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At 1 atm pressure, for a reaction $\Delta H = + 30.558$ kJ $mol^{-1}$ and $\Delta S = 0.066$ kJ $mol^{-1}$. The temperature at which free energy is equal to to zero and the nature of the reaction below this temperature are


(a) 483 K, spontaneous
(b) 443 K, non-spontaneous
(c) 443 K, spontaneous
(d) 463 K, non-spontaneous
Can you answer this question?
 
 

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Answer: 463 K, non-spontaneous
$\Delta G = \Delta H - T \Delta S = 0$
$\therefore T = \frac{\Delta H}{\Delta S} = \frac{+30.558 \text{ kJ } mol^{-1}}{0.066 \text{ kJ }K^{-1} mol^{-1}}$ = 463 K
$\Delta G = \Delta H - T \Delta S = (30.558 - T \times 0.066)$ kJ $mol^{-1}$
At T<463 K,
0.066 T < 463 $\times$ 0.066
or, 0.066 T < 30.558
or, 0 < 30.558 - 0.066 T
or, 0 < $\Delta G$
Thus, at T < 463 K, $\Delta G$ > 0 i.e. , process is non-spontaneous.
answered Feb 26, 2014 by mosymeow_1
 

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