The heat of transition `(Delta H_(t))` of graphite into diamond would be, where
C (graphite) `+O_(2)(g)to CO_(2)(g) , Delta H =` x kJ
C (diamond) `+O_(2)(g)to CO_(2)(g) , Delta H =` y kJ

One gram mole of graphite and diamond were burnt to form CO_(2) gas: C_(("graphite"))+O_(2)(g)toCO_(2)(g)," "DeltaH^(@)=-399.5kJ C_(("diamond"))+O_(2)(g)toCO_(2)(g)," "DeltaH^(@)=-395.4 kJ

Give that C_(s) +O_(2_(g))to CO_(2_((g))), Delta H^(0)=-xkJ 2CO_((g))+O_(2_((g)))to 2CO_(2_((g))), Delta H^(0)=-ykJ .

Calculate the heat of formation of carbon monoxide from the following data. . (i) C(s) + O_(2)(g) to CO_(2) (g) , Delta H =- 393.5 kJ (ii) CO(g) + (1)/(2) O_(2)(g) to CO_(2)(g) , Delta H = - 282.8 kJ

On the basis of thermochemical equation (a), (b) and (c ), which of the algebric relationship is correct. ( a) C("graphite") + O_(2)(g) rarr CO_(2)(g) , Delta_(r)H = x kJ mol^(-1) (b) C("graphite") + (1)/(2)O_(2)(g) rarr CO(g) , Delta_(r)H = y kJ mol^(-1) ( c) CO(g) + (1)/(2)O_(2)(g) rarr CO_(2)(g) , Delta_(r)H = z kJ mol^(-1)