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Gibbs Free Energy & Spontaneity

Ka for CH3COO...

Question

$K_{a} f o r C H_{3} C O O H a t 25^{o} C i s 1.754 \times 10^{- 5} .$ What is the value of $△ S^{o}$ in $J / m o l K$ for the ionisation of $C H_{3} C O O H$ if $△ H^{o}$ is $1.55 k J / m o l$ ?
log(1.754 ) = 0.244

-96.44

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96.26

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50.56

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17.54

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Solution

The correct option is A -96.44
we know the relation
$(△ G^{o})_{298} = - 2.303 R T l o g K$
$= - 2.303 \times 8.314 \times 298 \times l o g (1.754 \times 10^{- 5})$
$= - 2.303 \times 8.314 \times 298 \times (- 5 + l o g (1.754)$
$= - 2.303 \times 8.314 \times 298 \times (- 5 + 0.244)$
$= - 2.303 \times 8.314 \times 298 \times - 4.756$
$= 27137 J$
$△ G^{o} = △ H^{o} - T △ S^{o}$
$27137.01 = 1550 - 298 △ S^{o}$
$△ S^{0} = - 85.86 J / m o l K$
Ionisation of Acetic acid:

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