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 group having isoelectronic species is:

$$\Delta $$U is equal to :

The freezing point of benzene decreases by 0.45⁰C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (K_f for benzene = 5.12 K kg mol^–1)