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The enthalpy of vaporization of diethyl ether (l) is 25 kJ mol−1 at its boiling point, 35°C. Calculate ∆S° for the conversion of :

(i) liquid to vapour and

(ii) vapour to liquid at 35°C.

1 Answer

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Best answer

(i) For vaporization of diethyl ether,

vapHΘ = 26 kJ mol−1 = 26 × 103 J mol−1

T = 35 + 273= 308 K

\(\therefore\) ΔvapSΘ = \(\frac{\Delta_{vap}H^{\ominus}}{T}\)

\(\frac{26\times10^3\,J\,mol^{-1}}{35K}\)

= 84.4 JK-1 mol-1

(ii) The conversion of vapour into liquid is condensation. The enthalpy of condensation is negative of enthalpy of vaporization.

\(\therefore\) ∆condHΘ = −26 kJ mol−1

= −26 × 103 J mol−1

\(\therefore\) condSΘ = \(\frac{\Delta_{cond}H^{\ominus}}{T}\)

\(\frac{-26\times10^3\,J\,mol^{-1}}{308K}\)

= -84.4 J K-1 mol-1

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