Heat of formation of 2 moles of `NH_(3)(g) " is" -90 kJ`l bond energies of H-H and N-H bonds are 435 kJ and `390 kJ mol^(-1)` respectively. The value of the bond energy of `N -= N` is ?

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

The bond energies of H - H and I - I bonds are 435 kJ mol^-1 and 150 kJ mol^-1 respectively. If DeltaH_f^o for HI is 26.5 kJ mol^-1 then bond enthalpy of H-I bond is

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)

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 of H - H and Cl - Cl are 430 kJ mol^(-1) and 240 kJ mol^(-1) , respectively, and Delta_(f) H^(@) for HCl is - 90 kJ mol^(-1) . Bond enthalpy of HCl is

At standard conditions, if the change in the enthalpy for the following reaction is -109KJ mol^(-1) H_2(g) + Br_2(g) rarr 2HBr(g) Given that bond energy of H_2 and Br_2 is 435kj mol^(-1) and 192kj mol^(-1) , respectively. What is the bond energy of HBr ?

Enthalpy of formation of NH_(3) is -X kJ and Delta H_(H-H) , Delta H_(N-H) are respectively Y kJ mol^(-1) and Z kj mol^(-1) . The value of Delta H_(N = N) is

Calculate the enthalpy of formation of water, given that the bond energies of H-H, O=O and O-H bond are 433 kJ mol^(-1), 492 kJ mol^(-1) , and 464 kJ mol^(-1) , respectively.

Bond energies of H-H and Cl-Cl are "430 kJ mol"^(-1) and "242 kJ mol"^(-1) respectively. DeltaH_(f) for HCl is "91 kJ mol"^(-1) . What will be the bond energy of H-Cl ?

H_(2)(g) +(1)/(2)O_(2)(g) rarr H_(2)O(g) DeltaH =- 242 kJ mol^(-1) Bond energy of H_(2) and O_(2) is 436 and 500 kJ mol^(-1) , respectively. What is bond energy of O-H bond?