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Gibbs Free Energy & Spontaneity

When one mole...

Question

When one mole of univalent salt AB is dissolved in water at $300 K$ , $3.7 k J$ of heat is absorbed and $△ G$ of the process is $- 8.3 k J$ . What is the value of $△ S$ for the process?

40 J K^{- 1}

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30 J K^{- 1}

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20 J K^{- 1}

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- 15.3 J K^{- 1}

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Solution

The correct option is A $40 J K^{- 1}$
Given :
$T = 300 K, △ H = 3.7 k J △ G = - 8.3 k J$
To calculate $△ S = ?$
We know,
$△ G = △ H - T △ S$
putting values
$\Rightarrow - 8.3 = 3.7 - 300 \times △ S$
$\Rightarrow △ S = 0.04 k J K^{- 1}$
$= 40 J K^{- 1}$

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Gibbs Free Energy & Spontaneity

Standard XIII Chemistry

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When one mole of univalent salt AB is dissolved in water at 300 K, 3.7 kJ of heat is absorbed and △G of the process is −8.3 kJ. What is the value of △S for the process?

The correct option is A 40 JK−1Given : T=300 K, △H=3.7 kJ△G=−8.3kJ To calculate △S=? We know, △G=△H−T△S putting values ⇒−8.3=3.7−300×△S ⇒△S=0.04 kJK−1 =40 JK−1

When one mole of univalent salt AB is dissolved in water at $300 K$ , $3.7 k J$ of heat is absorbed and $△ G$ of the process is $- 8.3 k J$ . What is the value of $△ S$ for the process?

The correct option is A $40 J K^{- 1}$
Given :
$T = 300 K, △ H = 3.7 k J △ G = - 8.3 k J$
To calculate $△ S = ?$
We know,
$△ G = △ H - T △ S$
putting values
$\Rightarrow - 8.3 = 3.7 - 300 \times △ S$
$\Rightarrow △ S = 0.04 k J K^{- 1}$
$= 40 J K^{- 1}$