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(C) The combustion reaction for methane is:$CH_{4_{(g)}} + 2O_{2_{(g)}} 	o CO_{2_{(g)}} + 2H_2O_{(l)}$; $\Delta H = -210.8 \, kcal$The enthalpy of re

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$-20.0$

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(C) The combustion reaction for methane is:$CH_{4_{(g)}} + 2O_{2_{(g)}} 	o CO_{2_{(g)}} + 2H_2O_{(l)}$; $\Delta H = -210.8 \, kcal$The enthalpy of reaction is given by:$\Delta H = [\Delta H_f(CO_{2_{(g)}}) + 2 	imes \Delta H_f(H_2O_{(l)})] - [\Delta H_f(CH_{4_{(g)}}) + 2 	imes \Delta H_f(O_{2_{(g)}})]$Substituting the given values (where $\Delta H_f(O_{2_{(g)}}) = 0$):$-210.8 = [-94.2 + 2 	imes (-68.3)] - \Delta H_f(CH_{4_{(g)}})$$-210.8 = -94.2 - 136.6 - \Delta H_f(CH_{4_{(g)}})$$-210.8 = -230.8 - \Delta H_f(CH_{4_{(g)}})$$\Delta H_f(CH_{4_{(g)}}) = -230.8 + 210.8 = -20.0 \, kcal$

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Given the following thermochemical equations:$C_{(s)} + O_{2_{(g)}} \to CO_{2_{(g)}} + 94.2 \, kcal$$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \to H_2O_{(l)} + 68.3 \, kcal$$CH_{4_{(g)}} + 2O_{2_{(g)}} \to CO_{2_{(g)}} + 2H_2O_{(l)} + 210.8 \, kcal$Calculate the heat of formation of methane in $kcal$.