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(I) Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero? 

(II) Derive the relationship between standard free energy (∆G°) and equilibrium constant (keq)

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(I) A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.

(II)

1. In a reversible process, system is at all times in perfect equilibrium with its surroundings. 

2. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up. 

3. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible. 

4. It is possible only if at equilibrium, the free energy of a systepi is minimum. 

5. Lets consider a general equilibrium reaction, A + B ⇄ C + D 

The free energy change of the above reaction in any state (∆G) is related to the standard free energy change of the reaction (∆G°) according to the following equation.

∆G = ∆G° +RTIn Q ………………. (1)

where Q is reaction quotient and is defined as the ratio of concentrajion of the products to the concentration of the reactants under nonequilibrium condition.

6. When equilibrium is attained, there is no further free energy change i.e. ∆G = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes, ∆G° =-RTln Keq ……(2) 

This equation is known as Van’t Hoff equation. 

∆G° = -2.303 RTlogKeq ……… (3) 

We also know that, 

∆G° = ∆H° -T∆S° 

= – RT In K …….. (4)

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