Bond energy of `(N-H)` bond is `yKJ mol^(-1)` under standard state. Thus,change in internal energy in the following process is
`NH_(3)(g)rarrN(g)+3H(g)`

Bond energies of N-=N, H-H and N-H bonds are 945,463 & 391 kJ mol^(-1) respectively, the enthalpy of the following reactions is : N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)

Give the bond energies of N equiv N, H-H and N-H bonds as 945, 436 and 391 kJ mol^-1 respectively, the enthalpy of the following reaction N 2(g) ​ +3H 2(g) ​ →2NH 3(g) ​ is:

Given the bond energies N-N , N-H and H-H bond are 945, 436 and 391KJmol^(-1) respectively, the enthalpy change of the reaction N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g) is

Given the bond energies N=N , H and H-H bond are 945, 436 and 391KJmol^(-1) respectively, the enthalpy change of the reaction N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g) is

Bond energies of H - H bond is 80 kJ/mol, I - I bond is 100 kJ/mol and for H - I bond is 200 kJ/mol, the enthalpy of the reaction : H_(2)(g) + I_(2)(g) rarr 2HI(g) is

Bond energies of H - H bond is 80 kJ/mol, I - I bond is 100 kJ/mol and for H - I bond is 200 kJ/mol, the enthalpy of the reaction : H_(2)(g) + I_(2)(g) rarr 2HI(g) is

Calculate the bond energy of O-H bond in H_(2)O(g) at the standard state from the following data: (1) H_(2)(g) to 2H(g),DeltaH^(0)=436kJ*mol^(-1) (2) (1)/(2)O_(2)(g)toO(g),DeltaH^(0)=249k*mol^(-1) (3) H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(g),DeltaH_(f)^(0)[H_(2)O(g)]=-241.8kJ*mol^(-1) .