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iren [92.7K]
3 years ago
5

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 mol

es of liquid Co vaporizes at 3097 °C, 1 atm is _______ J/K.
Chemistry
1 answer:
tino4ka555 [31]3 years ago
8 0

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.

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