Calculate the heat of formation of ammonia from the following data:
`N_(2)(g)+3H_(2)(g)to2NH_(3)(g)`
The bond energies of `N-=N,H-H and N-H ` bonds are 226,104 and 92 kcal respectively.

Calculate the enthalpy of the following reaction: H_(2)C =CH_(2)(g) +H_(2)(g) rarr CH_(3)-CH_(3)(g) The bond energies of C-H, C-C,C=C , and H-H are 99,83,147 ,and 104kcal respectively.

Calculate the enthalpy change for the reaction CH_2 = CH_2(g) + H_2(g0 to CH_3 - CH_3(g) . The bond energies of C - C , C - H, H- H and C = C bonds are 83 kcal, 99 kcal 104 kcal and 145 kcal respectivley.

Heat of formation of 2 mol of NH_(3)(g) is =90 kJ , bond energies of H-H and N-H bonds are 435kJ and 390 kJ mol^(-1) , respectively. The value of the bond enegry of N-=N will be

Heat of formation of 2 mol of NH_(3)(g) is =90 kJ , bond energies of H-H and N-H bonds are 435kJ and 390 kJ mol^(-1) , respectively. The value of the bond enegry of N-=N will be

Enthalpy change for the reaction 2N_((g))+6H_((g))to2NH_(3(g))is-900 KJ The dissociation energy of N - H bond is

Enthalpy change for the reaction 2N_((g))+6H_((g))to2NH_(3(g))is-900 KJ The dissociation energy of N - H bond is