A reaction is at equilibrium at $153^{o} C$ and the enthalpy change for the reaction is $42.6 k J m o l^{- 1} .$ What will be the value of $△ S$ ?

120

J K^{- 1} m o l^{- 1}

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426.2

J K^{- 1} m o l^{- 1}

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373.1

J K^{- 1} m o l^{- 1}

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100

J K^{- 1} m o l^{- 1}

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Solution

The correct option is D 100 $J K^{- 1} m o l^{- 1}$
We know that ,
$△ G = △ H - T △ S$
At equilibrium, $△ G = 0$
So, $△ H = T △ S$
$△ S = \frac{△ H}{T}$
$T = 153 + 273 = 426 K$
putting the values we get :
$△ S = \frac{42600 J m o l^{- 1}}{426 K} = 100 J K^{- 1} m o l^{- 1}$

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