1.

Calculate the internal energy at 298 K 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-1mol-1)

Answer»

Given : 

ΔH = - 42.0 kJ mol-1,

T = 298 K, 

ΔU = ?

\(\frac{1}{2}\)N2(g)\(\frac{3}{2}\)H2(g) ⟶ NH3(g)

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

ΔH = ΔU + ΔnRT

∴ ΔU = ΔH – ΔnRT

= - 42 – (- 1) × 8.314 × 298 × 10-3

= - 42 + 2.477

= - 39.523 kJ

∴ ΔU = -39.523 kJ



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