Given below are two statements.
Statement I : The choice of reducing agents for metals extraction can be made by using Ellingham diagram, a plot of \(\Delta G\) vs temperature.
Statement II : The value of \(\Delta S\) increases from left to right in Ellingham diagram.
In the light of the above statements, choose the most appropriate answer from the options given below :
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In Ellingham diagrams, the slope is \(-\Delta S\). A sudden change in slope indicates a phase transition (melting or boiling) of one of the reactants or products.
To determine the most appropriate answer to the given statements regarding the Ellingham diagram, let's analyze each statement:
Statement I: "The choice of reducing agents for metals extraction can be made by using Ellingham diagram, a plot of \( \Delta G \) vs temperature."
The Ellingham diagram is a graph that plots the Gibbs free energy change \( \Delta G \) against temperature for the formation of oxides of various elements. It helps in determining which reducing agent can reduce a particular metal oxide by observing the relative positions of the lines in the diagram.
If the line for a particular metal is lower than that of a reducing agent, it means that the reducing agent can effectively reduce the metal oxide to the metal.
Since the Ellingham diagram indeed assists in selecting reducing agents based on thermodynamic feasibility, Statement I is true.
Statement II: "The value of \( \Delta S \) increases from left to right in Ellingham diagram."
The slope of the lines in the Ellingham diagram corresponds to the change in entropy (\( \Delta S \)) during the oxidation process. The relationship between \( \Delta G \), \( \Delta H \) (change in enthalpy), and \( \Delta S \) is given by the equation:
\(\Delta G = \Delta H - T\Delta S\).
An increase in temperature may lead to a change in the slope of the plot due to entropy considerations, but this does not imply that \( \Delta S \) itself continually increases as you move from left to right.
The characteristic slopes of the lines are specific to each reaction and do not universally imply an increase in \( \Delta S \) from left to right on the diagram. Hence, Statement II is false.
Based on the above analysis, the correct answer is: Statement I is true but Statement II is false.
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