Two reactions are given below:
$2 Fe_{(s)} + \frac{3}{2} O_{2_{(g)}} \rightarrow Fe_2 O_{3_{(s)}}, \Delta H^{o} = -822 \ kJ/mol$
$C_{(s)} + \frac{1}{2} O_{2_{(g)}} \rightarrow CO_{(g)}, \Delta H^{o} = -110 \ kJ/mol$
Then,the enthalpy change for the following reaction is:
$3 C_{(s)} + Fe_2 O_{3_{(s)}} \rightarrow 2 Fe_{(s)} + 3 CO_{(g)}$

Enthalpy of a compound is equal to its

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 $(N-H) = 389 \ kJ/mol$?

For the reaction $OF_{2(g)} \to O_{(g)} + 2F_{(g)}$,$\Delta_{rxn}H$ is $368 \ kJ$. What is the average $O-F$ bond energy in $kJ/mol$?

The enthalpy change $(\Delta H)$ for the neutralisation of $1 \ M \ HCl$ by caustic potash in dilute solution at $298 \ K$ is ..... $kJ$.

Consider the following reactions:
$(i)$ $H_{(aq)}^{+} + OH^{-}_{(aq)} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_1 \ kJ \ mol^{-1}$
$(ii)$ $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$
$(iii)$ $CO_{2_{(g)}} + H_{2_{(g)}} \longrightarrow CO_{(g)} + H_2O_{(l)}$,$\Delta H = -X_3 \ kJ \ mol^{-1}$
$(iv)$ $C_2H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + H_2O_{(l)}$,$\Delta H = -X_4 \ kJ \ mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is