Question

Calculate the internal energy at 298K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is – 42.0 kJ mol^-1.

(Given: R = 8.314 J K^-1 mol^-1)

Answer 1

Formation of 1 mole of ammonia

`1/2 N_(2(g)) + 3/2 H_(2(g)) -> NH_(3(g))`

Δn = (no.of moles of gaseous product) - (no.of moles of gaseous reactant)

`= 1 - (1/2 + 3/2)`

= −1

ΔH = ΔU + PΔV

ΔU = ΔH − ΔnRT

`= 42-(-1) xx 8.314 xx 298 xx 10^(-3)`

= −42 + 2.477 

= −39.523 kJ

∴ ΔU = −39.523 kJ