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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) |
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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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