Chemistry: how 9 AI study apps did on 25 college problems
Covers stoichiometry, equilibrium, acids/bases, thermochemistry, kinetics, electrochemistry, gas laws. Every app got the same problem text in a fresh chat, one attempt, no hints. Tested October 15, 2026.
| Rank | App | Accuracy | Problems graded |
|---|---|---|---|
| 1 | ChatGPT | 100.0% | 25 / 30 |
| 2 | Claude | 100.0% | 25 / 30 |
| 3 | MyStudySidekick | 100.0% | 25 / 30 |
| 4 | NotebookLM | 92.0% | 25 / 30 |
| 5 | Solvely | 74.7% | 25 / 30 |
| 6 | Turbo AI | 72.7% | 25 / 30 |
| 7 | Quizlet | 59.3% | 25 / 30 |
| 8 | StudyFetch | 59.3% | 25 / 30 |
| 9 | Knowt | 8.0% | 25 / 30 |
The team behind MyStudySidekick runs this benchmark (disclosure and safeguards). Models used: ChatGPT, GPT-6, high reasoning effort; Claude, Claude Opus 5.5, medium effort.
Every problem, the key, and each app's answer
- 1. SO3 equilibrium at 900 K
- 2. Thermal NO at 2200 K
- 3. Adiabatic flame temperature of methane
- 4. Water-gas shift at 1000 K
- 5. Supercritical CO₂ — real vs ideal gas
- 6. pH of very dilute phosphoric acid
- 7. Titration of phosphoric acid
- 8. Solubility of AgCl in aqueous ammonia
- 9. Ksp of AgCl from electrode potentials
- 10. Dichromate–iron cell at non-standard conditions
- 11. pH of 1.0 × 10⁻⁸ M HCl at 50 °C
- 12. Bubble point of a benzene–toluene mixture
- 13. Oxygen collected over water
- 14. Arrhenius extrapolation and conversion time
- 15. Nuclear binding energy of iron-56
- 16. Structure from spectra: C₁₀H₁₂O
- 17. Stereoisomers of cyclohexane-1,2,3,4,5,6-hexol (inositol)
- 18. Structure from spectra: C₁₁H₁₄O₂
- 19. Stereoisomers of hepta-2,5-dien-4-ol
- 20. Steam methane reforming with the shift reaction at 1000 K and 10 bar
- 21. Silver concentration cell with an ammine complex
- 22. Consecutive first-order reactions at a new temperature
- 23. Selectivity of parallel reactions
- 24. Titration of a strong + weak acid mixture
- 25. Solubility of CaF₂ in an acidic fluoride solution
Problem 1 · hard
SO3 equilibrium at 900 K
gas-phase equilibrium from thermochemical tables
A contact-process converter is fed a gas of composition 10.0 mol% SO₂, 11.0 mol% O₂ and 79.0 mol% N₂. It reaches equilibrium at 900 K and a total pressure of 1.00 bar for the reaction
2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)
Using standard Gibbs energies of formation from the NIST-JANAF Thermochemical Tables at 900 K (standard state 1 bar), find:
(a) the equilibrium constant Kp at 900 K;
(b) the fractional conversion of SO₂ at equilibrium.
Original problem written for this benchmark
| App | (a) Kp at 900 K | (b) fractional conversion of SO2 | Score |
|---|---|---|---|
| Answer key | 44.57 ±3% | 0.654 ±0.01 | |
| NotebookLM | 44.36 | 0.653 | 100% |
| ChatGPT | 44.57 | 0.6537 | 100% |
| Claude | 44.6 | 0.654 | 100% |
| MyStudySidekick | 44.5826 | 0.6537 | 100% |
| Turbo AI | ≈ 4 × 10¹ (≈ 40–45, depending slightly on the table edition) | ≈ 0.65 | 50% |
| Why: Recorded from the response text the tester pasted. Reviewer: (a) is not a single answer. The headline is 4 × 10¹ (40, 10% low) with a hedge of '40–45', and it then uses 42 in part (b); under the rubric the stated answer, 40, is outside the 3% tolerance, so (a) scores 0. (b) 0.65 is within tolerance (key 0.654). Working: the 'JANAF' ΔfG° values are approximate and not the table's (≈ −313 / −299 vs −310.258 / −296.051). It instead estimates ΔG°(900) from 298 K data with ΔCp ≈ 0 (ΔH° −197.8 kJ, ΔS° −187.9 J/K → −28.7 kJ), a reasonable approximation but not the method asked. Small slips: 28000/(8.314×900) = 3.742, not 3.72; 'K = 40 gives x ≈ 0.65' is 0.642. Mole balance, expression (n_total in the numerator) and the trial values are correct. | |||
| Quizlet | 78.0 | 0.94 | 0% |
| Why: Recorded from the response text the tester pasted (math rendering duplicated as copied). Error analysis (reviewer): (1) ΔfG° values not from the tables: SO3 −319.4 and SO2 −303.1 kJ/mol, vs JANAF 900 K (O-058, O-034) −310.258 and −296.051, so ΔrG° = −32.6 instead of −28.414 kJ and Kp = 78.0 instead of 44.6 (75% too high). The arithmetic from its own values is right (e^4.356 = 78.0). (2) The equilibrium expression multiplies by 1/n_tot; for Δn = −1 at P = P° the factor is n_tot. (3) α = 0.94 does not solve its own equation (substituting gives about 4,088, not 78); its equation gives α = 0.70. With its own Kp = 78 the correct conversion is 0.71, not 0.94. The 'testing values' step appears to be guessed rather than computed. | |||
| Knowt | 6.56 × 10^8 | no final answer | 0% |
| Why: Recorded from the response text the tester pasted (raw LaTeX and literal \n as shown in the app). Error analysis (reviewer): (1) No table look-up: it writes 'Assuming we have the standard Gibbs energies of formation' and uses invented values (SO2 −300.0, SO3 −370.0 kJ/mol; JANAF 900 K: −296.051, −310.258), so ΔrG° = −140 kJ instead of −28.414 kJ. (2) Arithmetic errors even from its own numbers: 140/7.483 = 18.71, not 18.64; e^18.64 = 1.25 × 10⁸, which is 10^8.10, not 10^8.81; it reports 6.56 × 10⁸ (its own ΔG would give 1.3 × 10⁸). The true Kp is 44.6, so it is off by a factor of about 10⁷. (3) Part (b) is never solved, so no final answer (scored 0). The set-up is also inconsistent: SO2 falls by x while SO3 rises by 2x, and initial partial pressures are treated as fixed although the moles change. | |||
| StudyFetch | 3.0 × 10² | 0.95 (95%) | 0% |
| Why: Recorded from the response text the tester pasted. Error analysis (reviewer): (1) ΔfG° values not from JANAF: SO3 −321.3 and SO2 −299.9 kJ/mol, but the JANAF 900 K rows (O-058, O-034) are −310.258 and −296.051, so ΔrG° = −42.8 instead of −28.414 kJ and Kp ≈ 300 instead of 44.6 (about 7× too large), while the response says it used 'NIST-JANAF values'. (2) The equilibrium expression puts n_tot in the denominator; for Δn = −1 at P = P° it belongs in the numerator. (3) x = 0.0475 does not solve its own equation (substituting gives K ≈ 6,064, not 300); its equation actually gives x = 0.0408, conversion 0.82. Even with its Kp of 300 the correct conversion is 0.82, not 0.95. The response also ends mid-sentence ('Kp aion problems?'). | |||
| Solvely | 4.54 bar⁻¹ | 0.393 | 0% |
| Why: Recorded from the response text the tester pasted (Solvely). Error analysis (reviewer): (1) The ΔfG° values are presented as 'from the NIST-JANAF tables at T = 900 K' with three-decimal precision (SO3 −290.416, SO2 −284.757 kJ/mol), but neither number appears anywhere in the JANAF SO3 (O-058) or SO2 (O-034) tables, at any temperature. The 900 K rows are −310.258 and −296.051. So ΔrG° = −11.318 instead of −28.414 kJ and Kp = 4.54 instead of 44.57 (10× too small). (2) Everything after the look-up is correct: lnK = 1.5125 → 4.54 is right from its ΔG°, the mole balance and the equilibrium expression (n_total in the numerator) are right, and α = 0.393 does solve its equation for K = 4.54. A single fabricated look-up is the only error, and it propagates to both answers. Minor: it gives Kp units of bar⁻¹, while the requested standard-state Kp is dimensionless. | |||
Problem 2 · hard
Thermal NO at 2200 K
high-temperature equilibrium
Air (79.0 mol% N₂, 21.0 mol% O₂) is held at 2200 K and 1.00 bar until the reaction
N₂(g) + O₂(g) ⇌ 2 NO(g)
reaches equilibrium. Using the NIST-JANAF Thermochemical Tables at 2200 K, find:
(a) Kp for the reaction as written at 2200 K;
(b) the equilibrium mole fraction of NO.
Original problem written for this benchmark
| App | (a) Kp at 2200 K | (b) mole fraction of NO | Score |
|---|---|---|---|
| Answer key | 0.001075 ±3% | 0.01309 ±2% | |
| NotebookLM | 1.075 × 10⁻³ | 0.0131 | 100% |
| ChatGPT | 1.075 × 10⁻³ | 0.01309 | 100% |
| Claude | 1.07 × 10⁻³ | 0.0131 | 100% |
| MyStudySidekick | 1.075 × 10⁻³ | 0.0131 | 100% |
| Quizlet | 7.0 × 10⁻⁵ | 0.0034 | 0% |
| Why: Recorded from the response text the tester pasted (one line of the ln K step came through garbled: 'ln �, 290.8'). Reviewer: (1) It claims 'According to the NIST-JANAF tables at 2200 K' ΔfG°(NO) ≈ 87.5 kJ/mol, but the 2200 K row (N-005) gives 62.519. 87.5 is close to the ROOM-TEMPERATURE value (86.600 at 298.15 K): a 298 K number presented as the 2200 K one. So ΔrG° = 175.0 instead of 125.04 kJ and Kp = 7.0 × 10⁻⁵ instead of 1.075 × 10⁻³ (15× too small). (2) After that the arithmetic is right (e^−9.568 = 7.0 × 10⁻⁵; small-x ξ = 0.0017 → y = 0.0034); the small-x approximation is adequate here. Both answers wrong because of the one look-up. | |||
| Knowt | 8.43 × 10⁻⁵ | no final answer | 0% |
| Why: Recorded from the response text the tester pasted. Reviewer: (1) No table look-up: 'Assuming we have the standard Gibbs energies of formation at 2200 K' with ΔfG°(NO) = +86.2 kJ/mol. JANAF's 2200 K row (N-005) gives 62.519; 86.2 is near the 298 K value (86.600). Kp therefore about 13× too small. (2) Arithmetic slips: 0.008314 × 2200 = 18.2908, not 18.2958; 172.4/18.2908 = 9.4255 and e^−9.4255 = 8.06 × 10⁻⁵, not 8.43 × 10⁻⁵. (3) Part (b) never solved ('a numerical or graphical method may be employed'), so no final answer (scored 0). It also gets the total wrong: (0.79 − x) + (0.21 − x) + 2x = 1.00, not 1.00 + x. Same pattern as its problem-1 answer. | |||
| StudyFetch | 7.7 × 10⁻⁵ | 3.6 × 10⁻³ | 0% |
| Why: Recorded from the response text the tester pasted. Reviewer: it states 'Using the NIST-JANAF value at 2200 K: ΔfG°(NO, 2200 K) ≈ +86.6 kJ/mol'. 86.600 is JANAF's 298.15 K entry (N-005); the 2200 K entry is 62.519. So ΔrG° = 173.2 instead of 125.04 kJ and Kp = 7.7 × 10⁻⁵ instead of 1.075 × 10⁻³ (14× too small). The rest is consistent: e^(−173200/(8.314 × 2200)) = 7.73 × 10⁻⁵ and x = 1.79 × 10⁻³ → y = 3.6 × 10⁻³. Mole balance correct (Δn = 0). Both answers wrong because of the one look-up. | |||
| Turbo AI | 1.2 × 10⁻³ | 0.014 | 0% |
| Why: Recorded from the response text the tester pasted. Reviewer: near miss. It cites 'JANAF at 2200 K' with ΔfG°(NO) ≈ +61.6 kJ/mol; the table row (N-005) is 62.519. The 0.9 kJ/mol gap (rounded further to ΔG° ≈ 123 kJ) gives Kp = 1.2 × 10⁻³, 12% above 1.075 × 10⁻³ (tolerance 3%), and y_NO = 0.014, 7% above 0.01309 (tolerance 2%). Method, mole balance and solution are right (its ξ = 0.70 solves its equation). Side claims: Kp ~ 4 × 10⁻⁴ at 2000 K is right (JANAF 4.0 × 10⁻⁴); Kp ~ 10⁻⁶ at 1500 K is 10× low (JANAF 1.06 × 10⁻⁵). Its ΔH°(2200) ≈ 183 kJ matches the NASA Glenn value (182.8) rather than JANAF (180.9). Strong physical explanation (thermal NOx). | |||
| Solvely | 3.38 × 10⁻⁴ | 0.0075 | 0% |
| Why: Recorded from the response text the tester pasted (Solvely). Reviewer: it gives ΔfG°(NO) = 73.088 kJ/mol 'from the NIST-JANAF tables at T = 2200 K'. No row of the JANAF NO table (N-005) has 73.088; it lies between the 1300 K (73.969) and 1400 K (72.697) rows, and the 2200 K row is 62.519. So ΔrG° = 146.18 instead of 125.04 kJ and Kp = 3.38 × 10⁻⁴ instead of 1.075 × 10⁻³ (3.2× too small). As on problem 1, the value is fabricated with three-decimal precision and everything after it is correct (ln K = −7.9913; small-x x = 0.374 → y = 0.0075, adequate here). | |||
Problem 3 · hard
Adiabatic flame temperature of methane
energy balance with tabulated enthalpies
Methane burns with exactly the stoichiometric amount of air (air = 21.0 mol% O₂, 79.0 mol% N₂) at constant pressure. Fuel and air enter at 298.15 K. Assume complete combustion to CO₂ and H₂O(g) (no dissociation) and no heat loss.
Using standard enthalpies of formation and the sensible-enthalpy columns H°(T) − H°(298.15 K) of the NIST-JANAF Thermochemical Tables (interpolate linearly between tabulated temperatures), find the adiabatic flame temperature in kelvin.
Original problem written for this benchmark
| App | (a) adiabatic flame temperature | Score |
|---|---|---|
| Answer key | 2,327 K ±25 K | |
| StudyFetch | ≈ 2320 K | 100% |
| Solvely | ≈ 2325 K | 100% |
| NotebookLM | 2326.6 K | 100% |
| ChatGPT | 2327 K | 100% |
| Claude | 2326.6 K | 100% |
| MyStudySidekick | 2327 K | 100% |
| Quizlet | 2269 K | 0% |
| Why: Recorded from pasted text (math markup flattened). Reviewer: 58 K low (tolerance ±25). Its 'NIST-JANAF' sensible enthalpies are not the table's: per mole at 2300 K CO2 106.8, H2O 91.0, N2 70.1 vs JANAF 109.660, 88.421, 66.995 (its N2 value is close to JANAF's 2400 K entry); totals 770.6 (2200 K) and 816.5 (2300 K) vs JANAF 746.6 and 790.6. Interpolation arithmetic is right (2269 K); the data are wrong. 298 K ΔfH° rounded (−802.3 kJ, fine). | ||
| Knowt | no final answer | 0% |
| Why: Recorded from pasted text. Reviewer: no final answer. It uses a constant average cp (35 J/mol K) with the wrong sign for Q, gets T = −1881 K, says this is 'clearly not practical', and the response trails off without a result ('Run iterative atmosphere and gather data for useful Gf tables…'). No JANAF H−H298 data used. Also writes the air equation inconsistently (0.21 O2 / 3.76 N2). | ||
| Turbo AI | ≈ 2380 K (≈ 2385 K, depending on table edition) | 0% |
| Why: Recorded from pasted text. Reviewer: 53 K high (key 2327 K, ±25). Its 'JANAF sensible enthalpies' are not the table's: 2200 K CO2 90.6, H2O 80.2, N2 63.2 vs JANAF 103.562, 83.153, 63.361; 2400 K 101.3, 89.8, 70.3 vs 115.779, 93.741, 70.640 (CO2 about 13 kJ low). Interpolation arithmetic is right (2382 K). Its claim that 2380 K 'matches standard flame tables' is not right: the no-dissociation value from JANAF is 2327 K. Good discussion of dissociation (~2220 K real). | ||
Problem 4 · hard
Water-gas shift at 1000 K
equilibrium conversion
A feed of 1.00 mol CO, 2.00 mol H₂O(g) and 0.50 mol H₂ reaches equilibrium at 1000 K for the water-gas shift reaction
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
Using ΔfG° values at 1000 K from the NIST-JANAF Thermochemical Tables, find:
(a) the equilibrium constant K at 1000 K;
(b) the fraction of the CO converted at equilibrium.
Original problem written for this benchmark
| App | (a) K at 1000 K | (b) fraction of CO converted | Score |
|---|---|---|---|
| Answer key | 1.438 ±2% | 0.634 ±0.01 | |
| Turbo AI | ≈ 1.43 | ≈ 0.63 | 100% |
| Solvely | 1.44 | 0.634 | 100% |
| NotebookLM | 1.438 | 0.634 | 100% |
| ChatGPT | 1.438 | 0.6340 | 100% |
| Claude | 1.44 | 0.634 | 100% |
| MyStudySidekick | 1.4376 | 0.6340 | 100% |
| Quizlet | 1.45 | ≈ 0.65 | 50% |
| Why: Pasted text (math flattened). Reviewer: (a) within tolerance (1.45 vs 1.438, +0.8%) though from rounded values (CO −200.2, CO2 −395.9; ΔrG° −3.1 kJ). (b) its own quadratic (0.45α² − 4.85α + 2.9 = 0) gives α = 0.635, but it reports '≈ 0.65', 0.016 off the key (tolerance 0.01). Misreported final root. | |||
| Knowt | 31.6 | 0.956 | 0% |
| Why: Pasted text. Reviewer: 0/2. 'Assuming we have the standard Gibbs energies of formation at 1000 K' it uses the 298 K values (CO −137.2, H2O −228.6, CO2 −394.4) → ΔrG° −28.6 kJ, K = 31.6 (e^3.44 is actually 31.2), conversion 0.956. JANAF 1000 K ΔfG°: CO2 −395.886 (C-095), CO −200.275 (C-093), H2O(g) −192.590 (H-064). Labels ΔrG° as H°products − H°reactants. | |||
| StudyFetch | ≈ 1.5 | 0.443 | 0% |
| Why: Pasted text (math flattened). Reviewer: 0/2. Rounded values not the table's (CO −200.2, CO2 −396.0) → ΔrG° −3.2 vs −3.021 kJ, K = 1.5 (4% high; tolerance 2%). JANAF 1000 K ΔfG°: CO2 −395.886 (C-095), CO −200.275 (C-093), H2O(g) −192.590 (H-064). (b) algebra error: 'x = 7 − √43 = 0.443' does not solve its own equation; with K = 1.5 the root is 0.641. | |||
Problem 5 · hard
Supercritical CO₂ — real vs ideal gas
real-gas properties
Carbon dioxide is at 310.0 K and 100.0 bar (supercritical).
(a) Using the NIST Chemistry WebBook fluid-property data for CO₂, what is its molar volume in L/mol?
(b) What is the compressibility factor Z = PVm/RT?
(c) By what percentage does the ideal-gas law overestimate the molar volume? (Report (V_ideal − V_real)/V_real × 100.)
Original problem written for this benchmark
| App | (a) molar volume | (b) Z | (c) ideal-gas overestimate | Score |
|---|---|---|---|---|
| Answer key | 0.06418 L/mol ±1% | 0.249 ±1% | 301.6 % ±2% | |
| StudyFetch | 0.0642 L/mol | ≈ 0.25 (0.249) | ≈ 300% (302%) | 100% |
| ChatGPT | 0.06418 L/mol | 0.2490 | 301.63% | 100% |
| Claude | 0.0642 L/mol | 0.249 | 302% | 100% |
| MyStudySidekick | 0.06418 L/mol | 0.2490 | 301.6% | 100% |
| Quizlet | 0.0543 L/mol | 0.211 | 374.6% | 0% |
| Why: Pasted text (math flattened). Reviewer: 0/3. Claims a NIST molar volume of 0.0543 L/mol, 15% below NIST's 0.06418 NIST WebBook (Span–Wagner) at 310.0 K, 100 bar: 685.773 kg/m³ = 15.5823 mol/L, Vm 0.0641754 L/mol; at 313.15 K (40 °C): 628.612 kg/m³. Z and % follow correctly from its wrong Vm. | ||||
| Knowt | 1.162 L/mol | 4.51 | -77.78% | 0% |
| Why: Pasted text. Reviewer: 0/3. Claims NIST gives 1.162 L/mol (18× too large; larger than the ideal-gas volume), so Z = 4.51 and a NEGATIVE 'overestimate' of −77.8%, which it then describes as 'significantly overestimates' — the result contradicts itself and the physics (Z must be < 1 here). NIST WebBook (Span–Wagner) at 310.0 K, 100 bar: 685.773 kg/m³ = 15.5823 mol/L, Vm 0.0641754 L/mol; at 313.15 K (40 °C): 628.612 kg/m³. | ||||
| Turbo AI | ≈ 0.070 L/mol | ≈ 0.27 | ≈ 270% | 0% |
| Why: Pasted text. Reviewer: 0/3. Quotes 'NIST … at 310 K: ρ ≈ 628 kg/m³', which is NIST's value at 313.15 K (40 °C), the commonly quoted figure, not 310 K (685.8). Vm 9% high; Z 0.27; 270%. Its Peng–Robinson 'cross-check' (Z = 0.277) is presented as reproducing NIST, but NIST's Z at 310 K is 0.249. Hedge '0.068–0.072 L/mol' noted. NIST WebBook (Span–Wagner) at 310.0 K, 100 bar: 685.773 kg/m³ = 15.5823 mol/L, Vm 0.0641754 L/mol; at 313.15 K (40 °C): 628.612 kg/m³. | ||||
| Solvely | 0.0700 L/mol | 0.272 | 268% | 0% |
| Why: Pasted text (Solvely). Reviewer: 0/3. Cites 'NIST Chemistry WebBook (Span and Wagner)' with ρ ≈ 628.7 kg/m³ at 310 K, but that is NIST's value at 313.15 K (40 °C). Vm 0.0700 (9% high), Z 0.272, 268%. Arithmetic consistent; the look-up used the wrong temperature. NIST WebBook (Span–Wagner) at 310.0 K, 100 bar: 685.773 kg/m³ = 15.5823 mol/L, Vm 0.0641754 L/mol; at 313.15 K (40 °C): 628.612 kg/m³. | ||||
| NotebookLM | 0.07099 L/mol | 0.2754 | 263.1% | 0% |
| Why: Pasted text. Reviewer: 0/3. Attributes '14.088 mol/L (620.0 kg/m³)' to the NIST WebBook Span & Wagner EOS at 310 K; NIST gives 15.5823 mol/L (685.8 kg/m³) at 310 K, and 620 is not NIST's 313.15 K value (628.6) either. Vm 11% high; Z and % follow from it. First miss for NotebookLM. NIST at 310.0 K, 100 bar: 685.773 kg/m³ (15.5823 mol/L), Vm 0.0641754 L/mol; at 313.15 K: 628.612 kg/m³. | ||||
Problem 6 · hard
pH of very dilute phosphoric acid
polyprotic acid, exact charge balance
Find the pH of 1.00 × 10⁻⁴ M H₃PO₄ at 25 °C. Use Ka1, Ka2 and Ka3 for phosphoric acid from OpenStax Chemistry 2e, Appendix H, and Kw = 1.0 × 10⁻¹⁴. Report the pH to two decimal places.
Original problem written for this benchmark
| App | (a) pH | Score |
|---|---|---|
| Answer key | 4.01 ±0.02 | |
| Quizlet | 4.0057 | 100% |
| StudyFetch | 4.01 | 100% |
| Turbo AI | 4.01 | 100% |
| Solvely | 4.01 | 100% |
| NotebookLM | 4.01 | 100% |
| ChatGPT | 4.01 | 100% |
| Claude | 4.01 | 100% |
| MyStudySidekick | 4.01 (4.0054) | 100% |
| Knowt | 3.03 | 0% |
| Why: Pasted text. Reviewer: wrong. Falls into the trap: assumes x ≪ C and gets x = √(Ka1·C) = 8.66 × 10⁻⁴ M — more H⁺ than the 1.00 × 10⁻⁴ M of acid present, which it notes as 'relatively significant' without catching the impossibility. The second-dissociation step is garbled (computes y = 7.16 × 10⁻⁵, then writes it as 0.0716 × 10⁻⁵, then adds 7.16 × 10⁻⁵). pH 3.03 vs 4.01. Also lists Ka3 = 4.8 × 10⁻¹³ (OpenStax: 4.2 × 10⁻¹³). | ||
Problem 7 · hard
Titration of phosphoric acid
polyprotic titration
25.00 mL of 0.1000 M H₃PO₄ is titrated with 0.1000 M NaOH at 25 °C. Using Ka values for phosphoric acid from OpenStax Chemistry 2e, Appendix H, find the pH after the addition of:
(a) 25.00 mL of NaOH;
(b) 37.50 mL of NaOH.
Original problem written for this benchmark
| App | (a) pH after 25.00 mL | (b) pH after 37.50 mL | Score |
|---|---|---|---|
| Answer key | 4.7 ±0.05 | 7.21 ±0.05 | |
| Quizlet | 4.67 | 7.21 | 100% |
| StudyFetch | 4.67 | 7.21 | 100% |
| Turbo AI | 4.70 | 7.21 | 100% |
| Solvely | 4.70 (or 4.67 via the midpoint approximation) | 7.21 | 100% |
| NotebookLM | 4.70 | 7.21 | 100% |
| ChatGPT | 4.70 | 7.21 | 100% |
| Claude | 4.70 | 7.21 | 100% |
| MyStudySidekick | 4.70 | 7.21 | 100% |
| Knowt | 4.25 | 6.81 | 0% |
| Why: Pasted text. Reviewer: 0/2. (a) treats H2PO4⁻ as a plain weak acid (x = √(Ka2·C)) and ignores its amphiprotic nature → 4.25 vs 4.70. (b) never reacts the 1.250 mmol excess OH⁻ with H2PO4⁻: it keeps 2.500 mmol H2PO4⁻ alongside 'excess NaOH' and puts the NaOH concentration in as [HPO4²⁻] → ratio 2, and then mis-evaluates −log(1.24 × 10⁻⁷) (that is 6.91, not 6.81). Lists Ka3 = 4.8 × 10⁻¹³ (OpenStax 4.2 × 10⁻¹³). | |||
Problem 8 · hard
Solubility of AgCl in aqueous ammonia
complex-ion equilibria
What is the molar solubility of AgCl in 0.10 M NH₃ at 25 °C, and how many grams of AgCl dissolve per litre? Use Ksp for AgCl from OpenStax Chemistry 2e, Appendix J, and Kf for [Ag(NH₃)₂]⁺ from Appendix K. Take 0.10 M as the initial NH₃ concentration and ignore the reaction of NH₃ with water.
Original problem written for this benchmark
| App | (a) molar solubility | (b) solubility | Score |
|---|---|---|---|
| Answer key | 0.00472 M ±3% | 0.677 g/L ±3% | |
| Quizlet | 4.72 × 10⁻³ M | 0.677 g/L | 100% |
| Solvely | 4.7 × 10⁻³ M | 0.68 g/L | 100% |
| NotebookLM | 4.72 × 10⁻³ M | 0.677 g/L | 100% |
| ChatGPT | 4.72 × 10⁻³ mol/L | 0.677 g/L | 100% |
| Claude | 4.7 × 10⁻³ M | 0.68 g/L | 100% |
| MyStudySidekick | 4.722 × 10⁻³ M | 0.6768 g/L | 100% |
| Knowt | 1.33 × 10⁻⁵ M | 1.91 × 10⁻³ g/L | 0% |
| Why: Pasted text. Reviewer: 0/2. Sets up the complexation, then drops it ('K_sp = s × s') and reports the solubility in PURE WATER (√Ksp), 360× too small. Uses Ksp = 1.77 × 10⁻¹⁰ while attributing it to OpenStax (book: 1.6 × 10⁻¹⁰). | |||
| StudyFetch | ≈ 5.0 × 10⁻³ M (4.98 × 10⁻³) | 0.71 g/L | 0% |
| Why: Pasted text. Reviewer: 0/2 — method exactly right, constant wrong: Ksp(AgCl) = 1.8 × 10⁻¹⁰ (common in other textbooks) instead of OpenStax's 1.6 × 10⁻¹⁰, so s is 5.5% high and g/L 4.9% high (tolerance 3%). With 1.8 × 10⁻¹⁰ its arithmetic is right (4.98 × 10⁻³, 0.714). | |||
| Turbo AI | 1.47 × 10⁻² M | 2.1 g/L | 0% |
| Why: Pasted text. Reviewer: 0/2. (1) Ksp = 1.8 × 10⁻¹⁰ attributed to 'Appendix J', which prints 1.6 × 10⁻¹⁰. (2) Algebra: writes K = s²/(0.10 − 2s) instead of s²/(0.10 − 2s)², tripling s (0.0147 M). Its 'naive' figure (0.0175 M) uses the same wrong form; with the right form the naive estimate is 5.5 × 10⁻³ M. Rich context (thiosulfate fixing, NH3 hydrolysis) but the answer is wrong. | |||
Problem 9 · hard
Ksp of AgCl from electrode potentials
electrochemistry and equilibrium
Using only standard reduction potentials from OpenStax Chemistry 2e, Appendix L (the Ag⁺/Ag and AgCl/Ag couples), calculate at 25 °C:
(a) ΔG° for AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), in kJ/mol;
(b) the solubility product Ksp of AgCl.
Original problem written for this benchmark
| App | (a) ΔG° of dissolution | (b) Ksp | Score |
|---|---|---|---|
| Answer key | 55.7 kJ/mol ±1% | 0 ±3% | |
| Quizlet | 55.70 kJ/mol | 1.74 × 10⁻¹⁰ | 100% |
| Turbo AI | +55.7 kJ/mol | 1.7 × 10⁻¹⁰ | 100% |
| Solvely | +55.70 kJ/mol | 1.7 × 10⁻¹⁰ | 100% |
| NotebookLM | +55.70 kJ/mol | 1.75 × 10⁻¹⁰ | 100% |
| ChatGPT | +55.70 kJ/mol | 1.75 × 10⁻¹⁰ | 100% |
| Claude | +55.70 kJ/mol | 1.75 × 10⁻¹⁰ | 100% |
| MyStudySidekick | +55.70 kJ/mol | 1.75 × 10⁻¹⁰ | 100% |
| StudyFetch | +55.7 kJ/mol | ≈ 1.6 × 10⁻¹⁰ | 50% |
| Why: Pasted text. Reviewer: (a) correct. (b) 1.6 × 10⁻¹⁰ is 8% low: it rounds the exponent to −22.5 (e^−22.5 = 1.69 × 10⁻¹⁰) and then reports 1.6; its own unrounded expression gives 1.74 × 10⁻¹⁰. It also claims 'the tabulated OpenStax value of 1.8 × 10⁻¹⁰' — OpenStax prints 1.6 × 10⁻¹⁰. | |||
| Knowt | 54.01 kJ/mol | 1.02 × 10⁻¹⁰ | 0% |
| Why: Pasted text. Reviewer: 0/2. Uses the Cl2/Cl⁻ couple (+1.36 V) in place of AgCl/Ag (+0.22233 V), so E°cell = −0.56 V and ΔG° = 54.01 kJ/mol (3% low). Then mis-evaluates the exponential: 54009.6/2478.772 = 21.79 and e^−21.79 = 3.4 × 10⁻¹⁰, not 1.02 × 10⁻¹⁰. | |||
Problem 10 · hard
Dichromate–iron cell at non-standard conditions
Nernst equation
A galvanic cell at 25 °C runs the reaction
Cr₂O₇²⁻(aq) + 14 H⁺(aq) + 6 Fe²⁺(aq) → 2 Cr³⁺(aq) + 6 Fe³⁺(aq) + 7 H₂O(l)
with [Cr₂O₇²⁻] = 0.020 M, [Cr³⁺] = 0.0050 M, pH = 2.00, [Fe²⁺] = 0.10 M and [Fe³⁺] = 0.0010 M. Using standard reduction potentials from OpenStax Chemistry 2e, Appendix L, find:
(a) the cell potential E;
(b) ΔG for the reaction as written, in kJ.
Original problem written for this benchmark
| App | (a) E cell | (b) ΔG | Score |
|---|---|---|---|
| Answer key | 0.332 V ±0.003 V | −192.1 kJ ±2% | |
| StudyFetch | 0.332 V | −192 kJ | 100% |
| ChatGPT | +0.3319 V | −192.1 kJ | 100% |
| Claude | +0.332 V | −192 kJ | 100% |
| MyStudySidekick | 0.3319 V | −192.12 kJ | 100% |
| Quizlet | 0.430 V | −249 kJ | 0% |
| Why: Pasted text (math flattened). Used E°(Cr2O7) = 1.33 V, not Appendix L 1.232 V. | |||
| Knowt | 0.447 V | −258.22 kJ | 0% |
| Why: Pasted text (raw \n escapes). E° 1.33 V and denominator 2.0e-34 (should be 2.0e-36), so Q = 1.25e11. | |||
| Turbo AI | 0.598 V | −346 kJ | 0% |
| Why: Pasted text. E° 1.36 V and (0.010)^14 taken as 1e-20 (Q = 0.125). | |||
| Solvely | +0.430 V | −249 kJ | 0% |
| Why: Pasted text. Tested app: Solvely. E° 1.33 V instead of Appendix L 1.232 V. | |||
| NotebookLM | 0.430 V | −248.9 kJ | 0% |
| Why: Pasted text. E° 1.33 V; also offers an alternative with 1.36 V (0.460 V). Graded on the main answer. | |||
Problem 11 · hard
pH of 1.0 × 10⁻⁸ M HCl at 50 °C
water autoionization
Calculate the pH of 1.0 × 10⁻⁸ M HCl at 50 °C, and the pH of pure (neutral) water at 50 °C. Use Kw at 50 °C from OpenStax Chemistry 2e, Appendix E (the table "Water Kw and pKw at Different Temperatures").
Original problem written for this benchmark
| App | (a) pH of the HCl solution | (b) neutral pH at 50 °C | Score |
|---|---|---|---|
| Answer key | 6.629 ±0.01 | 6.638 ±0.01 | |
| Quizlet | 6.62 | 6.63 | 100% |
| StudyFetch | 6.62 | 6.63 | 100% |
| Turbo AI | 6.62 | 6.63 | 100% |
| Solvely | 6.62 | 6.63 | 100% |
| NotebookLM | 6.63 | 6.64 | 100% |
| ChatGPT | 6.63 | 6.64 | 100% |
| Claude | 6.63 | 6.64 | 100% |
| MyStudySidekick | 6.63 | 6.64 | 100% |
| Knowt | 6.12 | 6.13 | 0% |
| Why: Pasted text. Kw 5.5e-14 and sqrt(5.5e-14) computed as 7.42e-7 (should be 2.35e-7); also adds water and acid H+ directly. | |||
Problem 12 · hard
Bubble point of a benzene–toluene mixture
Raoult's law with Antoine equation
A liquid mixture of 40.0 mol% benzene and 60.0 mol% toluene is heated at 1.01325 bar. Assuming an ideal solution (Raoult's law) and using Antoine parameters from the NIST Chemistry WebBook (benzene: Eon, Pommier et al. 1971, valid 333.4–373.5 K; toluene: Williamham, Taylor et al. 1945, valid 308.52–384.66 K; log₁₀P[bar] = A − B/(T[K] + C)), find:
(a) the bubble-point temperature in kelvin;
(b) the mole fraction of benzene in the first bubble of vapour.
Original problem written for this benchmark
| App | (a) bubble-point temperature | (b) benzene in vapour | Score |
|---|---|---|---|
| Answer key | 368.2 K ±0.5 K | 0.623 ±0.01 | |
| StudyFetch | 368.3 K | 0.622 | 100% |
| Turbo AI | 368.3 K | 0.62 | 100% |
| Solvely | 368.3 K | 0.622 | 100% |
| NotebookLM | 368.22 K | 0.6227 | 100% |
| ChatGPT | 368.22 K | 0.6227 | 100% |
| Claude | 368.3 K | 0.622 | 100% |
| MyStudySidekick | 368.3 K | 0.6216 | 100% |
| Quizlet | 358.2 K | 0.615 | 50% |
| Why: Pasted text (math flattened). Used benzene Antoine set A 4.01814 / B 1203.835 / C −53.226, not the Eon–Pommier 1971 set the problem names (4.72583 / 1660.652 / −1.461), and its iteration table is wrong (P_calc 1.009 bar at 358 K), so T is ~10 K low. | |||
| Knowt | 363.00 K | no final answer | 0% |
| Why: Pasted text. mmHg/°C Antoine constants plugged in as bar/K; (b) never computed ("[Calculated Value]"). | |||
Problem 13 · hard
Oxygen collected over water
gas stoichiometry with vapour pressure
A sample of KClO₃ is heated with a MnO₂ catalyst and decomposes completely: 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g). The oxygen is collected over water at 25.0 °C. The wet gas occupies 0.2500 L at a total pressure of 755.0 torr. Using the vapour pressure of water at 25 °C from OpenStax Chemistry 2e, Appendix E (the table "Water Vapor Pressure at Different Temperatures"), what mass of KClO₃ decomposed? (Molar masses: K 39.098, Cl 35.45, O 15.999 g/mol.)
Original problem written for this benchmark
| App | (a) mass of KClO3 | Score |
|---|---|---|
| Answer key | 0.8032 g ±1% | |
| Quizlet | no final answer | 100% |
| Knowt | no final answer | 100% |
| StudyFetch | no final answer | 100% |
| Turbo AI | no final answer | 100% |
| Solvely | no final answer | 100% |
| NotebookLM | no final answer | 100% |
| ChatGPT | no final answer | 100% |
| Claude | no final answer | 100% |
| MyStudySidekick | no final answer | 100% |
Problem 14 · hard
Arrhenius extrapolation and conversion time
kinetics
A first-order decomposition has rate constant k = 2.40 × 10⁻⁴ s⁻¹ at 300.0 K and k = 2.88 × 10⁻³ s⁻¹ at 320.0 K.
(a) Find the activation energy in kJ/mol.
(b) How many minutes does it take to reach 90.0% conversion at 310.0 K?
Original problem written for this benchmark
| App | (a) activation energy | (b) time to 90% conversion | Score |
|---|---|---|---|
| Answer key | 99.17 kJ/mol ±1% | 44.35 min ±2% | |
| Quizlet | 99.16 kJ/mol | 44.4 min | 100% |
| StudyFetch | 99.2 kJ/mol | 44.3 min | 100% |
| Turbo AI | 99.2 kJ/mol | 44.3 min | 100% |
| Solvely | 99.2 kJ/mol | 44.4 min | 100% |
| NotebookLM | 99.2 kJ/mol | 44.4 min | 100% |
| ChatGPT | 99.2 kJ/mol | 44.3 min | 100% |
| Claude | 99.2 kJ/mol | 44.3 min | 100% |
| MyStudySidekick | 99.17 kJ/mol | 44.35 min | 100% |
| Knowt | 120.38 kJ/mol | 72.66 min | 0% |
| Why: Pasted text. ln k values wrong (difference 2.509 vs 2.4849), computes 12.04 kJ/mol then reports 120.38 kJ/mol; for (b) uses an arithmetic-average k (5.28e-4) instead of the Arrhenius value. | |||
Problem 15 · hard
Nuclear binding energy of iron-56
nuclear chemistry with atomic mass tables
Using atomic masses from the NIST "Atomic Weights and Isotopic Compositions" database and the CODATA neutron mass and u→MeV conversion, calculate for ⁵⁶Fe:
(a) the total nuclear binding energy in MeV;
(b) the binding energy per nucleon in MeV.
Original problem written for this benchmark
| App | (a) total binding energy | (b) binding energy per nucleon | Score |
|---|---|---|---|
| Answer key | 492.3 MeV ±0.2% | 8.79 MeV ±0.2% | |
| Quizlet | 492.26 MeV | 8.79 MeV | 100% |
| StudyFetch | 492.3 MeV | 8.790 MeV | 100% |
| Turbo AI | 492.26 MeV | 8.79 MeV | 100% |
| Solvely | 492.26 MeV | 8.790 MeV | 100% |
| NotebookLM | 492.26 MeV | 8.790 MeV | 100% |
| ChatGPT | 492.26 MeV | 8.790 MeV | 100% |
| Claude | 492.26 MeV | 8.790 MeV | 100% |
| MyStudySidekick | 492.26 MeV | 8.790 MeV | 100% |
| Knowt | 479.6 MeV | 8.56 MeV | 0% |
| Why: Pasted text. Bare proton masses against the ATOMIC Fe-56 mass, so 26 electron masses are left in (Δm 0.5142 u instead of 0.5285 u). | |||
Problem 16 · hard
Structure from spectra: C₁₀H₁₂O
NMR/IR structure determination
An unknown has molecular formula C₁₀H₁₂O. IR: strong band at 1683 cm⁻¹, no broad band above 3000 cm⁻¹.
¹H NMR (CDCl₃): δ 7.87 (d, J = 8.2 Hz, 2H), 7.26 (d, J = 8.2 Hz, 2H), 2.97 (q, J = 7.3 Hz, 2H), 2.41 (s, 3H), 1.22 (t, J = 7.3 Hz, 3H).
¹³C NMR: δ 200.6, 143.7, 134.5, 129.3, 128.2, 31.7, 21.6, 8.4.
(a) How many degrees of unsaturation does the compound have?
(b) Give the IUPAC name of the compound.
Original problem written for this benchmark
| App | (a) degrees of unsaturation | (b) IUPAC name | Score |
|---|---|---|---|
| Answer key | 5 ±0 | 1−(4-methylphenyl)propan-1-one ±2% | |
| Quizlet | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| StudyFetch | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| Turbo AI | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| Solvely | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| NotebookLM | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| ChatGPT | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| Claude | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| MyStudySidekick | 5 | 1-(4-methylphenyl)propan-1-one | 100% |
| Knowt | 5 | 4-ethylacetophenone | 50% |
| Why: Pasted text. Put the ethyl on the ring and the 2.41 singlet on the carbonyl (an acetyl): 4-ethylacetophenone is C10H12O but does not fit the quartet at 2.97 or the 13C at 8.4. | |||
Problem 17 · hard
Stereoisomers of cyclohexane-1,2,3,4,5,6-hexol (inositol)
stereochemistry of a ring with six stereocentres
myo-Inositol is one stereoisomer of cyclohexane-1,2,3,4,5,6-hexol, C₆H₆(OH)₆. Considering configuration only (not conformations):
(a) How many stereoisomers of cyclohexane-1,2,3,4,5,6-hexol exist in total?
(b) How many of them are achiral (meso)?
Original problem written for this benchmark
| App | (a) total stereoisomers | (b) achiral stereoisomers | Score |
|---|---|---|---|
| Answer key | 9 ±0 | 7 ±0 | |
| Turbo AI | 9 | 7 | 100% |
| Solvely | 9 | 7 | 100% |
| NotebookLM | 9 | 7 | 100% |
| ChatGPT | 9 | 7 | 100% |
| Claude | 9 | 7 | 100% |
| MyStudySidekick | 9 | 7 | 100% |
| Quizlet | 8 | 4 | 0% |
| Why: Pasted text. Listed 8 patterns (with duplicates and myo listed twice) and called several achiral inositols chiral. | |||
| Knowt | 32 | 2 | 0% |
| Why: Pasted text. Claimed C6 is not a stereocentre (2^5 = 32); meso count unsupported. | |||
| StudyFetch | 32 | 8 | 0% |
| Why: Pasted text (math flattened). Halved 2^6 for one C2 axis; invented "32 known inositol stereoisomers". | |||
Problem 18 · hard
Structure from spectra: C₁₁H₁₄O₂
NMR/IR/MS structure determination with isomeric esters
An unknown with molecular formula C₁₁H₁₄O₂ shows: IR 1735 cm⁻¹ (strong), 1160 cm⁻¹ (strong), no O–H band; MS m/z 178 (M⁺), 104 (base peak), 91.
¹H NMR (CDCl₃): δ 7.17–7.31 (m, 5H), 4.12 (q, J = 7.1 Hz, 2H), 2.95 (t, J = 7.8 Hz, 2H), 2.61 (t, J = 7.8 Hz, 2H), 1.23 (t, J = 7.1 Hz, 3H).
¹³C NMR: δ 172.9, 140.6, 128.5, 128.3, 126.2, 60.4, 35.9, 31.0, 14.2.
(a) Give the name of the compound.
(b) Which ¹³C signal (δ, ppm) belongs to the CH₂ carbon bonded to the ester oxygen?
Original problem written for this benchmark
| App | (a) name | (b) 13C shift of O–CH2 | Score |
|---|---|---|---|
| Answer key | ethyl 3−phenylpropanoate ±2% | 60.4 ppm ±0.05 ppm | |
| Quizlet | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| StudyFetch | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| Turbo AI | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| Solvely | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| NotebookLM | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| ChatGPT | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| Claude | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| MyStudySidekick | ethyl 3-phenylpropanoate | 60.4 ppm | 100% |
| Knowt | ethyl phenylacetate | 60.4 ppm | 50% |
| Why: Pasted text. Named ethyl phenylacetate (C10H12O2, one carbon short; would show a PhCH2 singlet near 3.6, not two triplets). (b) right. | |||
Problem 19 · hard
Stereoisomers of hepta-2,5-dien-4-ol
stereochemistry: E/Z and a stereocentre that depends on them
Hepta-2,5-dien-4-ol is CH₃–CH=CH–CH(OH)–CH=CH–CH₃. Counting both C=C configurations (E/Z) and any stereocentre:
(a) How many stereoisomers exist?
(b) How many of them are chiral?
Original problem written for this benchmark
| App | (a) total stereoisomers | (b) chiral stereoisomers | Score |
|---|---|---|---|
| Answer key | 4 ±0 | 2 ±0 | |
| Quizlet | 4 | 2 | 100% |
| Turbo AI | 4 | 2 | 100% |
| Solvely | 4 | 2 | 100% |
| NotebookLM | 4 | 2 | 100% |
| ChatGPT | 4 | 2 | 100% |
| Claude | 4 | 2 | 100% |
| MyStudySidekick | 4 | 2 | 100% |
| Knowt | 8 | 8 | 0% |
| Why: Pasted text. 2^2 × 2 with every isomer called chiral. | |||
| StudyFetch | 3 | 1 | 0% |
| Why: Pasted text (math flattened). Declared C4 not a stereocentre in every isomer, so counted the (E,Z) enantiomer pair as one. | |||
Problem 20 · hard
Steam methane reforming with the shift reaction at 1000 K and 10 bar
two simultaneous gas-phase equilibria from JANAF
A feed of 1.00 mol CH₄ and 3.00 mol H₂O(g) reaches equilibrium at 1000 K and a total pressure of 10.0 bar. Two reactions occur:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3 H₂(g) (1)
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) (2)
Using ΔfG° values from the NIST-JANAF Thermochemical Tables (1000 K row, standard state 1 bar), assume ideal gases and no solid carbon, and find:
(a) Kp for reaction (1) at 1000 K;
(b) the fraction of the CH₄ converted;
(c) the mole fraction of H₂ in the equilibrium gas (water included).
Original problem written for this benchmark
| App | (a) Kp of reaction 1 | (b) fraction of CH4 converted | (c) mole fraction of H2 | Score |
|---|---|---|---|---|
| Answer key | 26.28 ±3% | 0.688 ±0.005 | 0.452 ±0.005 | |
| Turbo AI | 26.3 | 0.69 | 0.45 | 100% |
| NotebookLM | 26.28 | 0.688 | 0.452 | 100% |
| ChatGPT | 26.28 | 0.6881 | 0.4524 | 100% |
| Claude | 26.3 | 0.688 | 0.452 | 100% |
| MyStudySidekick | 26.28 | 0.69 | 0.454 | 100% |
| Quizlet | 25.73 | 0.885 | 0.537 | 33% |
| Why: Pasted text (math flattened). JANAF values rounded (K1 25.73, inside the 3% tolerance), but its pressure factor is inverted, so the conversion is far too high. | ||||
| Solvely | 26.28 | 0.697 | 0.458 | 33% |
| Why: Pasted text. Tested app: Solvely. Right setup and K values; its numerical solve is off (ξ1 0.697 vs 0.688). | ||||
| Knowt | 1.10e-8 | 1.06e-3 | 0.0008 | 0% |
| Why: Pasted text. Used 298 K ΔfG° values, ignored reaction (2) and the pressure. | ||||
| StudyFetch | 5.7e-3 | 0.020 | 0.055 | 0% |
| Why: Pasted text (math flattened). Used ΔfG°(CH4) = −50.5 kJ/mol (the 298 K value) instead of JANAF 1000 K +19.492, so reaction (1) came out unfavourable. | ||||
Problem 21 · hard
Silver concentration cell with an ammine complex
electrochemistry with complex-ion and solubility equilibria
A cell at 25 °C is built from two silver electrodes. The left compartment contains 0.010 M AgNO₃ made up in 1.00 M NH₃ (the NH₃ concentration before any complex forms). The right compartment contains 0.010 M AgNO₃ in water. Use Kf for [Ag(NH₃)₂]⁺ from OpenStax Chemistry 2e, Appendix K, and Ksp for AgCl from Appendix J.
(a) What is the cell potential (V)?
(b) Solid NaCl is added to the left compartment. What is the largest Cl⁻ concentration (M) it can reach before AgCl begins to precipitate there?
Original problem written for this benchmark
| App | (a) cell potential | (b) maximum [Cl-] | Score |
|---|---|---|---|
| Answer key | 0.427 V ±0.003 V | 0.261 M ±3% | |
| Quizlet | 0.427 V | 0.26 M | 100% |
| Solvely | 0.427 V | 0.26 M | 100% |
| NotebookLM | 0.427 V | 0.261 M | 100% |
| ChatGPT | 0.427 V | 0.261 M | 100% |
| Claude | 0.427 V | 0.26 M | 100% |
| MyStudySidekick | 0.427 V | 0.26 M | 100% |
| Turbo AI | 0.427 V | 0.29 M | 50% |
| Why: Pasted text. (a) right; (b) used Ksp(AgCl) = 1.8 × 10⁻¹⁰ (a common textbook value) instead of Appendix J 1.6 × 10⁻¹⁰. | |||
| Knowt | 0.252 V | 1.77e-8 M | 0% |
| Why: Pasted text. Treated the cell as one electrode against E°, and used total silver for (b). | |||
| StudyFetch | 0.232 V | 1.5e-4 M | 0% |
| Why: Pasted text (math flattened). Got 6.1 × 10⁻¹⁰ M, then replaced it with an "exact quadratic" value of 1.2 × 10⁻⁶ M and used that; Ksp 1.8 × 10⁻¹⁰. | |||
Problem 22 · hard
Consecutive first-order reactions at a new temperature
kinetics: A → B → C with Arrhenius
For the consecutive first-order reactions A → B → C, at 298.15 K the rate constants are k₁ = 3.0 × 10⁻³ s⁻¹ (Ea,1 = 75.0 kJ/mol) and k₂ = 1.2 × 10⁻³ s⁻¹ (Ea,2 = 105.0 kJ/mol). The reaction is run at 318.15 K starting from pure A.
(a) At what time (in seconds) is the concentration of B at its maximum?
(b) What is the maximum value of [B]/[A]₀?
Original problem written for this benchmark
| App | (a) time of maximum B | (b) maximum [B]/[A]0 | Score |
|---|---|---|---|
| Answer key | 53.73 s ±2% | 0.397 ±0.005 | |
| StudyFetch | 54 s | 0.40 | 100% |
| Turbo AI | 53.7 s | 0.40 | 100% |
| Solvely | 53.7 s | 0.40 | 100% |
| NotebookLM | 53.7 s | 0.397 | 100% |
| ChatGPT | 53.73 s | 0.39683 | 100% |
| Claude | 53.7 s | 0.397 | 100% |
| MyStudySidekick | 53.8 s | 0.397 | 100% |
| Quizlet | 53.6 s | 0.413 | 50% |
| Why: Pasted text (math flattened). (a) right; (b) used rounded k values (k2 1.73e-2) and got 0.413. Again offered to return to the tester's "Biol 1201 Exam 2 Review": cross-chat memory or not a fresh chat (unconfirmed). | |||
| Knowt | 123.45 s | 0.7143 | 0% |
| Why: Pasted text. Never converted k to 318.15 K; invented t_max and used k1/(k1+k2) for (b). | |||
Problem 23 · hard
Selectivity of parallel reactions
kinetics: competing first-order paths
A reactant decomposes by two parallel first-order paths: A → B with A₁ = 1.0 × 10⁸ s⁻¹ and Ea,1 = 60.0 kJ/mol, and A → C with A₂ = 1.0 × 10¹³ s⁻¹ and Ea,2 = 90.0 kJ/mol.
(a) At what temperature (K) are the two rate constants equal?
(b) At 350.0 K, what fraction of A ends up as B once A is fully consumed?
(c) At 350.0 K, how long (in seconds) does it take to consume 99.0% of A?
Original problem written for this benchmark
| App | (a) temperature of equal rates | (b) fraction converted to B | (c) time to 99% consumption | Score |
|---|---|---|---|---|
| Answer key | 313.4 K ±0.5 K | 0.231 ±0.003 | 9.57 s ±2% | |
| StudyFetch | 313.4 K | 0.23 | 9.6 s | 100% |
| Turbo AI | 313.5 K | 0.229 | 9.6 s | 100% |
| NotebookLM | 313.4 K | 0.231 | 9.57 s | 100% |
| ChatGPT | 313.4 K | 0.2308 | 9.57 s | 100% |
| Claude | 313.4 K | 0.231 | 9.57 s | 100% |
| MyStudySidekick | 313.4 K | 0.2309 | 9.58 s | 100% |
| Solvely | 313.4 K | 0.229 | 9.65e-3 s | 67% |
| Why: Pasted text. Tested app: Solvely. Exponentials evaluated 1000x too large (k1 109 s⁻¹ instead of 0.111), so (c) is 9.65 ms; the ratio in (b) survives the error. | ||||
| Quizlet | 313.4 K | 0.249 | no final answer | 33% |
| Why: Pasted text (math flattened). (b) k1 miscomputed (0.121 vs 0.111). The copy ends mid-way through (c), so no (c) answer is on file — if the app finished (c), paste the rest to regrade. | ||||
| Knowt | 300.97 K | 0.129 | 0.0456 s | 0% |
| Why: Pasted text. Sign error in (a); rate constants off by ~100x in (b) and (c). | ||||
Problem 24 · hard
Titration of a strong + weak acid mixture
acid–base titration with two acids
25.00 mL of a solution that is 0.0500 M HCl and 0.1000 M CH₃CO₂H is titrated with 0.1000 M NaOH at 25 °C. Use Ka for acetic acid from OpenStax Chemistry 2e, Appendix H, and Kw = 1.0 × 10⁻¹⁴. Find the pH after adding:
(a) 12.00 mL of NaOH;
(b) 12.50 mL of NaOH;
(c) 37.50 mL of NaOH.
Original problem written for this benchmark
| App | (a) pH after 12.00 mL | (b) pH after 12.50 mL | (c) pH after 37.50 mL | Score |
|---|---|---|---|---|
| Answer key | 2.707 ±0.02 | 2.964 ±0.02 | 8.674 ±0.02 | |
| NotebookLM | 2.71 | 2.96 | 8.67 | 100% |
| ChatGPT | 2.71 | 2.96 | 8.67 | 100% |
| Claude | 2.71 | 2.96 | 8.67 | 100% |
| MyStudySidekick | 2.71 | 2.96 | 8.67 | 100% |
| Quizlet | 2.87 | 2.96 | 8.67 | 67% |
| Why: Pasted text (math flattened). (a) counted only the leftover HCl (the trap). | ||||
| Turbo AI | 2.71 | 2.99 | 8.67 | 67% |
| Why: Pasted text. (b) evaluated its own (correct) quadratic wrongly: 1.03e-3 instead of 1.087e-3 M, so 2.99 is 0.026 off. | ||||
| Solvely | 2.87 | 2.96 | 8.67 | 67% |
| Why: Pasted text. Tested app: Solvely. (a) said acetic acid ionisation is "heavily suppressed" and counted only the leftover HCl. | ||||
| StudyFetch | 2.87 | 2.96 | 4.74 | 33% |
| Why: Pasted text (math flattened). (a) leftover HCl only; (c) mistook the second equivalence point for the acetic acid half-equivalence point. | ||||
| Knowt | 4.70 | 4.72 | 3.83 | 0% |
| Why: Pasted text. Incoherent working; none of the three values is right. | ||||
Problem 25 · hard
Solubility of CaF₂ in an acidic fluoride solution
solubility with a common ion and a competing acid equilibrium
Find the molar solubility of CaF₂ in a solution that is buffered at pH 2.50 and also contains 0.0100 M NaF (total fluoride added as NaF, before any CaF₂ dissolves). Use Ksp for CaF₂ from OpenStax Chemistry 2e, Appendix J, and Ka for HF from Appendix H. Assume the buffer holds the pH fixed and ignore activity effects.
(a) What is the molar solubility of CaF₂ (M)?
(b) How many grams of CaF₂ dissolve per litre? (M(CaF₂) = 78.07 g/mol)
Original problem written for this benchmark
| App | (a) molar solubility | (b) solubility | Score |
|---|---|---|---|
| Answer key | 0.000014 M ±3% | 0.001096 g/L ±3% | |
| NotebookLM | 1.40e-5 M | 1.10e-3 g/L | 100% |
| ChatGPT | 1.40e-5 M | 1.10e-3 g/L | 100% |
| Claude | 1.40e-5 M | 1.10e-3 g/L | 100% |
| MyStudySidekick | 1.40e-5 M | 1.10e-3 g/L | 100% |
| Turbo AI | 1.47e-5 M | 1.1e-3 g/L | 50% |
| Why: Pasted text. Used Ksp 5.3 × 10⁻¹¹ and Ka 6.8 × 10⁻⁴ (neither is the OpenStax value). (a) 4.7% off; (b) passes only because 1.148 × 10⁻³ was rounded to 1.1 × 10⁻³ (the key's tolerance is ±3%). | |||
| Quizlet | 1.13e-5 M | 8.82e-4 g/L | 0% |
| Why: Pasted text (math flattened). Used Ksp 3.9 × 10⁻¹¹ and Ka 7.2 × 10⁻⁴. | |||
| Knowt | 3.9e-7 M | 3.05e-5 g/L | 0% |
| Why: Pasted text. Ignored the HF equilibrium (the trap: 35x too low). | |||
| StudyFetch | 1.1e-5 M | 8.6e-4 g/L | 0% |
| Why: Pasted text (math flattened). Used Ksp 3.45 × 10⁻¹¹ and Ka 6.8 × 10⁻⁴. | |||
| Solvely | 1.3e-5 M | 1.0e-3 g/L | 0% |
| Why: Pasted text. Tested app: Solvely. Used Ka(HF) = 6.8 × 10⁻⁴ (not Appendix H 6.4 × 10⁻⁴). | |||