Question 5.3.4: Given the following information for mercury, Hg (at 1 atm), ......
Given the following information for mercury, Hg (at 1 atm), calculate the amount of heat needed (at 1 atm) to vaporize a 30.0-g sample of liquid mercury at its normal boiling point of 357 ºC.
boiling point = 357 ºC ∆H_{vap}(357 ºC) = 59.3 kJ/mol
melting point = −38.9 ºC ∆H_{fus}(−38.9 ºC) = 2.33 kJ/mol
specific heat (liquid) = 0.139 J/g · ºC
You are asked to calculate the amount of heat needed to vaporize a sample of liquid mercury at its normal boiling point.
You are given the mass of the mercury, its boiling and melting points, its heat of vaporization and heat of fusion, and the specific heat of the liquid.
This is a constant-temperature process in which the liquid is vaporized at its normal boiling point. Use the heat of vaporization of mercury to calculate the amount of heat needed to vaporize the liquid.
30.0 g \times \frac{1 \text{ mol Hg}}{200.6 \text{ g}}\times \frac{59.3 \text{ kJ}}{1\text{ mol Hg}} = 8.87 kJ
Is your answer reasonable? The quantity of liquid mercury being vaporized is less than one mol, so it should require less energy to vaporize than the heat of vaporization of mercury, the amount of energy required to vaporize one mole of liquid.