Calculate the internal energy change at 298 K for the following reacti

1.	Calculate the internal energy change at 298 K for the following reaction :\(\frac{1}{2}\)N2(g) + \(\frac{3}{2}\)H2(g) → NH3(g)The enthalpy change at constant pressure is -46.0 kJ mol-1. (R = 8.314 JK-1 mol-1).1/2N2(g) + 3/2H2(g) → NH3(g)
Answer» 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

Calculate the internal energy change at 298 K for the following reaction :\(\frac{1}{2}\)N2(g) + \(\frac{3}{2}\)H2(g) → NH3(g)The enthalpy change at constant pressure is -46.0 kJ mol-1. (R = 8.314 JK-1 mol-1).1/2N2(g) + 3/2H2(g) → NH3(g)

Answer»

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

Discussion

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