0.3 g of ethane undergoes combustion at 27^{ | vedclass

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Question

(B) $1$. Bomb calorimeter measures heat at constant volume,so the heat released is $\Delta U$.$2$. Moles of ethane $(C_2H_6)$ = $\frac{0.3 \ g}{30 \ g

Vedclass · Accepted Answer

$1006$

Vedclass · Answer

(B) $1$. Bomb calorimeter measures heat at constant volume,so the heat released is $\Delta U$.$2$. Moles of ethane $(C_2H_6)$ = $\frac{0.3 \ g}{30 \ g \ mol^{-1}} = 0.01 \ mol$.$3$. Heat released for $0.01 \ mol$ = $C 	imes \Delta T = 20 \ kJ \ K^{-1} 	imes 0.5 \ K = 10 \ kJ$.$4$. Heat released for $1 \ mol$ $(\Delta U)$ = $\frac{10 \ kJ}{0.01 \ mol} = -1000 \ kJ \ mol^{-1}$.$5$. Combustion reaction: $C_2H_6(g) + 3.5 O_2(g) ightarrow 2CO_2(g) + 3H_2O(l)$.$6$. Change in gaseous moles $(\Delta n_g)$ = $2 - (1 + 3.5) = -2.5$.$7$. Relation between $\Delta H$ and $\Delta U$: $\Delta H = \Delta U + \Delta n_g RT$.$8$. $\Delta H = -1000 \ kJ \ mol^{-1} + (-2.5 	imes 8.3 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1} 	imes 300 \ K)$.$9$. $\Delta H = -1000 - 6.225 = -1006.225 \ kJ \ mol^{-1}$.$10$. The heat evolved is $1006 \ kJ \ mol^{-1}$ (nearest integer).

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