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If a compound has reasonance (Delta H is...

If a compound has reasonance (`Delta H` is usually negative for formation reactions) then.

A

The magnitude of experimental value of its heat of formation decreases.

B

The magnitude of experimental value of its heat of formation increases.

C

The stability of the compound decreases

D

The stability of the compound remains unaffected

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The correct Answer is:
B
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In which of the following changes, Delta H is always negative :

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 ?

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 which of the process, Delta S is negative?

Assertion (A): The enthalpy of formation of H_(2)O(l) is greater than that of H_(2)O(g) . Reason (R ) : Enthalpy change is negative for the condensation reaction H_(2)O(g) rarr H_(2)O(l) .

Assertion : DeltaH for an exothermic reaction is negative nad for an endothermic reaction is positive. Reason : Enthalpy is an extensive property.

For a chemical reaction, DeltaH is negative and DeltaS is negative and DeltaS is negative . This reaction is

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

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