Question

The enthalpy of formation of NH₃(g) as per the following reaction is -46.11 kJ/mol

at 298 K.

N₂(g) + 3H₂ (g) --> 2NH₃(g)

Calculate the value of enthalpy of formation of NH₃(g) at 100 °C. The C_p

values are given as: N₂(g) = 29.12; H₂(g) = 28.82; NH₃(g) = 35.06. Assume the

given heat capacity values to be temperature independent in the range.

Answer 1

$\Delta$_rCp = 2*35.06 - 3*28.82 - 29.12 = -45.46

$\Delta$_fH_373К = 1/2(2$\Delta$_rH_298К + $\Delta$_rCp*(373-298)) = 1/2(- 2*46110 - 45.46*75) = - 47814.75 J/mol = - 47.81 kJ/mol