Catalonia PAU Chemistry common mistakes
Catalonia PAU Chemistry · Understand the traps and how to avoid them.
When balancing redox equations in acidic or basic medium, the candidate balances atoms and charges without adding H2O and H+ (or OH-) to close the oxygen and hydrogen balance, then accepts an equation whose charge is unequal on the two sides.
The half-reaction method is taught as three separate steps (mass balance, oxygen balance, charge balance) and candidates collapse them into one, forgetting that in basic medium the added H+ must be neutralised with an equal number of OH- on both sides. Under exam pressure they stop as soon as the main species look balanced.
What to do instead
Adopt a fixed checklist for each half-reaction: (1) balance the element that changes oxidation state, (2) add H2O to balance O, (3) add H+ to balance H, (4) add electrons to balance charge, (5) in basic medium add OH- to both sides equal to the H+ count and cancel the resulting H2O. Before combining, verify numerically that the total charge on each side is identical.
The candidate writes the equilibrium constant for a heterogeneous system including pure solids or pure liquids, or inverts Kc and Kp (products over reactants versus reactants over products) depending on how the equation happens to be written in the prompt.
Kc is memorised as a formula rather than as a statement about activities. Because pure condensed phases have unit activity, they must be omitted, but this rule is easily forgotten when a solid appears in the equation. Inversion happens because the same reaction may be quoted in either direction and the candidate does not re-derive K from the quoted direction.
What to do instead
Always write K from the law of mass action using activities, then set a(solid)=1 and a(pure liquid)=1 explicitly, crossing them out. Immediately after writing the equation, annotate the arrow direction and confirm that the product of the reaction appears in the numerator. Practise converting: if the equation is reversed, K becomes 1/K; if multiplied by n, K becomes K^n.
In weak acid–base problems the candidate substitutes the analytical concentration directly into pH = -log c, or uses the simple square-root approximation for [H+] without checking whether the degree of dissociation is small enough to justify neglecting the subtracted term.
Strong-acid formulas are over-learned and the weak-acid ICE table is skipped. Even when the quadratic is set up, candidates assume x is negligible compared with c without verifying the 5% condition, which fails for moderately weak acids at low concentration.
What to do instead
Write the full equilibrium expression Ka = x^2/(c - x) each time, then test x/c < 0.05 before dropping x from the denominator. If the test fails, solve the quadratic. For bases, compute pOH first and convert with pH = 14 - pOH at 25 degrees Celsius, and state the temperature whenever the value 14 is used.
When predicting the pH of a salt solution, the candidate labels every salt of a strong acid and strong base as neutral and every other salt as acidic, ignoring which ion hydrolyses and whether the conjugate acid or conjugate base is the stronger hydrolysing species.
The rule is remembered as a two-way table (strong/strong neutral, weak/strong acidic) without the underlying idea that the spectator ion is the one from the strong partner. Salts of weak acids with weak bases, and acid salts such as hydrogencarbonates, break the table and are misclassified.
What to do instead
For each salt, split it into cation and anion, identify the parent acid and parent base, and ask which conjugate is strong enough to react with water. Compare Ka and Kb for salts of weak acid plus weak base; the larger constant determines the sign. Treat amphiprotic anions with pH = (pKa1 + pKa2)/2 and note the approximation.
The candidate computes cell potentials by adding the standard potentials of the two half-reactions as written, or changes the sign of E when a half-reaction is multiplied by a coefficient, producing a cell emf that is wrong although the redox equation is correct.
E is an intensive property, so it does not scale with stoichiometry, and this contrasts with enthalpy and Gibbs energy, which are extensive. Candidates transfer the extensive-additivity habit from thermochemistry and also forget that the anode potential must be subtracted, not added.
What to do instead
Compute Ecell = Ecathode - Eanode using the reduction potentials exactly as tabulated, without multiplying or reversing signs. If half-reactions are reversed, remember the sign of E flips but the magnitude does not change with coefficients. Cross-check with Delta G = -nFE, where n does scale, and confirm that a positive Ecell corresponds to a negative Delta G.
In kinetics the candidate reads the overall order from the stoichiometric coefficients of the balanced equation, or determines the rate law from a single experiment instead of comparing experiments in which one concentration varies while the others are held constant.
The balanced equation describes the overall transformation, not the rate-determining step, and this distinction is the central idea of mechanism-based kinetics. Candidates also confuse the rate constant units with the order and cannot use them as a consistency check.
What to do instead
State explicitly that the rate law must be determined experimentally and that orders are the exponents in the rate equation, not the stoichiometric numbers. To find each order, select two runs that differ only in that reactant and use the ratio method. Verify the units of k against the overall order (mol dm^-3 s^-1 for zero order, s^-1 for first order, dm^3 mol^-1 s^-1 for second order).
The candidate decides spontaneity from the sign of the enthalpy change alone, or uses the sign of Delta S without considering temperature, and so misclassifies endothermic reactions that are entropy-driven or reactions whose spontaneity reverses at a crossover temperature.
Everyday experience equates exothermic with spontaneous, and the Gibbs equation is memorised as a formula without interpreting the competition between enthalpy and entropy. The temperature dependence of the T Delta S term is not treated as a variable.
What to do instead
Build the four-case table for the signs of Delta H and Delta S and state the temperature range in which each gives Delta G < 0. When both quantities are known, solve T = Delta H / Delta S for the crossover and discuss the result in kelvin. Always report Delta G in kJ per mole consistently and check that Delta S is converted to kJ when combined with Delta H.
When using Hess's law or bond energies, the candidate adds the given enthalpies without reversing the sign of reactions that were inverted in the manipulation, or forgets that bond energies are defined for bond breaking and so must be summed with reversed signs for bonds formed.
The arithmetic manipulation of thermochemical equations is treated as algebra on numbers rather than on reactions, so the sign convention is lost. Bond energy tables list positive dissociation energies, and candidates add them as if they were formation enthalpies.
What to do instead
Rewrite each given equation with its Delta H beside it, mark inverted equations with a struck-through arrow and an explicit sign change, then sum. For bond energies use Delta H = sum(bonds broken) - sum(bonds formed), all values taken as positive dissociation energies. Verify the result against an independent route such as formation enthalpies when data allow.
In organic nomenclature and isomerism the candidate names a compound by choosing the longest chain without regard to the principal functional group, or counts stereoisomers without checking for a plane of symmetry, so a molecule with a stereocentre is reported as having two enantiomers when it is meso.
The priority rules (principal group, then multiple bonds, then substituents, then lowest locants) are applied in the wrong order, and stereochemistry is treated as a counting exercise rather than a symmetry question. Cis/trans and E/Z descriptors are also confused with each other.
What to do instead
Apply the IUPAC priority sequence in order and write the chosen parent chain before numbering. For stereochemistry, draw the molecule, locate stereocentres, then test for an internal mirror plane before stating the number of stereoisomers; a meso compound has stereocentres but is achiral. Distinguish cis/trans (two substituents) from E/Z (CIP priorities on each carbon).
In VSEPR and polarity questions the candidate assigns molecular geometry from the electron-pair geometry without removing lone pairs from the count, or declares a molecule with polar bonds to be non-polar without checking whether the bond dipoles cancel by symmetry.
The two-stage procedure (count domains, then place atoms only) is compressed into one step, so water is described as linear or ammonia as trigonal planar. Polarity is then judged from electronegativity differences alone, and vector cancellation in symmetric arrangements such as CO2 or BF3 is overlooked.
What to do instead
Draw the Lewis structure, count bonding domains and lone pairs, name the electron-domain arrangement, then derive the molecular shape from atom positions only and quote the ideal angle with the lone-pair compression noted. For polarity, draw the individual bond dipoles as arrows and add them as vectors: identical dipoles opposing in a linear or trigonal planar or tetrahedral symmetric arrangement cancel, whereas a lone pair breaks the symmetry.