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For the gaseous reaction involving the complete combustion of isobutane

(a) $\Delta H = \Delta E$
(b) $\Delta H > \Delta E$
(c) $\Delta H = \Delta E = 0$
(d) $\Delta H < \Delta E$
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  • When, $\Delta n_g = +ve, \Delta H > \Delta E$
  • When, $\Delta n_g = -ve, \Delta H < \Delta E$
Answer: $\Delta H > \Delta E$
The gas phase combustion of isobutane ($C_4H_{10}$) represented as
$C_4H_10 (g) + \frac{13}{2} O_2 (g)\longrightarrow 4CO_2 (g) + 5H_2O(g)$
Since, $\Delta n_g = n_p -n_r$
$\Delta n_g = 9 - (\frac{13}{2} + 1) = 9 - \frac{15}{2} = + \frac{3}{2} $
Since, $\Delta n_g = +ve, \Delta H > \Delta E$
answered Feb 24, 2014 by mosymeow_1

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