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

\Delta

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

Quizplus · Accepted Answer

The standard enthalpy of formation of NH3(g) can be calculated using the bond enthalpy of the N-H bond and the standard enthalpies of formation of H(g) and N(g).$\Delta$H_fS1U1P1$\circ$S1S1P0(NH3(g))= 3$\times$390 kJ.mol^-1+ $\Delta$H_fS1U1P1$\circ$S1S1P0(H(g)) + $\Delta$H_fS1U1P1$\circ$S1S1P0(N(g))$\Delta$H_fS1U1P1$\circ$S1S1P0(NH3(g))= 3$\times$390 kJ.mol^-1+ 217.9 kJ.mol^-1+ 472.6 kJ.mol^-1$\Delta$H_fS1U1P1$\circ$S1S1P0(NH3(g))= -44 kJ.mol^-1

Use the Following Information to Determine the Standard Enthalpy of Formation