3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
For the reaction 2 SO₂(g) + O₂(g) → 2 SO₃(g), ΔH° = −198 kJ/molrxn. How much heat is released when 1.20 mol of SO₂(g) is consumed in this reaction at constant pressure?
Answer and reasoning
A238 kJ A student who applies ΔH per mole of SO₂, ignoring its coefficient of 2, picks this: 1.20 × 198 = 238 kJ. 198 kJ is released for every 2 mol of SO₂ consumed.
B475 kJ A student who multiplies the moles of SO₂ by its coefficient picks this: 1.20 × 2 × 198 = 475 kJ. Moles of reaction = moles of SO₂ ÷ 2, not × 2.
C119 kJCorrect One mole of reaction consumes 2 mol of SO₂, so 1.20 mol of SO₂ is 0.600 mol of reaction. The heat released is 0.600 molrxn × 198 kJ/molrxn = 119 kJ.
D198 kJ A student who thinks the ΔH value is the heat released whatever amount reacts picks this. 198 kJ is released per mole of reaction; only 0.600 mol of reaction occurs here.
Two solutions, both at 21.0°C, are mixed in an open foam cup, and an exothermic reaction occurs. The graph shows the temperature of the mixture over time. Which statement best explains the change in temperature after the maximum is reached?
Answer and reasoning
AThe energy released by the reaction is stored in the bonds of the products A student who thinks energy is stored in chemical bonds picks this. Once the reaction is over, the products do not take back energy; the mixture cools because it transfers energy to the surroundings.
BCold from the room flows into the mixture until it reaches room temperature A student who thinks cold flows into warm objects picks this. Nothing flows in; the warm mixture loses energy to the cooler room.
CThe energy released by the reaction is used up once the reaction has stopped A student who thinks energy is used up picks this. Energy is conserved; the energy leaving the mixture is gained by the surroundings.
DThermal energy is transferred from the warmer mixture to the cooler surroundingsCorrect The reaction released thermal energy, so the products and solution became warmer than the room. They then transfer thermal energy to the cooler surroundings, and the temperature falls slowly toward room temperature as they approach thermal equilibrium.
When aqueous solutions of an acid and a base are mixed, an exothermic reaction occurs and the temperature of the solution rises. Which statement best explains the rise in temperature at the particle level?
Answer and reasoning
ABreaking the bonds in the reactants releases the energy stored in them, which makes all of the particles move faster A student who thinks energy is stored in bonds and released when they break picks this. Breaking bonds requires energy; the energy comes out when the new bonds in the products form.
BThe products have lower potential energy than the reactants, and the difference becomes kinetic energyCorrect Bonds broken and formed in the reaction leave the products with less chemical potential energy than the reactants. The energy difference increases the average kinetic energy of the particles, which is measured as a higher temperature.
CThe products have higher potential energy than the reactants, and that extra energy is what makes the solution hotter A student who thinks more chemical potential energy means a higher temperature picks this. If the products had more potential energy, that energy would come from the particles' kinetic energy and the solution would cool.
DThe reaction produces particles of heat, which spread out among the water molecules and raise the solution's temperature A student who thinks heat is made of particles picks this. The energy released makes the existing particles move faster; no particles of heat are produced.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.6.A.1 Enthalpy change of reaction (ΔH) Fix
Enthalpy change of reaction (ΔH)
The heat released or absorbed by a chemical reaction carried out at constant pressure. A negative ΔH means heat is released (exothermic); a positive ΔH means heat is absorbed (endothermic).
Molar enthalpy of reaction (kJ/molrxn)
ΔH for the amounts shown in the balanced equation as written. The heat for a given amount of reaction is q = nrxn × ΔH, where nrxn = moles of a reactant or product ÷ its coefficient in the equation.
Mole of reaction
The reaction taking place once in the molar amounts given by the coefficients: for 2 SO₂ + O₂ → 2 SO₃, 1 mol of reaction consumes 2 mol of SO₂ and 1 mol of O₂.
ΔH from solution calorimetry
The heat absorbed or released by the solution (q = mcΔT) is the negative of the heat of the reaction, so ΔH = −qsolution ÷ nrxn.
Students often think A ΔH given in kJ/mol is the heat for each mole of any reactant, so the moles of any reactant can be multiplied directly by ΔH. In fact No. kJ/molrxn is per mole of reaction as written. For 2 SO₂ + O₂ → 2 SO₃ with ΔH = −198 kJ/molrxn, 198 kJ is released for each 2 mol of SO₂ consumed, which is 99 kJ per mole of SO₂.
Students often think The moles of a reactant are multiplied by its coefficient to find the amount of reaction, so a reactant with coefficient 2 gives twice as much reaction. In fact No. Moles of reaction = moles of the species ÷ its coefficient. Because 2 mol of SO₂ are consumed per mole of reaction, 1.20 mol of SO₂ corresponds to 0.600 mol of reaction.
6.6.A.2 Thermal equilibrium after a reaction Fix
Thermal equilibrium after a reaction
Products formed at a temperature different from that of their surroundings exchange energy with the surroundings until both are at the same temperature.
Exothermic reaction
A reaction in which thermal energy is transferred to the surroundings as reactants convert to products; in a calorimeter the temperature of the mixture rises.
Endothermic reaction
A reaction in which thermal energy is transferred from the surroundings as reactants convert to products; in a calorimeter the temperature of the mixture falls.
Students often think Cold flows from cool surroundings into a warm object, lowering its temperature. In fact No. A warm mixture cools because it transfers thermal energy to the cooler surroundings. Cold is not a substance or a form of energy that moves.
Students often think The energy released by a reaction is used up once the reaction stops, so the temperature of the mixture falls. In fact No. Energy is conserved. The warm mixture transfers thermal energy to its cooler surroundings until the mixture and the surroundings reach the same temperature.
6.6.A.3 Chemical potential energy Fix
Chemical potential energy
Energy associated with the arrangement of atoms and the bonds between them. Reactants and products differ in chemical potential energy because bonds are broken and formed in the reaction.
Potential energy to kinetic energy in a reaction
When products have lower chemical potential energy than reactants, the energy difference increases the average kinetic energy of the particles, which is observed as a rise in temperature; the reverse gives a fall in temperature.
Students often think Energy is stored in chemical bonds: breaking bonds releases the stored energy, and energy can be put back into the bonds of the products. In fact No. Breaking a bond always requires energy; energy is released when bonds form. A reaction releases energy overall when the bonds formed in the products release more energy than is needed to break the bonds in the reactants.
Students often think Products with more chemical potential energy than the reactants make the mixture hotter, because higher energy means a higher temperature. In fact No. Temperature reflects the average kinetic energy of the particles. When products have more chemical potential energy than the reactants (endothermic), that energy came from the kinetic energy of the particles and the surroundings, so the mixture cools.
5 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 5
The energy diagram shown is for the hypothetical reaction X(g) + 2 Y(g) → 2 Z(g), carried out at constant pressure, with the energies given for the amounts in the equation as written. Based on the diagram, how much heat is released when 0.60 mol of Y(g) reacts?
Answer and reasoning
A36 kJCorrect ΔH = energy of products − energy of reactants = 40 kJ − 160 kJ = −120 kJ per mole of reaction. 0.60 mol of Y is 0.30 mol of reaction, so 0.30 × 120 kJ = 36 kJ is released.
B72 kJ A student who applies ΔH per mole of Y, ignoring its coefficient of 2, picks this: 0.60 × 120 = 72 kJ. 120 kJ is released for every 2 mol of Y.
C48 kJ A student who takes the height of the peak above the reactants (160 kJ) as the heat of reaction picks this: 0.30 × 160 = 48 kJ. That height is the activation energy; ΔH is the product level minus the reactant level.
D12 kJ A student who reads the energy level of the products (40 kJ) as the heat of reaction picks this: 0.30 × 40 = 12 kJ. ΔH is the difference between the product and reactant levels, −120 kJ.
Working From the diagram, ΔH = 40 kJ − 160 kJ = −120 kJ/molrxn. nrxn = 0.60 mol Y ÷ 2 = 0.30 molrxn. Heat released = 0.30 × 120 kJ = 36 kJ.
A student mixes the two solutions described in the table in a foam-cup calorimeter. Assume that the mixture has a density of 1.00 g/mL and a specific heat capacity of 4.18 J/(g·°C) and that no heat is lost to the surroundings. Based on the data, what is ΔH for the reaction H⁺(aq) + OH⁻(aq) → H₂O(l)?
Answer and reasoning
A−37.3 kJ/molrxn A student who counts only the 50.0 g of acid solution as the mass that warms picks this. Both solutions are mixed and warm together, so the mass is 75.0 g.
B−28.0 kJ/molrxn A student who divides by the total moles of HCl and NaOH, 0.100 mol, picks this. One mole of reaction uses 1 mol of H⁺ and 1 mol of OH⁻, so 0.0500 mol of each is 0.0500 mol of reaction.
C−2.80 kJ/molrxn A student who thinks ΔH is the heat released in the experiment picks this, without dividing by the moles of reaction. ΔH is per mole of reaction: −2.80 kJ ÷ 0.0500 molrxn.
D−56.0 kJ/molrxnCorrect The 75.0 g of mixture warms by 8.93°C, so qsolution = (75.0)(4.18)(8.93) J = 2.80 kJ. HCl and NaOH are each 0.0500 mol, so 0.0500 mol of reaction releases 2.80 kJ, and ΔH = −2.80 kJ ÷ 0.0500 molrxn = −56.0 kJ/molrxn.
Working Mass of mixture = 75.0 mL × 1.00 g/mL = 75.0 g. ΔT = 29.03 − 20.10 = 8.93°C. qsolution = (75.0 g)(4.18 J/(g·°C))(8.93°C) = 2.80 × 10³ J = 2.80 kJ. Moles: HCl 0.0500 L × 1.00 M = 0.0500 mol; NaOH 0.0250 L × 2.00 M = 0.0500 mol, so 0.0500 molrxn. ΔH = −2.80 kJ ÷ 0.0500 molrxn = −56.0 kJ/molrxn.
In experiment 1, 50.0 mL of 1.00 M HCl(aq) and 50.0 mL of 1.00 M NaOH(aq), both at 21.0°C, are mixed in a foam-cup calorimeter, and the temperature change ΔT and the value of ΔH in kJ/molrxn are determined. In experiment 2, the procedure is repeated with 100.0 mL of each solution, both at 21.0°C. How do the results of experiment 2 compare with those of experiment 1?
Answer and reasoning
AΔT is about doubled, and ΔH is about the same A student who thinks ΔT depends only on how much reaction occurs picks this. The doubled heat is shared by a doubled mass of solution, so ΔT stays about the same.
BΔT is about the same, and ΔH is twice as negative A student who thinks ΔH is the heat released in the experiment picks this. ΔH is per mole of reaction, so doubling the amounts doubles q but not ΔH.
CΔT is about the same, and ΔH is about the sameCorrect Twice as much reaction releases twice the heat, but that heat warms twice the mass of solution, so ΔT is about the same. ΔH is heat per mole of reaction; both the heat and the moles of reaction double, so ΔH is unchanged.
DΔT is about doubled, and ΔH is twice as negative A student who thinks every result doubles when the amounts double picks this. ΔT (heat per mass of solution) and ΔH (heat per mole of reaction) are both ratios whose parts double together, so neither changes.
Working Experiment 2 has twice the moles of reaction, so twice the heat is released, but twice the mass of solution absorbs it: ΔT = q/(mc) is about the same. ΔH = −q/nrxn: q and nrxn both double, so ΔH is the same.
A student mixes 50.0 mL of 1.00 M HCl(aq) with 50.0 mL of 1.00 M NaOH(aq) in a foam-cup calorimeter to determine ΔH for the reaction that occurs. The student's value is −51 kJ/molrxn, less negative than the accepted value of −56 kJ/molrxn. The student's calculations contain no errors and use the amounts that were actually mixed. Which of the following could account for the difference?
Answer and reasoning
AThe thermometer read 0.5°C higher than the true temperature for each reading taken A student who thinks any thermometer error changes the measured temperature change picks this. A constant offset is the same in the initial and final readings, so it cancels in ΔT and does not affect ΔH.
BThe cup let thermal energy from the reaction mixture pass through it to the surroundingsCorrect In an exothermic reaction, thermal energy is transferred to the surroundings. If some of it leaves the cup instead of warming the solution, the measured ΔT is too small, so the calculated heat and the magnitude of ΔH are too small: ΔH is less negative.
CThe solutions mixed were less concentrated than those used to find the accepted value A student who thinks ΔH is the total heat released, so that less reactant gives a smaller ΔH, picks this. Less concentrated solutions release less heat, but ΔH is the heat per mole of reaction, which does not depend on how much reacts.
DThe temperatures were measured in degrees Celsius rather than in kelvins for each reading A student who thinks a temperature change in °C is smaller than in kelvins picks this. A change of 1°C is a change of 1 K, so ΔT and ΔH are the same on either scale.
A student mixes 50.0 mL of HCl(aq) with 50.0 mL of NaOH(aq) in a foam-cup calorimeter; the NaOH is in excess. From the mass and specific heat capacity of the mixture and its initial and maximum temperatures, the student calculates the heat released, q. Which additional quantity does the student need in order to determine ΔH for the reaction H⁺(aq) + OH⁻(aq) → H₂O(l) in kJ/molrxn?
Answer and reasoning
AThe concentration of the NaOH(aq) solution A student who thinks the amount of any reactant gives the moles of reaction picks this. The NaOH is in excess, so some of it is left over, and the amount present does not show how much reaction took place.
BThe concentration of the HCl(aq) solutionCorrect ΔH in kJ/molrxn is the heat released divided by the moles of reaction. The HCl is completely used up, so the moles of reaction equal the moles of HCl, which are found from its concentration and its volume, 50.0 mL.
CThe temperature change converted to kelvins A student who thinks a temperature change in degrees Celsius is smaller than the same change in kelvins picks this. A temperature change has the same size on both scales, so q is already correct.
DNothing more, as the heat released equals ΔH A student who thinks ΔH is the heat released in the experiment, whatever amount reacts, picks this. The heat released depends on how much HCl reacted, so it must be divided by the moles of reaction to give ΔH in kJ/molrxn.
Working ΔH = −q/(moles of reaction). The HCl is the limiting reactant, so moles of reaction = moles of H⁺ = (concentration of HCl)(0.0500 L). The volume is known, so the missing quantity is the concentration of the HCl(aq).
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
