Changes of State: Molar Enthalpy of Vaporization and Fusion

Recall: What is meant by the molar enthalpy (heat) of vaporization, ΔHvap?

• The heat required to raise the temperature of any sample by 1 °C

• The heat released when one mole of gas condenses to a liquid

• The heat required to vaporize one mole of a liquid

• The heat required to melt one mole of a solid

Skill/Concept: According to the phase-change diagram for water, which numerical relationship is always true?

• ΔHvap = −ΔHcond

• ΔHfus = +ΔHcond

• ΔHsolid = +ΔHvap

• ΔHfus = −ΔHvap

Refer to the table titled "Standard Enthalpies of Combustion". Which substance has the least exothermic standard enthalpy of combustion (largest value closest to zero) under standard state conditions?

• Sucrose (C12H22O11, s)

• Octane (C8H18, l)

• Glucose (C6H12O6, s)

• Propane (C3H8, g)

• Methane (CH4, g)

A sample has mass m and molar mass MM. The reaction has molar enthalpy change ΔH (kJ/mol). According to the formula shown under the diagrams, which single expression gives the heat q released or absorbed by this sample?

• q = (m ÷ MM) × ΔH

• q = (MM ÷ m) × ΔH

• q = m × ΔH

• q = ΔH ÷ (m ÷ MM)

The worked example converts 54.0 g of C6H12O6 to moles using the molar mass 180.18 g/mol. What is the calculated amount, to three significant figures?

• 0.300 mol C6H12O6

• 0.180 mol C6H12O6

• 3.34 mol C6H12O6

• 0.540 mol C6H12O6

Given the balanced reaction C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) with ΔH_comb = −2808 kJ per mole of glucose, what is the heat released when 0.300 mol of glucose is burned? Report the magnitude in kJ.

• 842 kJ

• 936 kJ

• 2808 kJ

• 468 kJ

Before calculating exactly, the analysis predicts that the energy released by burning 54.0 g of glucose will be less than one-third of |ΔH_comb|. Which reasoning best supports this prediction?

• Because 54.0 g is less than one-third of glucose’s molar mass, fewer than one-third of a mole will burn.

• Because the coefficients of O2 and CO2 are both six, the energy must be divided by six.

• Because bomb calorimeters at constant volume reduce the heat by a factor of three.

• Because ΔH_comb applies per mole of oxygen, and only two moles of oxygen react.

A student needs to calculate the heat required to melt a sample using q = m × ΔH_fus. For a 25.7 g sample of solid methanol at its melting point, which expression correctly sets up the calculation before inserting the tabulated molar heat of fusion value from a data table? Assume ΔH_fus is given per gram or must be converted from kJ/mol as needed.

• q = 25.7 g × ΔH_fus

• q = 25.7 g ÷ ΔH_fus

• q = ΔH_fus ÷ 25.7 g

When 275 g of ammonia condenses at its boiling point, which statement about the heat involved is correct if ΔH_cond is the enthalpy of condensation (a negative of ΔH_vap)?

• Heat is released and equals q = m × ΔH_cond.

• Heat is absorbed and equals q = m ÷ ΔH_cond.

• No heat is exchanged because temperature is constant.

• Heat is released and equals q = m × c × ΔT.

Which conversion factor correctly converts grams of CH3OH to moles of CH3OH in the melting calculation?

• 1 mol CH3OH / 32.04 g CH3OH

• 32.04 g CH3OH / 1 mol CH3OH

• 1 g CH3OH / 32.04 mol CH3OH

• 3.22 kJ / 1 mol CH3OH