Which of the following relations is correct?

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(D) $1$. The standard enthalpy of formation $(\Delta H_f^o)$ of a compound is equal to the standard enthalpy of combustion $(\Delta H_c^o)$ of its con

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All of the above

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(D) $1$. The standard enthalpy of formation $(\Delta H_f^o)$ of a compound is equal to the standard enthalpy of combustion $(\Delta H_c^o)$ of its constituent elements when the reaction produces one mole of the compound.$2$. For option $A$: The combustion of $H_2(g)$ is $H_2(g) + \frac{1}{2}O_2(g) ightarrow H_2O(l)$,which is the formation reaction of $H_2O(l)$. Thus,$\Delta H_f^o(H_2O, l) = \Delta H_c^o(H_2, g)$.$3$. For option $B$: The combustion of $C(	ext{graphite})$ is $C(	ext{graphite}) + O_2(g) ightarrow CO_2(g)$,which is the formation reaction of $CO_2(g)$. Thus,$\Delta H_f^o(CO_2, g) = \Delta H_c^o(C, 	ext{graphite})$.$4$. For option $C$: The combustion of $CO(g)$ is $CO(g) + \frac{1}{2}O_2(g) ightarrow CO_2(g)$. Using Hess's Law,$\Delta H_c^o(CO, g) = \Delta H_f^o(CO_2, g) - \Delta H_f^o(CO, g)$. Since $\Delta H_f^o(CO_2, g) = \Delta H_c^o(C, 	ext{graphite})$,we get $\Delta H_c^o(CO, g) = \Delta H_c^o(C, 	ext{graphite}) - \Delta H_f^o(CO, g)$.$5$. Therefore,all the given relations are correct.

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