Calculate the entropy change ( $\triangle_{vap} S^{∘}$ , J K^-1 mol^-1) when 1.00 mol of mercury at its boiling point (T_b = 356.55 °C) vaporizes (in the process the temperature does not change, $\triangle_{vap} H^{∘}$ for mercury is + 59.3 kJ mol^-1).

Question

Quizplus · Accepted Answer

The equation for entropy change during vaporization is:

$\triangle_{vap} S^{∘} = \frac{\triangle_{vap} H^{∘}}{T}$

Where $\triangle_{vap} H^{∘}$ is the enthalpy of vaporization and T is the temperature at which the vaporization occurs.

Plugging in the given values, we get:

$\triangle_{vap} S^{∘} = \frac{59.3\,kJ\,mol^{-1}}{629.7\,K}$

$\triangle_{vap} S^{∘} = 0.094\,kJ\,K^{-1}\,mol^{-1}$

Note that the units are kJ/K/mol, which can be simplified to J/K/mol. Therefore:

$\triangle_{vap} S^{∘} = 94\,J\,K^{-1}\,mol^{-1}$

The answer is B, +94.

Calculate the Entropy Change △vapS∘\triangle_{vap} S^{∘}△vap​S∘ , J K-1 Mol-1) When 1

Calculate the Entropy Change $\triangle_{vap} S^{∘}$ , J K^-1 Mol^-1) When 1