+...} = \dfrac{ n_i}{\sum n_i}=\dfrac{n_i}{n_\gamma}$

$\underset{i}{\sum}X_i = 1$

$p$


### Partial and Integral Molar Properties

Q' = n~T~Q

- Q' is value of Q for entire solution
- Q is an extensive property
- Q is the integral molar property in the solution
- At constant T and P, Q' is the function of number of component.. Q^'^=f(n~1~,n~2~,n~3~,....n~i~)


### Extention to Gibbs free energy

(dG^'^)


### T vs Cp,H,S

Phase transformation in Metal and Alloys ,Porter and sterling

Variation of H and G with temperature for solid and liquid

BCC to gamma iron the volume decreases
Important

Transforming stable state back to original state will consume energy

## Variation of Equilibrim constant with temperature


Gibs-Helmon Equation


$\Delta G^0 = -RT \text{ ln }K$

In Materials Processing

Requires calculation of composition

Pressure of gases are not generally too high

If pressure is not too high

### Equilibrium involving ideal gases

$X_i = \dfrac{p_i}{P_T}$

At elevated temperature, diffusion among gases happens very fast and chemical equilibrium is achieved very easily

Therefore actual composition of gas mixture inside a high temperature furnace is generally assumed same as that of equilibrium

Example:

H_2(g)+1/2O_2(g)=H2O

### Chemical equilibrium involving pure condense and gas phase

A(liquid) $\rightarrow$ A(gas)

### Oxidation-Reducation quilibrium involving pure metal M and pure Metal oxide

2M(s,l) + O~2~(g) $\rightarrow$ 2MO(s,l)

$\Delta G_8^0 = 2\Delta G_f^0 = -RT ln K_8 = -RT ln {\right[\dfrac{a_{MO}^2}{a_M^2\cdot p_{O_2}} \left]}_{eq}$

Ellingham Diagram

## Phase Stability Diagram

Thermodynamic

## Ellingham Diagram

Graphical representation of $\Delta G°$ vs T

- Appliable to formation of oxides, sulphides, chloride


Basic Features

- delta G_f_ for per mol of O2

> Tutorial - 7
>>Q1.
>>Ans. $\Delta G_f^0 = -RT\text{ ln }K$

- Everything is in pure, standard state

$\Delta G° = \Delta H° - T\Delta S°$

$\Delta H°$ is intercept
$\Delta S°$ is slope

Change in slope

- We can say the lines lying lower in the digram denote a more stable oxide
the '2Fe+O~2~=2FeO'lines lies above '2CO+O~2~=2CO~2~')

^away from Aq. Media - Don't add water to DRM Iron as it is highly active it will spearate out hydrogen and oxygen thus forming iron oxide and hydrogen, it being an exothermic reaction heat will be produce and hydrogen will catch fire

CO/CO~2~ Ratio
H~2~/H~2~O Ratio

Production of Hydrogen
- Electrolysis
- Storage

#### Determination of Partial Pressure

- If $\Delta G°$ for reaction is known
- Drop a line at a temperature - connect the point to (0,0)
Hall Heroult processs
- metallic solution
- non metallic solution
- Produced aluminium (earlier concidered to be precious metal)
- Oxide Solution
- Sulphide Solution
- Halide solution
- Molten Al~2~O~3~ + Na~3~AlF~4~ - Dissolve
- from past 130 years this process is used to produce aluminia

#### Purity

- No material is pure
- trace quantity-parts per billion
- It may effect the properties significantly
- Example Hydrogen in steel may cause issues
Why pure metal is not stable:

Mixing -


Ideal Binary Solution
- Graph is symmetric for $\Delta G_m$, $\Delta H_m$, $\Delta S_m$
- $\Delta S_m$ is positive
- $\Delta G_m$ is negative
- X~A~ = X~B~ = 0.5
- There is no variation in $\Delta H_m$ because for an ideal solution $\Delta H_m = 0$

## Regular Solution

$\bar S^{M,id}_$

### Chemical Potential
- Gibbs defination mu is termed as "chemical potential"
- Physio-chemical equilibrium chemical potential must be uniform and uniformity of Temperature and pressure

## Multicomponent solution

Situation I
- Binary Solution of A-B, interaction type $\rightarrow$ A-A, B-B, A-B
- Activity coefficient of Bin Binary A-B is $\gamma_B$

Situation II
- Consider Ternary soltuion A-B-C, interaction type $\rightarrow$
A-B, A-C, B-C, A-A, B-B, C-C
- Activity coefficient of Bin Binary A-B is $\gamma_B'$ - due to addition of C

CaSi - Reduction - Deals with O~2~

# Kinetics of Metallurgical Process

- Kinetics deals with the study of chemical and other physio-chemical process/reaction rates
- Thermodynamics dea;s with feasibility of process via quantitative prediction
- At certain T and P
- Knowledge of rates of reaction is required to understand and control the process
- Metallurgical and materials processing involves more than one phase i.e h

Steps in heterogeneous Reaction
- Transport of the reactants from bulk phase to reaction sit(interface) -> Diffusion Step
- The Phase boundary raction -> Chemical reaction step
- Transport of reaction products from the interdace into the bulk -> Diffusion step

## Kinetics of Heterogeneous reaction

- Rate of chemical reaction depends on
- Concentration
- Temperature
- Pressure

Chemical reaction
- Component specific
- rate to concentration, temperature and pressure

$A\rightarrow B$

| Time | A | B |
| -------- | -------- | -------- |
| Initially | C~0~ | 0 |
| T = t | C | x |

1st Order reaction
$k = \dfrac {2.303}t \text{ log }\dfrac{C_0}{C_0-x}$

2nd Order reaction

A+B$\rightarrow$x+y
$\dfrac{-dC_a}{dt} = kC_aC_b$
$k = \dfrac{1}{t} \dfrac x{C(C-x)}$

Generalization

$\dfrac{d\text{ ln }k_e}{dt} = \dfrac{\Delta H°}{RT^2}$

$\dfrac{d\text{ ln }k_f}{dt} - \dfrac{d\text{ ln }k_b}{dt} = \dfrac{\Delta H°}{RT^2}$

k = A exp$\left[\dfrac{-E}{RT}\right]$

$\text{ln }k = \text{ln }A + \dfrac{-E}{RT}$

## Theories of Reactions Kinetics

- Collision Theory
- Absolute Reaction rate Theory

$r = Z_{AB} \text{ exp }\left(\dfrac{-E}{RT}\right)$

$Z_{AB} = n_A n_B d^2_{AB} \left[8\pi kT \dfrac{M_A+M_B}{M_AM_B}\right]^{1/2}$

n~A~ -
M~A~ - Molecular Mass of A
d~AB~ - Average diameter of A and B

### Thermodynamics of Electrochemical Cells

- Electrochemical cell involve coupling of chemical reaction and electrical current
- Electrolytic extraction
- Electrolytic plating
- Corrosion
- Slag-Metal reaction
- Batteries
- Device for thermodynamic measurement
- Oxygen-Free High Conduction Copper

### Cell Equilibirum

Galvanic Cell

In Daniel Cell
V~ext~ < V~cell~
Zn(s) + CuSo~4~ -> Cu(s) + ZnSO~4~