The enthalpy change on freezing of 1 mol of water at 5^oC to ice at −5^oC is :

(Given $$\Delta $$_fusH = 6 kJ mol^$$-$$1 at 0^oC,
C_p(H₂O, $$\ell $$ = 75.3J mol^$$-$$1 K^$$-$$1)
C_p(H₂O s) =36.8 J mol^$$-$$1 K^$$-$$1)

$$\Delta $$U is equal to :

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 :

If 100 mole of H₂O₂ decompose at 1 bar and 300 K, the work done (kJ) by one mole of O₂(g) as it expands against 1 bar pressure is :

2H₂O₂(l)    $$\rightleftharpoons$$    2H₂O(l) + O₂(g)

(R = 8.3 J K ^$$-$$1 mol^$$-$$1)