$$\Delta $$U is equal to :

The group having isoelectronic species is:

Given, $${C_{(graphite)}} + {O_2} \to C{O_2}(g)$$;

$${\Delta _r}{H^o}$$ = - 393.5 kJ mol^-1

$${{\rm H}_2}(g)$$ + $${1 \over 2}{O_2}(g)$$$$\to {{\rm H}_2}{\rm O}(l)$$

$${\Delta _r}{H^o}$$ = - 285.8 kJ mol^-1

$$C{O_2}(g)$$ + $$2{{\rm H}_2}{\rm O}(l) \to$$ $$C{H_4}(g)$$ + $$2{O_2}(g)$$

$${\Delta _r}{H^o}$$ = + 890.3 kJ mol^-1

Based on the above thermochemical equations, the value of $${\Delta _r}{H^o}$$ at 298 K for the reaction

$${C_{(graphite)}}$$ + $$2{{\rm H}_2}(g) \to$$ $$C{H_4}(g)$$ will be :

The radius of the second Bohr orbit for hydrogen atom is:
(Planck’s Const. h = 6.6262 × 10^-34 Js; mass of electron = 9.1091 × 10^-31 kg; charge of electron (e) = 1.60210 × 10^-19 C; permittivity of vacuum ($${\varepsilon _0}$$) = 8.854185 × 10^-12 kg^-1 m^-3 A²)