The total enthalpies of reactants and products are $H_R$ and $H_P$ respectively. For an exothermic reaction,which of the following is true?

At $1 \ bar$ and $298 \ K$,the standard molar enthalpy of formation of which substance is zero?

$C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$; $\Delta H = -3267.7 \ kJ \ mol^{-1}$. Given that the standard enthalpies of formation of $CO_{2(g)}$ and $H_2O_{(g)}$ are $-393.5 \ kJ \ mol^{-1}$ and $-285.85 \ kJ \ mol^{-1}$ respectively,calculate the standard enthalpy of formation of benzene $(C_6H_{6(l)})$.

Enthalpy of sublimation of a substance is equal to

Determine the enthalpy of formation for $H_2O_2(\ell)$,using the listed enthalpies of reaction:
$N_2H_{4(\ell)} + 2H_2O_{2(\ell)} \to N_{2(g)} + 4H_2O_{(\ell)}; \Delta _r H_1^o = -818 \, kJ/mol$
$N_2H_{4(\ell)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(\ell)}; \Delta _r H_2^o = -622 \, kJ/mol$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(\ell)}; \Delta _r H_3^o = -285 \, kJ/mol$
Calculate the value in $kJ/mol$.

From the following data:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is: