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X gm of ethanal was subjected to combustion in a bomb calorimeter. The heat produced is Y joules. Then

(a) $\Delta E_{(combustion)} = -XJ$
(b) $\Delta E_{(combustion)} = - YJ$
(c) $\Delta E_{(combustion)} = - \frac{44Y}{X} J mol^{-1}$
(d) $\Delta H_{(combustion)} = - \frac{44Y}{X} J mol^{-1}$
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Answer: $\Delta E_{(combustion)} = - \frac{44Y}{X} J mol^{-1}$
$CH_3CHO + \frac{5}{2} O_2 + \longrightarrow 2CO_2 + 2H_2O$
Heat of combustion when one burns to give desirable products is
X gm ethanal give heat = Y joules
$\therefore$ 44 gm ethanal gives heat = $\frac{44Y}{X} J mol^{-1}$
As reaction is exothermic,
$\therefore \Delta E_{(combustion)} = - \frac{44Y}{X} J mol^{-1}$
answered Feb 23, 2014 by mosymeow_1

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