At temperatures above 85 K, decarboxylation of acetic acid becomes a spontaneous process under
standard state conditions. The standard entropy change ( in J / K - mol ) of the reaction
$C H_{3} C O O H (a q) \to C H_{4} (g) + C O_{2} (g)$ is :
Given : $Δ H_{f}^{0} [C H_{3} C O O H (a q)] = - 484$ kJ / mole
$Δ H_{f}^{0} [C O_{2} (g)] = - 392$ kJ / mole
$Δ H_{f}^{0} [C H_{4} (g)] = - 75$ kJ / mole

200 J / K mole

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300J / K mole

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400 J / K mole

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100J / K mole

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Solution

The correct option is A 200 J / K mole
$Δ_{r} H^{\circ}$ = $Δ H_{f}^{0} [C O_{2} (g)]$ + $Δ H_{f}^{0} [C H_{4} (g)]$ - $Δ H_{f}^{0} [C H_{3} C O O H (a q)]$
$Δ_{r} H^{\circ}$ = -392-75 + 484 = 17 kJ/mol
It is given that above 85 K reaction becomes spontaneous which implies at 85K $G$ = 0
Therefore at 85 K, $Δ_{r} H^{\circ}$ = $T Δ_{r} S^{\circ}$
$Δ_{r} S^{\circ}$ = (17X1000)/85 J/K mol
$Δ_{r} S^{\circ}$ = 200 J/K mol (Correct Option:A)

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