AP Chemistry
HOW IS THE AP CHEMISTRY EXAM STRUCTURED? • THE MULTIPLE-CHOICE SECTION • 60 QUESTIONS WITH 4 ANSWER CHOICES EACH (USED TO BE 5) • 1 HOUR 30 MINUTES • WORTH 50% OF YOUR SCORE • NO CALCULATOR!
• SOME OF THESE QUESTIONS WILL EXIST AS A PART OF QUESTION GROUPS (WHICH CONSIST OF A FEW QUESTIONS THAT ASK ABOUT ONE SET OF DATA) WHILE OTHERS WILL STAND ON THEIR OWN.
• THE FREE-RESPONSE SECTION • 4 SHORT RESPONSE QUESTIONS • 3 LONG RESPONSE QUESTIONS • 1 HOUR 45 MINUTES • WORTH 50% OF YOUR SCORE
• CALCULATOR ALLOWED! • QUESTIONS TOPICS • • • • •
EXPERIMENTAL DESIGN ANALYZING DATA AND IDENTIFYING PATTERNS OR EXPLAINING PHENOMENA CREATING OR ANALYZING ATOMIC AND MOLECULAR VIEWS TO EXPLAIN OBSERVATIONS ARTICULATING AND THEN TRANSLATING BETWEEN REPRESENTATIONS OF DATA FOLLOWING LOGICAL/ANALYTICAL PATHWAYS TO SOLVE A PROBLEM
6 BIG IDEAS • BIG IDEA 1: THE CHEMICAL ELEMENTS ARE FUNDAMENTAL BUILDING MATERIALS OF MATTER, AND ALL MATTER CAN BE UNDERSTOOD IN TERMS OF ARRANGEMENT OF ATOMS. THESE ATOMS RETAIN THEIR IDENTITIES IN CHEMICAL REACTIONS. • THIS COVERS: • • • • • • • • •
COMPOUNDS IN ATOMIC MOLECULAR THEORY USING MASS DATA TO IDENTIFY THE COMPOSITION OR IDENTITY OF A CHEMICAL SUBSTANCE UNITS IN CHEMISTRY: PARTICLES VS. MOLES VS. MASS VS. VOLUME ELECTRONS (DISTRIBUTION IN ATOMS AND IONS, ENERGY LEVELS, COULOMB’S LAW, CLASSIC SHELL MODEL VS. QUANTUM MECHANICAL MODEL) STRUCTURE OF THE PERIODIC TABLE MASS SPECTROMETRY SPECTROSCOPY/LIGHT ABSORPTION AND CHEMICAL COMPOSITION OF A SOLUTION THE LAW OF CONSERVATION OF MASS GRAVIMETRIC ANALYSIS AND TITRATION
6 BIG IDEAS • BIG IDEA 2: CHEMICAL AND PHYSICAL PROPERTIES OF MATERIALS CAN BE EXPLAINED BY THE STRUCTURE AND ARRANGEMENT OF ATOMS, IONS, OR MOLECULES AND THE FORCES BETWEEN THEM • THIS COVERS: • • • • • • • • • • • • • • • •
MOLECULAR STRUCTURE AND ITS CONNECTION TO A SUBSTANCE’S PROPERTIES PHASE CHANGES (SOLIDS, LIQUIDS, GASSES) CHROMATOGRAPHY INTERACTIONS BETWEEN SOLUTES AND SOLVENTS LONDON DISPERSION FORCES POLARITY OF ATOMS AND MOLECULES COULOMB’S LAW AND THE INTERACTIONS OF IONS CHEMICAL BONDS AND WHY CERTAIN ATOMS FORM CERTAIN TYPES OF BONDS BOND POLARITY PROPERTIES OF METALLIC ELEMENTS LEWIS DIAGRAMS AND VSEPR PROPERTIES OF IONIC SOLIDS PROPERTIES OF METAL ALLOYS METALLIC BONDING AND THE ELECTRON SEA MODEL PROPERTIES OF COVALENT SOLIDS PROPERTIES OF MOLECULAR SOLIDS
6 BIG IDEAS • BIG IDEA 3: CHANGES IN MATTER INVOLVE THE REARRANGEMENT AND/OR REORGANIZATION OF ATOMS AND/OR THE TRANSFER OF ELECTRONS • THIS COVERS: • • • • • • •
WRITING BALANCED CHEMICAL EQUATIONS DIFFERENCE BETWEEN PHYSICAL CHANGE, CHEMICAL CHANGE, OR AMBIGUOUS CHANGE STOICHIOMETRIC CALCULATIONS TO PREDICT OUTCOME OF REACTIONS BRONSTED-LOWRY ACIDS AND BASES REDOX REACTIONS GALVANIC OR ELECTROLYTIC REACTIONS HALF-CELL REACTIONS/POTENTIALS AND FARADAY’S LAWS
6 BIG IDEAS • BIG IDEA 4: RATES OF CHEMICAL REACTIONS ARE DETERMINED BY DETAILS OF THE MOLECULAR COLLISIONS. • THIS COVERS: • FACTORS THAT IMPACT RATES OF CHEMICAL REACTIONS (TEMPERATURE, CONCENTRATION, SURFACE AREA) • CALCULATING RATE OF A ZEROTH-, FIRST-, OR SECOND-ORDER REACTION • HOW HALF-LIFE RELATES TO RATE CONSTANT IN A FIRST-ORDER REACTION • RATE LAW AND MOLECULAR COLLISIONS • CATALYSTS AND THEIR IMPACT ON CHEMICAL REACTIONS
6 BIG IDEAS • BIG IDEA 5: THE LAWS OF THERMODYNAMICS DESCRIBE THE ESSENTIAL ROLE OF ENERGY AND EXPLAIN AND PREDICT THE DIRECTION OF CHANGES IN MATTER. • THIS COVERS: • THERMAL ENERGY TRANSFER • RELATING THE MAGNITUDES OF ENERGY CHANGES IN TWO INTERACTING SYSTEMS (DIRECTION OF ENERGY FLOW, TYPE OF ENERGY) • HOW ENERGY CHANGES RELATE TO HEAT CAPACITY, ENTHALPY OF FUSION, ENTHALPY OF REACTION, PV WORK • CALORIMETRY • ENTHALPY OF REACTIONS AND CONNECTION TO CHEMICAL BONDS • NONCOVALENT INTERACTIONS BETWEEN MOLECULES
• WHETHER REACTIONS ARE THERMODYNAMICALLY FAVORED OR NOT • GIBBS FREE ENERGY CALCULATIONS • LE CHATELIER’S PRINCIPLE • EQUILIBRIUM CONSTANTS
6 BIG IDEAS • BIG IDEA 6: ANY BOND OR INTERMOLECULAR ATTRACTION THAT CAN BE FORMED CAN BE BROKEN. THESE TWO PROCESSES ARE IN A DYNAMIC COMPETITION, SENSITIVE TO INITIAL CONDITIONS AND EXTERNAL PERTURBATIONS. • THIS COVERS: • • • • • • • • • • • • • •
HOW CHANGES TO CHEMICAL REACTIONS AFFECT Q AND K RELATIVE RATES OF FORWARD AND REVERSE REACTIONS EQUILIBRIUM CONSTANT (K) AND HOW TO CALCULATE IT CALCULATING EQUILIBRIUM CONDITIONS OF A SYSTEM DIRECTION OF THE SHIFT RESULTING FROM STRESSES PLACED ON A SYSTEM AT CHEMICAL EQUILIBRIUM DESIGN A SET OF CONDITIONS THAT WILL OPTIMIZE A CERTAIN REACTION OUTCOME THE DISTINCTION BETWEEN STRONG AND WEAK ACID SOLUTIONS WITH SIMILAR PH VALUES INTERPRETING TITRATION DATA TO DETERMINE THE CONCENTRATION OF THE TITRANT/PKA/PKB CALCULATING THE PH OF A SOLUTION AND ITS MAKEUP ACID/BASE REACTIONS BUFFER SOLUTIONS - DESIGN, IDENTIFICATION, REACTIONS LABILE PROTONS AND HOW THEY AFFECT PH SOLUBILITY OF SALTS AND KSP EQUILIBRIUM CONSTANT IN TERMS OF △G° AND RT
SAMPLE MULTIPLE-CHOICE QUESTION • MANY QUESTIONS ON THE AP CHEMISTRY EXAM ASK YOU TO MAKE PREDICTIONS ABOUT CHEMICAL PROPERTIES OR REACTIONS BASED ON DATA LIKE THIS. IN THIS CASE, THE ANSWER IS A. THE COULOMBIC ATTRACTIONS ARE WEAKER IN NACL THAN THEY ARE IN NAF BECAUSE THE IONIC RADIUS OF FIS SMALLER THAN THAT OF CL-. THE ATTRACTION BETWEEN MOLECULES WILL BE GREATER IN NAF, AND THE BONDS WILL BE HARDER TO BREAK.
SAMPLE SHORT FREE-RESPONSE QUESTION • PART A REQUIRES AN UNDERSTANDING OF WHY OR WHY NOT REACTIONS MIGHT OCCUR BETWEEN MOLECULES. FIRST, YOU NEED TO EXPLAIN HOW COLLISION ENERGY AFFECTS WHETHER TWO MOLECULES WILL REACT WITH EACH OTHER. ONLY COLLISIONS WITH ENOUGH ENERGY TO OVERCOME THE ACTIVATION ENERGY BARRIER (TYPICALLY REPRESENTED BY THE VARIABLE EA) WILL REACH THE TRANSITION STATE AND BREAK THE F-F BOND. • THEN, YOU NEED TO IDENTIFY ONE OTHER FACTOR BESIDES COLLISION ENERGY THAT INFLUENCES THE LIKELIHOOD OF A REACTION BETWEEN TWO COLLIDING MOLECULES. YOU COULD SAY THAT IN ORDER FOR A COLLISION TO BE SUCCESSFUL, THE MOLECULES MUST HAVE THE CORRECT ORIENTATION. YOU WOULD NEED TO MENTION THE SPECIFIC BONDS BEING FORMED AND BROKEN. ONLY MOLECULES WITH THE CORRECT ORIENTATION CAN START TO FORM THE N-F BOND AND BREAK THE F-F BOND. THE MOLECULES HAVE TO CONTACT EACH OTHER IN VERY SPECIFIC PLACES FOR THE TRANSITION TO TAKE PLACE.
SAMPLE SHORT FREE-RESPONSE QUESTION • PART B IS ABOUT RATE LAWS, AND THE FIRST PART IS PRETTY STRAIGHTFORWARD. YOU HAVE A 50/50 CHANCE OF CIRCLING THE RIGHT ONE EVEN IF YOU HAVE NO IDEA WHAT THE ANSWER IS. FOR THE RECORD, IT'S THE SECOND OPTION, RATE = K[NO2][F2]. THEN YOU NEED TO EXPLAIN WHY YOU MADE YOUR CHOICE TO GET THE LAST POINT ON THE QUESTION. THE SECOND RATE LAW IS THE CORRECT ANSWER BECAUSE STEP I IS THE SLOWER, RATE-DETERMINING STEP IN THE REACTION MECHANISM. STEP I IS AN ELEMENTARY REACTION, SO ITS RATE LAW COMES FROM THE STOICHIOMETRY OF THE REACTION MOLECULES, NO2 AND F2.
SAMPLE LONG FREE-RESPONSE QUESTION • IN PART A OF THIS QUESTION, YOU’RE ASKED TO WRITE TWO NET-IONIC EQUATIONS. WRITING BALANCED EQUATIONS BASED ON EXPERIMENTAL SCENARIOS IS AN IMPORTANT SKILL FOR THE TEST. FOR PART I, THE NEUTRALIZATION REACTION IS H+ + OH- = H2O (LIQUID). FOR PART II, THE PRECIPITATION REACTION IS BA2+ + SO42- = BASO4 (SOLID).
• IN PART B, YOU NEED AN UNDERSTANDING OF WHAT CAUSES ELECTRICAL CONDUCTIVITY IN CHEMICAL SUBSTANCES AND WHY THE CONDUCTIVITY DECREASES AT FIRST IN THE SITUATION DESCRIBED. FOR PART I, THE SOLUTION IS CONDUCTING ELECTRICITY AS THE FIRST 30 ML OF THE H2SO4 ARE ADDED DUE TO THE PRESENCE OF BA2+AND/OR OH- IONS THAT HAVEN’T YET BEEN SCOOPED UP FOR THE REACTIONS (YOU COULD MENTION EITHER ONE AND STILL GET A POINT). FOR PART II, YOU COULD SAY THAT THE CONDUCTIVITY DECREASES BECAUSE THESE TWO TYPES OF IONS ARE STEADILY REMOVED BY THE PRECIPITATION AND NEUTRALIZATION REACTIONS (BA2+ IONS ARE TAKEN TO FORM BASO4, AND OH- IONS ARE TAKEN TO FORM WATER). •
SIDE NOTE: THE CONDUCTIVITY GOES BACK UP AFTER THE EQUIVALENCE POINT BECAUSE OF THE ADDITIONAL H-AND SO42- IONS THAT NOW EXIST IN SOLUTION AFTER ALL THE BA2+ AND OH- IONS HAVE BEEN USED UP BY THE REACTIONS.
SAMPLE LONG FREE-RESPONSE QUESTION • PART C REQUIRES SOME ATTENTION TO DETAIL IN UNIT CONVERSION AS WELL AS A LOGICAL ASSESSMENT OF THE INFORMATION YOU’RE GIVEN. MOLARITY IS MOLES PER LITER, SO THE QUESTION IS HOW MANY MOLES OF BA(OH)2 WERE THERE PER LITER IN THE ORIGINAL SOLUTION WITHOUT THE ADDED H2SO4. SINCE THE CONDUCTIVITY STARTS GOING BACK UP AFTER 30 ML OF H2SO4 ARE ADDED, THAT MEANS THAT AT THAT POINT THE NUMBER OF MOLES OF H2SO4 IS EQUAL TO THE NUMBER OF MOLES OF BAOH2 IN THE ORIGINAL SOLUTION. WE CAN CALCULATE THAT 30 ML OF 0.10 M H2SO4 IS EQUIVALENT TO 0.0030 MOLES (0.10 MOLES/LITER MULTIPLIED BY 0.030 LITERS). THERE SHOULD BE THE SAME NUMBER OF MOLES OF BAOH2 IN THE ORIGINAL SOLUTION, SO WE CAN DIVIDE 0.0030 MOLES BY THE ORIGINAL 0.025 L (25 ML) TO ARRIVE AT OUR ANSWER OF 0.12 MOLES/LITER OR A MOLARITY OF 0.12 M.
SAMPLE LONG FREE-RESPONSE QUESTION • PART D REQUIRES YOU TO USE KSP (THE SOLUBILITY PRODUCT CONSTANT) TO DETERMINE THE AMOUNT OF BA2+ IONS THAT REMAIN IN SOLUTION AT THE EQUIVALENCE POINT. THE QUESTION TELLS US THAT FOR BASO4, KSP = 1.0 X 10-10. THE SOLUBILITY PRODUCT CONSTANT EQUALS THE PRODUCT OF THE NUMBER OF IONS OF EACH COMPONENT OF THE PRECIPITATE. EACH OF THESE IS RAISED TO THE POWER OF ITS COEFFICIENT IN THE ORIGINAL NET IONIC EQUATION, WHICH IN THIS CASE IS 1 FOR BOTH: • KSP = [BA2+] X [SO42-] • AT THE EQUIVALENCE POINT, THE AMOUNT OF EACH OF THESE IONS IS EQUAL. THIS MEANS THAT [BA2+] X [SO42-] = [BA2+]2 AND [BA2+]2 = 1.0 X 10-10. THE NUMBER OF BA2+ IONS WOULD BE THE SQUARE ROOT OF KSP, WHICH IS 1.0 X 10-5 M.
SAMPLE LONG FREE-RESPONSE QUESTION • PART E ASKS YOU TO EXPLAIN WHY THERE IS A LOWER CONCENTRATION OF BA2+ IONS IN SOLUTION AS THE AMOUNT OF H2SO4 ADDED INCREASES PAST THE EQUIVALENCE POINT. IN THIS CASE, YOU WOULD NEED TO MENTION THE COMMON ION EFFECT AND THE FACT THAT IF YOU ADD SULFATE IONS TO AN EQUILIBRIUM REACTION INVOLVING OTHER SULFATE IONS, THE REACTION WILL CONSUME THE ADDED IONS TO REACH A NEW EQUILIBRIUM. THIS MEANS THAT MORE OF THE PRECIPITATE (BASO4) IS FORMED, AND MORE BA2+ IONS ARE TAKEN OUT OF SOLUTION TO CONTRIBUTE TO IT.
AP EXAM – JUSTIFY ANSWERS • SO, AS YOU CAN SEE, THE QUESTIONS ON THE TEST RANGE FROM SHORT AND SWEET TO LONG AND MODERATELY EVIL. AN IMPORTANT THREAD THAT RUNS THROUGH ALL OF THEM IS THAT YOU NEED TO KNOW BASIC BACKGROUND INFORMATION ABOUT WHY CERTAIN SUBSTANCES ACT THE WAY THEY DO (WHY DO SOME SUBSTANCES HAVE HIGHER BOILING POINTS THAN OTHERS? WHAT DOES COLLISION ENERGY HAVE TO DO WITH MOLECULAR REACTIONS? WHY DO SOME CHEMICAL SUBSTANCES CONDUCT ELECTRICITY?). BEING ABLE TO JUSTIFY YOUR ANSWERS IS ALWAYS VERY IMPORTANT. MAKE SURE YOU NEVER LOSE SIGHT OF THE FUNDAMENTALS AS YOU GET INTO MORE COMPLEX CALCULATIONS AND CONCEPTS.
HOW IS THE AP CHEMISTRY EXAM SCORED? • TO CALCULATE YOUR MULTIPLE-CHOICE RAW SCORE, ADD UP ALL THE QUESTIONS YOU ANSWERED CORRECTLY. THIS MEANS YOU CAN EARN A MAXIMUM OF 60 POINTS ON THE MULTIPLE CHOICE SECTION. THE FREE-RESPONSE SECTION IS A LITTLE MORE COMPLICATED, BUT IF YOU HAVE SCORING GUIDELINES, YOU SHOULD BE ABLE TO FIGURE OUT HOW MANY POINTS YOU’VE EARNED. SHORT RESPONSE QUESTIONS ARE WORTH 4 POINTS, AND LONG RESPONSE QUESTIONS ARE WORTH 10 POINTS, SO YOU CAN EARN A MAXIMUM OF 46 POINTS TOTAL ON THIS SECTION. • CONVERT EACH OF THESE SCORES INTO NUMBERS OUT OF 50 SO THAT THEY EACH MAKE UP HALF OF YOUR FINAL RAW SCORE. SAY YOU GOT 40 OUT OF 60 MULTIPLE-CHOICE QUESTIONS CORRECT. YOU WOULD CONVERT THIS SCORE TO THE EQUIVALENT FRACTION OF 33 OUT OF 50. THEN, IF YOU GOT 30 OUT OF 46 POINTS ON THE FREE-RESPONSE SECTION, YOU WOULD CONVERT THAT TO THE EQUIVALENT FRACTION OF 32 OUT OF 50 POINTS. ADD THE TWO SCORES OUT OF 50 TOGETHER TO GET YOUR FINAL RAW SCORE OUT OF 100. YOU CAN USE THE CONVERSION CHART BELOW TO GIVE YOURSELF A ROUGH ESTIMATE OF HOW YOUR RAW SCORE MIGHT TRANSLATE TO A CERTAIN AP SCORE. IN THIS CASE, YOUR RAW SCORE OF 65 WOULD BE RIGHT IN THE MIDDLE OF THE 4 RANGE.
HOW IS THE AP CHEMISTRY EXAM SCORED?
We can’t be absolutely sure that these raw score ranges will correlate exactly with these AP scores because the curve is slightly different every year. If you find you’re close to the bottom of your goal score range in practice testing, don't get complacent! You should probably put in a little more studying so you can feel more secure.
5 STUDY TIPS FOR AP CHEMISTRY • #1: ALWAYS ASK WHY • DON’T GLOSS OVER QUESTIONS THAT YOU GOT RIGHT THROUGH LUCKY GUESSES. IF YOU DON’T UNDERSTAND EXACTLY WHY THE CORRECT ANSWER IS CORRECT, YOU NEED TO REVIEW THE CONCEPT UNTIL YOU DO. CHEMISTRY BUILDS ON ITSELF, SO IF YOU DON’T GET THE FUNDAMENTAL REASON YOUR ANSWER WAS CORRECT OR INCORRECT, YOU COULD BE IN FOR A WHOLE MESS OF TROUBLE IN THE FUTURE. • FOR EXAMPLE, YOU MAY HAVE MEMORIZED THAT A CERTAIN MOLECULAR COMPOUND HAS A HIGHER BOILING POINT THAN ANOTHER, BUT THAT DOESN’T MEAN YOU NECESSARILY KNOW WHY THIS IS THE CASE. MAKE SURE YOU ALWAYS KNOW WHY CERTAIN PROPERTIES OCCUR BASED ON MOLECULAR AND ATOMIC STRUCTURE SO THAT YOU CAN JUSTIFY YOUR ANSWERS AND ADAPT YOUR KNOWLEDGE TO A VARIETY OF DIFFERENT SCENARIOS.
5 STUDY TIPS FOR AP CHEMISTRY • #2: MEMORIZE FORMULAS • YOU SHOULD MEMORIZE ALL THE FORMULAS YOU NEED TO KNOW FOR THE TEST. EVEN THOUGH YOU’LL GET A FORMULA SHEET, IT’S GOING TO BE MUCH EASIER TO GET THROUGH THE QUESTIONS IF YOU DON’T HAVE TO KEEP CONSULTING IT. FOR EACH FORMULA, MAKE SURE YOU KNOW THE TYPES OF QUESTIONS IT WILL HELP YOU ANSWER AND HOW ELSE IT COULD COME INTO PLAY ON THE TEST.
• #3: REVIEW YOUR LABS • LABS ARE CRITICAL IN AP CHEMISTRY BECAUSE THEY SHOW YOU THE REAL-LIFE IMPLICATIONS OF THE FACTS YOU’VE BEEN STUDYING. YOU’LL SEE MANY QUESTIONS ON THE EXAM THAT DEAL WITH LAB SCENARIOS, AND IT’S MUCH EASIER TO UNDERSTAND THESE TYPES OF QUESTIONS IF YOU’RE SOMEWHAT FAMILIAR WITH THE SETUP. IT’S CRUCIAL TO UNDERSTAND WHY YOU GOT THE RESULTS YOU DID FOR EACH LAB AND CONNECT THEM TO FACTS ABOUT CHEMICAL REACTIONS AND PROPERTIES OF DIFFERENT SUBSTANCES.
5 STUDY TIPS FOR AP CHEMISTRY • #4: LEARN TO ESTIMATE • THE MULTIPLE-CHOICE SECTION OF THE AP CHEMISTRY EXAM DOESN’T LET YOU USE A CALCULATOR. THIS IS KIND OF SCARY FOR SOME PEOPLE, BUT IT SHOULDN’T BE A BIG OBSTACLE IF YOU’RE WELL-PREPARED. YOU’LL SAVE YOURSELF A LOT OF TIME IF YOU PRACTICE DOING MULTIPLE-CHOICE QUESTIONS AND ESTIMATING LOGICAL ANSWERS WITHOUT GOING THROUGH LONG CALCULATIONS. IT BECOMES EASY TO ESTIMATE THE ANSWERS TO THESE PROBLEMS IF YOU'RE VERY FAMILIAR WITH THE MECHANICS OF CHEMICAL REACTIONS.
• #5: GET A REVIEW BOOK • THIS IS ONE OF THE AP CLASSES WHERE IT’S EXTREMELY HELPFUL TO HAVE A REVIEW BOOK TO GUIDE YOUR STUDYING. SINCE THE MATERIAL IS COMPLICATED, AND THERE ARE A LOT OF DIFFERENT THINGS YOU NEED TO KNOW HOW TO DO, A REVIEW BOOK CAN HELP GROUND YOU AND GET A BETTER HANDLE ON HOW YOU WANT TO STRUCTURE YOUR REVIEW OVERALL. YOU’LL ALSO GET A LOT OF ADDITIONAL PRACTICE PROBLEMS AND ANSWER EXPLANATIONS. YOU SHOULD ALSO USE YOUR LABS AND NOTES FROM CLASS, BUT THE REVIEW BOOK WILL HELP YOU ORGANIZE YOUR THOUGHTS.
• SOME OF THESE QUESTIONS WILL EXIST AS A PART OF QUESTION GROUPS (WHICH CONSIST OF A FEW QUESTIONS THAT ASK ABOUT ONE SET OF DATA) WHILE OTHERS WILL STAND ON THEIR OWN.
• THE FREE-RESPONSE SECTION • 4 SHORT RESPONSE QUESTIONS • 3 LONG RESPONSE QUESTIONS • 1 HOUR 45 MINUTES • WORTH 50% OF YOUR SCORE
• CALCULATOR ALLOWED! • QUESTIONS TOPICS • • • • •
EXPERIMENTAL DESIGN ANALYZING DATA AND IDENTIFYING PATTERNS OR EXPLAINING PHENOMENA CREATING OR ANALYZING ATOMIC AND MOLECULAR VIEWS TO EXPLAIN OBSERVATIONS ARTICULATING AND THEN TRANSLATING BETWEEN REPRESENTATIONS OF DATA FOLLOWING LOGICAL/ANALYTICAL PATHWAYS TO SOLVE A PROBLEM
6 BIG IDEAS • BIG IDEA 1: THE CHEMICAL ELEMENTS ARE FUNDAMENTAL BUILDING MATERIALS OF MATTER, AND ALL MATTER CAN BE UNDERSTOOD IN TERMS OF ARRANGEMENT OF ATOMS. THESE ATOMS RETAIN THEIR IDENTITIES IN CHEMICAL REACTIONS. • THIS COVERS: • • • • • • • • •
COMPOUNDS IN ATOMIC MOLECULAR THEORY USING MASS DATA TO IDENTIFY THE COMPOSITION OR IDENTITY OF A CHEMICAL SUBSTANCE UNITS IN CHEMISTRY: PARTICLES VS. MOLES VS. MASS VS. VOLUME ELECTRONS (DISTRIBUTION IN ATOMS AND IONS, ENERGY LEVELS, COULOMB’S LAW, CLASSIC SHELL MODEL VS. QUANTUM MECHANICAL MODEL) STRUCTURE OF THE PERIODIC TABLE MASS SPECTROMETRY SPECTROSCOPY/LIGHT ABSORPTION AND CHEMICAL COMPOSITION OF A SOLUTION THE LAW OF CONSERVATION OF MASS GRAVIMETRIC ANALYSIS AND TITRATION
6 BIG IDEAS • BIG IDEA 2: CHEMICAL AND PHYSICAL PROPERTIES OF MATERIALS CAN BE EXPLAINED BY THE STRUCTURE AND ARRANGEMENT OF ATOMS, IONS, OR MOLECULES AND THE FORCES BETWEEN THEM • THIS COVERS: • • • • • • • • • • • • • • • •
MOLECULAR STRUCTURE AND ITS CONNECTION TO A SUBSTANCE’S PROPERTIES PHASE CHANGES (SOLIDS, LIQUIDS, GASSES) CHROMATOGRAPHY INTERACTIONS BETWEEN SOLUTES AND SOLVENTS LONDON DISPERSION FORCES POLARITY OF ATOMS AND MOLECULES COULOMB’S LAW AND THE INTERACTIONS OF IONS CHEMICAL BONDS AND WHY CERTAIN ATOMS FORM CERTAIN TYPES OF BONDS BOND POLARITY PROPERTIES OF METALLIC ELEMENTS LEWIS DIAGRAMS AND VSEPR PROPERTIES OF IONIC SOLIDS PROPERTIES OF METAL ALLOYS METALLIC BONDING AND THE ELECTRON SEA MODEL PROPERTIES OF COVALENT SOLIDS PROPERTIES OF MOLECULAR SOLIDS
6 BIG IDEAS • BIG IDEA 3: CHANGES IN MATTER INVOLVE THE REARRANGEMENT AND/OR REORGANIZATION OF ATOMS AND/OR THE TRANSFER OF ELECTRONS • THIS COVERS: • • • • • • •
WRITING BALANCED CHEMICAL EQUATIONS DIFFERENCE BETWEEN PHYSICAL CHANGE, CHEMICAL CHANGE, OR AMBIGUOUS CHANGE STOICHIOMETRIC CALCULATIONS TO PREDICT OUTCOME OF REACTIONS BRONSTED-LOWRY ACIDS AND BASES REDOX REACTIONS GALVANIC OR ELECTROLYTIC REACTIONS HALF-CELL REACTIONS/POTENTIALS AND FARADAY’S LAWS
6 BIG IDEAS • BIG IDEA 4: RATES OF CHEMICAL REACTIONS ARE DETERMINED BY DETAILS OF THE MOLECULAR COLLISIONS. • THIS COVERS: • FACTORS THAT IMPACT RATES OF CHEMICAL REACTIONS (TEMPERATURE, CONCENTRATION, SURFACE AREA) • CALCULATING RATE OF A ZEROTH-, FIRST-, OR SECOND-ORDER REACTION • HOW HALF-LIFE RELATES TO RATE CONSTANT IN A FIRST-ORDER REACTION • RATE LAW AND MOLECULAR COLLISIONS • CATALYSTS AND THEIR IMPACT ON CHEMICAL REACTIONS
6 BIG IDEAS • BIG IDEA 5: THE LAWS OF THERMODYNAMICS DESCRIBE THE ESSENTIAL ROLE OF ENERGY AND EXPLAIN AND PREDICT THE DIRECTION OF CHANGES IN MATTER. • THIS COVERS: • THERMAL ENERGY TRANSFER • RELATING THE MAGNITUDES OF ENERGY CHANGES IN TWO INTERACTING SYSTEMS (DIRECTION OF ENERGY FLOW, TYPE OF ENERGY) • HOW ENERGY CHANGES RELATE TO HEAT CAPACITY, ENTHALPY OF FUSION, ENTHALPY OF REACTION, PV WORK • CALORIMETRY • ENTHALPY OF REACTIONS AND CONNECTION TO CHEMICAL BONDS • NONCOVALENT INTERACTIONS BETWEEN MOLECULES
• WHETHER REACTIONS ARE THERMODYNAMICALLY FAVORED OR NOT • GIBBS FREE ENERGY CALCULATIONS • LE CHATELIER’S PRINCIPLE • EQUILIBRIUM CONSTANTS
6 BIG IDEAS • BIG IDEA 6: ANY BOND OR INTERMOLECULAR ATTRACTION THAT CAN BE FORMED CAN BE BROKEN. THESE TWO PROCESSES ARE IN A DYNAMIC COMPETITION, SENSITIVE TO INITIAL CONDITIONS AND EXTERNAL PERTURBATIONS. • THIS COVERS: • • • • • • • • • • • • • •
HOW CHANGES TO CHEMICAL REACTIONS AFFECT Q AND K RELATIVE RATES OF FORWARD AND REVERSE REACTIONS EQUILIBRIUM CONSTANT (K) AND HOW TO CALCULATE IT CALCULATING EQUILIBRIUM CONDITIONS OF A SYSTEM DIRECTION OF THE SHIFT RESULTING FROM STRESSES PLACED ON A SYSTEM AT CHEMICAL EQUILIBRIUM DESIGN A SET OF CONDITIONS THAT WILL OPTIMIZE A CERTAIN REACTION OUTCOME THE DISTINCTION BETWEEN STRONG AND WEAK ACID SOLUTIONS WITH SIMILAR PH VALUES INTERPRETING TITRATION DATA TO DETERMINE THE CONCENTRATION OF THE TITRANT/PKA/PKB CALCULATING THE PH OF A SOLUTION AND ITS MAKEUP ACID/BASE REACTIONS BUFFER SOLUTIONS - DESIGN, IDENTIFICATION, REACTIONS LABILE PROTONS AND HOW THEY AFFECT PH SOLUBILITY OF SALTS AND KSP EQUILIBRIUM CONSTANT IN TERMS OF △G° AND RT
SAMPLE MULTIPLE-CHOICE QUESTION • MANY QUESTIONS ON THE AP CHEMISTRY EXAM ASK YOU TO MAKE PREDICTIONS ABOUT CHEMICAL PROPERTIES OR REACTIONS BASED ON DATA LIKE THIS. IN THIS CASE, THE ANSWER IS A. THE COULOMBIC ATTRACTIONS ARE WEAKER IN NACL THAN THEY ARE IN NAF BECAUSE THE IONIC RADIUS OF FIS SMALLER THAN THAT OF CL-. THE ATTRACTION BETWEEN MOLECULES WILL BE GREATER IN NAF, AND THE BONDS WILL BE HARDER TO BREAK.
SAMPLE SHORT FREE-RESPONSE QUESTION • PART A REQUIRES AN UNDERSTANDING OF WHY OR WHY NOT REACTIONS MIGHT OCCUR BETWEEN MOLECULES. FIRST, YOU NEED TO EXPLAIN HOW COLLISION ENERGY AFFECTS WHETHER TWO MOLECULES WILL REACT WITH EACH OTHER. ONLY COLLISIONS WITH ENOUGH ENERGY TO OVERCOME THE ACTIVATION ENERGY BARRIER (TYPICALLY REPRESENTED BY THE VARIABLE EA) WILL REACH THE TRANSITION STATE AND BREAK THE F-F BOND. • THEN, YOU NEED TO IDENTIFY ONE OTHER FACTOR BESIDES COLLISION ENERGY THAT INFLUENCES THE LIKELIHOOD OF A REACTION BETWEEN TWO COLLIDING MOLECULES. YOU COULD SAY THAT IN ORDER FOR A COLLISION TO BE SUCCESSFUL, THE MOLECULES MUST HAVE THE CORRECT ORIENTATION. YOU WOULD NEED TO MENTION THE SPECIFIC BONDS BEING FORMED AND BROKEN. ONLY MOLECULES WITH THE CORRECT ORIENTATION CAN START TO FORM THE N-F BOND AND BREAK THE F-F BOND. THE MOLECULES HAVE TO CONTACT EACH OTHER IN VERY SPECIFIC PLACES FOR THE TRANSITION TO TAKE PLACE.
SAMPLE SHORT FREE-RESPONSE QUESTION • PART B IS ABOUT RATE LAWS, AND THE FIRST PART IS PRETTY STRAIGHTFORWARD. YOU HAVE A 50/50 CHANCE OF CIRCLING THE RIGHT ONE EVEN IF YOU HAVE NO IDEA WHAT THE ANSWER IS. FOR THE RECORD, IT'S THE SECOND OPTION, RATE = K[NO2][F2]. THEN YOU NEED TO EXPLAIN WHY YOU MADE YOUR CHOICE TO GET THE LAST POINT ON THE QUESTION. THE SECOND RATE LAW IS THE CORRECT ANSWER BECAUSE STEP I IS THE SLOWER, RATE-DETERMINING STEP IN THE REACTION MECHANISM. STEP I IS AN ELEMENTARY REACTION, SO ITS RATE LAW COMES FROM THE STOICHIOMETRY OF THE REACTION MOLECULES, NO2 AND F2.
SAMPLE LONG FREE-RESPONSE QUESTION • IN PART A OF THIS QUESTION, YOU’RE ASKED TO WRITE TWO NET-IONIC EQUATIONS. WRITING BALANCED EQUATIONS BASED ON EXPERIMENTAL SCENARIOS IS AN IMPORTANT SKILL FOR THE TEST. FOR PART I, THE NEUTRALIZATION REACTION IS H+ + OH- = H2O (LIQUID). FOR PART II, THE PRECIPITATION REACTION IS BA2+ + SO42- = BASO4 (SOLID).
• IN PART B, YOU NEED AN UNDERSTANDING OF WHAT CAUSES ELECTRICAL CONDUCTIVITY IN CHEMICAL SUBSTANCES AND WHY THE CONDUCTIVITY DECREASES AT FIRST IN THE SITUATION DESCRIBED. FOR PART I, THE SOLUTION IS CONDUCTING ELECTRICITY AS THE FIRST 30 ML OF THE H2SO4 ARE ADDED DUE TO THE PRESENCE OF BA2+AND/OR OH- IONS THAT HAVEN’T YET BEEN SCOOPED UP FOR THE REACTIONS (YOU COULD MENTION EITHER ONE AND STILL GET A POINT). FOR PART II, YOU COULD SAY THAT THE CONDUCTIVITY DECREASES BECAUSE THESE TWO TYPES OF IONS ARE STEADILY REMOVED BY THE PRECIPITATION AND NEUTRALIZATION REACTIONS (BA2+ IONS ARE TAKEN TO FORM BASO4, AND OH- IONS ARE TAKEN TO FORM WATER). •
SIDE NOTE: THE CONDUCTIVITY GOES BACK UP AFTER THE EQUIVALENCE POINT BECAUSE OF THE ADDITIONAL H-AND SO42- IONS THAT NOW EXIST IN SOLUTION AFTER ALL THE BA2+ AND OH- IONS HAVE BEEN USED UP BY THE REACTIONS.
SAMPLE LONG FREE-RESPONSE QUESTION • PART C REQUIRES SOME ATTENTION TO DETAIL IN UNIT CONVERSION AS WELL AS A LOGICAL ASSESSMENT OF THE INFORMATION YOU’RE GIVEN. MOLARITY IS MOLES PER LITER, SO THE QUESTION IS HOW MANY MOLES OF BA(OH)2 WERE THERE PER LITER IN THE ORIGINAL SOLUTION WITHOUT THE ADDED H2SO4. SINCE THE CONDUCTIVITY STARTS GOING BACK UP AFTER 30 ML OF H2SO4 ARE ADDED, THAT MEANS THAT AT THAT POINT THE NUMBER OF MOLES OF H2SO4 IS EQUAL TO THE NUMBER OF MOLES OF BAOH2 IN THE ORIGINAL SOLUTION. WE CAN CALCULATE THAT 30 ML OF 0.10 M H2SO4 IS EQUIVALENT TO 0.0030 MOLES (0.10 MOLES/LITER MULTIPLIED BY 0.030 LITERS). THERE SHOULD BE THE SAME NUMBER OF MOLES OF BAOH2 IN THE ORIGINAL SOLUTION, SO WE CAN DIVIDE 0.0030 MOLES BY THE ORIGINAL 0.025 L (25 ML) TO ARRIVE AT OUR ANSWER OF 0.12 MOLES/LITER OR A MOLARITY OF 0.12 M.
SAMPLE LONG FREE-RESPONSE QUESTION • PART D REQUIRES YOU TO USE KSP (THE SOLUBILITY PRODUCT CONSTANT) TO DETERMINE THE AMOUNT OF BA2+ IONS THAT REMAIN IN SOLUTION AT THE EQUIVALENCE POINT. THE QUESTION TELLS US THAT FOR BASO4, KSP = 1.0 X 10-10. THE SOLUBILITY PRODUCT CONSTANT EQUALS THE PRODUCT OF THE NUMBER OF IONS OF EACH COMPONENT OF THE PRECIPITATE. EACH OF THESE IS RAISED TO THE POWER OF ITS COEFFICIENT IN THE ORIGINAL NET IONIC EQUATION, WHICH IN THIS CASE IS 1 FOR BOTH: • KSP = [BA2+] X [SO42-] • AT THE EQUIVALENCE POINT, THE AMOUNT OF EACH OF THESE IONS IS EQUAL. THIS MEANS THAT [BA2+] X [SO42-] = [BA2+]2 AND [BA2+]2 = 1.0 X 10-10. THE NUMBER OF BA2+ IONS WOULD BE THE SQUARE ROOT OF KSP, WHICH IS 1.0 X 10-5 M.
SAMPLE LONG FREE-RESPONSE QUESTION • PART E ASKS YOU TO EXPLAIN WHY THERE IS A LOWER CONCENTRATION OF BA2+ IONS IN SOLUTION AS THE AMOUNT OF H2SO4 ADDED INCREASES PAST THE EQUIVALENCE POINT. IN THIS CASE, YOU WOULD NEED TO MENTION THE COMMON ION EFFECT AND THE FACT THAT IF YOU ADD SULFATE IONS TO AN EQUILIBRIUM REACTION INVOLVING OTHER SULFATE IONS, THE REACTION WILL CONSUME THE ADDED IONS TO REACH A NEW EQUILIBRIUM. THIS MEANS THAT MORE OF THE PRECIPITATE (BASO4) IS FORMED, AND MORE BA2+ IONS ARE TAKEN OUT OF SOLUTION TO CONTRIBUTE TO IT.
AP EXAM – JUSTIFY ANSWERS • SO, AS YOU CAN SEE, THE QUESTIONS ON THE TEST RANGE FROM SHORT AND SWEET TO LONG AND MODERATELY EVIL. AN IMPORTANT THREAD THAT RUNS THROUGH ALL OF THEM IS THAT YOU NEED TO KNOW BASIC BACKGROUND INFORMATION ABOUT WHY CERTAIN SUBSTANCES ACT THE WAY THEY DO (WHY DO SOME SUBSTANCES HAVE HIGHER BOILING POINTS THAN OTHERS? WHAT DOES COLLISION ENERGY HAVE TO DO WITH MOLECULAR REACTIONS? WHY DO SOME CHEMICAL SUBSTANCES CONDUCT ELECTRICITY?). BEING ABLE TO JUSTIFY YOUR ANSWERS IS ALWAYS VERY IMPORTANT. MAKE SURE YOU NEVER LOSE SIGHT OF THE FUNDAMENTALS AS YOU GET INTO MORE COMPLEX CALCULATIONS AND CONCEPTS.
HOW IS THE AP CHEMISTRY EXAM SCORED? • TO CALCULATE YOUR MULTIPLE-CHOICE RAW SCORE, ADD UP ALL THE QUESTIONS YOU ANSWERED CORRECTLY. THIS MEANS YOU CAN EARN A MAXIMUM OF 60 POINTS ON THE MULTIPLE CHOICE SECTION. THE FREE-RESPONSE SECTION IS A LITTLE MORE COMPLICATED, BUT IF YOU HAVE SCORING GUIDELINES, YOU SHOULD BE ABLE TO FIGURE OUT HOW MANY POINTS YOU’VE EARNED. SHORT RESPONSE QUESTIONS ARE WORTH 4 POINTS, AND LONG RESPONSE QUESTIONS ARE WORTH 10 POINTS, SO YOU CAN EARN A MAXIMUM OF 46 POINTS TOTAL ON THIS SECTION. • CONVERT EACH OF THESE SCORES INTO NUMBERS OUT OF 50 SO THAT THEY EACH MAKE UP HALF OF YOUR FINAL RAW SCORE. SAY YOU GOT 40 OUT OF 60 MULTIPLE-CHOICE QUESTIONS CORRECT. YOU WOULD CONVERT THIS SCORE TO THE EQUIVALENT FRACTION OF 33 OUT OF 50. THEN, IF YOU GOT 30 OUT OF 46 POINTS ON THE FREE-RESPONSE SECTION, YOU WOULD CONVERT THAT TO THE EQUIVALENT FRACTION OF 32 OUT OF 50 POINTS. ADD THE TWO SCORES OUT OF 50 TOGETHER TO GET YOUR FINAL RAW SCORE OUT OF 100. YOU CAN USE THE CONVERSION CHART BELOW TO GIVE YOURSELF A ROUGH ESTIMATE OF HOW YOUR RAW SCORE MIGHT TRANSLATE TO A CERTAIN AP SCORE. IN THIS CASE, YOUR RAW SCORE OF 65 WOULD BE RIGHT IN THE MIDDLE OF THE 4 RANGE.
HOW IS THE AP CHEMISTRY EXAM SCORED?
We can’t be absolutely sure that these raw score ranges will correlate exactly with these AP scores because the curve is slightly different every year. If you find you’re close to the bottom of your goal score range in practice testing, don't get complacent! You should probably put in a little more studying so you can feel more secure.
5 STUDY TIPS FOR AP CHEMISTRY • #1: ALWAYS ASK WHY • DON’T GLOSS OVER QUESTIONS THAT YOU GOT RIGHT THROUGH LUCKY GUESSES. IF YOU DON’T UNDERSTAND EXACTLY WHY THE CORRECT ANSWER IS CORRECT, YOU NEED TO REVIEW THE CONCEPT UNTIL YOU DO. CHEMISTRY BUILDS ON ITSELF, SO IF YOU DON’T GET THE FUNDAMENTAL REASON YOUR ANSWER WAS CORRECT OR INCORRECT, YOU COULD BE IN FOR A WHOLE MESS OF TROUBLE IN THE FUTURE. • FOR EXAMPLE, YOU MAY HAVE MEMORIZED THAT A CERTAIN MOLECULAR COMPOUND HAS A HIGHER BOILING POINT THAN ANOTHER, BUT THAT DOESN’T MEAN YOU NECESSARILY KNOW WHY THIS IS THE CASE. MAKE SURE YOU ALWAYS KNOW WHY CERTAIN PROPERTIES OCCUR BASED ON MOLECULAR AND ATOMIC STRUCTURE SO THAT YOU CAN JUSTIFY YOUR ANSWERS AND ADAPT YOUR KNOWLEDGE TO A VARIETY OF DIFFERENT SCENARIOS.
5 STUDY TIPS FOR AP CHEMISTRY • #2: MEMORIZE FORMULAS • YOU SHOULD MEMORIZE ALL THE FORMULAS YOU NEED TO KNOW FOR THE TEST. EVEN THOUGH YOU’LL GET A FORMULA SHEET, IT’S GOING TO BE MUCH EASIER TO GET THROUGH THE QUESTIONS IF YOU DON’T HAVE TO KEEP CONSULTING IT. FOR EACH FORMULA, MAKE SURE YOU KNOW THE TYPES OF QUESTIONS IT WILL HELP YOU ANSWER AND HOW ELSE IT COULD COME INTO PLAY ON THE TEST.
• #3: REVIEW YOUR LABS • LABS ARE CRITICAL IN AP CHEMISTRY BECAUSE THEY SHOW YOU THE REAL-LIFE IMPLICATIONS OF THE FACTS YOU’VE BEEN STUDYING. YOU’LL SEE MANY QUESTIONS ON THE EXAM THAT DEAL WITH LAB SCENARIOS, AND IT’S MUCH EASIER TO UNDERSTAND THESE TYPES OF QUESTIONS IF YOU’RE SOMEWHAT FAMILIAR WITH THE SETUP. IT’S CRUCIAL TO UNDERSTAND WHY YOU GOT THE RESULTS YOU DID FOR EACH LAB AND CONNECT THEM TO FACTS ABOUT CHEMICAL REACTIONS AND PROPERTIES OF DIFFERENT SUBSTANCES.
5 STUDY TIPS FOR AP CHEMISTRY • #4: LEARN TO ESTIMATE • THE MULTIPLE-CHOICE SECTION OF THE AP CHEMISTRY EXAM DOESN’T LET YOU USE A CALCULATOR. THIS IS KIND OF SCARY FOR SOME PEOPLE, BUT IT SHOULDN’T BE A BIG OBSTACLE IF YOU’RE WELL-PREPARED. YOU’LL SAVE YOURSELF A LOT OF TIME IF YOU PRACTICE DOING MULTIPLE-CHOICE QUESTIONS AND ESTIMATING LOGICAL ANSWERS WITHOUT GOING THROUGH LONG CALCULATIONS. IT BECOMES EASY TO ESTIMATE THE ANSWERS TO THESE PROBLEMS IF YOU'RE VERY FAMILIAR WITH THE MECHANICS OF CHEMICAL REACTIONS.
• #5: GET A REVIEW BOOK • THIS IS ONE OF THE AP CLASSES WHERE IT’S EXTREMELY HELPFUL TO HAVE A REVIEW BOOK TO GUIDE YOUR STUDYING. SINCE THE MATERIAL IS COMPLICATED, AND THERE ARE A LOT OF DIFFERENT THINGS YOU NEED TO KNOW HOW TO DO, A REVIEW BOOK CAN HELP GROUND YOU AND GET A BETTER HANDLE ON HOW YOU WANT TO STRUCTURE YOUR REVIEW OVERALL. YOU’LL ALSO GET A LOT OF ADDITIONAL PRACTICE PROBLEMS AND ANSWER EXPLANATIONS. YOU SHOULD ALSO USE YOUR LABS AND NOTES FROM CLASS, BUT THE REVIEW BOOK WILL HELP YOU ORGANIZE YOUR THOUGHTS.
