Which substance has a standard molar enthalpy of formation equal to zero at $298 \ K$?

If the enthalpies of formation of $Al_2O_3$ and $Cr_2O_3$ are $-1596 \, kJ/mol$ and $-1134 \, kJ/mol$ respectively,then calculate $\Delta H$ for the following reaction in $kJ$:
$2Al + Cr_2O_3 \to Al_2O_3 + 2Cr$

Calculate the $N-N$ bond energy in $N_2H_4$ from the given bond enthalpy data.
$\varepsilon_{N-H} = 393 \ kJ/mol$
$\varepsilon_{H-H} = 436 \ kJ/mol$
$\Delta H_{vap}[N_2H_{4(l)}] = 18 \ kJ/mol$
$N_2H_{4(l)} + H_{2(g)} \to 2NH_{3(g)} : \Delta H = -142 \ kJ/mol$
....... $kJ/mol$

If the bond energies of $H-H$,$Br-Br$ and $H-Br$ are $433$,$192$ and $364 \ kJ \ mol^{-1}$ respectively,the $\Delta H^o$ for the reaction,$H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$ is.....$kJ$

Calculate the standard enthalpy change for the reaction,$C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \rightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$. Given: $\Delta_{f}H^{\circ}(C_2H_5OH) = -280 \ kJ \ mol^{-1}$,$\Delta_{f}H^{\circ}(CO_2) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f}H^{\circ}(H_2O) = -285 \ kJ \ mol^{-1}$.

Combustion of $1$ $mol$ of benzene is expressed as:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of combustion of $2$ $mol$ of benzene is $-x$ $kJ$.
$x = . . . . . . . . . .$
$(1)$ Standard enthalpy of formation of $1$ $mol$ of $C_6H_{6(l)}$ is $48.5$ $kJ \ mol^{-1}$.
$(2)$ Standard enthalpy of formation of $1$ $mol$ of $CO_{2(g)}$ is $-393.5$ $kJ \ mol^{-1}$.
$(3)$ Standard enthalpy of formation of $1$ $mol$ of $H_2O_{(l)}$ is $-286$ $kJ \ mol^{-1}$.