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Question

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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) $-55{ JK^{ -1 } }{ mol }^{ -1 }$

Answer

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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 $-50{ JK^{ -1 } }{ mol }^{ -1 }$.