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Chapter 10: Problem 99

The molar heat of fusion of sodium metal is 2.60 kJ/mol,whereas its heat of vaporization is 97.0 kJ/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 g sodium at its normal melting point? c. What quantity of heat would be needed to vaporize 1.00 g sodium at its normal boiling point? d. What quantity of heat would be evolved if 1.00 g sodium vapor condensed at its normal boiling point

Short Answer

Expert verified
a. The heat of vaporization is larger than the heat of fusion because breaking intermolecular bonds in a liquid state requires more energy than in a solid state. b. 0.113 kJ of heat is needed to melt 1.00 g of sodium at its normal melting point. c. 4.22 kJ of heat is needed to vaporize 1.00 g of sodium at its normal boiling point. d. 4.22 kJ of heat would be evolved if 1.00 g of sodium vapor condensed at its normal boiling point.

Step by step solution

01

a. Comparing heat of fusion and heat of vaporization

Heat of fusion refers to the amount of heat needed to convert a substance from solid to liquid state without changing temperature. Heat of vaporization is the amount of heat required to convert a substance from liquid to gas state, also without changing temperature. The heat of vaporization is larger than the heat of fusion because breaking the intermolecular bonds in a liquid state requires more energy than breaking those in a solid state. In a liquid state, the particles have more freedom to move, so there are more bonds to break before they can become gas particles.
02

b. Calculate the heat needed to melt 1.00 g sodium

To calculate the amount of heat needed, we'll use the formula: Heat (q) = n * ΔH_f, where n is the number of moles, and ΔH_f is the molar heat of fusion. First, we need to convert mass (1.00 g) into moles using the molar mass of sodium (Na) which is 22.99 g/mol: n = mass / molar_mass n = 1.00 g / 22.99 g/mol ≈ 0.0435 mol Now substitute the values into the formula: q = 0.0435 mol * 2.60 kJ/mol ≈ 0.113 kJ So, 0.113 kJ of heat is needed to melt 1.00 g of sodium at its normal melting point.
03

c. Calculate the heat needed to vaporize 1.00 g sodium

Again, we'll use the formula Heat (q) = n * ΔH_v, where this time ΔH_v is the molar heat of vaporization. We have already calculated the number of moles, now substitute the values into the formula: q = 0.0435 mol * 97.0 kJ/mol ≈ 4.22 kJ So, 4.22 kJ of heat is needed to vaporize 1.00 g of sodium at its normal boiling point.
04

d. Quantity of heat evolved during condensation

When sodium vapor condenses, it releases the same amount of heat as it takes to vaporize it. Therefore, the amount of heat evolved in this process is the same as the amount needed to vaporize 1.00 g of sodium at its normal boiling point. So, 4.22 kJ of heat would be evolved if 1.00 g of sodium vapor condensed at its normal boiling point.

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