Answers __exclusive__ — 1972 Ap Chemistry Free Response
1972 AP Chemistry Free-Response section followed a different structure than modern exams, consisting of 18 questions with a total time of 110 minutes
Problem 3: Thermodynamics – The ΔH Calculation
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 ap chemistry free response answers
* 2024. 2024 1-7. 2024, 1. 2024, 2. 2024, 3. 2024, 4. 2024, 5. 2024, 6. 2024, 7. * 2023. 2023, 1. 2023, 2. 2023, 3. 2023, 4. 2023, Adrian Dingle's Chemistry Pages 16.17 ap chemistry frq 1972 energy 1972 AP Chemistry Free-Response section followed a different
- Anode (oxidation): Zn → Zn²⁺ + 2e⁻, E° = +0.76 V (reverse sign)
- Cathode (reduction): Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V
- E°_cell = 0.76 + 0.34 = 1.10 V
To the remaining AgCl solid, add 1M NH3 → AgCl dissolves forming [Ag(NH3)2]⁺ complex. Anode (oxidation): Zn → Zn²⁺ + 2e⁻, E° = +0
Use stoichiometry to find moles of $H_2$: From part (a), moles of $M = 0.00880\text mol$. Ratio of $M$ to $H_2$ is $2:3$. $$ \textMoles H_2 = \frac32 \times \textMoles M $$ $$ \textMoles H_2 = 1.5 \times 0.00880\text mol = 0.0132\text mol $$