$H^{+}_{(aq)} + OH^{-}_{(aq)} \rightarrow H_2O_{(l)} + 56 \text{ kJ/mol}$. The heat of neutralization for the complete neutralization of $1 \text{ mole}$ of $H_2SO_4$ will be ...... $\text{kJ}$.

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

Consider the following reactions:
$(i)$ $H_{(aq)}^{+} + OH^{-}_{(aq)} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_1 \ kJ \ mol^{-1}$
$(ii)$ $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$
$(iii)$ $CO_{2_{(g)}} + H_{2_{(g)}} \longrightarrow CO_{(g)} + H_2O_{(l)}$,$\Delta H = -X_3 \ kJ \ mol^{-1}$
$(iv)$ $C_2H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + H_2O_{(l)}$,$\Delta H = -X_4 \ kJ \ mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is

The standard enthalpies of formation for $NH_3$,$CO_2$,$HI$,and $SO_2$ are $-46.19$,$-393.4$,$+24.94$,and $-296.9 \, kJ/mol$ respectively. What is the increasing order of their stability?

Calculate the enthalpy change for the reaction
$H_2 + F_2 \longrightarrow 2HF$
given that
Bond energy of $H-H$ bond $= 434 \ kJ/mol$
Bond energy of $F-F$ bond $= 158 \ kJ/mol$
Bond energy of $H-F$ bond $= 565 \ kJ/mol$
Result in $kJ$.

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is: