
What does the Ellingham diagram represent ?
A. G with temperature
B. h with temperature
C. G with pressure
D. G-TS with temperature
Answer
541.5k+ views
Hint: In the Ellingham diagram, a graph is drawn which has the Gibbs energy change in the y axis and temperature , plotted in the x axis.
Complete step-by-step answer: In order to answer the question, we need to learn abut Ellingham diagrams. Gibbs free energy for a reaction is given by the equations $\Delta G=\Delta H-T\Delta S$and $\Delta G=-RT\ln K$Negative value of $\Delta G$and positive value of K favour the formation of products.
Ellingham had drawn and observed graphs of formation of sulphides, oxides and halides versus temperature. The plots of $\Delta G$ metal oxide, per mole of oxygen versus temperature are useful in deciding the reducing agent and temperature for reducing a specific metal oxide. All these graphs have positive slopes showing decrease of entropy except for the formation of CO(g) from coke that shows increase in entropy with increase in temperature.
A sudden increase in slope, as in case of Zn or Mg, indicates melting. The temperature at which two graphs intersect give $\Delta G=0$ for the reaction of one oxide with other element The element of the lower graph works as reducing agent for the oxide of the other, for example, Below temperature 1623 K (approx) Mg metal will reduce $A{{l}_{2}}{{O}_{3}}$ (but not economical) and above this temperature it is All that reduces MgO (but not economical).
So, we can determine whether a reaction is feasible or not, with the help of Ellingham diagram. As it consists of $\Delta G$ in the y axis, so it helps In determining the Gibbs energy change of a particular reaction.
So, we get the correct answer as option A.
Note: It is to be noted that without the help of Ellingham diagrams, it will not be possible to determine whether a hypothetical reaction would be possible or not and whether the compound will get reduced.
Complete step-by-step answer: In order to answer the question, we need to learn abut Ellingham diagrams. Gibbs free energy for a reaction is given by the equations $\Delta G=\Delta H-T\Delta S$and $\Delta G=-RT\ln K$Negative value of $\Delta G$and positive value of K favour the formation of products.
Ellingham had drawn and observed graphs of formation of sulphides, oxides and halides versus temperature. The plots of $\Delta G$ metal oxide, per mole of oxygen versus temperature are useful in deciding the reducing agent and temperature for reducing a specific metal oxide. All these graphs have positive slopes showing decrease of entropy except for the formation of CO(g) from coke that shows increase in entropy with increase in temperature.
A sudden increase in slope, as in case of Zn or Mg, indicates melting. The temperature at which two graphs intersect give $\Delta G=0$ for the reaction of one oxide with other element The element of the lower graph works as reducing agent for the oxide of the other, for example, Below temperature 1623 K (approx) Mg metal will reduce $A{{l}_{2}}{{O}_{3}}$ (but not economical) and above this temperature it is All that reduces MgO (but not economical).
So, we can determine whether a reaction is feasible or not, with the help of Ellingham diagram. As it consists of $\Delta G$ in the y axis, so it helps In determining the Gibbs energy change of a particular reaction.
So, we get the correct answer as option A.
Note: It is to be noted that without the help of Ellingham diagrams, it will not be possible to determine whether a hypothetical reaction would be possible or not and whether the compound will get reduced.
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