The heat of neutralization for $2 \ moles$ of $\text{LiOH}$ and $1 \ mole$ of $H_2SO_4$ at $25^{\circ} C$ is $-69.6 \ kJ$. The heat of ionisation of $\text{LiOH}$ will be nearly $:-$

What is the enthalpy change for the reaction $NaOH_{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$ called?

The enthalpy of combustion of methane,graphite and dihydrogen at $298 \, K$ are $-890.3 \, kJ \, mol^{-1}$,$-393.5 \, kJ \, mol^{-1}$ and $-285.8 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of $CH_{4(g)}$ will be:
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$

The enthalpy of the reaction,$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$ is $\Delta H_1$ and that of $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$ is $\Delta H_2$. Then:

For the reaction $N_2 + 3 X_2 \longrightarrow 2 NX_3$,where $X = F, Cl$ (the average bond energies are $F-F = 155 \ kJ \ mol^{-1}$,$N-F = 272 \ kJ \ mol^{-1}$,$Cl-Cl = 242 \ kJ \ mol^{-1}$,$N-Cl = 200 \ kJ \ mol^{-1}$ and $N \equiv N = 941 \ kJ \ mol^{-1}$),the heats of formation of $NF_3$ and $NCl_3$ in $kJ \ mol^{-1}$,respectively,are closest to

The heat of neutralization of $HCl$ by $NaOH$ under certain conditions is $-55.9 \, kJ \, mol^{-1}$ and that of $HCN$ by $NaOH$ is $-12.1 \, kJ \, mol^{-1}$. The heat of ionization of $HCN$ is .............. $kJ \, mol^{-1}$.