Based on the values of $B.E.$ given,calculate the $\Delta_fH^o$ of $N_2H_4\, (g)$ in $kJ\, mol^{-1}$.
Given : $N-N = 159\, kJ\, mol^{-1};\, H-H = 436\, kJ\, mol^{-1}$
$N \equiv N = 941\, kJ\, mol^{-1};\, N-H = 398\, kJ\, mol^{-1}$

The heat of transition $(\Delta H_t)$ of graphite into diamond would be,where
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$

The enthalpy of combustion of carbon disulfide $(CS_2)$ is $-110.2 \, kJ \, mol^{-1}$. If the enthalpies of formation of $SO_2$ and $CO_2$ are $-297.4 \, kJ \, mol^{-1}$ and $-394.5 \, kJ \, mol^{-1}$ respectively,the enthalpy of formation of carbon disulfide is:

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 $kJ$.

The enthalpy of solution of $BaCl_{2(s)}$ and $BaCl_2 \cdot 2H_2O_{(s)}$ are $-20.6 \, kJ \, mol^{-1}$ and $8.8 \, kJ \, mol^{-1}$ respectively. The enthalpy change for the reaction $BaCl_{2(s)} + 2H_2O_{(l)} \to BaCl_2 \cdot 2H_2O_{(s)}$ is $...... \, kJ \, mol^{-1}$.

The average $C-H$ bond energy is $416 \ kJ \ mol^{-1}$. Which of the following equations correctly represents the bond dissociation of $CH_4$?