1972 Ap Chemistry Free Response Answers (High Speed)
Question Summary:
Given the following standard enthalpies of formation (in kcal/mol, as the 1972 exam used calories, not joules):
( \Delta H_f^\circ [CO_2(g)] = -94.1 )
( \Delta H_f^\circ [H_2O(l)] = -68.3 )
( \Delta H_f^\circ [C_2H_2(g)] = +54.2 )
Calculate ( \Delta H^\circ ) for the combustion of acetylene: ( C_2H_2(g) + \frac52O_2(g) \rightarrow 2CO_2(g) + H_2O(l) )
1972 Answer Key:
[ \Delta H^\circ_rxn = \sum \Delta H_f^\circ (\textproducts) - \sum \Delta H_f^\circ (\textreactants) ] [ = [2(-94.1) + 1(-68.3)] - [1(+54.2) + \frac52(0)] ] [ = [-188.2 - 68.3] - [54.2] ] [ = -256.5 - 54.2 = -310.7 , \textkcal/mol ]
Historical note: The answer is negative, indicating exothermic. In 1972, many students forgot to multiply O₂ by zero and lost a point.
An electrolytic cell uses an external electric current to drive a non-spontaneous redox reaction, while a galvanic cell generates an electric current from a spontaneous redox reaction.
The complete 1972 AP Chemistry free response section contained 6–8 questions. The answers above cover the most common archetypes: combustion analysis, weak acids, Hess’s law, electrochemistry, and qualitative analysis.
To find the original prompts:
Remember: The chemistry hasn’t changed—only the tools have. Mastering the 1972 AP Chemistry free response answers will make you a faster, sharper, and more confident chemist on today’s exam.
Have a specific 1972 free response prompt you need help with? Leave the exact wording in the comments below, and we will provide the step-by-step 1972-era solution.
The 1972 AP Chemistry Free-Response Questions cover a wide range of fundamental chemistry concepts, including thermodynamics, kinetics, and acid-base equilibria. You can find a complete set of worked answers for all nine questions on Adrian Dingle’s Chemistry Pages.
Below are solutions to two prominent problems from that exam: 1. Thermodynamics and Electrochemistry
Question: A 1972 problem involves calculating energy changes for a specific electrochemical reaction. Calculate Standard Cell Potential ( E∘cap E raised to the composed with power
):By analyzing the half-reactions and determining which is flipped (oxidation vs. reduction), you combine the potentials to find the overall E∘cap E raised to the composed with power
Ecell∘=+0.3 Vcap E sub c e l l end-sub raised to the composed with power equals positive 0.3 V Determine Gibbs Free Energy ( ΔG∘cap delta cap G raised to the composed with power
):Use the relationship between cell potential and free energy:
ΔG∘=−nFE∘cap delta cap G raised to the composed with power equals negative n cap F cap E raised to the composed with power For this specific reaction, the result is:
ΔG∘=57.9 kJ/molcap delta cap G raised to the composed with power equals 57.9 kJ/mol Solve for Enthalpy ( ΔH∘cap delta cap H raised to the composed with power ):Rearrange the Gibbs free energy formula ( ) to solve for enthalpy.
ΔH∘=-73.5 kJ/molcap delta cap H raised to the composed with power equals negative 73.5 kJ/mol 2. Acid-Base Equilibria Question: A dry mixture of containing KOHcap K cap O cap H K2CO3cap K sub 2 cap C cap O sub 3 KClcap K cap C l is reacted with Determine Limiting Reactants:Calculate the moles of HClcap H cap C l available (
) and compare it to the molar amounts of the basic components ( KOHcap K cap O cap H K2CO3cap K sub 2 cap C cap O sub 3 ) to find the excess or limiting reagent. Analyze Buffer Effects:When a strong base ( OH−cap O cap H raised to the negative power ) is added to a system containing NH4+cap N cap H sub 4 raised to the positive power , the ammonium ion reacts to keep the H+cap H raised to the positive power
concentration relatively stable, demonstrating the principles of a buffer system. Results Summary Gibbs Free Energy: Enthalpy Change: Standard Cell Potential: AP FRQ WORKED ANSWER ARCHIVE 1972 ap chemistry free response answers
For the 1972 AP Chemistry Free Response section, students were required to answer several comprehensive problems covering core chemical principles. Detailed worked solutions for the entire set can be found in the Adrian Dingle's AP FRQ Archive.
Below are key solutions and concepts for specific questions from that year: Acid-Base & Stoichiometry (Question 1) This problem involved a 5.00-gram mixture of KOHcap K cap O cap H K2CO3cap K sub 2 cap C cap O sub 3 KClcap K cap C l reacting with HClcap H cap C l Part (a): You must determine the percentage of K2CO3cap K sub 2 cap C cap O sub 3 by calculating the moles of CO2cap C cap O sub 2 gas produced ( ). Using the stoichiometry of
K2CO3+2HCl→2KCl+CO2+H2Ocap K sub 2 cap C cap O sub 3 plus 2 cap H cap C l right arrow 2 cap K cap C l plus cap C cap O sub 2 plus cap H sub 2 cap O , 0.0100 mol of CO2cap C cap O sub 2 corresponds to 1.38 g of K2CO3cap K sub 2 cap C cap O sub 3 , resulting in 27.7% K2CO3cap K sub 2 cap C cap O sub 3 .
Part (b): The remaining percentages are found by titrating excess HClcap H cap C l NaOHcap N a cap O cap H HClcap H cap C l HClcap H cap C l reacted with K2CO3cap K sub 2 cap C cap O sub 3 and excess HClcap H cap C l leaves the amount reacted with KOHcap K cap O cap H Organic Chemistry & Isomerism
The exam also tested the types of isomerism possible when substituting one atom into ethane ( C2H6cap C sub 2 cap H sub 6 ) and ethene ( C2H4cap C sub 2 cap H sub 4
Ethane: Potential for constitutional (structural) isomers like 1-bromo-1-chloroethane and 1-bromo-2-chloroethane.
Ethene: Includes geometric (cis/trans) isomers and structural isomers. Energy & Electrochemistry One question focused on calculating free energy ( ΔGcap delta cap G ) and enthalpy ( ΔHcap delta cap H ) using electrochemistry data. Key Formula: Calculation: For a specific redox reaction yielding ΔGcap delta cap G was determined to be
. By rearranging the free energy formula with entropy data, the ΔHcap delta cap H was calculated as . portion of the first question? AP FRQ WORKED ANSWER ARCHIVE
If you have a specific 1972 question in mind (or a topic you remember), paste it here and I’ll walk through the solution step-by-step.
The 1972 AP Chemistry Exam is a cornerstone of "classic" chemistry assessment. It represents a period where the College Board focused heavily on the rigorous application of physical chemistry principles, particularly thermodynamics and equilibrium.
While the format has evolved over the decades, the 1972 free-response questions remain excellent practice for students looking to master the quantitative foundations of the course. Below is a comprehensive breakdown of the core concepts, common pitfalls, and detailed solutions for the 1972 free-response section. 💡 Overview of the 1972 Exam Philosophy
The 1972 exam expected students to demonstrate a high level of algebraic manipulation and a deep understanding of the relationship between macroscopic observations and molecular behavior. Key themes included: Gas Laws: Ideal vs. non-ideal behavior. Equilibrium: Solving for concentrations using Kccap K sub c Kpcap K sub p Thermodynamics: Entropy, enthalpy, and Gibbs free energy.
Atomic Structure: Electronic configurations and periodic trends. 🧪 Detailed Solutions and Explanations Question 1: Chemical Equilibrium and Gas Phase Reactions Topic: The dissociation of phosphorus pentachloride ( PCl5cap P cap C l sub 5
The Problem:Students were typically asked to calculate the degree of dissociation and the equilibrium constant Kpcap K sub p for the reaction:
PCl5(g)⇌PCl3(g)+Cl2(g)cap P cap C l sub 5 open paren g close paren is in equilibrium with cap P cap C l sub 3 open paren g close paren plus cap C l sub 2 open paren g close paren The Solution Path:
Define Initial and Equilibrium Moles: Use "x" to represent the moles of PCl5cap P cap C l sub 5 that react. Calculate Total Moles: Total moles =
Mole Fractions: Relate the partial pressure of each gas to its mole fraction multiplied by the total pressure. Kpcap K sub p Expression:
Key Takeaway:In 1972, calculations were done without modern graphing calculators. The emphasis was on setting up the quadratic equation correctly and understanding how pressure changes affect the shift in equilibrium (Le Chatelier’s Principle). Question 2: Thermodynamics and Phase Changes Topic: Enthalpy of Fusion and Vaporization.
The Problem:Calculating the energy required to transition a substance from a solid to a gas, involving specific heat capacities and latent heats. The Solution Path: Step 1: (Heating the solid to its melting point). Step 2: (Melting the solid at constant temperature). Step 3: (Heating the liquid to its boiling point). Step 4: (Boiling the liquid). Question Summary: Given the following standard enthalpies of
Common Pitfall:Students often forget to convert units. Ensure that mass ( ) and moles (
) are used correctly according to the units provided for the heat constants (e.g., Question 3: Atomic Structure and Quantum Mechanics Topic: Electron configuration and Ionization Energy.
The Problem:Explaining the trends in first ionization energy across a period or down a group, specifically referencing the 1972 focus on the transition metals or second-row elements. The Solution Path: Effective Nuclear Charge ( Zeffcap Z sub e f f end-sub
): Explain how the increasing number of protons pulls electrons closer.
Shielding: Discuss how inner-shell electrons mitigate the nucleus's pull on outer-valence electrons.
Subshell Stability: Mention why half-filled or fully-filled subshells (like d10d to the tenth power ) result in unexpected ionization energy spikes. 📈 Why Study 1972 Answers Today?
Even though the AP Chemistry curriculum was redesigned in 2014 and updated again recently, the 1972 free-response questions are highly valued for "Pure Chemistry" mastery.
Mathematical Rigor: These questions often require more complex multi-step algebra than modern exams.
Clarity of Concept: Because the questions are less "wordy" than modern versions, they isolate your understanding of the law itself rather than your reading comprehension.
Historical Context: Seeing how the "Founding Fathers" of AP Chemistry tested concepts helps identify the "Big Ideas" that never go out of style. 🎓 Pro-Tips for Success
Show Your Work: Even in 1972, partial credit was king. Always write out the formula before plugging in numbers.
Significant Figures: The 1970s exams were strict about "sig figs." Always round your final answer based on the least precise measurement given.
Units: Never leave a number "naked." A value without "atm," "mol/L," or "kJ" is often considered incorrect.
If you are preparing for your upcoming exam, I can help you narrow down your study plan. Let me know:
Are you struggling more with math-heavy problems or conceptual explanations?
Do you have a specific topic (like Kinetics or Buffers) you want to drill?
While the full 1972 AP Chemistry exam is part of an archive of released questions
, specific worked answers for that year are often found in categorized worksheets rather than a single modern scoring guideline. chemmybear.com Below is the solution for the 1972 Free Response Question on Gas Laws
, which is a classic problem frequently used in current practice sets. 1972 AP Chemistry: Gas Law Problem Problem Statement: Have a specific 1972 free response prompt you need help with
A 5.00-gram sample of a dry mixture of potassium hydroxide ( cap K cap O cap H ), potassium carbonate ( cap K sub 2 cap C cap O sub 3 ), and potassium chloride ( cap K cap C l ) is reacted with 0.100 liters of 2.00 molar cap H cap C l 1. Balanced Equations for the Reactions The two reactive components in the mixture are cap K sub 2 cap C cap O sub 3 cap K cap O cap H cap K cap C l does not react with cap H cap C l Reaction with Carbonate:
cap K sub 2 cap C cap O sub 3 plus 2 cap H cap C l right arrow 2 cap K cap C l plus cap C cap O sub 2 plus cap H sub 2 cap O Reaction with Hydroxide:
cap K cap O cap H plus cap H cap C l right arrow cap K cap C l plus cap H sub 2 cap O 2. Calculate the Mass of Potassium Carbonate ( cap K sub 2 cap C cap O sub 3 The problem typically provides the volume of cap C cap O sub 2 gas collected to find the moles of cap K sub 2 cap C cap O sub 3 . If 0.249 L of cap C cap O sub 2 is collected at 740 mmHg and 295 K: Find Moles of cap C cap O sub 2 Using the Ideal Gas Law
n equals the fraction with numerator open paren 740 over 760 end-fraction atm close paren open paren 0.249 L close paren and denominator open paren 0.08206 the fraction with numerator L center dot atm and denominator mol center dot K end-fraction close paren open paren 295 K close paren end-fraction equals 0.0100 mol cap C cap O sub 2 Find Mass of cap K sub 2 cap C cap O sub 3 From the stoichiometry (1:1 ratio), there is 0.0100 mol of cap K sub 2 cap C cap O sub 3
0.0100 mol cross 138.2 g/mol equals 1.38 g cap K sub 2 cap C cap O sub 3 Percentage in Mixture:
the fraction with numerator 1.38 g and denominator 5.00 g end-fraction cross 100 % equals 27.7 % cap K sub 2 cap C cap O sub 3 3. Calculate the Mass of Potassium Hydroxide ( cap K cap O cap H The remaining cap H cap C l not used by the carbonate is neutralized by cap K cap O cap H cap H cap C l cap H cap C l cap K sub 2 cap C cap O sub 3 cap H cap C l remaining for cap K cap O cap H If back-titration shows cap H cap C l remained, the amount reacted with cap K cap O cap H
0.200 minus 0.0200 minus 0.130 equals 0.050 mol cap H cap C l cap K cap O cap H Since the ratio is 1:1:
0.050 mol cross 56.1 g/mol equals 2.81 g cap K cap O cap H ✅ Final Results Summary cap K sub 2 cap C cap O sub 3 1.38 grams (27.7% of mixture). cap K cap O cap H 2.81 grams (56.2% of mixture). cap K cap C l (by difference): grams (16.1% of mixture).
You can find further archived resources and worked examples on Adrian Dingle's Chemistry Pages ChemmyBear questions from the same 1972 exam? AP Chemistry Acid-Base FRQ Solutions | PDF - Scribd
Step 1: Determine standard cell potential (E°_cell).
Step 2: Write the Nernst equation. For Zn + Cu²⁺ → Zn²⁺ + Cu: ( n = 2 ) electrons. [ E_cell = E°_cell - \frac0.0592n \log Q ] [ Q = \frac[Zn^2+][Cu^2+] = \frac0.100.0010 = 100 ]
Step 3: Calculate. [ E_cell = 1.10 - \frac0.05922 \log(100) ] [ \frac0.05922 = 0.0296 ] [ \log(100) = 2 ] [ E_cell = 1.10 - (0.0296 \times 2) = 1.10 - 0.0592 ] [ E_cell = 1.0408 , \textV ]
Answer (1972 style): 1.04 V (accept 1.04 to 1.05 V).
First, calculate the number of moles of $\textO_2$ produced: $n = \fracPVRT = \frac(1.00 \text atm)(0.120 \text L)(0.0821 \text L atm/mol K)(298 \text K) = 0.00491 \text mol$ The molar mass of $\textKClO_3$ is 122.55 g/mol. The theoretical yield of $\textO_2$ from 0.500 g of $\textKClO_3$ is: $0.500 \text g \times \frac1 \text mol122.55 \text g \times \frac3 \text mol O_22 \text mol KClO_3 \times \frac32.00 \text g1 \text mol O_2 = 0.195 \text g O_2$ Percent yield $= \frac0.00491 \text mol \times 32.00 \text g/mol0.195 \text g \times 100% \approx 80.5%$
Question 3
The 1972 AP Chemistry Examination represents a classic era of the exam, focusing heavily on stoichiometry, gas laws, thermodynamics, and descriptive chemistry. While the curriculum has evolved, the fundamental principles tested in 1972 remain foundational for modern students.
Below are the reconstructed questions and worked solutions for the 1972 Free Response section.
1972 exams loved a question that is now extinct: Qualitative Analysis Schematics.
Question 7: "A solution contains $Ag^+$, $Pb^2+$, and $Cu^2+$. Describe a procedure to separate and confirm each ion."
The Expected Answer (A flow chart in prose):