In the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}; \Delta H = 80 \ kJ$,$\Delta H$ is known as

Calculate the enthalpy change in $kJ$ for the reaction: $2C_{(graphite)} + 2H_{2(g)} \to C_2H_{4(g)}$
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
$C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H = -1410.9 \ kJ$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(l)} \quad \Delta H = -285.8 \ kJ$

If $\Delta H_f (H_2O) = X$,then the heat of neutralization of $CH_3COOH$ and $NaOH$ will be:

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

If the standard heat of the reaction $Fe_2O_{3(s)} + 3CO_{(g)} = 2Fe_{(s)} + 3CO_{2(g)}$ is $-6.6 \, kcal$,then $\Delta H_f^o$ for $Fe_2O_{3(s)}$ is $...... \, kcal/mol$. [Given: $\Delta H_f^o$ of $CO_{(g)} = -26.4 \, kcal$ and $\Delta H_f^o$ of $CO_{2(g)} = -94 \, kcal$]

Based on the bond enthalpy $(B.E.)$ values given,the standard enthalpy of formation $(\Delta_fH^o)$ of $N_2H_{4(g)}$ is ...... $kJ\ mol^{-1}$.
Given: $B.E.(N-N) = 159\ kJ\ mol^{-1}$,$B.E.(H-H) = 436\ kJ\ mol^{-1}$,$B.E.(N \equiv N) = 941\ kJ\ mol^{-1}$,$B.E.(N-H) = 398\ kJ\ mol^{-1}$.