Answer to Question #176130 in Physical Chemistry for Vikram

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_p⁰ (Jk^-1mol^-1) values are given as: N₂(g)=29.12;H₂(g)=28.82; NH₃(g)=35.06• Assume the given heat capacity value to be temperature independent in the range.

Answer 1

Enthalpy for NH₃$=\frac{-46.11×373}{298}=-57.71 kj$