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…… = Delta H - T Delta S...

…… `= Delta H - T Delta S`

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The correct Answer is:
`Delta G`
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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. For the conversion of Ca(s) to Ca(s) which of the following represent the Delta G vs. T ?

A porcess is nonspontaneous at evey temperature if (i) Delta H gt 0, Delta S = 0 (ii) Delta H lt 0, Delta S gt 0 (iii) Delta H gt 0, Delta S lt 0 (iv) Delta H = 0, Delta S lt 0

For reaction 2Cl_((s)) to Cl_(2(s)), the signs of Delta H "and " Delta S respectively are _______ .

For a given reaction, Delta H = 35.5 kJ mol^(-1) and Delta S = 83.6 JK^(-1) mol(-1) . The reaction is spontaneous at : (Assume that Delta H and Delta S do not vary with temperature)