Enthalpy change for a reaction does not depend upon

Calculate the enthalpy change for the following reaction, using the given bond energies $(\text{kJ/mol})$: $C-H = 414$, $H-O = 463$, $H-Cl = 431$, $C-Cl = 326$, and $C-O = 335$.
$CH_3OH(g) + HCl(g) \rightarrow CH_3Cl(g) + H_2O(g)$

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

For the reaction $H_{2(g)} + C_{2}H_{4(g)} \rightarrow C_{2}H_{6(g)}$,the enthalpy change is ....... $Kcal \, mol^{-1}$. Given bond energies: $H-H = 103$,$C-H = 99$,$C-C = 80$,and $C=C = 145 \, Kcal \, mol^{-1}$.

The heat of neutralization of $HCl$ by $NaOH$ is $-57.3 \, kJ/mol$. If the heat of neutralization of $HCN$ by $NaOH$ is $-12.1 \, kJ/mol$,then the enthalpy of dissociation of $HCN$ is $...... \, kJ$.

The $H-H$ bond energy is $430 \ kJ \ mol^{-1}$ and $Cl-Cl$ bond energy is $240 \ kJ \ mol^{-1}$. $\Delta H$ for the formation of $HCl$ is $-90 \ kJ \ mol^{-1}$. The $H-Cl$ bond energy is about: