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100+ Free Berlin Abitur Chemistry 3. PF (GK) Practice Questions
Berlin Abitur Chemistry — third examination subject (3. Prüfungsfach, Grundkurs / Pool) practice questions are available now; exam metadata is being verified.
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2026 Statistics
Key Facts: Berlin Abitur Chemistry 3. PF (GK) Exam
255 min
GK Chemie Arbeitszeit incl. ~30 min Lese-/Auswahlzeit (AV Anlage 5b; ps_chemie_2026_gk)
SenBJF ps_chemie_2026_gk.pdf; AV Prüfungen Anlage 5b
3 of 4
Tasks worked; candidate alone selects (teacher does not pre-select)
SenBJF Prüfungsschwerpunkte Chemie 2026 Grundkurs
+≤60 min
Possible Zeitverlängerung if fachpraktische Teilaufgabe chosen
SenBJF ps_chemie_2026_gk.pdf; AV Prüfungen
24.04.2026
Chemie LF + 3. PF Pool main written date (Berlin Prüfungsplan)
SenBJF pruefungsplan-2026.pdf
Q1–Q4
Each Prüfungsaufgabe spans ≥3 Qualifikationsphase Kurshalbjahre
SenBJF ps_chemie_2026_gk.pdf; RLP Teil C Chemie
Free 100-question English MCQ study bank for Berlin Abitur Chemie 3. Prüfungsfach (Grundkurs). Official exam: German constructed-response, 3 of 4 tasks, 255 min (+ optional practical extension), Notenpunkte 0–15. Not an official format or language simulation; distinct from Leistungsfach.
Sample Berlin Abitur Chemistry 3. PF (GK) Practice Questions
Try these sample questions to test your Berlin Abitur Chemistry 3. PF (GK) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1According to the official Prüfungsschwerpunkte Chemie 2026 (Grundkurs) of the SenBJF, what is the working time for the written Abitur in Chemie as 3. Prüfungsfach?
A.225 minutes of pure writing time plus a separate oral part
B.255 minutes, which already includes an individual reading and task-selection time of 30 minutes
C.315 minutes, the same as for the Leistungsfach exams
D.180 minutes, because Grundkurs exams are always shortened
Explanation: The official Hinweise zur Vorbereitung state: 'Die Arbeitszeit beträgt 255 Minuten und beinhaltet eine individuelle Lese- und Auswahlzeit für die Prüflinge von 30 Minuten.' The 30-minute reading and selection period is part of the 255 minutes, not added on top.
2How is a Berlin central Abitur Prüfungsaufgabe in Chemie (Grundkurs 2026) structured?
A.Six tasks, all of which must be answered
B.Two tasks chosen by the teacher from a school list
C.Four tasks, of which the candidate alone selects and works on three
D.One compulsory task plus a fixed experimental protocol
Explanation: The Prüfungsschwerpunkte state that one Prüfungsaufgabe consists of four Aufgaben, of which three must be worked on in the given time. The selection is made exclusively by the candidate; the teacher does not choose tasks.
3A peptide bond between two amino acids is formed by which type of reaction?
A.Condensation: the carboxyl group of one amino acid reacts with the amino group of another, releasing water
B.Hydrolysis: a water molecule splits the two amino acids apart
C.Radical chain polymerization across C=C double bonds
D.Redox reaction transferring electrons between the side chains
Explanation: Amino acids link via a Kondensationsreaktion: the -COOH of one amino acid and the -NH2 of another form an amide (peptide) bond -CO-NH- while one water molecule is eliminated. Repetition of this step builds the polypeptide chains of Proteine, the Q1 topic of the Berlin Rahmenlehrplan.
4What is the pH of a hydrochloric acid solution with c(HCl) = 0.01 mol/L at 25 °C?
A.pH = 1
B.pH = 12
C.pH = 7
D.pH = 2
Explanation: Hydrochloric acid is a strong acid, so it is fully protolyzed and c(H3O+) = 0.01 mol/L = 10^-2 mol/L. Therefore pH = -lg(10^-2) = 2. Strong-acid pH calculations of this kind are standard Grundkurs material in the Säure-Base-Reaktionen Kurshalbjahr (Q3).
5In redox chemistry, an Oxidation is best defined as which of the following?
A.Gain of electrons by a particle
B.Loss of electrons (Elektronenabgabe), i.e. an increase in oxidation number
C.Gain of protons by a particle
D.Release of heat to the surroundings
Explanation: Oxidation is the donation of electrons: the oxidation number of the species increases. Its partner process, Reduction, is the uptake of electrons; together they form a Redoxreaktion, the core of the Donator-Akzeptor-Prinzip half-year (Q3).
6Which statement about the Gibbs free energy ΔG = ΔH − T·ΔS is correct for judging whether a reaction runs freiwillig (spontaneously)?
A.A reaction is spontaneous whenever ΔH is positive
B.A reaction is spontaneous whenever ΔS is negative
C.A reaction can run spontaneously only when ΔG is negative under the given conditions
D.ΔG is always zero at any temperature
Explanation: The Gibbs-Helmholtz equation combines enthalpy and entropy: only ΔG < 0 marks an exergonic, spontaneous reaction; ΔG > 0 means endergonic, and ΔG = 0 marks equilibrium. Whether ΔG is negative depends on both ΔH and the T·ΔS term, so neither ΔH nor ΔS alone decides.
7Which reagent is used to detect Stärke (starch), and what is the positive result?
A.Iodine-potassium iodide solution (Lugolsche Lösung), giving a blue-black coloration
B.Fehling's solution, giving a brick-red precipitate
C.Ninhydrin, giving a blue-violet coloration
D.Bromthymolblau, turning yellow
Explanation: Iodine molecules slip into the helical amylose chains of starch and form an inclusion complex that appears blue-black; this is the classic starch test with Lugolsche Lösung. Both starch and the reagent appear on the official Berlin material list for experimental tasks.
8How does a Katalysator accelerate a chemical reaction, and what does it NOT change?
A.It increases the average kinetic energy of all particles and is consumed in the process
B.It shifts the equilibrium toward the products and increases the yield
C.It raises the activation energy so that fewer collisions succeed
D.It lowers the activation energy via an alternative reaction path, without changing the equilibrium position
Explanation: A catalyst opens an alternative pathway with lower Aktivierungsenergie, so a larger fraction of collisions leads to reaction; it emerges unchanged. Because it accelerates forward and reverse reactions equally, the Lage des Gleichgewichts and the equilibrium constant K stay the same.
9In the Haber-Bosch synthesis N2 + 3 H2 ⇌ 2 NH3 (ΔH < 0), why is the industrial process run at about 450 °C although lower temperatures would give a higher equilibrium yield of ammonia?
A.Because ammonia decomposes explosively below 450 °C
B.Because the chosen temperature is a compromise: the reaction rate would be too low at lower temperatures, so a moderate yield reached quickly is preferred
C.Because high temperature increases the equilibrium constant of exothermic reactions
D.Because nitrogen is inert above 450 °C and cannot react
Explanation: For an exothermic equilibrium, lowering the temperature raises the ammonia yield but slows the reaction drastically, even with the iron catalyst. Industry therefore accepts a smaller equilibrium yield at roughly 450 °C (plus high pressure around 200-350 bar) to reach equilibrium fast — a classic Le-Chatelier trade-off examined at Grundkurs level.
10What is the pH of a sodium hydroxide solution with c(NaOH) = 0.001 mol/L at 25 °C (pKw = 14)?
A.pH = 3
B.pH = 9
C.pH = 11
D.pH = 14
Explanation: NaOH dissociates completely, so c(OH-) = 10^-3 mol/L and pOH = 3. With pH + pOH = 14 at 25 °C, the pH is 14 − 3 = 11. The pOH detour is the standard route for strong bases.
About the Berlin Abitur Chemistry 3. PF (GK) Practice Questions
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