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If $\Delta H^\circ _f$ and $\Delta G ^\circ _f$ of $CH_3OH(g)$ at 298 K is -200.7 kJ/mol and -162 kJ/mol respectively, then absolute entropy of $CH_3OH(g)$ at 298 K is (given absolute entropies of C(s), $H_2(g)$ and $O_2(g)$ at 298 K are 5.7, 130.60 and 205.00 J/mol respectively).

(a) $239.6 JK^{-1}mol^{-1}$
(b) $369.36 JK^{-1}mol^{-1}$
(c) $-369.36 JK^{-1}mol^{-1}$
(d) none of these
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Answer: 239.6 $JK^{-1}mol^{-1}$
$\Delta S^\circ _f[CH_3OH(l)] = \frac{\Delta H^\circ _f - \Delta ^\circ _f}{T} = \frac{(-200.7 + 162.0)kJ/mol}{298 K} = -129.8 JK^{-1}mol^{-1}$
$C(s)+2H_2(g)+\frac{1}{2}O_2(g) \rightarrow CH_3OH(g)$
For this reaction, $\Delta G ^\circ _f = -129.8 JK^{-1}mol^{-1}$
$\Longrightarrow1 mol.[-129.8 JK^{-1}mol^{-1}] = 1 mol. [S^\circ (CH_3OH)]$
$\Longrightarrow1-1mol. [S^\circ(C)]-2mol. [S^\circ (H_2)]-\frac{1}{2} mol. [S^\circ (O_2)]$
$\Longrightarrow1-129.8 JK^{-1} = 1 mol. [S^\circ (CH_3OH)]-[5.7 + 2(130.6)+\frac{1}{2}(205.0)] JK^{-1}$
$\therefore S^\circ (CH_3OH)$ = 239.6 $JK^{-1}mol^{-1}$
answered Feb 28, 2014 by mosymeow_1

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