From the given reaction,$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H = -92.6 \ kJ$,the enthalpy of formation of $NH_3$ is (in $kJ$):

The heat of neutralization of a strong acid with a strong base is constant and is equal to:

When $2 \, mol$ of $C_2H_6$ is completely combusted,$3129 \, kJ$ of heat is released. The heat of formation of $C_2H_6$ is ..... $kJ/mol$. Given that $\Delta H_f$ for $CO_2$ and $H_2O$ are $-395 \, kJ/mol$ and $-286 \, kJ/mol$ respectively.

The heat evolved in the combustion of benzene is given by the equation $C_6H_{6(l)} + 7.5 O_{2(g)} \to 3H_2O_{(l)} + 6CO_{2(g)}$,$\Delta H = -781.0 \ kcal \ mol^{-1}$. Which of the following quantities of heat energy will be evolved when $39 \ g$ of benzene is burnt in an open container?

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

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 ..... $kJ/mol$.