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`

Calculate the DeltaH in joules for C("graphite")toC("diamond") from the following data: C("graphite")+O_(2)(g)toCO_(2)(g):" "DeltaH^(@)=-393.5kJ C("diamond")+O_(2)(g)toCO_(2)(g)," "DeltaH^(@)=-395.4kJ

Calculate DeltaH in Joules for C_(("graphite")) rarr C_(("Diamond")) by using the following data C_(("graphiter")) + O_(2(g)) rarr CO_(2(g)) , DeltaH^@ =-393.5KJ C_(("Diamond")) + O_(2(g)) rarr CO_(2(g)) , DeltaH^@ =-395.4KJ

Why are the standard reaction enthalpies of the following two reactions different? (1) C(graphite, s) +O_(2)(g)toCO_(2)(g),DeltaH^(0)=-393.5kJ (2) C(diamond, s) +O_(2)(g)toCO_(2)(g),DeltaH^(0)=-395.4kJ

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

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

C_("graphite")+O_(2)(g)rarrCO_(2)(g) deltaH=-94.05Kcalmol^(-1) C_("diamond")+O_(2)(g),DeltaH=-94.50Kcalmol^(-1) Therefore,

Calculate the heat of formation (DeltaH) of CO from the following data. (I) C_(graph ite)+2H_(2)(g)toCH_(4)(g)," " DeltaH=-94kcal (II) CO(g)+(1)/(2)O_(2)toCO_(2)(g)," "DeltaH=-68kcal