(to get 3H2 on left). [ 3H2 + 1.5O2 → 3H2O \quad ΔH = -857.4 , \textkJ ]
[ 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 ]
[ [H^+] = \sqrt(0.100)(1.8 \times 10^-5) ] [ [H^+] = \sqrt1.8 \times 10^-6 ] 1972 ap chemistry free response answers
For decades, the Advanced Placement (AP) Chemistry exam has been the gold standard for high school college-level chemistry. While modern students focus on quantum mechanics and organic spectroscopy, the 1972 exam represents a fascinating "retro" challenge—a test built in an era of slide rules, log tables, and heavy emphasis on stoichiometry, equilibrium, and descriptive chemistry.
(to get 2C on left). [ 2C + 2O2 → 2CO2 \quad ΔH = -787.0 , \textkJ ] (to get 3H2 on left)
By: AP Curriculum Historians
(since we want ethanol as product, not reactant). [ 2CO2 + 3H2O → C2H5OH + 3O2 \quad ΔH = +1367 , \textkJ ] (to get 2C on left)
pH = 2.87