The enthalpy change during a reaction does not depend on .......

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

$S + \frac{3}{2} O_2 \to SO_3 + 2x \ kcal$
$SO_2 + \frac{1}{2} O_2 \to SO_3 + y \ kcal$
Find out the heat of formation of $SO_2$.

Given the reactions:
$C + \frac{1}{2}O_2 \to CO : \Delta H = -12 \ kJ$
$CO + \frac{1}{2}O_2 \to CO_2 : \Delta H = -10 \ kJ$
For the reaction $C + O_2 \to CO_2 : \Delta H = x \ kJ$,the value of $x$ is: (in $kJ$)

What is the heat of formation of $HCl_{(g)}$ from the following equation (in $kJ$)?
$H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)} \quad \Delta_{r}H = -194 \ kJ$

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.