The reaction for the formation of $NaCl$ is:

Given: $2Fe + \frac{3}{2} O_2 \to Fe_2O_3$,$\Delta H = -193.4 \ kJ$;
$Mg + \frac{1}{2} O_2 \to MgO$,$\Delta H = -140.2 \ kJ$.
What is the $\Delta H$ of the reaction $3Mg + Fe_2O_3 \to 3MgO + 2Fe$ in $kJ$?

Calculate the enthalpy of formation of nitric oxide $(NO)$ in $kJ \, mol^{-1}$ from the following data:
$NO_{(g)} + CO_{(g)} \rightarrow \frac{1}{2} N_{2(g)} + CO_{2(g)}; \Delta H^o = -372.2 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO) = -110.5 \, kJ \, mol^{-1}$
$\Delta H_f^o (CO_2) = -393.5 \, kJ \, mol^{-1}$

Hess's law is applicable for the determination of heat of

The standard heats of formation for $CH_4$, $C_2H_4$, and $C_3H_8$ are $-17.9$, $12.5$, and $-24.8 \ kcal/mol$ respectively. The heat of reaction $(\Delta H)$ for the reaction $CH_4 + C_2H_4 \rightarrow C_3H_8$ in $kcal$ is:

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 $..... \, kJ$.