"The resultant heat change in a reaction is the same whether it takes place in one or several stages." This statement is called

Diborane is formed from the elements as shown in equation $(i)$:
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)} \dots (i)$
Given that:
$H_2O_{(l)} \longrightarrow H_2O_{(g)}, \quad \Delta H_1^{\circ} = 44 \, kJ$
$2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$B_2H_{6(g)} + 3 O_{2(g)} \longrightarrow B_2O_{3(s)} + 3 H_2O_{(g)}, \quad \Delta H_3^{\circ} = -2035 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}, \quad \Delta H_4^{\circ} = -286 \, kJ$
The $\Delta H^{\circ}$ for the reaction $(i)$ is $..... \, kJ$.

For the reaction $2H_{2(g)} + O_{2(g)} \to 2H_2O_{(l)}$,the enthalpy change is $\Delta H = -571 \ kJ$. If the $H-H$ bond energy is $435 \ kJ \ mol^{-1}$ and the $O=O$ bond energy is $498 \ kJ \ mol^{-1}$,calculate the average bond energy of the $O-H$ bond in $kJ \ mol^{-1}$.

The following two reactions are known:
$Fe_2O_{3(s)} + 3CO_{(g)} \rightarrow 2Fe_{(s)} + 3CO_{2(g)}; \Delta H = -26.8 \ kJ$
$FeO_{(s)} + CO_{(g)} \rightarrow Fe_{(s)} + CO_{2(g)}; \Delta H = -16.5 \ kJ$
The value of $\Delta H$ for the following reaction:
$Fe_2O_{3(s)} + CO_{(g)} \rightarrow 2FeO_{(s)} + CO_{2(g)}$ is ............. $kJ$.

Calculate the enthalpy of formation of ethylene in $kJ\,mol^{-1}$ from the following data:
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -393.5\,kJ$
$H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.2\,kJ$
$C_2H_{4(g)} + 3 O_{2(g)} \to 2 CO_{2(g)} + 2 H_2O_{(l)} ; \Delta H = -1410.8\,kJ$

Given the following thermochemical equations:
$1) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -787 \ kJ$
$2) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)}, \Delta H = -286 \ kJ$
$3) \ C_2H_{2(g)} + \frac{5}{2} O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(l)}, \Delta H = -1310 \ kJ$
Calculate the enthalpy of formation of acetylene $(C_2H_{2(g)})$ in $kJ \ mol^{-1}$.