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Berlin Abitur Chemie LK practice questions

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Question 1Physical chemistry, amounts and reactions

In thermochemistry, what does a negative reaction enthalpy ΔRH < 0 mean for the system under standard discussion in the Leistungskurs?

  1. The reaction is endothermic and absorbs heat from the surroundings
  2. The reaction is exothermic; the system releases heat to the surroundings Correct answer.
  3. The reaction cannot occur spontaneously under any conditions
  4. The activation energy is necessarily zero

Explanation

By convention ΔRH < 0 denotes an exothermic reaction: enthalpy of the system decreases as heat is released to the surroundings. Endothermic reactions have ΔRH > 0. Sign of ΔRH alone does not set Ea to zero or forbid spontaneity (Gibbs energy also matters).

Question 2Physical chemistry, amounts and reactions

The Satz von Hess states that the overall reaction enthalpy depends only on initial and final states. Which calculation correctly applies Hess's law?

  1. ΔRH(overall) equals the sum of ΔRH of a sequence of steps that add to the same net reaction Correct answer.
  2. ΔRH(overall) equals the product of the step enthalpies
  3. ΔRH(overall) equals only the activation energy of the slowest step
  4. ΔRH(overall) equals the difference of boiling points of products and reactants

Explanation

Hess's law: enthalpy is a state function, so ΔRH for a net reaction equals the algebraic sum of ΔRH values of any pathway connecting the same reactants and products-used constantly in LK calorimetry and formation-enthalpy cycles.

Question 3Physical chemistry, amounts and reactions

In a coffee-cup calorimeter, 50.0 mL of 1.00 mol·L−1 HCl is mixed with 50.0 mL of 1.00 mol·L−1 NaOH. The temperature rises by 6.5 K. Assume c = 4.18 J·g−1·K−1, density ≈ 1.00 g·mL−1, and neglect the calorimeter's heat capacity. Approximate q for the solution and the molar neutralisation enthalpy.

  1. q ≈ −2.7 kJ for the solution; ΔRH ≈ −54 kJ·mol−1 for the neutralisation (exothermic) Correct answer.
  2. q ≈ +2.7 kJ; ΔRH ≈ +54 kJ·mol−1 (endothermic)
  3. q ≈ −27 kJ; ΔRH ≈ −540 kJ·mol−1
  4. q ≈ 0 because acids and bases cancel thermally

Explanation

m ≈ 100 g; q_sol = m·c·ΔT ≈ 100·4.18·6.5 ≈ 2717 J ≈ +2.72 kJ absorbed by the solution, so q_rxn ≈ −2.72 kJ. Amount of H2O formed = 0.050 mol, so ΔRH ≈ −2.72/0.050 ≈ −54 kJ·mol−1 - a typical strong-acid/strong-base estimate.

Question 4Physical chemistry, amounts and reactions

Which statement correctly distinguishes Reaktionenthalpie ΔRH from Aktivierungsenergie EA?

  1. ΔRH is the net heat change between reactants and products; EA is the energy barrier that must be surmounted for reaction to proceed at a measurable rate Correct answer.
  2. EA equals ΔRH for all exothermic reactions
  3. ΔRH is always larger than EA
  4. Catalysts change ΔRH but never change EA

Explanation

Thermodynamics (ΔRH) compares reactant and product energy levels. Kinetics (EA) describes the barrier height on the reaction coordinate. Catalysts lower EA (and may open a new path) without changing the net ΔRH of the balanced reaction.

Question 5Physical chemistry, amounts and reactions

For a reaction that is first order in reactant A, which statement about the rate law is correct?

  1. Rate = k; concentration of A does not appear
  2. Rate = k·c(A); doubling c(A) doubles the rate Correct answer.
  3. Rate = k·c(A)2 always, for every reaction labelled 'first order'
  4. Rate is independent of temperature for first-order reactions

Explanation

First order in A means the rate is proportional to c(A). Doubling concentration doubles rate. Temperature still affects k via the Arrhenius factor; order is about concentration dependence, not about ignoring T.

Question 6Physical chemistry, amounts and reactions

Which change increases the rate of a homogeneous reaction in solution without changing the stoichiometric equation?

  1. Lowering the temperature while keeping concentrations fixed
  2. Increasing the concentration of a reactant that appears in the rate law, or raising temperature (Arrhenius) Correct answer.
  3. Removing a catalyst that was present
  4. Diluting all reactants equally so every concentration falls

Explanation

Rates rise when reactant concentrations in the rate law increase or when T rises (more molecules exceed EA). Cooling, removing catalysts, or diluting reactants generally slow the reaction.

Question 7Physical chemistry, amounts and reactions

For the equilibrium N2(g) + 3 H2(g) ⇌ 2 NH3(g), which expression is the correct concentration equilibrium constant Kc?

  1. Kc = c(N2)·c(H2)3 / c(NH3)2
  2. Kc = c(NH3)2 / [c(N2)·c(H2)3] Correct answer.
  3. Kc = c(NH3) / [c(N2)·c(H2)]
  4. Kc = c(N2) + 3 c(H2) − 2 c(NH3)

Explanation

Kc is products over reactants with stoichiometric exponents: Kc = c(NH3)2 / [c(N2)·c(H2)3]. Sums of concentrations are not equilibrium constants.

Question 8Physical chemistry, amounts and reactions

Le Chatelier: for N2 + 3 H2 ⇌ 2 NH3 (exothermic forward), which stress shifts equilibrium toward more ammonia?

  1. Increasing temperature at constant pressure
  2. Decreasing the total pressure by expanding the volume at constant temperature
  3. Increasing total pressure (or decreasing volume) at constant temperature, favouring the side with fewer gas moles Correct answer.
  4. Adding a catalyst that only speeds the reverse reaction

Explanation

The forward reaction converts 4 gas moles to 2. Higher pressure (lower volume) favours NH3. Higher T favours the endothermic reverse. Catalysts accelerate both directions equally and do not shift K.

Question 9Physical chemistry, amounts and reactions

At equilibrium, which statement is always true for a reversible reaction in a closed system?

  1. All reactants have been completely converted to products
  2. Forward and reverse rates are equal, so macroscopic concentrations stay constant Correct answer.
  3. The activation energy of the forward reaction is zero
  4. Kc must equal 1

Explanation

Dynamic equilibrium means equal forward and reverse rates; concentrations are constant but not necessarily equal, and Kc can take any positive value.

Question 10Physical chemistry, amounts and reactions

A reaction has Kc = 2.5×10−6 at 298 K. What does this indicate qualitatively?

  1. Products strongly dominate at equilibrium
  2. Reactants strongly dominate at equilibrium; the equilibrium lies far to the left Correct answer.
  3. The reaction is necessarily explosive
  4. The reaction cannot have an activation energy

Explanation

Very small Kc means the numerator (products) is tiny relative to reactants at equilibrium - the position lies toward reactants. Kinetics (explosiveness, EA) are separate from Kc.

Question 11Physical chemistry, amounts and reactions

For the esterification equilibrium carboxylic acid + alcohol ⇌ ester + water, removing water as it forms typically:

  1. Decreases the yield of ester by Le Chatelier's principle
  2. Shifts the equilibrium toward ester, increasing conversion of the acid/alcohol Correct answer.
  3. Changes the stoichiometric coefficients of the balanced equation
  4. Makes Kc smaller because water is a product

Explanation

Removing a product shifts equilibrium toward products (Le Chatelier), raising ester yield. Kc itself at fixed T is unchanged; the system adjusts concentrations to restore Q = Kc.

Question 12Physical chemistry, amounts and reactions

Calculate the amount of heat released when 2.00 mol of methane burns completely if ΔRH = −890 kJ·mol−1 for CH4 + 2 O2 → CO2 + 2 H2O.

  1. 445 kJ absorbed
  2. 890 kJ released
  3. 1780 kJ released Correct answer.
  4. 1780 kJ absorbed

Explanation

q = n·ΔRH = 2.00 mol · (−890 kJ·mol−1) = −1780 kJ. The negative sign means 1780 kJ is released to the surroundings.

Question 13Physical chemistry, amounts and reactions

Using average bond enthalpies is an approximate way to estimate ΔRH. Which limitation is important in LK reasoning?

  1. Bond-enthalpy estimates ignore that tabulated averages are not molecule-specific and neglect phase/state subtleties compared with calorimetric ΔRH Correct answer.
  2. Bond enthalpies always give exact ΔRH to 0.01 kJ·mol−1
  3. Bond enthalpies can only be used for ionic lattice energies
  4. Bond-enthalpy methods forbid writing thermochemical equations

Explanation

Average bond enthalpies are approximate mean values; real bonds vary with molecular environment. Calorimetry / Hess cycles with formation enthalpies are preferred when precise ΔRH is required.

Question 14Physical chemistry, amounts and reactions

For an elementary bimolecular collision A + B → products, how does increasing pressure in the gas phase (at constant T) typically affect rate if concentrations rise with pressure?

  1. Rate usually increases because collision frequency rises with concentration Correct answer.
  2. Rate always decreases because molecules move slower at high pressure
  3. Rate is unaffected by concentration for bimolecular steps
  4. Rate becomes zero above 1 bar

Explanation

Higher gas pressure at fixed T increases concentrations and collision frequency, so bimolecular rates typically increase. Temperature, not pressure alone, primarily sets molecular speeds.

Question 15Physical chemistry, amounts and reactions

The equilibrium constant Kc for a reaction is 4.0. If Q = 0.25 at some moment, which statement is correct?

  1. The system is at equilibrium
  2. The forward reaction is favoured until Q rises to equal Kc Correct answer.
  3. The reverse reaction is favoured until Q falls further
  4. Kc must immediately become 0.25

Explanation

When Q < Kc, the reaction quotient is too 'reactant-heavy'; net forward reaction proceeds until Q = Kc. Kc is fixed at constant T.

Question 16Physical chemistry, amounts and reactions

Stoichiometry: 2 H2 + O2 → 2 H2O. Starting with 4.0 mol H2 and 3.0 mol O2, what is the limiting reactant and maximum H2O amount?

  1. O2 limiting; 6.0 mol H2O
  2. H2 limiting; 4.0 mol H2O Correct answer.
  3. H2 limiting; 2.0 mol H2O
  4. Neither limiting; 7.0 mol H2O

Explanation

H2 needs 2.0 mol O2 to make 4.0 mol H2O; 3.0 mol O2 is excess. H2 is limiting and yields 4.0 mol H2O. (If O2 were limiting, 3.0 mol O2 → 6.0 mol H2O, but that would need 6.0 mol H2, which is unavailable.)

Question 17Physical chemistry, amounts and reactions

Which energy diagram feature corresponds to an endothermic reaction with a sizable activation barrier?

  1. Products lower than reactants and no barrier
  2. Products higher in enthalpy than reactants, with a transition-state maximum above both Correct answer.
  3. Products and reactants at identical enthalpy with EA = 0
  4. A barrier only on the product side with reactants higher than the transition state

Explanation

Endothermic: products lie above reactants in enthalpy. A kinetic barrier appears as a transition-state peak higher than both ends of the path.

Question 18Physical chemistry, amounts and reactions

Which statement about homogeneous vs heterogeneous catalysis is correct?

  1. Homogeneous: catalyst in the same phase as reactants; heterogeneous: catalyst in a different phase (e.g., solid metal with gas reactants) Correct answer.
  2. Heterogeneous catalysts must dissolve completely in the reaction mixture
  3. Homogeneous catalysts never interact chemically with reactants
  4. Only heterogeneous catalysis can lower activation energy

Explanation

Phase relationship defines the terms: same phase = homogeneous (e.g., acid in solution); different phase = heterogeneous (e.g., Pt surface). Both types work by providing lower-EA pathways.

Question 19Physical chemistry, amounts and reactions

For the gas-phase equilibrium 2 SO2 + O2 ⇌ 2 SO3, Kc = 2.8×102 at a given T. If all concentrations equal 0.10 mol·L−1, is the mixture at equilibrium?

  1. Yes, because all concentrations are equal
  2. No; Q = c(SO3)2 / [c(SO2)2·c(O2)] = 10, which is less than Kc, so net forward reaction proceeds Correct answer.
  3. No; Q is greater than Kc so only reverse reaction can occur forever
  4. Yes, because Kc is large

Explanation

Q = (0.10)2 / [(0.10)2·(0.10)] = 10. Since Q < Kc (280), the forward reaction is favoured until Q rises. Equal concentrations alone do not mean equilibrium.

Question 20Physical chemistry, amounts and reactions

Why does increasing temperature usually increase reaction rate even for an exothermic reaction?

  1. Because Kc must increase for all exothermic reactions when T rises
  2. Because a larger fraction of collisions exceed the activation energy (Arrhenius), even though equilibrium may shift toward reactants Correct answer.
  3. Because exothermic reactions have EA = 0 at high T
  4. Because molecules stop colliding at high T

Explanation

Kinetics: higher T raises the fraction of molecules with E ≥ EA. Thermodynamics: for exothermic equilibria, higher T decreases K - but that affects position, not the immediate rate acceleration.

Question 21Physical chemistry, amounts and reactions

In the protolysis of acetic acid in water, CH3COOH + H2O ⇌ CH3COO− + H3O+, which pair is a conjugate acid–base pair?

  1. CH3COOH and H2O
  2. CH3COOH and CH3COO− Correct answer.
  3. CH3COO− and H3O+
  4. H2O and CH3COO−

Explanation

A conjugate pair differs by one proton: CH3COOH (acid) / CH3COO− (conjugate base). H2O / H3O+ is the other conjugate pair in this equation.

Question 22Physical chemistry, amounts and reactions

For a weak acid HA with Ka = 1.0×10−5 and c0 = 0.10 mol·L−1, which approximate pH calculation is appropriate (x ≪ c0)?

  1. pH ≈ 1.0 using full dissociation
  2. c(H3O+) ≈ √(Ka·c0) = √(1.0×10−6) = 1.0×10−3; pH ≈ 3.0 Correct answer.
  3. pH = 14.0 exactly
  4. pH = Ka directly (1.0×10−5)

Explanation

Weak-acid approximation: Ka ≈ x2/c0 ⇒ x = √(Ka·c0). Here x = 1.0×10−3 mol·L−1, pH ≈ 3.0. Full dissociation would wrongly give pH 1.

Question 23Physical chemistry, amounts and reactions

A buffer contains comparable amounts of a weak acid and its conjugate base. Which statement is correct?

  1. Adding a small amount of strong acid or base changes pH only slightly because the conjugate pair consumes the added H3O+ or OH− Correct answer.
  2. Buffers keep pH exactly 7 under all conditions
  3. Buffers work only in organic solvents, never in water
  4. Buffers increase Ka of the weak acid permanently

Explanation

Buffer action: added H3O+ is largely consumed by A−; added OH− is consumed by HA. pH stays near pKa but is not magically fixed at 7 for every buffer.

Question 24Physical chemistry, amounts and reactions

In the Leistungskurs, Indikatorfarbstoffe are treated as weak acids HIn ⇌ H+ + In− with different colours for HIn and In−. When is the indicator mid-transition?

  1. Only when pH = 0
  2. Near pH ≈ pKa(HIn), where c(HIn) ≈ c(In−) Correct answer.
  3. Only when the solution is boiling
  4. When Ka of the analyte becomes infinite

Explanation

For HIn/In−, the colour transition centres around pH ≈ pKa, where both forms are present in comparable amounts. LK tasks often analyse indicator equilibria and suitable pH ranges.

Question 25Physical chemistry, amounts and reactions

Why must the indicator's transition interval match the steep pH jump of a titration's equivalence region?

  1. So the colour change occurs where a tiny volume of titrant causes a large pH change, marking equivalence accurately Correct answer.
  2. So the indicator permanently destroys the analyte
  3. So the titration never reaches equivalence
  4. Because indicators only work in non-aqueous media

Explanation

At equivalence the titration curve is steep; an indicator that flips in that window changes colour with negligible excess titrant, giving a sharp end point.

Question 26Physical chemistry, amounts and reactions

Oxidation numbers: what is the oxidation number of Mn in MnO4−?

  1. +2
  2. +4
  3. +7 Correct answer.
  4. −1

Explanation

Oxygen is −2 each (4×−2 = −8); the ion charge is −1, so Mn is +7. This is standard LK redox bookkeeping for permanganate oxidations.

Question 27Physical chemistry, amounts and reactions

In a redox reaction, which statement is correct?

  1. Oxidation is gain of electrons; reduction is loss of electrons
  2. Oxidation is loss of electrons; reduction is gain of electrons Correct answer.
  3. Both partners always lose electrons
  4. Redox reactions never involve electron transfer

Explanation

Oxidation = electron loss (oxidation number rises); reduction = electron gain (oxidation number falls). Electrons lost by the reductant equal electrons gained by the oxidant.

Question 28Physical chemistry, amounts and reactions

Balance the ionic skeleton in acidic solution: MnO4− + Fe2+ → Mn2+ + Fe3+. How many Fe2+ ions are oxidised per MnO4− reduced?

  1. 1
  2. 3
  3. 5 Correct answer.
  4. 7

Explanation

Mn(+7 → +2) gains 5 e−; each Fe2+ → Fe3+ loses 1 e−. Therefore 5 Fe2+ are required per MnO4−.

Question 29Physical chemistry, amounts and reactions

A titration uses 25.00 mL of 0.100 mol·L−1 NaOH to neutralise 20.00 mL of HCl. What was c(HCl)?

  1. 0.080 mol·L−1
  2. 0.100 mol·L−1
  3. 0.125 mol·L−1 Correct answer.
  4. 0.250 mol·L−1

Explanation

n(NaOH) = 0.02500 L · 0.100 mol·L−1 = 0.00250 mol. For HCl + NaOH → NaCl + H2O, n(HCl) = 0.00250 mol; c = 0.00250/0.02000 = 0.125 mol·L−1.

Question 30Physical chemistry, amounts and reactions

Which species is the oxidising agent in: Zn + Cu2+ → Zn2+ + Cu?

  1. Zn (it is oxidised)
  2. Cu2+ (it is reduced by gaining electrons) Correct answer.
  3. Zn2+
  4. Metallic copper product only

Explanation

The oxidising agent is the species that is reduced: Cu2+ gains electrons. Zn is the reducing agent (oxidised to Zn2+).

Question 31Physical chemistry, amounts and reactions

pKa of acetic acid is 4.76. For a buffer with c(CH3COOH) = c(CH3COO−), approximate pH is:

  1. 0
  2. 4.76 Correct answer.
  3. 7.00
  4. 14.00

Explanation

Henderson–Hasselbalch: pH = pKa + lg(c(A−)/c(HA)). When the ratio is 1, pH = pKa = 4.76.

Question 32Physical chemistry, amounts and reactions

Strong acid–strong base titration curve: where is the equivalence point pH for HCl with NaOH at typical concentrations?

  1. Near pH 3 because HCl is acidic
  2. Near pH 7 because the salt solution is neutral (strong–strong) Correct answer.
  3. Near pH 11 always, regardless of concentrations
  4. Exactly at pH = pKa of HCl

Explanation

Equivalence produces water + NaCl; for strong–strong titration the equivalence pH is about 7 (ideally). Weak-acid titrations equivalence points are higher.

Question 33Physical chemistry, amounts and reactions

Which equation correctly represents autoprotolysis of water?

  1. 2 H2O ⇌ H3O+ + OH− Correct answer.
  2. H2O → H2 + ½ O2
  3. H2O + CO2 → H2CO3 only as a redox process
  4. H2O → 2 H+ + O2− in all solvents identically

Explanation

Water's autoprotolysis equilibrium is 2 H2O ⇌ H3O+ + OH− with Kw = 1.0×10−14 at 25 °C.

Question 34Physical chemistry, amounts and reactions

In alkaline solution, balancing O and H for redox often uses OH− and H2O. Which statement is methodologically correct?

  1. You may never use H2O or OH− when balancing in base
  2. Balance atoms and charge using H2O, OH−, and electrons so both mass and charge are conserved Correct answer.
  3. Only oxidation numbers matter; atom balance is optional
  4. Charge balance is unnecessary if oxidation numbers look neat

Explanation

Whether acidic (H+/H2O) or alkaline (OH−/H2O), half-reaction balancing must conserve atoms and charge with the appropriate proton/hydroxide inventory plus electrons.

Question 35Physical chemistry, amounts and reactions

A weak base B with Kb = 1.0×10−5 and c0 = 0.10 mol·L−1 has approximate c(OH−) = √(Kb·c0) = 1.0×10−3. What is pH?

  1. 3.0
  2. 7.0
  3. 11.0 Correct answer.
  4. 14.0

Explanation

pOH = −lg(1.0×10−3) = 3.0; pH = 14.0 − 3.0 = 11.0 at 25 °C.

Question 36Physical chemistry, amounts and reactions

Which observation best indicates that a redox reaction has occurred in aqueous solution (conceptual LK check)?

  1. Oxidation numbers of elements change and electrons are transferred between partners Correct answer.
  2. Only the colour of an indicator changes without any element changing oxidation number
  3. The solution is stirred
  4. The container is made of glass

Explanation

Redox is defined by oxidation-number change / electron transfer. Indicator colour change alone can be a simple acid–base proton transfer without redox.

Question 37Physical chemistry, amounts and reactions

Calculate the volume of 0.200 mol·L−1 KMnO4 (acidic) needed to oxidise 0.0100 mol Fe2+ completely (MnO4− + 5 Fe2+ → Mn2+ + 5 Fe3+).

  1. 10.0 mL Correct answer.
  2. 25.0 mL
  3. 50.0 mL
  4. 100 mL

Explanation

n(MnO4−) required = 0.0100/5 = 0.00200 mol. V = n/c = 0.00200/0.200 = 0.0100 L = 10.0 mL.

Question 38Physical chemistry, amounts and reactions

Amphoteric species such as HCO3− can act as:

  1. Only oxidising agents in every reaction
  2. Either acids or bases depending on the reaction partner (proton donor or acceptor) Correct answer.
  3. Neither acids nor bases
  4. Only gases at room temperature

Explanation

Ampholytes can donate or accept protons: HCO3− + H3O+ → H2CO3 + H2O (base behaviour) or HCO3− + OH− → CO32− + H2O (acid behaviour).

Question 39Physical chemistry, amounts and reactions

Universal indicator shows different colours across pH ranges because:

  1. It is a mixture of indicators with staggered transition intervals covering a wide pH scale Correct answer.
  2. It contains only one indicator with a single sharp transition at pH 7
  3. It measures temperature instead of pH
  4. It works by Faraday electrolysis of water alone

Explanation

Universal indicators blend several acid–base indicators so successive colour changes map a broad pH range - a practical LK link to Indikatorfarbstoffe.

Question 40Physical chemistry, amounts and reactions

Disproportionation means:

  1. The same element is both oxidised and reduced in one reaction (e.g., parts of Cl2 to Cl− and ClO− in alkali) Correct answer.
  2. Only oxidation occurs, never reduction
  3. A precipitation without electron transfer
  4. An acid–base neutralisation with no oxidation-number change by definition of disproportionation

Explanation

In disproportionation one element undergoes simultaneous oxidation and reduction, producing species with higher and lower oxidation numbers (classic chlorine in alkali example).

Question 41Organic chemistry and macromolecules

Which statement correctly describes the structure of an α-amino acid with a nonionizable side chain in aqueous solution at neutral pH?

  1. It exists exclusively as an uncharged molecule with NH2 and COOH groups
  2. It exists predominantly as a Zwitterion carrying a protonated amino group (–NH3+) and a deprotonated carboxyl group (–COO−) Correct answer.
  3. It is always doubly positively charged
  4. It decomposes immediately into ammonia and carbon dioxide

Explanation

In aqueous solution an α-amino acid undergoes an internal proton transfer: the carboxyl group donates a proton to the amino group, forming a Zwitterion with –NH3+ and –COO−. This ionic character explains the high melting points and good water solubility of amino acids.

Question 42Organic chemistry and macromolecules

In the Biuret-Reaktion a few drops of dilute copper(II)-sulfate solution are added to a protein solution made alkaline with sodium hydroxide. Which observation proves peptide bonds, and which species causes it?

  1. A blue-violet (violet) colour caused by a copper(II) complex with at least two peptide bonds Correct answer.
  2. A brick-red precipitate of copper(I) oxide
  3. A blue-violet colour caused by free amino groups only
  4. A yellow colour caused by elemental sulfur

Explanation

The Biuret test detects peptide bonds: in alkaline solution Cu2+ ions form a violet complex with the nitrogen atoms of at least two peptide bonds. Free amino acids (single amino group, no peptide bond) do not give this reaction; the brick-red Cu2O precipitate belongs to the Fehling test for aldehydes.

Question 43Organic chemistry and macromolecules

Amino acids such as glycine are colourless solids with melting points above 230 °C that dissolve well in water but poorly in nonpolar solvents. Which structural feature explains these properties?

  1. Their zwitterionic structure, which creates strong ion-dipole interactions with water and ionic lattice forces in the solid Correct answer.
  2. Their long unbranched hydrocarbon chains
  3. Their ability to form extensive hydrogen-bond-free crystal lattices
  4. Their aromatic ring systems

Explanation

As Zwitterions, amino acid molecules carry full positive and negative charges. In the solid, strong electrostatic attraction between the ions produces a salt-like lattice with a high melting point; in water, ion-dipole interactions hydrate the ions and make them soluble.

Question 44Organic chemistry and macromolecules

Which statements about the Peptidbindung are correct? (1) It forms by Kondensation with elimination of water. (2) The C–N bond of the peptide unit is planar and not freely rotatable.

  1. Only statement (1) is correct
  2. Only statement (2) is correct
  3. Neither statement is correct
  4. Both statements are correct Correct answer.

Explanation

A peptide bond forms when the carboxyl group of one amino acid and the amino group of another condense, eliminating water. Because of partial double-bond character (mesomeric stabilisation) the atoms of the peptide unit lie in one plane and rotation about the C–N bond is hindered - while the bond at the tetrahedral α-carbon remains flexible.

Question 45Organic chemistry and macromolecules

In a FISCHER-Projektion of an α-L-Aminosäure (e.g. L-Alanin), how is the stereocentre drawn according to the convention taught in the Leistungskurs?

  1. The carboxyl group at the bottom, the amino group on the right
  2. The carboxyl group at the top, the amino group on the left Correct answer.
  3. The carboxyl group at the top, the amino group on the right
  4. The amino group at the top and the carboxyl group on the left

Explanation

In the Fischer convention for α-amino acids the carbon chain is vertical with the carboxyl group (the most oxidised group) at the top. In the L-series the amino group points to the left; in D-amino acids it points to the right. Proteins in living organisms are built almost exclusively from α-L-amino acids.

Question 46Organic chemistry and macromolecules

When egg white is heated or treated with heavy-metal ions, the protein denatures. What exactly happens during Denaturierung?

  1. The Primärstruktur is hydrolysed into free amino acids
  2. Only the peptide bonds are rearranged into new sequences
  3. The Sekundär-, Tertiär- and Quartärstrukturen are destroyed while the Primärstruktur (amino acid sequence) remains intact Correct answer.
  4. The protein is converted quantitatively into ammonia and carbon dioxide

Explanation

Denaturation disrupts the inter- and intramolecular interactions (hydrogen bonds, ionic and hydrophobic interactions) that hold the higher-order structures, so secondary, tertiary and quaternary structures collapse. The covalent peptide bonds of the primary structure stay intact - breaking them would be hydrolysis, not denaturation.

Question 47Organic chemistry and macromolecules

Which interaction is chiefly responsible for stabilising the α-helix of a protein's Sekundärstruktur?

  1. Hydrogen bonds between the C=O group of one peptide bond and the N–H group of another peptide bond along the backbone Correct answer.
  2. Covalent disulfide bridges between neighbouring turns of the helix
  3. Ionic bonds between the α-carboxyl and α-amino end groups only
  4. Van-der-Waals forces between the side chains and the solvent

Explanation

The α-helix is held together by a regular pattern of hydrogen bonds within the backbone: each carbonyl oxygen accepts an H-bond from the amide N–H four residues further along the chain. Side-chain interactions determine the tertiary structure; disulfide bridges can occur but are not the defining feature of the helix.

Question 48Organic chemistry and macromolecules

What characterises essenzielle Aminosäuren?

  1. They are the only amino acids that can form peptide bonds
  2. They are always D-amino acids
  3. They are acidic amino acids with two carboxyl groups
  4. They cannot be synthesised by the human body and must be supplied with food Correct answer.

Explanation

Essential amino acids (e.g. Leucin, Lysin, Methionin) cannot be produced by human metabolism in sufficient amounts and must be ingested with protein-containing food. The term refers to nutritional necessity, not to chemical structure or reactivity.

Question 49Organic chemistry and macromolecules

The complete cleavage of a protein into its amino acids by boiling with hydrochloric acid is which type of reaction?

  1. Kondensation
  2. Hydrolyse Correct answer.
  3. Polymerisation
  4. Denaturierung

Explanation

Splitting peptide bonds with water (acid- or enzyme-catalysed) is hydrolysis - the exact reverse of the Kondensation that formed them. Each cleaved peptide bond takes up one water molecule, regenerating a free amino and a free carboxyl group.

Question 50Organic chemistry and macromolecules

Which statement correctly distinguishes Thermoplaste, Duroplaste and Elastomere?

  1. Thermoplaste soften reversibly when heated because their chains are not cross-linked (or only weakly); Duroplaste are densely cross-linked and do not soften; Elastomere are loosely cross-linked and rubber-elastic Correct answer.
  2. Thermoplaste are densely cross-linked and heat-resistant; Duroplaste melt on heating
  3. All three classes melt reversibly, they differ only in colour
  4. Elastomere decompose below room temperature

Explanation

The classification follows from the degree of cross-linking: un-cross-linked chain molecules (Thermoplaste, e.g. PE) can slide past each other on heating and soften reversibly; densely cross-linked Duroplaste keep their shape until they decompose; loosely cross-linked Elastomere deform elastically and spring back.