1. Tardigrade
  2. Question
  3. Chemistry
  4. Bond dissociation enthalpies of H2(g) and N2(g) are 436.0 kJ mol-1 and 941.8 kJ mol-1, respectively, and enthalpy of formation of NH3(g) is -46 kJ mol-1. What are the enthalpy of atomisation of NH3(g) and the average bond enthalpy of N — H bond respectively (in kJ mol-1) ?

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Bond dissociation enthalpies of $H_{2(g)}$ and $N_{2(g)}$ are $436.0\, kJ\, mol^{-1}$ and $941.8 \,kJ\, mol^{-1}$, respectively, and enthalpy of formation of $NH_{3(g)}$ is $-46\, kJ\, mol^{-1}$. What are the enthalpy of atomisation of $NH_{3(g)}$ and the average bond enthalpy of $N — H$ bond respectively (in $kJ\, mol^{-1}$) ?

Thermodynamics

Solution:

$N_{2\left(g\right)}+3H_{2\left(g\right)} \to 2NH_{3\left(g\right)}$ ;
$\Delta H = -2 \times 46\,kJ\,mol^{-1}$
$\Delta H = \Sigma\left(BE\right)_{reactants} - \Sigma \left(BE\right)_{products}$
$= \left(941.8 + 3 \times 436\right) - \left(6x\right) = - 2 \times 46$
(here $x = BE$ of $N — H$ bonds)
$x = 390.3\, kJ\, mol^{-1}$
$NH_{3} \to N + 3\left(H\right)$
Heat of atomisation $= 3 \times 390.3 = 1170.9 \,kJ\, mol^{-1}$