2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
The heating curve shown is for a sample of water heated at a constant rate, starting as ice at −40°C. Which statement describes what happens during segment BC?
Answer and reasoning
ANo energy is absorbed, which is why the temperature of the sample stays constant A student who thinks a substance absorbing energy must rise in temperature picks this. The heater keeps supplying energy during BC; the ice absorbs it as it melts, at constant temperature.
BEnergy is absorbed as O–H bonds within the molecules break, while the temperature stays constant A student who thinks melting breaks covalent bonds within molecules picks this. Melting overcomes attractions between water molecules; the O–H bonds stay intact, and liquid water is still H₂O.
CEnergy is released as molecules leave the solid's structure, while the temperature stays constant A student who thinks breaking the attractions that hold particles together releases energy picks this. Separating attracting molecules requires energy, so melting absorbs energy.
DEnergy is absorbed as the solid's structure breaks down, while the temperature stays constantCorrect Segment BC is the flat segment at 0°C, where ice melts. Energy from the heater is absorbed to overcome attractions between the water molecules in the solid, so the system's energy increases while its temperature stays at the melting point.
The molar enthalpy of vaporization of water at 100°C is 40.7 kJ/mol. How much energy is released when 1.60 g of steam at 100°C condenses to liquid water at 100°C?
Answer and reasoning
A65.1 kJ A student who multiplies a molar enthalpy by a mass in grams picks this: (1.60)(40.7) = 65.1. The value is per mole, so the mass must first be converted to moles.
B3.61 kJCorrect n = 1.60 g ÷ 18.02 g/mol = 0.0888 mol. Condensing releases the same energy per mole that vaporizing absorbs, so the energy released is (0.0888 mol)(40.7 kJ/mol) = 3.61 kJ.
C40.7 kJ A student who thinks the molar enthalpy is the energy for the whole sample picks this. 40.7 kJ is released per mole; this sample is only 0.0888 mol.
D0.00 kJ A student who applies q = mcΔT to the phase change picks this: with ΔT = 0, q = 0. Condensation releases energy at constant temperature, found from n × ΔH.
Working n = 1.60 g ÷ 18.02 g/mol = 0.0888 mol. Condensation is the reverse of vaporization: molar enthalpy of condensation = −40.7 kJ/mol, so the energy released = (0.0888 mol)(40.7 kJ/mol) = 3.61 kJ.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.5.A.1 Phase transition Fix
Phase transition
A change of a substance from one phase to another (melting, freezing, vaporization or boiling, condensation) in which the substance itself does not change.
Energy change on melting and boiling
Energy must be transferred to a system to melt or boil a substance, so the energy of the system increases. Freezing and condensation release energy, so the energy of the system decreases.
Constant temperature during a phase change
While a pure substance melts or boils, its temperature stays constant: the energy transferred overcomes attractions between particles rather than increasing their average kinetic energy.
Heating curve
A graph of temperature against time or energy added for a substance heated at a steady rate. Sloping segments show one phase warming; flat segments show a phase change taking place at constant temperature.
Students often think A substance that absorbs energy must rise in temperature, so a substance whose temperature stays constant during heating is absorbing no energy. In fact No. While a pure substance melts or boils it absorbs energy at constant temperature; the energy overcomes attractions between particles instead of increasing their average kinetic energy, so the energy of the system increases.
Students often think When a molecular substance melts or boils, covalent bonds within its molecules break. In fact No. Melting and boiling a molecular substance such as water overcome attractions between molecules; the covalent bonds within each molecule stay intact, so the substance is still H₂O in all three phases.
6.5.A.2 Molar enthalpy of fusion (ΔHfus) Fix
Molar enthalpy of fusion (ΔHfus)
The energy absorbed when 1 mol of a substance melts at its melting point, in kJ/mol. The same amount of energy is released when 1 mol of the liquid freezes.
Molar enthalpy of vaporization (ΔHvap)
The energy absorbed when 1 mol of a liquid vaporizes, in kJ/mol; for water at 100°C it is 40.7 kJ/mol.
Complementary phase changes
A phase change and its reverse involve equal amounts of energy in opposite directions: the molar enthalpy of condensation equals −ΔHvap, and the molar enthalpy of freezing equals −ΔHfus.
Heat for a phase change
The heat absorbed or released when a sample changes phase is the number of moles that change phase multiplied by the molar enthalpy of that phase change (q = n × ΔH).
Students often think A molar enthalpy of a phase change can be multiplied directly by the mass of the sample in grams. In fact No. A molar enthalpy of vaporization or fusion is in kJ per mole, so it must be multiplied by the number of moles; a mass in grams is first divided by the molar mass.
Students often think The molar enthalpy of a phase change is the energy absorbed or released by the whole sample, whatever its amount, so the energy measured for a sample is its molar enthalpy. In fact No. A molar enthalpy is the energy per mole, so the energy for a sample is the number of moles that change phase multiplied by the molar enthalpy.
3 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 3
A 2.00 mol sample of a pure solid, X, is heated, and the graph shows its temperature as energy is added. Based on the graph, what is the molar enthalpy of fusion of X?
Answer and reasoning
A6.00 kJ/molCorrect The flat segment at 40°C, where X melts, runs from 4.0 kJ to 16.0 kJ, so melting absorbs 12.0 kJ. Dividing by the 2.00 mol that melt gives ΔHfus = 6.00 kJ/mol.
B8.00 kJ/mol A student who reads the energy at the end of the flat segment instead of the change across it picks this: 16.0 kJ ÷ 2.00 mol. The first 4.0 kJ warmed the solid; only 12.0 kJ was absorbed in melting.
C12.0 kJ/mol A student who takes the energy for the whole sample as the molar enthalpy picks this. The 12.0 kJ melted 2.00 mol, so the energy per mole is 6.00 kJ/mol.
D2.00 kJ/mol A student who thinks the solid melts while its temperature is rising picks the sloping first segment: 4.0 kJ ÷ 2.00 mol. That segment is the solid warming; melting is the flat segment at 40°C.
Working The flat segment (melting at 40°C) runs from 4.0 kJ to 16.0 kJ, so the energy absorbed in melting = 16.0 kJ − 4.0 kJ = 12.0 kJ. ΔHfus = 12.0 kJ ÷ 2.00 mol = 6.00 kJ/mol.
A pot of water is boiling steadily over a burner, and a thermometer in the water reads 100°C throughout. A student claims that the boiling water is not absorbing energy from the burner. Which statement best evaluates the claim?
Answer and reasoning
AThe claim is correct; absorbing energy would make the water hotter, but this water stays at 100°C A student who thinks absorbed energy must raise the temperature picks this. During boiling the temperature of a pure substance stays constant while it absorbs energy.
BThe claim is incorrect; the energy absorbed breaks the O–H bonds in the water molecules as they boil A student who thinks boiling breaks covalent bonds within molecules picks this. Boiling overcomes attractions between water molecules; the vapor is still H₂O, so its O–H bonds are intact.
CThe claim is incorrect; the water absorbs energy as it changes from liquid to gas, so its energy increasesCorrect Energy must be transferred to a system for a liquid to boil. The water absorbs energy from the burner as it changes from liquid to gas, so the energy of the water increases even though its temperature stays at 100°C.
DThe claim is correct; the burner's energy is carried away by bubbles of air rising out of the water A student who thinks the bubbles in boiling water are air picks this. The bubbles are water vapor, formed as the water absorbs energy from the burner and changes to gas.
A student wants to determine the molar enthalpy of vaporization of a pure liquid. The liquid is kept boiling by an electric heater, and the energy supplied by the heater is measured. Assuming that all of this energy is absorbed by the boiling liquid, which other measurement does the student need?
Answer and reasoning
AThe temperature change of the liquid while the energy is supplied A student who thinks q = mcΔT applies to a phase change picks this. The boiling liquid stays at its boiling point, so ΔT ≈ 0 even though energy is absorbed; the energy must be divided by the moles vaporized.
BThe volume of vapor formed, to divide by 22.4 L/mol A student who thinks every gas occupies 22.4 L per mole picks this. That value applies only to an ideal gas at STP, and the vapor forms at the liquid's boiling point; the moles are found from the mass vaporized.
CNone; the energy supplied is the molar enthalpy of vaporization A student who thinks a molar enthalpy is the energy for the whole sample picks this. The molar enthalpy is the energy per mole, so the moles vaporized must be known.
DThe mass of liquid that vaporizes while the energy is suppliedCorrect Molar enthalpy of vaporization = energy absorbed ÷ moles vaporized. The moles come from the mass of liquid that vaporizes and the molar mass, so that mass must be measured, for example from the loss in mass of the container.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
