The molar heat of fusion of sodium metal is \(2.60 \mathrm{~kJ} / \mathrm{mol}\), whereas its heat of vaporization is \(97.0 \mathrm{~kJ} / \mathrm{mol}\). a. Why is the heat of vaporization so much larger than the heat of fusion? b. What quantity of heat would be needed to melt \(1.00 \mathrm{~g}\) sodium at its normal melting point? c. What quantity of heat would be needed to vaporize \(1.00 \mathrm{~g}\) sodium at its normal boiling point? d. What quantity of heat would be evolved if \(1.00 \mathrm{~g}\) sodium vapor condensed at its normal boiling point?

The heat of fusion is the amount of heat required to change a substance from a solid to a liquid, while the heat of vaporization is the amount of heat required to change a substance from a liquid to a gas. For most substances, including sodium, the heat of vaporization is significantly larger than the heat of fusion. This is because when the substance transforms from solid to liquid, the bonds between the particles are weakened, allowing them to move more freely. However, when the substance vaporizes (transforms from liquid to gas), the bonds must be completely broken, which requires more energy. In this case, sodium's heat of vaporization is about 37 times higher than its heat of fusion, which indicates that it takes a lot more heat to convert sodium from a liquid to a gas than from a solid to a liquid.

To calculate the heat required to melt 1.00 g of sodium at its normal melting point, we'll use the molar heat of fusion and perform the following steps: 1. Find the moles (n) of sodium: \(n = \dfrac{mass}{molar ~mass}\), where the molar mass of sodium is 22.9897 g/mol. 2. Calculate the heat required: \(q = n * \Delta H_{fusion}\), where \(\Delta H_{fusion}\) is the molar heat of fusion (2.60 kJ/mol). Applying these steps, we get: \(n = \dfrac{1.00}{22.9897} = 0.0435 ~mol\) \(q = 0.0435 * 2.60 = 0.113 ~kJ\) Therefore, the heat required to melt 1.00 g of sodium at its normal melting point is 0.113 kJ.

To calculate the heat required to vaporize 1.00 g of sodium at its normal boiling point, we'll use the molar heat of vaporization and perform the following steps: 1. Use the moles (n) of sodium that we calculated in part b. 2. Calculate the heat required: \(q = n * \Delta H_{vaporization}\), where \(\Delta H_{vaporization}\) is the molar heat of vaporization (97.0 kJ/mol). Applying these steps, we get: \(q = 0.0435 * 97.0 = 4.219 ~kJ\) Therefore, the heat required to vaporize 1.00 g of sodium at its normal boiling point is 4.219 kJ.

The heat released during the condensation is equal in magnitude, but opposite in sign, to the heat required for vaporization. Therefore, the heat evolved when 1.00 g of sodium vapor condenses at its normal boiling point would be -4.219 kJ. In conclusion, we have explained why the heat of vaporization is larger than the heat of fusion and calculated the heat required for melting and vaporization of 1.00 g sodium, as well as the heat evolved during condensation.