The heats of formation of CO_{2(g)},H_2O_{(l) | vedclass

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(A) The combustion reaction for methane is: $CH_{4(g)} + 2O_{2(g)} 	o CO_{2(g)} + 2H_2O_{(l)}$.The heat of combustion $(\Delta H_c)$ is calculated us

Vedclass · Accepted Answer

$-212.9$

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(A) The combustion reaction for methane is: $CH_{4(g)} + 2O_{2(g)} 	o CO_{2(g)} + 2H_2O_{(l)}$.The heat of combustion $(\Delta H_c)$ is calculated using the formula: $\Delta H_c = \sum \Delta H_f(	ext{products}) - \sum \Delta H_f(	ext{reactants})$.$\Delta H_c = [\Delta H_f(CO_{2(g)}) + 2 	imes \Delta H_f(H_2O_{(l)})] - [\Delta H_f(CH_{4(g)}) + 2 	imes \Delta H_f(O_{2(g)})]$.Given $\Delta H_f(O_{2(g)}) = 0 \ kcal/mol$.$\Delta H_c = [-94.0 + 2 	imes (-68.4)] - [-17.9 + 2 	imes 0]$.$\Delta H_c = [-94.0 - 136.8] - [-17.9]$.$\Delta H_c = -230.8 + 17.9 = -212.9 \ kcal$.

The heats of formation of $CO_{2(g)}$,$H_2O_{(l)}$,and $CH_{4(g)}$ are $-94.0$,$-68.4$,and $-17.9 \ kcal$ respectively. The heat of combustion of methane is.....$kcal$.

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