Chemistry HL

1. 1.

   Plan an initial-rate investigation of the order with respect to A for an aqueous reaction A + B -> products. Concentrations quoted are initial concentrations after mixing. Give a suitable series of three A concentrations, explain how to keep B concentration and total mixed volume fixed while changing A, and state how the measured initial rates could determine the order in A. Keep temperature fixed.

   [3 marks] · no calculator

   Worked explanation

   1. The comparison must use concentrations after mixing, not the labels on stock bottles. At fixed total volume, varying A stock and compensating with solvent leaves the chosen B amount at the same concentration.
   2. Initial gradients minimise composition changes during measurement. For two trials at otherwise identical conditions, rate2/rate1 = ([A]2/[A]1)^m, allowing m to be calculated and checked with the third trial.

   Common mistake: Adding more A stock without compensating volume also dilutes B and cannot isolate the order in A.

   Marking points

   • A suitable series changes only initial A, for example 0.020, 0.040 and 0.060 mol/dm^3 at fixed temperature.
   • Use the same amount of B stock and same final total volume in all trials, replacing changes in A-stock volume with solvent.
   • Determine initial slopes from early product-time data and compare rate ratios with concentration ratios, or fit log(initial rate) against log[A] whose slope is the order.

   Examiner tip: Adding more A stock without compensating volume also dilutes B and cannot isolate the order in A.

2. 2.

   Hypothetical rate constants for the same reaction are 0.0100 s^-1 at 300 K and 0.0200 s^-1 at 310 K. Assume Arrhenius behaviour with constant A across this interval. Using ln(k2/k1) = Ea/R (1/T1 - 1/T2) and R = 8.314 J mol^-1 K^-1, calculate Ea in kJ/mol. Explain why two points cannot establish that this model holds over a wide temperature range.

   [3 marks]

   Worked explanation

   1. The reciprocal-temperature difference is 0.000107527 K^-1. Ea = 8.314 x ln(2) / 0.000107527 = approximately 53594 J/mol.
   2. The positive 53.6 kJ/mol is consistent with the rate constant increasing with temperature. Agreement at two points alone does not test curvature, changes of mechanism or repeatability.

   Common mistake: Using the Celsius temperatures in the reciprocal-temperature expression invalidates the Arrhenius calculation.

   Marking points

   • Substitutes ln(2) and (1/300 - 1/310) with temperatures in kelvin.
   • Ea = approximately 53600 J/mol = 53.6 kJ/mol.
   • Two points always define a straight line in an Arrhenius plot; more temperatures and repeat measurements are needed to assess linearity and uncertainty.

   Examiner tip: Using the Celsius temperatures in the reciprocal-temperature expression invalidates the Arrhenius calculation.

3. 3.

   A hypothetical dilution uses a 10.00 +/-0.02 cm^3 pipette and a 100.0 +/-0.1 cm^3 volumetric flask. Stock concentration uncertainty is negligible. Under a conservative additive relative-uncertainty rule, calculate the percentage uncertainty in diluted concentration. Then predict the direction of concentration bias if the flask is filled above the mark, and propose a prevention step.

   [3 marks]

   Worked explanation

   1. Since c(diluted) = c(stock) x V(pipette)/V(flask), add the specified relative bounds: 0.20% + 0.10% = 0.30%.
   2. Overfilling is a directional preparation error, not an extra random replicate fluctuation. At fixed transferred solute, a larger final volume lowers concentration; careful meniscus alignment prevents this specific bias.

   Common mistake: Adding 0.02 and 0.1 as if they were relative uncertainties ignores their different reference volumes.

   Marking points

   • Percentage uncertainty = (0.02/10.00 + 0.1/100.0) x 100 = 0.30%.
   • Filling above the mark gives a lower concentration because the fixed solute amount occupies too large a volume.
   • Approach the mark dropwise and view the bottom of the meniscus at eye level with the flask upright.

   Examiner tip: Adding 0.02 and 0.1 as if they were relative uncertainties ignores their different reference volumes.

4. 4.

   For written HL investigation preparation, assume rate=k[A][B] with k=0.50 dm^3 mol^-1 s^-1. The reaction consumes A and B in a 1:1 ratio. Initially [A]=0.010 mol dm^-3 and [B]=0.200 mol dm^-3 at constant volume and temperature. Calculate the initial rate and k_obs in the approximation rate=k_obs[A]. If A becomes 0.002 mol dm^-3, calculate B and the resulting k_obs, and evaluate whether the original constant-k_obs approximation is exact.

   [5 marks]

   Syllabus placement evidence: Chemistry guide, first assessment 2025, Reactivity 2.2.9-2.2.11 (HL); subject brief p.3 Inquiry process, Concluding and evaluating. Applies the supplied HL rate law and 1:1 stoichiometry to test an excess-reagent approximation. Reactivity 2.2.9-2.2.11 covers rate equations, reaction orders and rate-constant units; the pseudo-first-order approximation is explained through the supplied model. Written HL investigation preparation, not SL placement or an actual kinetic study.

   Worked explanation

   1. Holding B approximately constant converts the supplied second-order law to a first-order-looking dependence on A. The observed coefficient includes B and therefore has different units from k.
   2. The 1:1 stoichiometry at fixed volume removes equal concentrations of A and B. Although B remains in large excess, it is not unchanged; compare the calculated 4% coefficient drift with the intended measurement tolerance.

   Common mistake: An excess reagent is not an unconsumed reagent, and k_obs can change while the true k stays fixed at constant temperature.

   Marking points

   • Initial rate=0.50(0.010)(0.200)=0.00100 mol dm^-3 s^-1.
   • Initially k_obs=k[B]=0.100 s^-1.
   • A consumption=0.008 mol dm^-3, so [B]=0.192 mol dm^-3.
   • Later k_obs=0.0960 s^-1, a 4.0% decrease from its initial value.
   • The excess-B approximation is not exact; whether 4.0% drift is acceptable depends on precision/aim, and increasing B relative to A reduces its fractional depletion.

   Examiner tip: An excess reagent is not an unconsumed reagent, and k_obs can change while the true k stays fixed at constant temperature.

5. 5.

   A hypothetical fast-reaction investigation uses a sensor whose displayed signal after a sudden constant true signal S is S_display(t)=S(1-exp(-t/2.0)), with t in seconds. Treat this as a supplied response model, not a reaction law. Calculate the displayed fraction at t=1.0 s and the time to reach 95% of S. Explain why comparing first-second slopes from this sensor may misrepresent reactions with different fast time courses, and propose a targeted improvement.

   [4 marks]

   Syllabus placement evidence: Chemistry subject brief, first assessment 2025, p.3 Skills in the study of chemistry, Tools/Technology; Inquiry process, Exploring and designing / Concluding and evaluating. Evaluates measurement suitability for a rate investigation using a supplied response equation. HL preparation application, not a claim that this sensor model is prescribed recall, that a real reaction has been measured or that an IA is complete.

   Worked explanation

   1. The two-second constant belongs to the measurement device. Substitute t=1 into its response and solve exp(-t/2)=0.05 for the settling time.
   2. A device still far from a step response at one second cannot automatically resolve sub-second chemistry. The calibration model diagnoses a limitation; correcting a time-varying reaction would require a justified dynamic treatment, not merely dividing all slopes by 0.3935.

   Common mistake: The sensor's exponential response is not evidence that the chemical reaction is first order, and a single step-response fraction is not a universal rate correction.

   Marking points

   • Displayed fraction at 1.0 s is 1-exp(-0.5)=0.3935 approximately, or 39.35%.
   • For 95%, t=-2.0 ln(0.05)=5.991... s, approximately 6.0 s.
   • Instrument lag smooths/delays early changes, so first-second display slopes need not equal chemical initial rates or preserve their ratios.
   • Characterise response with an appropriate rapid calibration change and select a sufficiently faster sensor, or slow the reaction under controlled documented conditions and reassess response relative to reaction timescale.

   Examiner tip: The sensor's exponential response is not evidence that the chemical reaction is first order, and a single step-response fraction is not a universal rate correction.

6. 6.

   In written HL investigation preparation, hypothetical first-order rate constants are fitted as y=20.0-6000x, where y=ln(k/(1 s^-1)) and x=1/T in K^-1. Use ln(k/(1 s^-1))=ln(A/(1 s^-1))-Ea/(RT), with R=8.314 J mol^-1 K^-1. Identify the slope and intercept, calculate Ea in kJ mol^-1 and A with units, and predict k at 300 K. The fit was obtained at only two distinct temperatures: explain one additional measurement strategy needed before treating the relation as valid over a wider temperature range.

   [5 marks]

   Syllabus placement evidence: Chemistry guide, first assessment 2025, Reactivity 2.2.11-2.2.13 (additional higher level), printed p.54. Adds HL Arrhenius-factor and intercept analysis absent from the two-temperature activation-energy task. Hypothetical written preparation, not an SL requirement or completed kinetics investigation.

   Worked explanation

   1. Match y=b+mx to the supplied linear Arrhenius equation. A plot against 1/T gives m=-Ea/R, so the negative slope implies positive Ea. The slope has kelvin units because y is dimensionless.
   2. Multiply 6000 by 8.314 and divide by 1000 to get 49.884 kJ mol^-1. Exponentiate the intercept to recover A; first-order k and A both have s^-1 units.
   3. At 300 K the exponent is zero, not the rate constant: exp(0)=1. A line through two temperatures is a fit, not a test that A and the apparent activation energy remain constant beyond that interval.

   Common mistake: Reporting A=20 or k=0 confuses a logarithm with the quantity it describes.

   Marking points

   • Slope=-6000 K and intercept=20.0; the intercept is ln(A/(1 s^-1)), not A itself.
   • Ea=-slope x R=49884 J mol^-1=49.9 kJ mol^-1 approximately.
   • A=exp(20.0) s^-1=4.85 x 10^8 s^-1 approximately.
   • At 300 K, ln(k/(1 s^-1))=20.0-6000/300=0, giving k=1.00 s^-1.
   • Measure replicated rate constants at several additional temperatures across the proposed range and examine Arrhenius residuals/linearity and uncertainty; two temperatures alone cannot test curvature.

   Examiner tip: Reporting A=20 or k=0 confuses a logarithm with the quantity it describes.

7. 7.

   In hypothetical initial-rate trials at one temperature, concentrations are quoted after mixing and rates are in mol dm^-3 s^-1. Trial 1: [A]=0.0100, [B]=0.0200, rate=4.00 x 10^-5. Trial 2: [A]=0.0200, [B]=0.0200, rate=1.60 x 10^-4. Trial 3: [A]=0.0100, [B]=0.0400, rate=8.00 x 10^-5. Concentrations have units mol dm^-3. Assume rate=k[A]^m[B]^n over this range. Deduce m and n with evidence, give the rate equation and overall order, calculate k with units, and predict the initial rate for [A]=0.0300 and [B]=0.0100 at the same temperature. This is written investigation analysis, not actual measured data.

   [5 marks]

   Syllabus placement evidence: IB Chemistry guide, first assessment 2025, Reactivity 2.2.9-2.2.11 (additional higher level), printed p.54. Adds experimental-data rate-law inference absent from a supplied-law question or design-only initial-rate task. Written HL investigation preparation, with no completed experiment or canonical-area remapping claimed.

   Worked explanation

   1. Choose pairs that change only one initial concentration. The resulting ratios are 4=2^m and 2=2^n; the stoichiometric equation, which is not supplied, is unnecessary for inferring these empirical orders.
   2. Divide rate by the concentration product to obtain k. Its units are (mol dm^-3 s^-1)/(mol dm^-3)^3=dm^6 mol^-2 s^-1.
   3. Use the inferred law without changing k at the same temperature. This prediction assumes the same mechanism/rate-law regime and is not guaranteed at arbitrarily different concentrations.

   Common mistake: A larger raw rate does not determine an order unless the concentration ratio and controlled reactant are identified.

   Marking points

   • Trials 1 and 2 hold B fixed: doubling A quadruples rate, so m=2.
   • Trials 1 and 3 hold A fixed: doubling B doubles rate, so n=1.
   • rate=k[A]^2[B], overall order=3.
   • k=(4.00 x 10^-5)/((0.0100)^2 x 0.0200)=20.0 dm^6 mol^-2 s^-1; numerical value and units both required.
   • Predicted rate=20.0 x (0.0300)^2 x 0.0100=1.80 x 10^-4 mol dm^-3 s^-1.

   Examiner tip: A larger raw rate does not determine an order unless the concentration ratio and controlled reactant are identified.

8. 8.

   A hypothetical investigation establishes rate=k_obs[A][B]^2 for overall A+2B -> P. Candidate mechanism M has A+B <=> I (fast pre-equilibrium), then I+B -> P (slow). For this task use [I]=K[A][B] and rate_slow=k2[I][B]. Candidate N has A+B -> I (slow), then I+B -> P (fast), and its slow-step rate is k1[A][B]. Sum the steps for M, identify its intermediate, derive its predicted rate equation, and decide which candidate is consistent with the measured law. Explain why consistency does not prove M uniquely correct. Treat A, B, I and P as abstract species with the stated stoichiometry.

   [5 marks] · no calculator

   Syllabus placement evidence: IB Chemistry guide, first assessment 2025, Reactivity 2.2.6-2.2.10 (additional higher level), printed pp.53-54. Targets prescribed evaluation of mechanisms against kinetic and stoichiometric evidence, including a rate-determining step that is not first. The pre-equilibrium expression is supplied, not imposed as extra recall.

   Worked explanation

   1. Bookkeeping and kinetics are separate tests: both candidate sequences can give A+2B -> P, but that is not enough to explain the measured exponent of B.
   2. For M the slow collision uses I and B. The fast equilibrium makes I proportional to A times B, introducing a second B factor without requiring a termolecular slow collision.
   3. An intermediate has a finite chemical identity between steps, unlike a transition state at a barrier. Even a matching possible mechanism remains provisional because kinetic agreement is not unique identification.

   Common mistake: Reading overall coefficients directly as rate-law exponents bypasses the mechanism and the empirical evidence.

   Marking points

   • Adding the forward steps of M and cancelling I gives A+2B -> P.
   • I is an intermediate: formed in one step, consumed in another and absent from the net equation; it is not a transition state.
   • Substitution gives rate=k2*K[A][B]^2, with k_obs=k2*K under the supplied pre-equilibrium assumptions.
   • M is consistent; N predicts only first order in B and is inconsistent with the measured second-order B dependence under its stated slow-step model.
   • Other mechanisms can share a rate law and net stoichiometry. Independent evidence, such as detecting I or testing predictions under changed conditions, is needed to distinguish them.

   Examiner tip: Reading overall coefficients directly as rate-law exponents bypasses the mechanism and the empirical evidence.

9. 9.

   For written investigation analysis, an ideal-gas mixture undergoes A(g)+B(g) <=> C(g)+D(g) at constant temperature and volume. Kc=9.00 in the concentration convention used here. Initially [A]=[B]=0.600 mol dm^-3 and [C]=[D]=0.200 mol dm^-3. Calculate the initial reaction quotient Qc and predict the net direction. Let x mol dm^-3 react in that direction: write all four equilibrium concentrations, solve for x and give the equilibrium composition. Take the positive square root of Kc to avoid expanding a quadratic. Explain why initial concentrations must not simply be inserted as equilibrium concentrations.

   [5 marks]

   Syllabus placement evidence: IB Chemistry guide, first assessment 2025, Reactivity 2.3.5-2.3.6 (additional higher level), printed p.55. Adds general homogeneous equilibrium-composition work beyond the existing acid/base examples. The paired concentrations permit a linear solution after a square root, respecting the guide's no-quadratic expectation.

   Worked explanation

   1. The initial product/reactant concentration ratio is not Kc; it is a quotient to compare with Kc. Since it is too small, the forward direction raises it.
   2. The 1:1:1:1 stoichiometry and fixed volume give equal concentration changes. Identical paired concentrations let the equilibrium expression be written as a squared ratio; all concentrations must remain nonnegative.
   3. Solving the positive-ratio equation gives x=0.400, within the available 0.600. Checking the final values gives (0.600/0.200)^2=9.00. Kc and Qc are dimensionless for this equal-total-power concentration expression.

   Common mistake: Using 9=(0.200/0.600)^2 declares an initially non-equilibrium mixture equilibrated without allowing its composition to change.

   Marking points

   • Qc=(0.200 x 0.200)/(0.600 x 0.600)=1/9=0.111 approximately.
   • Qc<Kc, so net forward reaction initially forms more C and D.
   • [A]eq=[B]eq=0.600-x and [C]eq=[D]eq=0.200+x, in mol dm^-3.
   • (0.200+x)/(0.600-x)=sqrt(9.00)=3.00 gives x=0.400 mol dm^-3.
   • Equilibrium: A and B each 0.200, C and D each 0.600 mol dm^-3; initial concentrations are not equilibrium values because net change continues until Qc=Kc.

   Examiner tip: Using 9=(0.200/0.600)^2 declares an initially non-equilibrium mixture equilibrated without allowing its composition to change.

10. 10.

    For hypothetical written spectroscopy preparation, the short-wavelength convergence limit of the hydrogen series ending at n=1 is measured as 91.2 nm. Use E=hc/lambda, h=6.626 x 10^-34 J s, c=2.998 x 10^8 m s^-1 and NA=6.022 x 10^23 mol^-1. Convert the wavelength to metres, calculate the corresponding photon energy and first ionisation energy in kJ mol^-1, and explain why this limit rather than any one finite transition gives the ground-state ionisation threshold. Propose one specific instrumental check before trusting the inferred value.

    [5 marks]

    Syllabus placement evidence: IB Chemistry guide, first assessment 2025, Structure 1.3.6 (additional higher level), printed p.35; Inquiry 2 emission-spectrum measurements. Targets the explicit HL ionisation-energy calculation from a spectral convergence wavelength. Fits the scientific-investigation preparation topic despite lying outside kinetics; no live ultraviolet experiment is implied.

    Worked explanation

    1. Nanometres must become metres to match c. The result from hc/lambda is energy for one photon, which corresponds here to one atomic threshold event.
    2. Multiply by Avogadro's constant to obtain joules per mole, then divide by 1000. Do not multiply by NA twice or treat the photon result as already molar.
    3. The stated n=1 final level matters: a series ending at an excited level would not directly give first ionisation from ground-state hydrogen. A resolved spectral line close to the limit is still a finite transition, so instrument calibration and limit identification need separate scrutiny.

    Common mistake: Using any visible hydrogen line as the first-ionisation threshold confuses a finite excitation gap with removal of the ground-state electron.

    Marking points

    • 91.2 nm=9.12 x 10^-8 m.
    • Photon energy=(6.626 x 10^-34)(2.998 x 10^8)/(9.12 x 10^-8)=2.18 x 10^-18 J approximately.
    • Molar threshold=E*NA/1000=1.31 x 10^3 kJ mol^-1 approximately (unrounded about 1312).
    • At convergence the initial bound level tends to n=infinity; reversing that limiting transition corresponds to removing an electron from n=1 to the ionisation limit, not merely exciting it to another finite level.
    • Check/calibrate the wavelength scale against an independent appropriate reference spectrum in the measurement range, or verify sufficient spectral resolution to locate the convergence limit rather than treating the last resolved finite line as the limit.

    Examiner tip: Using any visible hydrogen line as the first-ionisation threshold confuses a finite excitation gap with removal of the ground-state electron.

Original, unofficial practice questions from SubjectScout — not official past-paper questions or complete syllabus coverage. Added English teaching explanations are AI-authored; independent teacher review is pending.