Heat of combustion of two isomers $x$ and $y$ are $17 \ kJ/mol$ and $12 \ kJ/mol$ respectively. From this information,it may be concluded that:

Given the thermochemical reactions: $C(\text{graphite}) + \frac{1}{2}O_2 \rightarrow CO; \Delta H = -110.5 \, kJ$ and $CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H = -283.2 \, kJ$. Calculate the heat of reaction for $C(\text{graphite}) + O_2 \rightarrow CO_2$ in $kJ$.

Enthalpy of formation of methane is $-75 \ kJ / mol$. What is the enthalpy change for formation of $24 \ g$ of methane (in $kJ$)?

At $25^{\circ}C$,the heats of combustion for $CH_{4(g)}$,$C_{(s)}$,and $H_{2(g)}$ are $-212.4 \, kcal$,$-94.0 \, kcal$,and $-68.4 \, kcal$ respectively. The heat of formation for $CH_{4(g)}$ in $kcal$ is:

If the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:

The reaction of methanol $(\Delta H_f^o = -238.7 \ kJ \ mol^{-1})$ with $2$-methylpropene produces methyl tert-butyl ether $(\Delta H_f^o = -313.6 \ kJ \ mol^{-1})$. Given the reaction: $(CH_3)_2C = CH_2 + CH_3OH \rightarrow (CH_3)_3C - OCH_3; \Delta H^o = -57.8 \ kJ \ mol^{-1}$,calculate the $\Delta H_f^o$ for $2$-methylpropene.