Question
Write a note on thermodynamic principle of metallurgy.
✓
Answer
Thermodynamic principle of metallurgy:
The extraction of metals from their oxides can be carried out by using different reducing agents.
For example, consider the reduction of a metal
oxide $M _{ x } O _{ y }$.
$
\frac{2}{y} M _{ x } O _{ y }( s ) \rightarrow \frac{2 x}{y} M ( s )+ O _2( g ) .........(1)
$
The above reduction may be carried out with carbon. In this case, the reducing agent carbon may be oxidised to either $CO$ or $CO _2$.
- $C + O _2 \rightarrow CO _{2( g )}$ ..........(2)
- $2 C + O _2 \rightarrow 2 CO _{( g )}$ .......(3)
If carbon monoxide is used as a reducing agent, it is oxidised to $CO _2$ as follows,
$
2 CO + O _2 \rightarrow 2 CO _{2( g )} .......(4)
$
A suitable reducing agent is selected based on the thermodynamic considerations. We know that for a spontaneous reaction, the change in free energy (AG) should be negative. Therefore, thermodynamically, the reduction of metal oxide [equation (1)] with a given reducing agent [Equation (2), (3) or (4)] can occur if the free energy change for the coupled reaction. [Equations (1) \& (2), (1) \& (3) or (1) \& (4)] is negative. Hence, the reducing agent is selected in such a way that it provides a large negative AG value for the coupled reaction.
Ellingham diagram:
The change in Gibbs free energy $(\Delta G)$ for a reaction is given by the expression.
$
\Delta G =\Delta H - T \Delta S \text { }........ (1)
$
where, $\Delta H$ is the enthalpy change, $T$ the temperature in kelvin and $\Delta S$ the entropy change. For an equilibrium process, $\Delta G^{\circ}$ can be calculated using the equilibrium constant by the following expression $\Delta G^{\circ}=-R T \operatorname{lnKp}$
Harold Ellingham used the above relationship to calculate the $\Delta G^{\circ}$ values at various temperatures for the reduction of metal oxides by treating the reduction as an equilibrium process. He has drawn a plot by considering the temperature in the $x$-axis and the standard free energy change for the formation of metal oxide in $y$-axis. The resultant plot is a straight line with $\Delta S$ as slope and $\Delta H$ as $y$-intercept. The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.Thermodynamic principle of metallurgy:
The extraction of metals from their oxides can be carried out by using different reducing agents.
For example, consider the reduction of a metal
oxide $M _{ x } O _{ y }$.
$
\frac{2}{y} M _{ x } O _{ y }( s ) \rightarrow \frac{2 x}{y} M ( s )+ O _2( g ) .........(1)
$
The above reduction may be carried out with carbon. In this case, the reducing agent carbon may be oxidised to either $CO$ or $CO _2$.
- $C + O _2 \rightarrow CO _{2( g )}$ ..........(2)
- $2 C + O _2 \rightarrow 2 CO _{( g )}$ .......(3)
If carbon monoxide is used as a reducing agent, it is oxidised to $CO _2$ as follows,
$
2 CO + O _2 \rightarrow 2 CO _{2( g )} .......(4)
$
A suitable reducing agent is selected based on the thermodynamic considerations. We know that for a spontaneous reaction, the change in free energy (AG) should be negative. Therefore, thermodynamically, the reduction of metal oxide [equation (1)] with a given reducing agent [Equation (2), (3) or (4)] can occur if the free energy change for the coupled reaction. [Equations (1) \& (2), (1) \& (3) or (1) \& (4)] is negative. Hence, the reducing agent is selected in such a way that it provides a large negative AG value for the coupled reaction.
Ellingham diagram:
The change in Gibbs free energy $(\Delta G)$ for a reaction is given by the expression.
$
\Delta G =\Delta H - T \Delta S \text { }........ (1)
$
where, $\Delta H$ is the enthalpy change, $T$ the temperature in kelvin and $\Delta S$ the entropy change. For an equilibrium process, $\Delta G^{\circ}$ can be calculated using the equilibrium constant by the following expression $\Delta G^{\circ}=-R T \operatorname{lnKp}$
Harold Ellingham used the above relationship to calculate the $\Delta G^{\circ}$ values at various temperatures for the reduction of metal oxides by treating the reduction as an equilibrium process. He has drawn a plot by considering the temperature in the $x$-axis and the standard free energy change for the formation of metal oxide in $y$-axis. The resultant plot is a straight line with $\Delta S$ as slope and $\Delta H$ as $y$-intercept. The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.
The extraction of metals from their oxides can be carried out by using different reducing agents.
For example, consider the reduction of a metal
oxide $M _{ x } O _{ y }$.
$
\frac{2}{y} M _{ x } O _{ y }( s ) \rightarrow \frac{2 x}{y} M ( s )+ O _2( g ) .........(1)
$
The above reduction may be carried out with carbon. In this case, the reducing agent carbon may be oxidised to either $CO$ or $CO _2$.
- $C + O _2 \rightarrow CO _{2( g )}$ ..........(2)
- $2 C + O _2 \rightarrow 2 CO _{( g )}$ .......(3)
If carbon monoxide is used as a reducing agent, it is oxidised to $CO _2$ as follows,
$
2 CO + O _2 \rightarrow 2 CO _{2( g )} .......(4)
$
A suitable reducing agent is selected based on the thermodynamic considerations. We know that for a spontaneous reaction, the change in free energy (AG) should be negative. Therefore, thermodynamically, the reduction of metal oxide [equation (1)] with a given reducing agent [Equation (2), (3) or (4)] can occur if the free energy change for the coupled reaction. [Equations (1) \& (2), (1) \& (3) or (1) \& (4)] is negative. Hence, the reducing agent is selected in such a way that it provides a large negative AG value for the coupled reaction.
Ellingham diagram:
The change in Gibbs free energy $(\Delta G)$ for a reaction is given by the expression.
$
\Delta G =\Delta H - T \Delta S \text { }........ (1)
$
where, $\Delta H$ is the enthalpy change, $T$ the temperature in kelvin and $\Delta S$ the entropy change. For an equilibrium process, $\Delta G^{\circ}$ can be calculated using the equilibrium constant by the following expression $\Delta G^{\circ}=-R T \operatorname{lnKp}$
Harold Ellingham used the above relationship to calculate the $\Delta G^{\circ}$ values at various temperatures for the reduction of metal oxides by treating the reduction as an equilibrium process. He has drawn a plot by considering the temperature in the $x$-axis and the standard free energy change for the formation of metal oxide in $y$-axis. The resultant plot is a straight line with $\Delta S$ as slope and $\Delta H$ as $y$-intercept. The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.Thermodynamic principle of metallurgy:
The extraction of metals from their oxides can be carried out by using different reducing agents.
For example, consider the reduction of a metal
oxide $M _{ x } O _{ y }$.
$
\frac{2}{y} M _{ x } O _{ y }( s ) \rightarrow \frac{2 x}{y} M ( s )+ O _2( g ) .........(1)
$
The above reduction may be carried out with carbon. In this case, the reducing agent carbon may be oxidised to either $CO$ or $CO _2$.
- $C + O _2 \rightarrow CO _{2( g )}$ ..........(2)
- $2 C + O _2 \rightarrow 2 CO _{( g )}$ .......(3)
If carbon monoxide is used as a reducing agent, it is oxidised to $CO _2$ as follows,
$
2 CO + O _2 \rightarrow 2 CO _{2( g )} .......(4)
$
A suitable reducing agent is selected based on the thermodynamic considerations. We know that for a spontaneous reaction, the change in free energy (AG) should be negative. Therefore, thermodynamically, the reduction of metal oxide [equation (1)] with a given reducing agent [Equation (2), (3) or (4)] can occur if the free energy change for the coupled reaction. [Equations (1) \& (2), (1) \& (3) or (1) \& (4)] is negative. Hence, the reducing agent is selected in such a way that it provides a large negative AG value for the coupled reaction.
Ellingham diagram:
The change in Gibbs free energy $(\Delta G)$ for a reaction is given by the expression.
$
\Delta G =\Delta H - T \Delta S \text { }........ (1)
$
where, $\Delta H$ is the enthalpy change, $T$ the temperature in kelvin and $\Delta S$ the entropy change. For an equilibrium process, $\Delta G^{\circ}$ can be calculated using the equilibrium constant by the following expression $\Delta G^{\circ}=-R T \operatorname{lnKp}$
Harold Ellingham used the above relationship to calculate the $\Delta G^{\circ}$ values at various temperatures for the reduction of metal oxides by treating the reduction as an equilibrium process. He has drawn a plot by considering the temperature in the $x$-axis and the standard free energy change for the formation of metal oxide in $y$-axis. The resultant plot is a straight line with $\Delta S$ as slope and $\Delta H$ as $y$-intercept. The graphical representation of variation of the standard Gibbs free energy of reaction for the formation of various metal oxides with temperature is called Ellingham diagram.
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