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(C) The heat of formation $(\Delta H_f)$ is defined as the enthalpy change when $1 \ mole$ of a compound is formed from its constituent elements in th

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$-65$

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(C) The heat of formation $(\Delta H_f)$ is defined as the enthalpy change when $1 \ mole$ of a compound is formed from its constituent elements in their standard states.The target reaction is: $2C_{(s)} + 3H_{2(g)} + \frac{1}{2}O_{2(g)} 	o C_2H_5OH_{(l)}, \Delta H_f = ?$Given equations:$(i) \ 2C_{(s)} + 2O_{2(g)} 	o 2CO_{2(g)}, \Delta H = 2 	imes (-94) = -188 \ kcal$$(ii) \ 3H_{2(g)} + \frac{3}{2}O_{2(g)} 	o 3H_2O_{(l)}, \Delta H = 3 	imes (-68) = -204 \ kcal$$(iii) \ 2CO_{2(g)} + 3H_2O_{(l)} 	o C_2H_5OH_{(l)} + 3O_{2(g)}, \Delta H = +327 \ kcal$Adding equations $(i)$,$(ii)$,and $(iii)$:$\Delta H_f = (-188) + (-204) + 327 = -392 + 327 = -65 \ kcal/mol$.

Calculate the heat of formation of ethyl alcohol in $kcal/mol$ from the following data:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}, \Delta H = -94 \ kcal$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}, \Delta H = -68 \ kcal$
$C_2H_5OH_{(l)} + 3O_{2(g)} \to 2CO_{2(g)} + 3H_2O_{(l)}, \Delta H = -327 \ kcal$

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Calculate the heat of formation of ethyl alcohol in $kcal/mol$ from the following data:$C_{(s)} + O_{2(g)} \to CO_{2(g)}, \Delta H = -94 \ kcal$$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}, \Delta H = -68 \ kcal$$C_2H_5OH_{(l)} + 3O_{2(g)} \to 2CO_{2(g)} + 3H_2O_{(l)}, \Delta H = -327 \ kcal$