K a for CH 3 COOH at 25- C is 1-754 - 10-5- What are S 0 in J - mol - K for the ionisation of CH 3 COOH if H 0 is -1-55 kJ - mol -log 1-754-0-244A- 96-26B- 50-56C- 17-54D- -96-44

The correct option is D 96.44we know the relation ( G^o)_298 = 2.303RT log K = 2.303 × 8.314 × 298 × log(1.754 × 10^ 5) = 2.303 × 8.314 × 298 ×( 5+ log(1.7 ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

K a for CH 3 ...

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 are $△ S^{o}$ 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
$(△ G^{o})_{298} = - 2.303 R T l o g K$
$= - 2.303 \times 8.314 \times 298 \times l o g (1.754 \times 10^{- 5})$
$= - 2.303 \times 8.314 \times 298 \times (- 5 + l o g (1.754)$
$= - 2.303 \times 8.314 \times 298 \times (- 5 + 0.244)$
$= - 2.303 \times 8.314 \times 298 \times - 4.756$
$= 27137 J .$
$△ G^{o} = △ H^{o} - T △ S^{o}$
$27137.01 = - 1550 - 298 △ S^{o}$
$△ S^{0} = - 96.265 J / m o l . K$

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Similar questions

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}

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for the ionisation of

C H_{3} C O O H

△ H^{o}

1.55 k J / m o l

?
log(1.754 ) = 0.244

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Q. What would be the ionisation constant

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and

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respectively. Calculate the equilibrium constant for the reaction:

{C H}_{3} C O O H + N a C N ⇌ {C H}_{3} C O O N a + H C N

Q. Specific conductance of 0.1 M

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Gibbs Free Energy & Spontaneity

Standard XII Chemistry

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Ka for CH3COOH at 25oC is 1.754×10−5. What are △So in J/mol.K for the ionisation of CH3COOH if △Ho is −1.55 kJ/mol? log(1.754 ) = 0.244

$K_{a} f o r C H_{3} C O O H a t 25^{o} C i s 1.754 \times 10^{- 5} .$ What are $△ S^{o}$ 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