Calculate the internal energy change at 298 K for the following reaction : 1/2N2(g) + 3/2H2(g) → NH3(g)

Question

Calculate the internal energy change at 298 K for the following reaction :

\(\frac{1}{2}\)N_2(g) + \(\frac{3}{2}\)H_2(g) → NH_3(g)

The enthalpy change at constant pressure is -46.0 kJ mol^-1. (R = 8.314 JK^-1 mol^-1).

1/2N_2(g) + 3/2H_2(g) → NH_3(g)

Answer 1

Given :

1/2N_2(g) + 3/2H_2(g) → NH_3(g)

ΔH= -46.0 kJ mol^-1 

ΔH = Heat of formation of NH₃ at constant pressure 

= -46.0 kJ mol^-1 = -4600 J mol^-1

ΔU = Change in internal energy = ? 

Δn = (Number of moles of ammonia) – (Number of moles of hydrogen + Number of moles of nitrogen)

= [1 – ( \(\frac{1}{2}\)+ \(\frac{3}{2}\))]= -1 mol

R = 8.314 JK^-1 mol^-1

T = Temperature in kelvin = 298 K

ΔH = Δ U + ΔnRT

∴ -46000 = ΔU + (-1 × 8.314 × 298)

∴ -46000 = ΔU – 2477.0

∴ ΔU = -46000 + 2477.0

= -43523 J 

= -43.523 kJ

∴ Change in internal energy = -43.523 kJ