The IB Chemistry HL Paper 1 Question Bank is an essential tool for students preparing for the IB Chemistry HL exam. The question bank provides a wide range of practice questions, covering all aspects of the IB Chemistry syllabus. The questions are designed to challenge and test students’ understanding of the concepts they have learned. The question bank is an invaluable resource for students, helping them to prepare fully for the rigours of the IB Chemistry HL Paper 1 exam.
Time: 1 hour
Instructions to candidates
A.) Atomic radius increases, Ionic radius decreases
B.) Atomic radius increases, Ionic radius increases
C.) Atomic radius decreases, Ionic radius increases
D.) Atomic radius decreases, Ionic radius decreases
Answer: B
The atomic and ionic radii increase down the group. This is due to the addition of a new shell at each successive element on moving down the group.
A.) K < Na < Si
B.) Si < Na < K
C.) Na < Si < K
D.) K < Si < Na
Answer: B
The atomic size increases down the group. Na and K belong to the same group. Hence Na<K. Also atomic size decreases along a period and Na and Si belong to the same period. ∴ Na>Si. On combining, the final order is Si<Na<K
A.) Atom gains an electron to form a negative ion
B.) Atom loses electron to form a positive ion
C.) Atom loses proton to form a negative ion
D.) Atom gains proton to form a positive ion
Answer: B
The most stable form of the sodium atom is Na+ and for it to come to this stable state, it needs to lose an electron from its outermost shell. Atoms do not lose protons to become ions as they are present inside the nucleus.
A.) High melting points
B.) Low densities
C.) Insulators of heat and electricity
D.) Soft in nature
Answer: D
Transition metals are known for having high melting points in addition to high densities, electrical and heat conductivity, variable oxidation states and their ability to form coloured compounds.
A.) Sigma bonds: 2 Pi bonds: 2
B.) Sigma bonds: 2 Pi bonds: 3
C.) Sigma bonds: 3 Pi bonds: 2
D.) Sigma bonds: 3 Pi bonds: 3
Answer: C
Each atom is connected by at least one single bond which will be counted as σ bonds and the rest are π bonds. Hence there are 3 sigma (σ) bonds and 2 pi (π) bonds in ethyne.
A.) Catalysts initiate reactions
B.) Catalysts increase the activation energy
C.) Catalysts increases the yield of the reaction
D.) Catalysts increase the rate of the reaction
Answer: D
Catalysts only increase the rate of the reaction by lowering the activation energy. They do not have the properties of increasing the yield of the reaction or initiating reactions.
A.) CH2O
B.) CH4
C.) C2H4
D.) CH4O
Answer: B
CH4O is the only molecule that has a hydrogen bonding between O and H. The other molecules have bonds between C and H which are not considered hydrogen bonds. Hydrogen bonds results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom
A.) Nitrogen oxides
B.) Sulfur dioxide
C.) Carbon dioxide
D.) Methane
Answer: C
Unpolluted rain has a slightly acidic pH of 5.6, because carbon dioxide and water in the air react together to form carbonic acid, a weak acid.
A.) Cl2
B.) CO2
C.) CH4
D.) H2O
Answer: A
Water (H2O) is polar because of the bent shape of the molecule. The shape means most of the negative charge from the oxygen on one side of the molecule and the positive charge of the hydrogen atoms is on the other side of the molecule.
A.) Decreasing temperature
B.) Increasing concentration
C.) Adding a catalyst
D.) Decreasing collision frequency
Answer: B
With an increase in concentration, the number of molecules with the minimum required energy will increase, and therefore the rate of the reaction will increase. Adding a catalyst, decreasing the temperature and collision frequency will reduce the activation energy.
A.) 0.05 M
B.) 0.10 M
C.) 0.15 M
D.) 0.20 M
Answer: D
The concentration of A after time t is given by the equation [A] = [A]0 * e(-kt), where [A]_0 is the initial concentration of A, k is the rate constant, and t is time. Plugging in the given values, we get [A] = 0.5 * e(-0.02*20) = 0.20 M.
A.) CH3CH2CH2OH
B.) (CH3)2CHCH2OH
C.) CH3CH(OH)CH3
D.) CH3CH2CH(OH)CH3
Answer: B
E1 elimination occurs when a carbocation is formed as an intermediate. The stability of the carbocation determines the rate of the reaction. In this case, (CH3)2CHCH2OH has a tertiary carbocation intermediate, which is more stable than the primary carbocation intermediate in the other options.
A.) -277 kJ/mol
B.) -290 kJ/mol
C.) -304 kJ/mol
D.) -328 kJ/mol
Answer: D
The standard enthalpy of formation of a compound is the enthalpy change when 1 mole of the compound is formed from its elements in their standard states. The relationship between standard enthalpies of combustion and standard enthalpies of formation is given by Hess’s law: ΔHf = ΣnΔHc(products) – ΣnΔHc(reactants). Plugging in the values, we get ΔHf = (-1367 kJ/mol) – [(3*(-393.5 kJ/mol)) + (2*(-285.8 kJ/mol))] = -328 kJ/mol.
A.) CH2O
B.) C2H5O
C.) C3H6O2
D.) C5H10O5
Answer: A
The empirical formula is the simplest whole number ratio of atoms in a compound, so divide each mole value by the smallest mole value, which is 0.10: C 0.30/0.10 = 3, H 0.60/0.10 = 6, O 0.10/0.10 = 1 . The empirical formula is therefore CH2O.
A.) +1
B.) +3
C.) +5
D.) +7
Answer: D
The sum of the oxidation states of all atoms in a molecule or ion is equal to the charge of the molecule or ion. Since H has an oxidation state of +1 and O has an oxidation state of -2, the oxidation state of Cl must be +7 to balance out the charge.
A.) A + B → C
B.) 2A → B
C.) A + 2B → 3C
D.) 2C → A + B
Answer: D
The rate of a chemical reaction is determined by the rate-determining step, which is usually the slowest step. In this case, the reaction with the slowest rate of reaction is likely to have the slowest rate-determining step. The correct answer is (d) because it is a reverse reaction, which is generally slower than forward reactions.
A.) 11.3 kJ/mol
B.) 16.5 kJ/mol
C.) 21.8 kJ/mol
D.) 27.1 kJ/mol
Answer: B
The activation energy (Ea) can be calculated using the Arrhenius equation, k = Ae(-Ea/RT). Taking the natural logarithm of this equation gives ln(k) = -Ea/RT + ln(A). By plotting ln(k) vs. 1/T, the slope of the line is equal to -Ea/R, where R is the gas constant. Using the data given, the slope of the line is (ln(3.0 x 10-3) – ln(2.5 x 10-3))/(1/308 – 1/298) = 2195.5 K. Multiplying this value by R (8.31 J/mol K) gives 18.2 kJ/mol. Therefore, the correct answer is (b).
A.) HF
B.) HCl
C.) HBr
D.) HI
Answer: D
The strength of an acid is determined by its tendency to donate a proton (H+). The strength of hydrogen halides increases down the group due to decreasing bond strength, so HI is the strongest acid.
A.) Dipole-dipole forces
B.) Hydrogen bonding
C.) London dispersion forces
D.) Ionic bonding
Answer: C
London dispersion forces result from the temporary uneven distribution of electrons in a molecule, which creates temporary dipoles that can induce dipoles in neighboring molecules.
A.) 0.06 M
B.) 0.15 M
C.) 0.25 M
D.) 0.35 M
Answer: C
The equation for the reaction between HCl and NaOH is HCl + NaOH → NaCl + H2O. At the equivalence point, the moles of HCl will equal the moles of NaOH. Therefore, (0.1 M) x (15 mL) = (C HCl) x (25 mL), where C HCl is the concentration of HCl. Solving for C HCl gives a concentration of 0.25 M.
A.) 0.71 L
B.) 1.43 L
C.) 2.86 L
D.) 4.67 L
Answer: A
This problem can be solved using Boyle’s Law, which states that the pressure and volume of a gas are inversely proportional, assuming constant temperature. Using the equation P1V1 = P2V2, where P1, V1, and P2 are the initial pressure, volume, and final pressure, respectively, we can solve for V2, the final volume. V2 = (P1V1)/P2 = (2.0 atm)(3.5 L)/(3.5 atm) = 2.0 L.
A.) CH2CH2CH2OH and CH3CH2OCH3
B.) CH3CH2COOH and CH3COCH3
C.) H2O and H2O2
D.) NaCl and NaBr
Answer: A
CH3CH2CH2OH and CH3CH2OCH3. Isomers are molecules that have the same molecular formula but different structures. In this case, the two molecules have the same number and types of atoms, but the arrangement of the atoms is different.
A.) 3.75 x 10-4 L mol-1 s-1
B.) 7.5 x 10-4 L mol-1 s-1
C.) 1.5 x 10-3 L mol-1 s-1
D.) 3.0 x 10-3 L mol-1 s-1
Answer: B
The rate law for this reaction is rate = k[A][B]. Therefore, the rate constant can be calculated by dividing the rate of reaction by the product of the initial concentrations of A and B: k = rate/([A][B]). Substituting the given values gives k = 1.5 x 10-2/(0.2 x 0.1) = 7.5 x 10-2 L mol-1 s-1 . Therefore, the correct answer is (b).
A.) It relates the rate constant of a reaction to the activation energy and temperature.
B.) It relates the equilibrium constant of a reaction to the standard Gibbs free energy change.
C.) It relates the solubility of a gas in a liquid to the partial pressure of the gas.
D.) It relates the rate of a reaction to the concentration of reactants.
Answer: A
The Arrhenius equation relates the rate constant (k) of a reaction to the activation energy (Ea), the temperature (T), and the frequency factor (A).
A.) +2
B.) +4
C.) +6
D.) +8
Answer: C
The oxidation state of hydrogen is +1 and the oxidation state of oxygen is -2. Therefore, the sum of the oxidation states in H2SO4 must
A.) Radio waves
B.) Microwaves
C.) X-rays
D.) Visible light
Answer: C
Electromagnetic radiation is classified by its wavelength and frequency. Higher frequency radiation has higher energy. X-rays have a higher frequency and therefore higher energy than visible light, microwaves, and radio waves.
A.) 0.625 L
B.) 1.25 L
C.) 2.50 L
D.) 5.00 L
Answer: B
This problem can be solved using the Ideal Gas Law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for V, we get V = (nRT)/P. Plugging in the given values, we get V = (0.025 mol)(0.0821 L·atm/K·mol)(273 K)/(1 atm) = 1.25 L.
A.) sp
B.) sp2
C.) sp3
D.) sp3d
Answer: D
The central atom in SF4 is sulfur, which has six valence electrons. To form four covalent bonds with four fluorine atoms, sulfur must hybridize its orbitals. The hybridization of the central atom is determined by the number of electron pairs around it. In this case, there are four electron pairs, giving a hybridization of sp3d.
A.) 24.2 min
B.) 27.2 min
C.) 30.2 min
D.) 33.2 min
Answer: A
The half-life of a first-order reaction can be calculated using the equation t1/2 = ln(2)/k. Substituting the given rate constant gives t1/2 = ln(2)/0.0286 = 24.2 min. Therefore, the correct answer is (a).
A.) Particles have zero volume
B.) Particles are in constant random motion
C.) Particles exert no attractive or repulsive forces on each other
D.) Particles have infinite energy
Answer: D
An ideal gas is a hypothetical gas composed of particles with zero volume, in constant random motion, and with no attractive or repulsive forces between them.
A.) 0
B.) 1
C.) 2
D.) 3
Answer: B
The electron configuration of a nitrogen atom is 1s22s22p3. The three unpaired electrons are in the 2p subshell.
A.) NaCl (s) → Na+ (aq) + Cl– (aq)
B.) CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
C.) H2O (l) → H2O (g)
D.) 2H2 (g) + O2 (g) → 2H2O (g)
Answer: C
This reaction involves breaking the hydrogen bonds between water molecules, which requires a significant amount of energy and therefore has the highest activation energy.
A.) NaOH
B.) HCl
C.) NH3
D.) KOH
Answer: B
HCl is a strong acid that donates a proton (H+) to a base according to the Bronsted-Lowry acid-base theory.
A.) ΔG is always negative for spontaneous reactions.
B.) ΔG is always positive for nonspontaneous reactions.
C.) ΔG can be negative or positive depending on temperature and pressure.
D.) ΔG is always zero at equilibrium.
Answer: C
Gibbs free energy (ΔG) depends on both the enthalpy (ΔH) and entropy (ΔS) of a system, as well as temperature and pressure, and can be positive or negative depending on these factors.
A.) Platinum wire
B.) Iron filings
C.) Enzyme
D.) Acid catalyst
Answer: A
Platinum wire is a heterogeneous catalyst because it is in a different phase (solid) from the reactants (gas or liquid) it is catalyzing.
A.) The molecule must have an odd number of pi electrons.
B.) The molecule must have a cyclic structure.
C.) The molecule must have a planar structure.
D.) The molecule must have at least one double bond.
Answer: B
A molecule must have a cyclic structure that is planar and follows Hückel’s rule (4n+2 pi electrons) to exhibit aromaticity.
A.) They result from the mixing of s and p orbitals.
B.) They are used to describe the shapes of molecules.
C.) They are used to explain the properties of metals.
D.) They are only found in the ground state of an atom.
Answer: A
Hybrid orbitals are formed by mixing atomic orbitals (usually s and p orbitals) in the valence shell of an atom to form new hybrid orbitals with different shapes and energies.
A.) The rate is proportional to the square of the concentration of one of the reactants.
B.) The rate is independent of the concentration of the reactants.
C.) The rate is proportional to the product of the concentrations of the reactants.
D.) The rate is proportional to the concentration of one of the reactants.
Answer: D
In a first-order reaction, the rate of the reaction is directly proportional to the concentration of one of the reactants.
A.) It is a measure of the solubility of a solute in a solvent.
B.) It is the equilibrium constant for a precipitation reaction.
C.) It is independent of temperature.
D.) It is the product of the concentrations of the dissolved ions in a solution.
Answer: B
The solubility product constant, Ksp, is the equilibrium constant for a precipitation reaction between a sparingly soluble salt and water, and it describes the extent to which the salt dissolves in water.
A.) H+
B.) Na+
C.) Br–
D.) Cl2
Answer:C
A nucleophile is an electron-rich species that is attracted to a positive or partially positive charge. Br– has an extra electron and can donate it to an electron-deficient atom or molecule, making it a nucleophile.
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