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

Step-by-step explanation:

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

Enthalpy of vaporization,
`\Delta H_(vap)` =
`1.85 mol * (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 * \Delta S_(vap)`

`\Delta S_(vap) = (719.8 * 10^(3) J)/(3370 K)`

= 213.6 J/K

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