Question

For cobalt, Co, the heat of vaporization at its normal boiling point of 3097 °C is 389.1 kJ/mol.The entropy change when 1.85 moles of liquid Co vaporizes at 3097 °C, 1 atm is _______ J/K.

Answer 1

Explanation:

First, for 1.85 mol Co the enthalpy will be calculated as follows.

   Enthalpy of vaporization, $\Delta H_{vap}$ = $1.85 mol \times \frac{389.1 kJ}{mol}$

                                 = 719.8 kJ

Now, we will convert the temperature into Kelvin as follows.

                  (3097 + 273) K

               = 3370 K

Therefore, entropy of vaporization will be calculated as follows.

        $\Delta H_{vap} = T \times \Delta S_{vap}$

        $\Delta S_{vap} = \frac{719.8 \times 10^{3} J}{3370 K}$

                      = 213.6 J/K

Thus, we can conclude that $\Delta S$ for vaporization is 213.6 J/K.