Which factor affects the heat of reaction in the Kirchhoff equation?

Enthalpy of formation of methane is $-75 \ kJ / mol$. What is the enthalpy change for formation of $24 \ g$ of methane (in $kJ$)?

In the reaction $H_2 + Cl_2 \rightarrow 2HCl$,heat is released. The bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $242 \ kJ \ mol^{-1}$ respectively. If the enthalpy of reaction is $-182 \ kJ \ mol^{-1}$,the bond energy of $H-Cl$ is . . . . . . $kJ \ mol^{-1}$.

When $1 \, \text{mol}$ of anhydrous salt $AB$ is dissolved in water,$21.0 \, J \, \text{mol}^{-1}$ of heat is released. The enthalpy of hydration of $AB$ is $-29.4 \, J \, \text{mol}^{-1}$. What is the enthalpy of solution of the hydrated salt $AB \cdot 2H_2O_{(s)}$ in $J \, \text{mol}^{-1}$ (in $.4$)?

If $(i)$ $C + O_2 \to CO_2$,$(ii)$ $C + 1/2 O_2 \to CO$,$(iii)$ $CO + 1/2 O_2 \to CO_2$,the heats of reaction are $Q$,$-12$,and $-10$ respectively. Then $Q =$ ?

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)