Free QCE Chemistry Exam Flashcards

Forward and reverse reaction rates are equal in a closed system under constant conditions. Reactions continue; concentrations remain constant but need not equal one another.

Kc = [C]^c[D]^d/([A]^a[B]^b), using equilibrium concentrations. Omit pure solids and pure liquids. Keep the balanced equation and temperature fixed when comparing Kc values.

Net forward reaction until equilibrium is restored. Q uses current concentrations; Q > Kc predicts net reverse reaction, while Q = Kc indicates equilibrium.

Adding one species without appreciably changing volume causes an immediate jump in that species. Compressing a gas mixture raises all gaseous concentrations immediately. Subsequent changes follow stoichiometry; a smooth change alone does not uniquely prove a temperature change.

It favours the side with fewer gas molecules; equal gas mole totals give no shift. Adding inert gas at constant volume changes total pressure but not reacting-species concentrations, so it does not shift the equilibrium.

It favours the endothermic reverse direction and decreases Kc. A catalyst accelerates both directions without changing the equilibrium composition or Kc at that temperature.

PbI2(s) ⇌ Pb2+(aq) + 2I−(aq). If molar solubility is s and other ion sources are negligible, Ksp = s(2s)^2 = 4s^3. A common ion changes this simple relationship; use actual equilibrium concentrations.

When Qsp, calculated with concentrations after mixing and dilution, exceeds Ksp. The common-ion effect generally reduces a sparingly soluble salt’s solubility at constant temperature; it does not change Ksp.

By one proton: NH4+/NH3 is an example. A Brønsted–Lowry acid donates H+ and a base accepts it. An amphiprotic species such as HCO3− can do either, depending on its reaction partner.

Strength describes ionisation in water; concentration describes amount per volume. Dilute HCl remains a strong acid. For sulfuric acid, distinguish its essentially complete first ionisation from the equilibrium of HSO4− in the second step.

pH = −log10[H+] = 3.00. At 298 K, Kw = 1.00 × 10⁻¹⁴ and pH + pOH = 14.00, so pOH = 11.00. Neutrality means [H+] = [OH−]; neutral pH varies with temperature.

For HA of initial concentration c, Ka = x²/(c−x), where x ≈ [H+] if water ionisation is negligible. Only when dissociation is small may x ≈ √(Ka c). Check the approximation against c; otherwise solve the equilibrium expression.

Its conjugate base consumes added H+ to form the weak acid. Added OH− is consumed by the weak acid. Capacity is finite. The 2025 syllabus requires this equilibrium explanation, not buffer calculations.

Use the balanced-reaction mole ratio, supported by the curve’s inflection or a conductometric change in slope. The indicator end point is an observed colour change, not the definition of equivalence. Select an indicator whose transition lies in the steep pH change.

The species that accepts electrons and is reduced. The reducing agent loses electrons and is oxidised. Follow oxidation-state changes rather than assuming that oxygen must be present.

Four oxygen atoms at −2 contribute −8. The ion totals −1, so manganese is +7. Elemental substances have oxidation state zero; a monatomic ion’s state equals its charge.

MnO4− + 8H+ + 5e− → Mn2+ + 4H2O. Balance non-O/H atoms, then O with water, H with H+, and charge with electrons. Both sides have charge +2.

Through the external circuit from the negative anode, where oxidation occurs, to the positive cathode, where reduction occurs. The salt bridge carries ions, not electrons, and maintains electrical neutrality.

A reduction half-cell potential relative to the standard hydrogen electrode. Use the data-book conditions: 298 K, 1 mol L⁻¹ aqueous species and 100 kPa gases. Nonstandard concentrations and kinetic barriers limit predictions from E°.

Subtract the anode’s tabulated reduction potential from the cathode’s. For Cu2+/Cu = +0.34 V and Zn2+/Zn = −0.76 V, E°cell = 0.34 − (−0.76) = +1.10 V. Do not multiply potentials when balancing electrons.

The power supply makes the anode positive and cathode negative, driving a nonspontaneous reaction. Oxidation remains at the anode and reduction at the cathode.

Na+ is reduced to sodium at the cathode; chloride is oxidised to chlorine at the anode: 2Cl− → Cl2 + 2e−. There is no water to compete, unlike aqueous electrolysis.

Water competes with dissolved ions, and concentration, electrode material and overpotential matter. Concentrated brine with suitable inert electrodes gives hydrogen at the cathode and chlorine at the anode; dilute solutions can favour oxygen instead.

Impure copper is the anode and pure copper the cathode in an appropriate Cu2+ electrolyte. Cu dissolves at the anode and deposits at the cathode; less readily oxidised impurities can collect as anode sludge.

q = It = 1930 C; n(e−) = q/96485 = 0.0200 mol. Cu2+ + 2e− → Cu gives 0.0100 mol Cu, or 0.636 g using 63.55 g mol⁻¹, assuming 100% current efficiency.

They share a functional group and general structural pattern; successive members differ by CH2. Similar chemistry follows from the functional group, while physical properties change with chain length and structure.

A molecular formula gives atom counts; a structural formula shows connectivity. C4H10 can describe butane or 2-methylpropane, so the molecular formula alone does not identify the compound.

Name the alcohol-derived alkyl group first, then the acid-derived alkanoate: CH3COOCH2CH3 is ethyl ethanoate. Trace atoms around the ester linkage instead of reading the chain as one continuous alkane.

For simple alkenes such as but-2-ene, each double-bond carbon has a hydrogen and a different substituent. Restricted rotation permits the matching groups to be on the same side (cis) or opposite sides (trans). Propene does not qualify because one double-bond carbon has two hydrogens; more general substituted alkenes may require E/Z nomenclature.

Look for a tetrahedral carbon bonded to four different groups. The resulting mirror-image arrangements may be non-superimposable optical isomers. A carbon in a C=C bond is not a tetrahedral chiral centre.

Alcohol molecules can hydrogen-bond; alkanes rely on dispersion forces. Compare similar size and branching, and use supplied data: no single functional-group ranking determines every boiling point.

Halogen substitution in an alkane replaces H under suitable UV conditions, forming HX as well as a haloalkane. Addition across an alkene C=C consumes the double bond; bromine decolourisation under test conditions supports unsaturation.

Under the ordinary addition conditions used in the syllabus, the major product is 2-bromopropane: H joins the double-bond carbon already bearing more hydrogens. Use stated reaction conditions rather than extending this rule to every mechanism.

Elimination with heated ethanolic hydroxide removes HX to form a C=C bond; aqueous hydroxide favours substitution to an alcohol. Reagents and solvent conditions are essential parts of a pathway answer.

Oxidise under controlled conditions and distil the aldehyde as it forms. Refluxing with excess oxidant instead favours the carboxylic acid. Secondary alcohols give ketones; tertiary alcohols resist these usual oxidation tests.

A carboxylic acid and alcohol form an ester and water under acid catalysis and heating. Removing product or using excess reactant can improve equilibrium yield; a catalyst speeds approach without changing Kc at fixed temperature.

Extensive branching in LDPE hinders close chain packing. Less-branched HDPE packs more closely and has greater crystallinity. Both contain crystalline and amorphous regions; LDPE is not wholly amorphous.

For a simple amino acid, the net charge is positive below its isoelectric point and negative above it. At the isoelectric point the average net charge is zero; the zwitterion contains both positive and negative groups.

Rf = distance travelled by spot/distance travelled by solvent front = 3.0/6.0 = 0.50. Compare values only under the same stationary-phase, solvent and other experimental conditions.

Its positive net charge drives it toward the negative electrode (cathode). Above pI it is negative and moves toward the positive electrode; at pI it has little net migration.

Use the QCAA June 2026 data book: a C=O absorption at 1700–1750 cm⁻¹ together with broad hydrogen-bonded acid O–H at 2500–3000 cm⁻¹. Combine peaks and other evidence; one carbonyl peak alone cannot identify a unique compound.

The molecular ion is the intact ion formed after electron loss; for charge +1, its m/z gives molecular mass. It may be weak or absent. The base peak is the most intense peak, normalised to 100%, and may be either a fragment or the molecular ion.

N2 + 3H2 ⇌ 2NH3 is exothermic. Lower temperature favours equilibrium ammonia but slows reaction. Increased pressure favours fewer gas molecules; an iron catalyst improves rate, and ammonia removal permits recycling of unreacted gases.

2SO2 + O2 ⇌ 2SO3, catalysed by V2O5. Temperature balances rate and equilibrium yield. SO3 is absorbed in sulfuric acid to form oleum, then diluted; adding SO3 directly to water creates a difficult acid mist.

Fermentation: C6H12O6 → 2C2H5OH + 2CO2 using yeast under suitable anaerobic conditions. Hydration: C2H4 + H2O ⇌ C2H5OH using steam and an acid catalyst. Compare feedstock, purity, rate and conditions.

Anode: 2H2 → 4H+ + 4e−. Cathode: O2 + 4H+ + 4e− → 2H2O. Overall: 2H2 + O2 → 2H2O. In alkaline cells use OH− in the balanced half-equations; the overall reaction is unchanged.

Actual product mass/theoretical product mass × 100%. Identify the limiting reagent from balanced mole ratios first. Recovering 7.2 g when 9.0 g is theoretically possible gives 80%; two successive 80% steps give 64% overall.

Open the C=C double bond to connect adjacent monomers, giving [–CH2–CH2–]n. Show continuation bonds through brackets and retain substituents for other alkene monomers; do not lose atoms from the repeat unit.

Their functional groups react to form ester links, eliminating water at each condensation linkage. Identify the linkage and reconstruct both monomers. Polyamides use amide links; PLA can arise from a hydroxycarboxylic-acid monomer.

Two peptide links form a linear tripeptide with loss of two water molecules. Read its amino-acid sequence from the N-terminus to the C-terminus. Joining two monosaccharides similarly forms a glycosidic linkage with loss of water.