Calculate the standard entropy of condensation of chloroform at its boiling point,335 K.The standard molar enthalpy of vaporization of chloroform at its boiling point is 31.4 J.K^-1.mol^-1. A)(-31.3 J.K^-1.mol^-1) B)+93.7 J.K^-1.mol^-1 C)(-93.7 J.K^-1.mol^-1) D)+31.4 J.K^-1.mol^-1 E)+506 J.K^-1.mol^-1

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Quizplus · Accepted Answer

The standard entropy of condensation is equal to the negative of the standard entropy of vaporization. The standard entropy of vaporization is given by the equation:$$\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$$where $\Delta H_{vap}$ is the standard molar enthalpy of vaporization and T is the boiling point. Substituting the given values into this equation, we get:$$\Delta S_{vap} = \frac{31.4 J.K^{-1}.mol^{-1}}{335 K} = 0.0937 J.K^{-1}.mol^{-1}$$Therefore, the standard entropy of condensation is:$$\Delta S_{cond} = -\Delta S_{vap} = -0.0937 J.K^{-1}.mol^{-1} = -93.7 J.K^{-1}.mol^{-1}$$

Calculate the Standard Entropy of Condensation of Chloroform at Its