`Delta_(f) G^(Ө)` vs `T` plot in the Ellingham diagram slopes downward for the reaction.
`Delta_(f) G^(Ө)` vs `T` plot in the Ellingham diagram slopes downward for the reaction.
A
`C+(1)/(2)O_(2)toCO`
B
`CO+(1)/(2)O_(2)toCO_(2)`
C
`2Ag+(1)/(2)O_(2)toAg_(2)O`
D
`Mg+(1)/(2)O_(2)toMgO`
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A
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DeltaG^(Ө) vs T plot in Ellingham diagram slopes downward for the reaction .
DeltaG^(@) vsT plot in the Ellingham diagram slopes down for the reaction.
The choice of reducing agent for metal oxide is decided by thermodynamic principles of metallurgy. Delta G must be negative. Ellingham diagram is a plot of Delta_(f)G^(0) vs T for the formation of oxides of various elements. All the plots slope upwards when temperature increases. Each plot is a straight line except when some change in phase takes place. A metal will reduce the oxide of other metals which lie above it in this diagram. At a temperature below 1623 K. Al cannot reduce
Ellingham diagrams are plots of DeltaG_(1)^(@) Vs T for the formation of
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. 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 chemical reaction R rarr P the variation in the concentration (R ) vs .Time (t) plot given as (i) predict the order of the reaction (ii) what is he slope of the curve?
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 ?
If Delta G^(@)gt 0 for a reaction then :
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