Organic Chemistry: Some Basic Principles and Techniques — Chemistry STD 11 Science — Question

$C_{(graphite)} +O_2(g) \rightarrow CO_2(g)\,;$

$\Delta _rH^o=-395.5 \, kJ\,mol^{-1}$

$H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O(l)\,;$   $\Delta _rH^o  =-285.8\, kJ\, mol^{-1}$

$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)\,;$

$\Delta _rH^o  = + 890.3\, kJ\, mol^{-1}$

Based on the above thermochemical equations, the value of, $\Delta _rH^o $ at $298\, K$ for the reaction

$C_{(graphite)} + 2H_2(g) \rightarrow CH_4(g) $ will be for $\Delta_{r} H^{\circ}$ ........... $kJ \,mol^{-1}$

Given $SO_3(g) + H_2O(l) \rightarrow H_2SO_4(l)$
$\Delta H = -130\,\, kcal\, mol^{-1}$
$SO_2(g) + 1/2O_2(g) \rightarrow SO_3(g)$
$\Delta H = -100 \,\,kcal\,\, mol^{-1}$
the enthalpy of formation of $H_2SO_4(l)$ would be ......$kcal\, mol^{-1}$

$C{O_{(g)}} + \frac{1}{2}{O_{2(g)}} \to C{O_{2(g)}}$

How are $\Delta E$ and $\Delta H$ releated for the reaction?