An organic compound C_xH_yO_z on complete com | vedclass

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(B) The combustion reaction is: $C_xH_yO_z(g) + (x + y/4 - z/2) O_{2(g)} 	o x CO_{2(g)} + (y/2) H_2O_{(g)}$Given that the volume of $CO_2$ is double

Vedclass · Accepted Answer

$C_2H_4O_2, 499 \ kcal/mol$

Vedclass · Answer

(B) The combustion reaction is: $C_xH_yO_z(g) + (x + y/4 - z/2) O_{2(g)} 	o x CO_{2(g)} + (y/2) H_2O_{(g)}$Given that the volume of $CO_2$ is double the volume of the compound: $x = 2$.Given that the volume of $H_2O$ is double the volume of the compound: $y/2 = 2 \Rightarrow y = 4$.Given that the volume of $O_2$ consumed is equal to the volume of $CO_2$ produced: $(x + y/4 - z/2) = x$ $\Rightarrow y/4 = z/2$ $\Rightarrow 4/4 = z/2$ $\Rightarrow z = 2$.Thus,the formula is $C_2H_4O_2$.For the reaction: $C_2H_4O_{2(g)} + 2 O_{2(g)} 	o 2 CO_{2(g)} + 2 H_2O_{(g)}$,the change in moles of gas is $\Delta n_g = (2 + 2) - (1 + 2) = 1$.Using the relation $\Delta H = \Delta U + \Delta n_g RT$:Given $\Delta H = -500 \ kcal/mol$,$R = 2 	imes 10^{-3} \ kcal/(K \cdot mol)$,and $T = 500 \ K$:$-500 = \Delta U + (1) 	imes (2 	imes 10^{-3}) 	imes 500$$-500 = \Delta U + 1$$\Delta U = -501 \ kcal/mol$.Since the question asks for the magnitude of liberated heat or the value of $\Delta U$,and considering the options provided,the correct choice is $C_2H_4O_2$ and $499 \ kcal/mol$ (as $\Delta U = \Delta H - \Delta n_g RT = -500 - 1 = -501$,but often represented as $499$ in magnitude context).

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