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Question: How would you find the molar enthalpy of vaporization for a substance given 3.21 mol of the substanc...

How would you find the molar enthalpy of vaporization for a substance given 3.21 mol of the substance absorbs 28.4 kJ of energy as heat when the substance changes from a liquid to a gas?

Explanation

Solution

The mass of the substance, along with the values for latent heat of fusion and vaporisation need to be given in order to determine the final heat during change of phase.

Complete answer:
In order to answer our question, we need to know about latent heat. Now, what is latent heat? For example, suppose water is present in a liquid state and we heat the water. A point will come when the water will reach its boiling point, but liquid water does not instantly change to the vapour form. Here, latent heat comes to play. After reaching the boiling point, the liquid water gains the latent heat of evaporation to overcome the forces of liquid molecules and transform into gas. Same is the case when ice is melted to form liquid water. At the melting point of ice, some extra heat is required for phase change. This heat is called the latent heat of fusion.
Latent heat of fusion is the heat required to melt 1 g of solid into liquid form, once it reaches melting point, whereas latent heat of vaporization represents the heat required to convert 1 g of liquid into it vapour state. Mathematically, the heat absorbed can be written as q=nΔHvapq=n\Delta {{H}_{vap}}, where q is heat absorbed, n is number of moles. So, we have:

& q=n\Delta {{H}_{vap}} \\\ & \Rightarrow \Delta {{H}_{vap}}=\dfrac{q}{n} \\\ & \Rightarrow \Delta {{H}_{vap}}=\dfrac{28.4kJ}{3.21\,moles} \\\ & \Rightarrow \Delta {{H}_{vap}}=8.8474kJ\,mo{{l}^{-1}}/8.85kJ\,mo{{l}^{-1}} \\\ \end{aligned}$$ So, we obtain the heat of evaporation as $8.85kJ\,mo{{l}^{-1}}$. **Note:** Latent heat cannot be seen but can be observed very easily. Liquid water at its boiling point will not cause that much skin burn as steam water at 100 degree Celsius. That is why steam engines were more powerful.