(a) The enthalpy of vaporisation of liquid ether at its boiling point

1.	(a) The enthalpy of vaporisation of liquid ether at its boiling point (35°C) is 26 kJ/mol. Calculate the entropy change for liquid ether to its vapour state.(b) At what temperature, will the following reaction be spontaneous?N2 + 3H2 → 2NH3, given ΔH = -95.4 kJ/mol and ΔS = -198.3 J k-1 mol-1(c) What are path functions?
Answer» (a) Enthalpy of vaporisation (ΔΗ) - 26 k/mol. Boiling point (T) = 35°C = (35 + 273) K = 308 K Now, at constant pressure Entropy change ∆S = \(\frac{-\Delta H}{T}\) ∆S = \(\frac{26K/mol}{308K}\) ∆S = 0.0844 kJ mol⁻¹ K⁻¹ (b) ∆_fH° = ∑∆H°_Products − ∑∆H°_Reacants −1323 = {2(−393.5) + 2(−249.0)} −{∆H°C₂H₄ + 0} −1323 = −1285 − ∆H°C₂H₄ + 0} ∆H°C₂H₄ = −1285 + 1323 = 38 kJ mol⁻¹ (c) Path functions: These are the functions whose magnitude depend on the path followed during a process as well as on the end states. E.g. work (w), heat (Q) are path functions.

(a) The enthalpy of vaporisation of liquid ether at its boiling point (35°C) is 26 kJ/mol. Calculate the entropy change for liquid ether to its vapour state.(b) At what temperature, will the following reaction be spontaneous?N2 + 3H2 → 2NH3, given ΔH = -95.4 kJ/mol and ΔS = -198.3 J k-1 mol-1(c) What are path functions?

Answer»

(a) Enthalpy of vaporisation (ΔΗ) - 26 k/mol.

Boiling point (T) = 35°C = (35 + 273)

K = 308 K

Now, at constant pressure

Entropy change ∆S = \(\frac{-\Delta H}{T}\)

∆S = \(\frac{26K/mol}{308K}\)

∆S = 0.0844 kJ mol⁻¹ K⁻¹

(b) ∆_fH° = ∑∆H°_Products − ∑∆H°_Reacants

−1323 = {2(−393.5) + 2(−249.0)}

−{∆H°C₂H₄ + 0}

−1323 = −1285 − ∆H°C₂H₄ + 0}

∆H°C₂H₄ = −1285 + 1323 = 38 kJ mol⁻¹

(c) Path functions: These are the functions whose magnitude depend on the path followed during a process as well as on the end states. E.g. work (w), heat (Q) are path functions.

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