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For a sponaneous reaction, the free ener...

For a sponaneous reaction, the free energy change must be negative, `Delta G=Delta H-T Delta S, Delta H` is the enthalpy change during the reaction. T is the absolute temperature, and `Delta S` is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide
`M+O_(2) to MO`
Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence `Delta S` is negative. Thus, if the temperature is raised then `T Delta S` becomes more negative,Since, `TDelta S` is substracted in the equation, then `Delta G` becomes less negative. Thus, the free energy change increases with the increase in temperature.
The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals.

For the conversion of Ca(s) to Ca(s) which of the following represent the `Delta G` vs. T ?

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
C
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For a sponaneous reaction, the free energy change must be negative, Delta G=Delta H-T Delta S, Delta H is the enthalpy change during the reaction. T is the absolute temperature, and Delta S is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide M+O_(2) to MO Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence Delta S is negative. Thus, if the temperature is raised then T Delta S becomes more negative,Since, TDelta S is substracted in the equation, then Delta G becomes less negative. Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals. As per the Ellingham diagram of oxides which of the following conclusion is true ?

For a sponaneous reaction, the free energy change must be negative, Delta G=Delta H-T Delta S, Delta H is the enthalpy change during the reaction. T is the absolute temperature, and Delta S is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide M+O_(2) to MO Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence Delta S is negative. Thus, if the temperature is raised then T Delta S becomes more negative,Since, TDelta S is substracted in the equation, then Delta G becomes less negative. Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals. Free energy change of Hg and Mg for the convertion to oxides the slpe of Delta G vsT has been changed above the boiling points of the given metal because :

For a sponaneous reaction, the free energy change must be negative, Delta G=Delta H-T Delta S, Delta H is the enthalpy change during the reaction. T is the absolute temperature, and Delta S is the change in entropy during the reaction. Consider a reaction such as the formation of an oxide M+O_(2) to MO Dioxygen is used up in the course of this reaction. Gases have a more random structure (less ordered) than liquid or solids. Consequently gases have a higher entropy than liquids and solids. In this reaction S (entropy or randomness) decreases, hence Delta S is negative. Thus, if the temperature is raised then T Delta S becomes more negative,Since, TDelta S is substracted in the equation, then Delta G becomes less negative. Thus, the free energy change increases with the increase in temperature. The free energy changes that occur when one mole of common reactant (in this case dioxygen) is used may be plotted graphically aginst temperature for a number of reactions of metals to their oxides. The following plot is called an Ellingham diagram for metal oxide. Understanding of Ellingham diagram is extremely important for the efficient extraction of metals. Which of the following elements can be prepared by heating the oxide above 400^(@)C ?

The free energy change, Delta G = 0, when

Which of the following reactions is Delta H less than Delta E ?

In which reaction(s), DeltaS in negative?

Which of the following reactions represents Delta H (hydration) :-

For an exothermic reaction to be spontaneous ( Delta S = negative)

In which of the following changes, Delta H is always negative :

For which reaction, (DeltaH > DeltaU) ? ( Delta H = change in enthalpy, Delta U = change in internal energy)