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(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?

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(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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