Use the following information to determine the standard enthalpy of formation of NH₃(g) .
N-H bond enthalpy = 390 kJ.mol^-¹
$\Delta$ H_f^$\omicron$(H(g) ) = 217.9 kJ.mol^-¹
$\Delta$ H_f^$\omicron$(N(g) ) = 472.6 kJ.mol^-¹

Question

Use the following information to determine the standard enthalpy of formation of NH₃(g) .
N-H bond enthalpy = 390 kJ.mol^-¹

\Delta

H_f^$\omicron$(H(g) ) = 217.9 kJ.mol^-¹

\Delta

H_f^$\omicron$(N(g) ) = 472.6 kJ.mol^-¹

Quizplus · Accepted Answer

The standard enthalpy of formation of NH3(g) can be calculated using the bond enthalpies of the N-H bonds and the enthalpies of formation of H(g) and N(g). The enthalpy of formation of NH3(g) is equal to the sum of the bond enthalpies of the N-H bonds minus the sum of the enthalpies of formation of H(g) and N(g).Therefore, the standard enthalpy of formation of NH3(g) is:$\Delta$H_f^{$\omicron $}(NH₃(g)) = 3(390 kJ.mol^-¹) - 217.9 kJ.mol^-¹ - 472.6 kJ.mol^-¹ = -44 kJ.mol^-¹

Use the Following Information to Determine the Standard Enthalpy of Formation