2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
Hydrogen and fluorine react exothermically: H₂(g) + F₂(g) → 2 HF(g). The reaction can be thought of as taking place in two stages: the bonds in the reactant molecules break, and then the bonds in the product molecules form. Which statement correctly describes the potential energy of the system during these stages?
Answer and reasoning
AIt falls as the H–H and F–F bonds break and rises as the H–F bonds form A student who thinks energy is stored in bonds and released when they break picks this. Breaking bonds requires energy and raises the potential energy; forming bonds releases energy and lowers it.
BIt rises as the H–H and F–F bonds break and rises as the H–F bonds form A student who thinks making bonds, like breaking them, needs an input of energy picks this. Bond formation releases energy, so the potential energy falls as the H–F bonds form.
CIt rises as the H–H and F–F bonds break and falls as the H–F bonds formCorrect Separating bonded atoms requires energy, so the potential energy of the system rises as the H–H and F–F bonds break. Forming the H–F bonds releases energy, so the potential energy falls, and it falls by more than it rose because the reaction is exothermic.
DIt falls as the H–H and F–F bonds break and falls as the H–F bonds form A student who thinks every stage of an exothermic reaction releases energy picks this. Breaking the H–H and F–F bonds requires energy even though the reaction is exothermic overall.
Hydrazine reacts with oxygen according to the equation N₂H₄(g) + O₂(g) → N₂(g) + 2 H₂O(g). The structural formulas of the reactants and products and a table of average bond enthalpies are shown. Based on the data, what is the estimated value of ΔH for the reaction?
Answer and reasoning
A−571 kJ/molrxnCorrect Breaking one N–N, four N–H and one O=O bond requires 163 + 4(391) + 495 = 2222 kJ. Forming one N≡N bond and four O–H bonds (two in each of two H₂O molecules) releases 941 + 4(463) = 2793 kJ. ΔH ≈ 2222 − 2793 = −571 kJ/molrxn.
B+571 kJ/molrxn A student who subtracts reactants from products picks this: 2793 − 2222 = +571. The energy released in forming the product bonds is subtracted from the energy required to break the reactant bonds, which gives −571 kJ/molrxn.
C+355 kJ/molrxn A student who counts the bonds in one molecule of each substance, ignoring the coefficient 2 of H₂O, picks this: 2222 − (941 + 2 × 463) = +355. Two H₂O molecules form, so four O–H bonds form.
D−818 kJ/molrxn A student who uses each bond energy once per molecule picks this: (163 + 391 + 495) − (941 + 2 × 463) = −818. N₂H₄ has four N–H bonds and each H₂O has two O–H bonds, and every one of them must be counted.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
6.7.A.1 Bond breaking and potential energy Fix
Bond breaking and potential energy
Separating two bonded atoms requires energy, so breaking a bond increases the potential energy of the system.
Bond formation and potential energy
When two atoms form a bond, energy is released, so forming a bond decreases the potential energy of the system.
Students often think Energy is stored in chemical bonds, so breaking bonds releases energy and forming bonds takes energy in. In fact No. Breaking a bond always requires energy, which raises the potential energy of the system; energy is released when bonds form. A reaction releases energy overall only when forming the product bonds releases more energy than breaking the reactant bonds requires.
Students often think Energy has to be put in both to break bonds and to make bonds, so the potential energy of the system rises in both stages. In fact No. Breaking a bond requires energy, but forming a bond releases energy and lowers the potential energy of the system.
6.7.A.2 Average bond energy (bond enthalpy) Fix
Average bond energy (bond enthalpy)
The energy required to break one mole of a particular bond, such as O–H, in gaseous molecules, averaged over many compounds that contain the bond. It is a positive quantity, in kJ/mol, and the same amount of energy is released when one mole of the bond forms.
Estimating ΔH from average bond energies
ΔH ≈ (sum of the bond energies of all bonds broken in the reactants) − (sum of the bond energies of all bonds formed in the products), counting every bond in every molecule and multiplying by the coefficients in the balanced equation. The result is an estimate because the tabulated values are averages.
Exothermic or endothermic from bond energies
If the energy released in forming the product bonds is greater than the energy required to break the reactant bonds, the reaction is exothermic (ΔH < 0). If the energy required is greater than the energy released, the reaction is endothermic (ΔH > 0).
Students often think ΔH from bond energies is found as products minus reactants: the sum of the bond energies of the products minus the sum of the bond energies of the reactants. In fact No. ΔH ≈ energy required to break the bonds in the reactants − energy released in forming the bonds in the products, so the bond energies of the products are subtracted from those of the reactants. Subtracting the other way round gives the right size with the wrong sign.
Students often think The bonds are counted in one molecule of each substance as drawn, without multiplying by the coefficients in the balanced equation. In fact No. The bonds in one molecule must be multiplied by that substance's coefficient. In N₂H₄ + O₂ → N₂ + 2 H₂O, two water molecules form, so four O–H bonds form, not two.
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 diagram shows the potential energy of the reactants, the separated atoms and the products for the hypothetical reaction X₂(g) + Y₂(g) → 2 XY(g), for the amounts in the equation as written. Based on the diagram, what is ΔH for the reaction?
Answer and reasoning
A+100 kJ/molrxn A student who takes the energy for the product bonds minus the energy for the reactant bonds picks this: 800 − 700 = +100. The 800 kJ is released and the 700 kJ is required, so ΔH = 700 − 800 = −100 kJ/molrxn.
B+700 kJ/molrxn A student who takes the energy change of the reaction to be the energy needed to break the reactant bonds picks this: 900 − 200 = 700 kJ. The 800 kJ released when the product bonds form must also be counted.
C−800 kJ/molrxn A student who takes ΔH to be the energy released when the product bonds form picks this: 900 − 100 = 800 kJ released. The 700 kJ required to break the reactant bonds must be subtracted from it.
D−100 kJ/molrxnCorrect Breaking the bonds in the reactants requires 900 − 200 = 700 kJ, and forming the bonds in the products releases 900 − 100 = 800 kJ. More energy is released than is required, so ΔH = 700 − 800 = −100 kJ/molrxn.
Working Energy required to break the bonds in X₂ and Y₂: 900 kJ − 200 kJ = 700 kJ. Energy released in forming the bonds in 2 XY: 900 kJ − 100 kJ = 800 kJ. ΔH = 700 kJ − 800 kJ = −100 kJ/molrxn (equivalently, 100 kJ − 200 kJ).
A student uses a table of average bond enthalpies and finds that, per mole of reaction between two gases, the bond enthalpies of all the bonds in the reactants add up to 590 kJ and the bond enthalpies of all the bonds in the products add up to 750 kJ. The student will carry out the reaction in an insulated container fitted with a temperature probe. Which prediction, with its justification, is consistent with these bond enthalpy totals?
Answer and reasoning
AThe temperature rises, as energy is released both when the bonds break and when the new bonds form A student who thinks energy is released at every stage of an exothermic reaction picks this. The temperature does rise, but breaking the reactant bonds requires 590 kJ; the rise comes from the 160 kJ by which the energy released exceeds the energy required.
BThe temperature rises, as more energy is released in forming bonds than is needed to break bondsCorrect Breaking the reactant bonds requires 590 kJ and forming the product bonds releases 750 kJ. The energy released is greater than the energy required, so the reaction is exothermic, with ΔH ≈ 590 − 750 = −160 kJ/molrxn, and the energy released raises the temperature of the contents of the insulated container.
CThe temperature falls, as energy has to be taken in to break all the bonds in the reactants A student who decides the energy change from bond breaking alone picks this. Energy is required to break the bonds, but more is released when the product bonds form, so the reaction is exothermic and the temperature rises.
DThe temperature falls, as more energy is taken in to form bonds than is released as bonds break A student who thinks breaking bonds releases energy and forming bonds takes it in picks this, treating the 750 kJ as absorbed when the product bonds form and the 590 kJ as released when the reactant bonds break. It is the other way round: 590 kJ is required to break bonds and 750 kJ is released as bonds form.
For the reaction H₂(g) + Cl₂(g) → 2 HCl(g), ΔH° = −184.6 kJ/molrxn. The bond enthalpy of the H–H bond is 436 kJ/mol, and the bond enthalpy of the Cl–Cl bond is 243 kJ/mol. Based on these data, what is the bond enthalpy of the H–Cl bond?
Answer and reasoning
A247 kJ/mol A student who writes ΔH as products minus reactants picks this: −184.6 = 2D − 679 gives D = 247 kJ/mol. The energy released in forming the H–Cl bonds is subtracted from the energy required to break the H–H and Cl–Cl bonds.
B864 kJ/mol A student who ignores the coefficient 2 of HCl, counting one H–Cl bond formed, picks this: D = 679 + 184.6 = 864 kJ/mol. Two moles of H–Cl bonds form per mole of reaction, so this energy is shared between two moles of bonds.
C524 kJ/mol A student who reads ΔH° as the energy change per mole of HCl, and so doubles it for 2 HCl, picks this: (679 + 2 × 184.6)/2 = 524 kJ/mol. The value −184.6 kJ/molrxn already applies to the equation as written, in which 2 mol of HCl form.
D432 kJ/molCorrect ΔH = energy required to break bonds − energy released in forming bonds: −184.6 = (436 + 243) − 2D, where D is the H–Cl bond enthalpy and two H–Cl bonds form. So 2D = 863.6 kJ and D = 432 kJ/mol.
The diagram represents a mixture of hydrogen and oxygen before and after the two gases react to form water. Each line between two atoms represents a bond, and a double line represents one double bond. Based on the diagram, how many bonds must be broken and how many bonds must be formed to convert the contents of the Before box into the contents of the After box?
Answer and reasoning
ABroken: 6 bonds; formed: 8 bondsCorrect Four H₂ molecules and two of the three O₂ molecules react, so 4 H–H bonds and 2 O=O bonds are broken, 6 in all. Each of the four H₂O molecules has two O–H bonds, so 8 bonds are formed. The O₂ molecule that remains is unchanged.
BBroken: 7 bonds; formed: 9 bonds A student who thinks every molecule present at the start reacts picks this, breaking the bonds in all seven molecules and then forming the O=O bond of the remaining O₂ molecule again. Only two O₂ molecules are needed; the third is unchanged, so its bond is neither broken nor formed.
CBroken: 6 bonds; formed: 4 bonds A student who counts each kind of bond once for each molecule that contains it picks this, counting one O–H bond for each of the four H₂O molecules. Each H₂O molecule has two O–H bonds, so 8 bonds are formed.
DBroken: 2 bonds; formed: 2 bonds A student who counts the bonds in one molecule of each substance picks this: one H–H bond and one O=O bond broken, and the two O–H bonds of one H₂O molecule formed. The count must include every molecule that reacts and every molecule that forms.
Working Before: 4 H₂ and 3 O₂. After: 4 H₂O and 1 O₂, so 4 H₂ and 2 O₂ react and 1 O₂ is unchanged. Bonds broken: 4 H–H + 2 O=O = 6. Bonds formed: 4 H₂O × 2 O–H = 8.
Using a table of average bond enthalpies, a student estimates ΔH for the hypothetical reaction X₂(g) + Y₂(g) → 2 XY(g) as −150 kJ/molrxn. The student then finds that the value used for the X–Y bond enthalpy was 15 kJ/mol too small; the other values were correct. What estimate of ΔH does the corrected X–Y bond enthalpy give?
Answer and reasoning
A−165 kJ/molrxn A student who counts the bonds in one XY molecule, without multiplying by the coefficient 2, picks this: −150 − 15. Two moles of X–Y bonds form per mole of reaction, so the change is 30 kJ.
B−120 kJ/molrxn A student who calculates ΔH as the bond energies of the products minus those of the reactants picks this, since a larger X–Y value then raises ΔH by 2 × 15 = 30 kJ. Forming stronger bonds releases more energy, which makes ΔH more negative.
C−150 kJ/molrxn A student who thinks the energy change of a reaction is set by the energy needed to break the bonds in the reactants picks this, because only a product bond enthalpy was changed. The energy released when the X–Y bonds form is part of ΔH, so the estimate changes.
D−180 kJ/molrxnCorrect Two moles of X–Y bonds form per mole of reaction, so the energy released in forming bonds is 2 × 15 = 30 kJ larger than the student first calculated. The energy required to break bonds is unchanged, so ΔH becomes −150 − 30 = −180 kJ/molrxn.
Working ΔH ≈ (energy to break X–X and Y–Y) − (energy released forming 2 X–Y). Raising the X–Y bond enthalpy by 15 kJ/mol raises the energy released by 2 × 15 = 30 kJ/molrxn, so ΔH = −150 − 30 = −180 kJ/molrxn.
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
