Given: $S + O_2 \rightarrow SO_2 : \Delta H_1 = -298.2 \ kJ$,$SO_2 + 1/2 \ O_2 \rightarrow SO_3 : \Delta H_2 = -98.7 \ kJ$,$SO_3 + H_2O \rightarrow H_2SO_4 : \Delta H_3 = -130.2 \ kJ$,and $H_2 + 1/2 \ O_2 \rightarrow H_2O : \Delta H_4 = -287.3 \ kJ$. Calculate the heat of formation of $H_2SO_4$ in $kJ$.

Calculate $\Delta H \ (kJ/mol)$ for the reaction:
$2FeO_{(s)} + \frac{1}{2} O_{2_{(g)}} \to Fe_2O_{3_{(s)}}$
Given $\Delta H$ values:
$(i)$ $Fe_2O_{3_{(s)}} + 3C_{(graphite)} \to 2Fe_{(s)} + 3CO_{(g)}$ : $492 \ kJ/mol$
$(ii)$ $FeO_{(s)} + C_{(graphite)} \to Fe_{(s)} + CO_{(g)}$ : $156 \ kJ/mol$
$(iii)$ $C_{(graphite)} + O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-393 \ kJ/mol$
$(iv)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-283 \ kJ/mol$

Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?

What is the enthalpy of formation of $NH_3$ if the bond enthalpies are $(N \equiv N) = 941 \ kJ/mol$,$(H-H) = 436 \ kJ/mol$,and $(N-H) = 389 \ kJ/mol$?

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: $B.E.(N-N) = 159\ kJ\ mol^{-1}$,$B.E.(H-H) = 436\ kJ\ mol^{-1}$,$B.E.(N \equiv N) = 941\ kJ\ mol^{-1}$,$B.E.(N-H) = 398\ kJ\ mol^{-1}$.

For the reaction $H_{2(g)} + I_{2(g)} \rightarrow 2HI_{(g)}; \Delta H = 12.40 \, Kcal$,the heat of formation of $HI$ is ....... $Kcal$.