`Delta H f^(0)` of diamond is

Delta HI^(@) of diamond is

Delta_(f) H of graphite is 0.23 kJ mol^(-1) and Delta_(f) H of diamond is 1.896 kh mol^(-1) . Delta H_(transition) from graphite to diamond is

Why Delta H_(f)^(@) of elements is zero?

In which of the following equations, Delta H ^(0) reaction equals to Delta H _(f) ^(0) for the product ?

DeltaH^(0) for a reaction F_(2) + 2HCl rarr 2HF + Cl_(2) is given as -352.8 kJ. Delta H_(f)^(0) for HF is -268.3 KJ mol^(-1) , then Delta H_(f)^(0) of HCl would be

Calorific value of H_2 is -143KJ g^(-1) . Hence, Delta H_(f)^(@) of H_2O is:

Calorific value of H_2 is -143KJ g^(-1) . Hence, Delta H_(f)^(@) of H_2O is:

Heat of formation, Delta H_(f)^(@) of an explosive compound like NCl_(3) is -

Heat of formation, Delta H_(f)^(@) of an explosive compound like NCl_(3) is -

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