Find the enthalpy of formation of Hydrogen fluorideon the basis of following data : Bond energy of H-H bond= 434kJ mol^(-1) Body energy of F-F bond = 158 kJ mol^(-1) Bond energyof H-Fbond = 565 kJ mol^(-1)

Find the enthalpy of formation of Hydrogen fluorideon the basis of fol

1.	Find the enthalpy of formation of Hydrogen fluorideon the basis of following data : Bond energy of H-H bond= 434kJ mol^(-1) Body energy of F-F bond = 158 kJ mol^(-1) Bond energyof H-Fbond = 565 kJ mol^(-1)
Answer» Solution :Aim `: (1)/(2) H_(2)+(1)/(2) F_(2) rarr HF, Delta_(r) H =?` `Delta_(r)H= Sigma B.E.` ( REACTANTS) `- Sigma`B.E. ( Products) `=(1)/(2)B.E. ( H_(2)) + (1)/(2) B.E. (F_(2))- B.E. (HF) = (1)/(2 ) xx434 +(1)/(2) XX 158 - 565 = - 269 kJ MOL^(-1)`