If the enthalpy of formation and enthalpy of solution of $HCl(g)$ are $-92.3 \ kJ/mol$ and $-75.14 \ kJ/mol$ respectively,then find the enthalpy of formation of $Cl^{-}(aq)$. [Assume $\Delta H_{f}(H^{+}) = 0 \ kJ/mol$]

Calculate $\Delta H$ for the reaction: $H_{2(g)} + O_{2(g)} \rightarrow H_2O_{2(g)}$ given the bond energies: $BE_{H-H} = 436 \ kJ/mol$,$BE_{O=O} = 499 \ kJ/mol$,$BE_{O-O} = 142 \ kJ/mol$,and $BE_{O-H} = 460 \ kJ/mol$. (in $kJ$)

"The quantity of heat which must be supplied to decompose a compound into its elements is equal to the heat evolved during the formation of that compound from the elements." This statement is known as

$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?
Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$

The bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio $1 : 1 : 0.5$. If the enthalpy of formation of $XY$ is $\Delta_fH = -200 \ kJ \ mol^{-1}$,find the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$.

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