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# The Enthalpy change for a reaction at equilibrium at $-20.5{ KJ }{ mol }^{ -1 }$. Then Entropy change for this equilibrium at 410 K is:A) $+50{ JK^{ -1 } }{ mol }^{ -1 }$B) $+55{ JK^{ -1 } }{ mol }^{ -1 }$C) $+75{ JK^{ -1 } }{ mol }^{ -1 }$D) $-50{ JK^{ -1 } }{ mol }^{ -1 }$E) $-55{ JK^{ -1 } }{ mol }^{ -1 }$  Verified
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Hint: At equilibrium change in Gibbs free energy is zero. By equating Gibbs free energy to zero we can find Enthalpy change for the reaction.

Formula Used: $\triangle G\quad =\quad \triangle H\quad -\quad T\triangle S$

The equilibrium rate of formation of products is equal to the rate of formation of reactants. At equilibrium spontaneity of reaction will be zero.

$\triangle G\quad =\quad \triangle H\quad -\quad T\triangle S$

Change in Gibbs Free energy is zero for equilibrium reaction. Equate Gibbs free change to
$0{ J }{ mol }^{ -1 }$.
Change in Enthalpy for equilibrium reaction is $-20.5{ KJ }{ mol }^{ -1 }$.
The temperature of equilibrium will be 410 K.
Substituting all these values we are left with one unknown which is the change in Entropy.

0 = $-20.5{ KJ }{ mol }^{ -1 }-{{410K}{\triangle\; S}}$

We know that $1{ KJ }{ mol }^{ -1 }$ = $1000{ J }{ mol }^{ -1 }$

$-20.5{ KJ }{ mol }^{ -1 }$= $-20.5\times1000{ J }{ mol }^{ -1 }$= $-20500{ J }{ mol }^{ -1 }$

0 = $-20500{ J }{ mol }^{ -1 }-{{410K}{\triangle\; S}}$

${{410K}{\triangle\; S}}$ = $-20500{ J }{ mol }^{ -1 }$

△S = ${ \dfrac { -20500 }{ 410 } J }{ mol }^{ -1 }$

△S =$-50{ JK^{ -1 } }{ mol }^{ -1 }$

Entropy change for this equilibrium at $410K$ is $-50{ JK^{ -1 } }{ mol }^{ -1 }$.

Therefore, option D is correct.

Note: At equilibrium change in Gibbs free is zero because the rate of formation of products and reactants will be the same so the reaction will not be spontaneous. We need to be careful with the sign of Enthalpy change and while solving the equation we need to reverse the sign when we send terms from R.H.S to L.H.S of an equation.