Chemistry 30 Diploma common mistakes
Chemistry 30 Diploma · Understand the traps and how to avoid them.
Stops at q = mcΔT and reports that joule value as the enthalpy change, leaving the answer in J rather than kJ/mol and dropping the sign, so dissolving an exothermic solid is reported with a positive ΔH.
The calculation is taught as a single formula, so students treat the calorimeter's heat gain as the reaction's enthalpy and never return to the definition of molar enthalpy or to the sign convention for the system versus the surroundings.
What to do instead
Force a three-line write-up every time: (1) q_surroundings = mcΔT with the sign of ΔT noted, (2) q_reaction = −q_surroundings, (3) ΔH = q_reaction ÷ moles of the limiting reactant, converted to kJ/mol. Check the final sign against whether the thermometer rose or fell before moving on.
Applies Hess's law to a set of given equations without reversing the sign of ΔH when an equation is flipped, and copies ΔH unchanged when the equation is multiplied by a factor.
ΔH gets treated as a property stuck to a substance rather than a property of the reaction exactly as written, so rearranging the equation feels like it should not touch the number.
What to do instead
Before summing, rewrite every step as a numbered line with the multiplier written beside the enthalpy, e.g. '2 × (equation b): ΔH = 2(−394)'. Sum the annotated lines in one column, carrying units of kJ throughout, and confirm the rewritten steps add to the target equation atom by atom.
In a galvanic cell, labels the positive electrode as the anode because 'positive = anode' was memorized from an electrolytic cell, or assigns oxidation to the cathode.
The two cell types are taught close together and the sign convention for electrodes flips between them, so a half-remembered rule from the other cell type overwrites the oxidation/reduction definition.
What to do instead
Write the two anchor statements before labelling anything: oxidation always occurs at the anode (electrons are lost there) and reduction always at the cathode. Then decide the sign from cell type only: anode is negative in a galvanic cell, positive in an electrolytic cell. Confirm with the E° table by identifying which species has the higher reduction potential and must be reduced.
Multiplies a standard reduction potential by the coefficient used to balance electrons, e.g. reports E° = 2(0.34 V) for a half-reaction doubled to match electron count.
The habit of scaling enthalpy in Hess's law transfers over, and the volt is not obviously an energy-per-charge quantity, so scaling the half-reaction seems to require scaling its potential.
What to do instead
State the definition each time: E° is joules per coulomb, an intensive property, so it is unchanged by multiplying a half-reaction. Only the number of electrons n changes, and n appears in ΔG° = −nFE°cell, which is where scaling legitimately enters.
Declares a cell reaction spontaneous when E°cell is negative, or reverses a half-reaction to place it at the anode but forgets to reverse the sign of its potential before adding.
The table of reduction potentials is used as a lookup of 'reaction energies' rather than as reduction data that must be converted to oxidation data, and the sign test is remembered as 'a cell always has a positive voltage anyway'.
What to do instead
Never reverse the table entries; keep every half-reaction as a reduction, then use E°cell = E°cathode − E°anode. Test spontaneity with E°cell > 0 and cross-check with ΔG° = −nFE°cell < 0. Also name the reducing and oxidizing agents from the species actually oxidized and reduced on the page.
Numbers the parent chain from the end that gives a substituent the lower locant even when a principal functional group or a multiple bond sits nearer the other end, e.g. naming an alcohol or alkene as if it were an alkane with a branch.
Introductory practice is dominated by simple substituted alkanes, so 'lowest substituent number' becomes the default rule and the IUPAC priority hierarchy is never applied once functional groups appear.
What to do instead
Run a fixed checklist before numbering: identify the principal functional group and make it the suffix, then the parent chain must contain it; give it the lowest possible locant; next give multiple bonds the lowest locants; only then minimize substituent locants and list them alphabetically. Write the candidate name from both ends and keep the one that wins at the highest-priority step.
Predicts the product of the wrong reaction type, for instance adding HBr across a C=C using the minor (anti-Markovnikov) regiochemistry, or treating bromine with benzene as an addition instead of a catalysed substitution.
Reagents are memorized as a list of 'things that react' without classifying the substrate's functional group or the role of the reagent and catalyst, so a familiar-looking reagent triggers the wrong mechanism.
What to do instead
Before writing any product, name the functional group in the reactant and classify the reagent as electrophile, nucleophile, oxidant or reductant, plus note the catalyst. Then apply the matching rule explicitly: Markovnikov addition with the more stable carbocation intermediate for HX across an alkene; substitution when an aromatic ring reacts with Br2/FeBr3; addition only when the ring is forced by H2/Pd or UV light with halogen.
Substitutes the initial concentration of a weak acid or base straight into Ka or Kb, as if it ionized completely, instead of solving for the equilibrium concentration.
Strong-acid practice drills the pattern [H+] = [acid], and the K expression looks like a one-step formula, so the ICE table is skipped whenever the numbers look easy.
What to do instead
Set up the ICE table as a required intermediate, however small the problem. Solve for x, then verify the assumption with x ÷ [HA]initial < 5%; if that fails, keep the −x in the denominator and solve the quadratic. Report [H+] at equilibrium, not the initial concentration.
States that adding reactant, compressing a gaseous system, or adding a catalyst changes the value of K or Ka, then uses the new K to find the final concentrations.
Le Châtelier's principle is learned as a list of disturbances that 'push' the reaction, and shifting the position of equilibrium is confused with changing the equilibrium constant itself.
What to do instead
Separate the two ideas explicitly in writing: a disturbance changes Q, so Q ≠ K momentarily and the system shifts until Q = K again; only a temperature change alters K. For each disturbance, write the Q expression, show which term changed, and state the direction that restores equality.
Calculates the pH of a buffer from Ka and the acid concentration alone, ignoring the conjugate base term, or assumes that diluting a buffer with water raises the pH roughly in proportion to the dilution.
Weak-acid pH problems were solved with [H+] ≈ √(Ka[HA]), and that template is reused even when a second source of the conjugate base is present, while dilution is assumed to act like dilution of a strong acid.
What to do instead
Use [H+] = Ka × [HA]/[A−] (or pH = pKa + log([A−]/[HA])) whenever both members of the conjugate pair are present, and note the ratio, not the absolute concentrations, sets the pH. For dilution of a buffer, show that [HA] and [A−] fall by the same factor, so the ratio and pH are essentially unchanged; reserve the square-root formula for a weak acid with no added conjugate base.