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Reduction of ZnO with carbon is done at ...

Reduction of ZnO with carbon is done at `1100^(@)C`.
Reason : `Delta G^(@)` is negative at this temperature thus, process is spontaneous.

A

If both assetion and reason are CORRECT, and reason is the CORRECT explanation of the assertion

B

If both assertion and reason are CORRECT, but reason is NOT the CORRECT explanation of the assertion

C

If assertion is CORRECT but reason is INCORRECT

D

If assertion is INCORRECT but reason is CORRECT

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The correct Answer is:
A
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Assertion : Reduction of ZnO with carbon is done at 1100^(@)C . Reason : At this temperature, DeltaG^(@) is negative and the process is spontaneous.

Reduction of Fe_(2)O_(3) with CO is done below 710^@C . Delta_(f) G^(Ө) is negative at this temperature , this process is spontaneous.

If Delta G gt 0, then the nature of the process will be (spontaneous/non-spontaneous)?

If Delta G gt 0 , then the nature of the process will be (spontaneous/non-spontaneous)?

Assertion : If both DeltaH^(@) and DeltaS^(@) are positive then reaction will be spontaneous at high temperature. Reason : All processes with positive entropy change are spontaneous.

When Delta H and TDeltaS both are negative , then for spontaneous process which option 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. For the conversion of Ca(s) to Ca(s) which of the following represent the Delta G vs. T ?

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 ?

P(g)+2Q(g)toPQ_(2)(g),DeltaH=18kJmol^(-1) The entropy change of the above reaction (DeltaS_("system") is 60 JK^(-1)mol^(-1)) . At what temperature the reaction becomes spontaneous?

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