MCAT Chem/Phys common mistakes
MCAT Chem/Phys · Understand the traps and how to avoid them.
Computes the pH of a solution of a weak acid alone (e.g., 0.10 M acetic acid) by plugging the acid concentration and its pKa straight into the Henderson–Hasselbalch equation.
The equation is retrieved by keyword association ('weak acid' → 'pH equation') rather than by checking whether a buffer actually exists; it is also the most recently memorized pH formula, so it crowds out the ICE-table method.
What to do instead
Before choosing a pH tool, ask one screening question: are HA and A⁻ both present in comparable amounts? If yes, use Henderson–Hasselbalch. If only HA is present, set up an ICE table with Ka and solve x ≈ √(Ka·C), then verify the small-x assumption.
Substitutes ΔH in kJ and ΔS in J/K directly into ΔG = ΔH − TΔS, producing a free-energy value off by a factor of 1000 and flipping the spontaneity conclusion.
Each quantity is copied into the equation in whatever unit it was given or remembered in; enthalpy is habitually thought of in kJ (biological convention) while entropy tables are almost always in J/K, so the mismatch is invisible unless units are written out.
What to do instead
On the scratch sheet, convert every thermodynamic quantity to joules and kelvin before writing the equation, and write the unit symbol next to each substituted number. Treat 'kJ' appearing anywhere in a ΔG calculation as a trigger to convert.
Assigns g = +9.8 m/s² while verbally treating upward as positive, then keeps a negative computed time, negative kinetic energy, or negative mass as if it were a real answer.
g is memorized as a positive magnitude rather than as a vector component, and the chosen coordinate axis exists only in the test-taker's head, never on paper, so the contradiction is never surfaced.
What to do instead
Draw the axis and list the sign of every vector (v₀, a, displacement) before writing any kinematics equation. Treat any negative time, mass, or kinetic energy as an automatic stop signal that a sign or the axis assignment is wrong.
Claims that adding another resistor in parallel increases the total resistance, or that the same current flows through each parallel branch.
Series rules are learned first and become the default mental model, and 'adding a component' intuitively feels additive; current is also intuitively treated as one conserved quantity rather than something that splits at a junction.
What to do instead
Use 1/R_eq = Σ(1/Rᵢ) and check the result against the invariant that parallel equivalent resistance is always smaller than the smallest branch. State explicitly which quantity is shared: voltage is the same across parallel elements, current is the same through series elements.
Performs exact long division or three-significant-figure multiplication on items whose four answer choices differ by a factor of 2 or by a power of ten.
Precision anxiety carried over from coursework where exact answers are graded, combined with distrust of approximation as 'guessing'.
What to do instead
Round every input to one or two significant figures and compare the rough result to the choices; only if two choices survive the estimate should the arithmetic be refined. Practise this until the estimate becomes the default first pass.
Predicts a substitution or elimination product without checking substrate class, nucleophile/base strength, solvent, and heat - for example assigning SN1 to a primary alkyl halide, or E2 to a tertiary substrate in aqueous ethanol with no strong base.
The mechanism is selected from the reaction's name or the chapter it appeared in rather than from the variables that actually determine it, and mixed cases (secondary substrates) are resolved by guessing.
What to do instead
Run a fixed four-item checklist before drawing any product: degree of substitution at the electrophilic carbon, strong vs weak nucleophile/base, protic vs aprotic solvent, and heat vs room temperature. Only assign SN1/SN2/E1/E2 after all four are explicitly named, and check stereochemistry (inversion vs racemization) as a final pass.
States that a noncompetitive inhibitor raises Km, or that a competitive inhibitor lowers Vmax.
The two inhibitor classes are memorized as 'the Km one' and 'the Vmax one' instead of being derived from where the inhibitor binds, so the direction of each effect is recalled by feel rather than by reasoning.
What to do instead
Derive both from the binding site every time: a competitive inhibitor occupies the active site, so excess substrate can outcompete it (Vmax unchanged, Km increases); a noncompetitive inhibitor binds elsewhere, so substrate cannot outcompete it (Vmax decreases, Km unchanged). Sketch the double-reciprocal lines to confirm.
Calculates the net charge of a peptide or free amino acid at a stated pH of 2 or 11 by assuming all acidic side chains are deprotonated and all basic ones protonated, as they would be at pH 7.
The pH 7 zwitterion picture is heavily overlearned and gets applied mechanically to problems that deliberately specify an extreme pH.
What to do instead
For every ionizable group, compare the stated pH with that group's pKa: pH < pKa means protonated, pH > pKa means deprotonated. Tally the charge group by group - α-carboxyl, α-amino, then each side chain - and sum only at the end.
Writes an mRNA sequence identical to the template strand, or copies a strand without tracking 5'→3' polarity and antiparallel orientation.
Both DNA strands are referred to loosely as 'the DNA sequence' in introductory teaching, and the coding strand is often the one printed in figures, so template and coding become interchangeable in memory.
What to do instead
Before transcribing, label the given strand as template or coding and write the 5' and 3' ends of both strands. Then apply the fixed rule: mRNA is complementary and antiparallel to the template strand, and identical to the coding strand except with U in place of T.
Uses a negative ΔG° as proof that a reaction proceeds forward under the cell's actual conditions, or asserts that an enzyme or catalyst drives the equilibrium toward more product.
Tabulated standard free energies are the values students actually see in tables, so standard-state and actual-state quantities collapse into one; and catalysts are associated with 'making the reaction happen', which gets misread as 'making more product'.
What to do instead
Keep the two symbols physically separate on paper: ΔG° applies at standard conditions (1 M, fixed T), while ΔG = ΔG° + RT ln Q applies at the conditions given. State in one sentence that a catalyst lowers activation energy and shortens the time to equilibrium but leaves K and ΔG° untouched.