If the ratio of the enthalpy of formation of $CO_2$ and $SO_2$ is $4:3$ and the enthalpy of formation of $CS_2$ is $26 \ kcal/mol$,then what will be the enthalpy of formation of $SO_2(g)$ based on the following reaction?
$CS_2(l) + 3O_2(g) \to CO_2(g) + 2SO_2(g)$

The enthalpies of dissolution of $BaCl_2(s)$ and $BaCl_2 \cdot 2H_2O(s)$ are $-20.6 \ kJ \ mol^{-1}$ and $8.8 \ kJ \ mol^{-1}$ respectively. Calculate the enthalpy of hydration for the given reaction: $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$

If the enthalpy of formation of $N_2O$ is $82 \, kJ \, mol^{-1}$,calculate the resonance energy of $N_2O$ in $kJ \, mol^{-1}$.
$N \equiv N \, (946 \, kJ \, mol^{-1}); \, N = N \, (418 \, kJ \, mol^{-1})$
$O = O \, (498 \, kJ \, mol^{-1}); \, N = O \, (607 \, kJ \, mol^{-1})$

Given $:$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$

Given the following thermochemical equations:
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)

"The quantity of heat which must be supplied to decompose a compound into its elements is equal to the heat evolved during the formation of that compound from the elements." This statement is known as