The enthalpy change for the reaction,$H_{2(g)} + C_2H_{4(g)} \to C_2H_{6(g)}$ is $......$ $kcal \ mol^{-1}$. The bond energies are,$[e_{H-H} = 103, e_{C-H} = 99, e_{C-C} = 80]$ and $[e_{C=C} = 145] \ kcal \ mol^{-1}$.

The heat of combustion of $C$,$S$ and $CS_2$ are $-393.3 \ kJ$,$-293.7 \ kJ$ and $-1108.76 \ kJ$ respectively. What will be the heat of formation of $CS_2$ in $kJ$?

Enthalpy of sublimation of a substance is equal to

The heat of combustion of ethanol into carbon dioxide and water is $-327 \ kcal$ at constant pressure. The heat evolved (in $cal$) at constant volume and $27^{\circ} C$ (assuming all gases behave ideally) is $\left( R = 2 \ cal \ mol^{-1} \ K^{-1} \right)$

The enthalpy of neutralization of $NaOH$ by $HCl$ is $-55.84 \, kJ/mol$ and that of $NH_4OH$ by $HCl$ is $-51.34 \, kJ/mol$. The enthalpy of ionization of $NH_4OH$ is ..... $kJ/mol$.

Based on the bond enthalpy $(B.E.)$ values given,the standard enthalpy of formation $(\Delta_fH^o)$ of $N_2H_{4(g)}$ is ...... $kJ\ mol^{-1}$.
Given: $B.E.(N-N) = 159\ kJ\ mol^{-1}$,$B.E.(H-H) = 436\ kJ\ mol^{-1}$,$B.E.(N \equiv N) = 941\ kJ\ mol^{-1}$,$B.E.(N-H) = 398\ kJ\ mol^{-1}$.