2-Day AP/Pre-AP Science Conference Corpus Christi Omni-Marina Hotel January 20 & 21, 2006 Misconceptions from the Perspective of AP Exam Grader and Chief Reader John I. Gelder Former Chief Reader AP Chemistry Department of Chemistry Oklahoma State University

* From Dr. Michael R Abraham talk on Misconceptions

Misconceptions or Alternate Conceptions Students have common, nonrandom conceptions concerning key scientific ideas that are at odds with scientific views Many of these conceptions are pervasive, stable, and resistant to change * From Dr. Michael R Abraham talk on Misconceptions

Comments Terms: Misconceptions, Preconceptions, Alternative Hypotheses, Naïve Theories, Alternative Frameworks, Children’s Science Chemistry as basis of focus Stages of interest in misconceptions amusement irritation fascination * From Dr. Michael R Abraham talk on Misconceptions

Enthalpy Enthalpy of Formation: 2Fe(s) + 3/2O2(g)  Fe2O3(s) ∆H˚f = -824 kJ mol-1 Fe(s) + 1/2O2(g)  FeO(s) ∆H˚f = -272 kJ mol-1 Calculate ∆H˚rxn for the reaction 2FeO(s) + 1/2O2(g)  Fe2O3(s)

Enthalpy Enthalpy of Formation: 2Fe(s) + 3/2O2(g)  Fe2O3(s) ∆H˚f = -824 kJ mol-1 Fe(s) + 1/2O2(g)  FeO(s) ∆H˚f = -272 kJ mol-1 Calculate ∆H˚rxn for the reaction 2FeO(s) + 1/2O2(g)  Fe2O3(s) ∆H˚rxn = n∆H˚f(products) - m∆H˚f(reactants) ∆H˚rxn = 1·∆H˚f(Fe2O3(s)) - [2·∆H˚f(FeO(s)) + 1/2· ∆H˚f(O2(g))] ∆H˚rxn = 1·-824 kJ mol-1 - (2·-272 kJ mol-1 + 1/2·0 kJ mol-1 ) ∆H˚rxn = -280 kJ mol-1

Enthalpy Enthalpy of reactions: 2Fe(s) + 3/2O2(g)  Fe2O3(s) ∆H˚f = -824 kJ mol-1 Fe(s) + 1/2O2(g)  FeO(s) ∆H˚f = -272 kJ mol-1 2FeO(s) + 1/2O2(g)  Fe2O3(s) ∆H˚rxn = -280 kJ mol-1 (per mol of reaction) ∆H˚rxn for 4FeO(s) + O2(g)  2Fe2O3(s) ?

Enthalpy Enthalpy of reactions: 2Fe(s) + 3/2O2(g)  Fe2O3(s) ∆H˚f = -824 kJ mol-1 Fe(s) + 1/2O2(g)  FeO(s) ∆H˚f = -272 kJ mol-1 2FeO(s) + 1/2O2(g)  Fe2O3(s) ∆H˚rxn = -280 kJ mol-1 ∆H˚rxn for 4FeO(s) + O2(g)  2Fe2O3(s) ∆H˚rxn = -560 kJ mol-1 Mol of reaction! ∆H˚rxn = -560 kJ per 2 mol Fe2O3 ∆H˚rxn = -560 kJ per 4 mol FeO ∆H˚rxn = -560 kJ per 1 mol O2

Free Energy (∆G˚) ∆G˚ has units of kJ mol-1 ∆G˚ = -RT ln K R = 8.314 J mol-1 K-1 T (K) To calculate ∆G˚ given K and the standard temperature the units on ∆G˚ are kJ mol-1. To calculate K given ∆G˚ and the standard temperature the units on ∆G˚ must be kJ mol-1 so K is unitless.

2002 AP Chemistry Exam Q7 An environmental concern is the depletion of O3 in Earth’s upper atmosphere, where O3 is normally in equilibrium with O2 and O. A proposed mechanism for the depletion of O3 in the upper atmosphere is shown below. Step I O3 + Cl  O2 + ClO Step II ClO + O  Cl + O2 b) Clearly identify the catalyst in the mechanism above. Justify your answer.

2002 AP Chemistry Exam Q7 An environmental concern is the depletion of O3 in Earth’s upper atmosphere, where O3 is normally in equilibrium with O2 and O. A proposed mechanism for the depletion of O3 in the upper atmosphere is shown below. Step I O3 + Cl  O2 + ClO Step II ClO + O  Cl + O2 Clearly identify the catalyst in the mechanism above. Justify your answer. Cl is the catalyst in the reaction. It is a reactant in Step I and reappears as a product in Step II. 1 point earned for identifying Cl as the catalyst 1 point earned for justification of Cl as the catalyst

2002 AP Chemistry Exam Q7 An environmental concern is the depletion of O3 in Earth’s upper atmosphere, where O3 is normally in equilibrium with O2 and O. A proposed mechanism for the depletion of O3 in the upper atmosphere is shown below. Step I O3 + Cl  O2 + ClO Step II ClO + O  Cl + O2 Clearly identify the catalyst in the mechanism above. Justify your answer. Most students correctly identified Cl as the catalyst. Their justification was often the textbook definition of a catalyst rather than connecting the catalyst to the steps in the mechanism(no point). A common misconception appears to be that students believe that if a catalyst does not appear in the final balanced equation, then it has no effect on the reaction. Cl is the catalyst in the reaction. It is a reactant in Step I and reappears as a product in Step II. 1 point earned for identifying Cl as the catalyst 1 point earned for justification of Cl as the catalyst

2002 AP Chemistry Exam Q6 Use the principles of atomic structure and/or chemical bonding to explain each of the following. In each part, your answer must include references to both substances. (a) The atomic radius of Li is larger than that of Be.

2002 AP Chemistry Exam Q6 Use the principles of atomic structure and/or chemical bonding to explain each of the following. In each part, your answer must include references to both substances. The atomic radius of Li is larger than that of Be. Both Li and Be have their outer electrons in the same shell (and/or they have the same number of inner core electrons shielding the valence electrons from the nucleus). However, Be has four protons and Li has only three protons. Therefore, the effective nuclear charge experienced (attraction experienced) by the valence (outer) electrons is greater in Be than in Li, so Be has a smaller atomic radius. 1 point earned for indicating that Be has more protons than Li. 1 point earned for indicating that since the electrons are at about the same distance from the nucleus, there is more attraction in Be as a result of the larger number of protons.

2002 AP Chemistry Exam Q6 Use the principles of atomic structure and/or chemical bonding to explain each of the following. In each part, your answer must include references to both substances. The atomic radius of Li is larger than that of Be. electron shielding or pairing energy is more important to atomic radius than nuclear charge (often also implying atoms with no nucleus; electron/proton ratio changes across the period; two electrons must have twice as much attraction to a nucleus as one; electronegativity is an explanation, not a result; electrons are point charges pulling the nucleus towards them atomic radius is determined by the size of the nucleus

2002 AP Chemistry Exam Q5 The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0 M NaOH. Indicate whether the value of q increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. The ∆T will be greater, so q  increases. There are more moles of  HCl and NaOH reacting so the final temperature of the mixture will be higher. 1 point for direction and explanation.

2002 AP Chemistry Exam Q5 The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0 M NaOH. Indicate whether the value of q increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. Arguments about increased mass are not acceptable because the total mass increase is negligible (the solutions have virtually the same density) and is not the driving force for increases in q. The ∆T will be greater, so q increases. There are more moles of HCl and  NaOH reacting so the final temperature of the mixture will be higher. 1 point for direction and explanation

Other Common Misconceptions 0. If it is not in the Acorn book, I do not touch it. Change in mass is a measurement. half filled or filled shells are happy Because of Le Chatelier's principle That just identifying a type of IMF is all that is necessary and that should imply relative strength. When IMF are the same that mass explains the stronger force. That the ionization energy increase going across a period because the atomic radius gets smaller. 7.  it is harder to remove the second electron in sodium, because filled shell is more stable.

Other Common Misconceptions 96,500 has 5 significant figures 9.   Phase change is the most important factor when determining change in entropy. That alcohols are strong bases. That all equivalence points are at pH=7. Who cares about Q! That the radius gets smaller going across a period just because there are more protons in the nucleus. The radius gets larger going down a group because there are more electrons. That restating the question is the answer. 16. If fluorine in in the problem that electronegativity is in the answer.

Other Common Misconceptions A molecule is nonpolar due to its molecular geometry. The dipoles due to the polar bonds cancel each other out due to their symmetric arrangement. That hydrogen bonding is an intramolecular attractive force. That there are more hydrogen bonds because there are more H's in the formula. Understanding 10–5 is larger than 10–6. Or writing 1.20 x 10–6 for 12.0 x 10–5. Write out general expressions.  For example, equilibrium expressions, Nernst equations. 21. ∆H, ∆S or ∆G for a reaction is per mol of reaction. 22. Everything is not at STP! 23. Write soluble ionic compounds as ions.

Other Common Misconceptions Have to know the symbols and charge of fundamental ions. Must omit spectator ions. MgCl2 is not Mg+ and Cl2–! Know the formula of all elements. 28. Someone must be oxidized and someone must be reduced. 29. Charges on ions do not have to be written in a chemical equation or in an equilibrium expression. 30. Pay attention to stoichiometry when setting up ICE table for solubility expressions. 31. (2x)2 ≠ 2x2 32. Be careful that proximity does not mean that the same calculation produces the correct answer (2004 Q2) calculate moles of Fe(s) and O2(g), one use the ideal gas law the other does not.