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Kerala Plus Two Chemistry study materials
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Kerala Plus Two Chemistry practice questions

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Question 1Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

In solution chemistry, molality is defined as the number of moles of solute dissolved per

  1. kilogram of the final solution
  2. litre of the final solution
  3. kilogram of solvent Correct answer.
  4. litre of pure solvent

Explanation

Molality (m) is the amount of solute in moles divided by the mass of the solvent in kilograms. Since it is based on the mass of solvent rather than the volume of solution, molality does not change when the solution is heated or cooled.

Question 2Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

For a solution of a non-volatile solute in a volatile solvent, Raoult's law expresses the vapour pressure of the solution as

  1. the sum of the vapour pressures of the pure solvent and the pure solute
  2. the mole fraction of the solute multiplied by the vapour pressure of the pure solvent
  3. the vapour pressure of the pure solvent, unchanged by the presence of solute
  4. the mole fraction of the solvent multiplied by the vapour pressure of the pure solvent Correct answer.

Explanation

Raoult's law states that the partial vapour pressure of the solvent equals its mole fraction in the solution times the vapour pressure of the pure solvent, p = x(solvent) × p°. Because the solute is non-volatile it contributes no vapour pressure of its own, so the solution pressure is lower than that of the pure solvent.

Question 3Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

Which of the following properties of a dilute solution is a colligative property?

  1. Refractive index of the solution
  2. Surface tension of the solution
  3. Viscosity of the solution
  4. Osmotic pressure of the solution Correct answer.

Explanation

Colligative properties depend only on the number of solute particles present and not on their chemical nature. The four colligative properties are relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Viscosity, refractive index and surface tension instead depend on the identity of the solute.

Question 4Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

A solution is prepared by dissolving 18 g of glucose (molar mass 180 g mol⁻¹) in 500 g of water. The molality of the solution is

  1. 0.100 mol kg⁻¹
  2. 0.200 mol kg⁻¹ Correct answer.
  3. 1.00 mol kg⁻¹
  4. 0.360 mol kg⁻¹

Explanation

Moles of glucose = 18 g ÷ 180 g mol⁻¹ = 0.10 mol, and the mass of solvent is 0.500 kg. Dividing gives molality = 0.10 mol ÷ 0.500 kg = 0.200 mol kg⁻¹. Note that the 500 g refers to water, the solvent, and not to the final solution.

Question 5Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

At 300 K the vapour pressure of pure benzene is 120 mm Hg. When 6.0 g of a non-volatile, non-electrolyte solute is dissolved in 78 g of benzene, the vapour pressure of the solution falls to 114 mm Hg. The molar mass of the solute is

  1. 342 g mol⁻¹
  2. 114 g mol⁻¹ Correct answer.
  3. 57 g mol⁻¹
  4. 228 g mol⁻¹

Explanation

The relative lowering of vapour pressure is (120 − 114)/120 = 0.050, which equals the mole fraction of solute. Benzene amounts to 78 g ÷ 78 g mol⁻¹ = 1.0 mol, so 0.050 = n₂/(1.0 + n₂), giving n₂ = 0.0526 mol. Hence molar mass = 6.0 g ÷ 0.0526 mol ≈ 114 g mol⁻¹.

Question 6Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

Acetic acid dimerises completely when dissolved in benzene. The van't Hoff factor for acetic acid in this solution is

  1. 1.0
  2. 0.25
  3. 2.0
  4. 0.5 Correct answer.

Explanation

Two molecules of acetic acid combine to form one dimer, so the number of particles is halved compared with the amount dissolved. For complete dimerisation the observed particle count is half the expected count, hence i = 0.5; in general i = 1 − α/2 for this association.

Question 7Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

The osmotic pressure of a 0.10 M aqueous glucose solution at 300 K is (R = 0.0821 L atm K⁻¹ mol⁻¹)

  1. 1.23 atm
  2. 24.6 atm
  3. 0.246 atm
  4. 2.46 atm Correct answer.

Explanation

For a non-electrolyte, osmotic pressure is given by π = cRT, with c in mol L⁻¹. Substituting gives π = 0.10 × 0.0821 × 300 ≈ 2.46 atm, so the solution supports a height equivalent to about 2.5 atm of pressure.

Question 8Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

The boiling point constant Kb for water is 0.52 K kg mol⁻¹. A 1.0 molal aqueous solution of a non-electrolyte will boil at (external pressure 1 atm)

  1. 99.48 °C
  2. 100.52 °C Correct answer.
  3. 100.052 °C
  4. 101.04 °C

Explanation

Elevation of boiling point is ΔTb = Kb × m, so ΔTb = 0.52 × 1.0 = 0.52 K. Pure water boils at 100.00 °C, therefore the solution boils at 100.52 °C; a non-electrolyte has i = 1 and no further correction is needed.

Question 9Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

A 0.10 molal aqueous solution of a weak monobasic acid HA freezes at −0.2046 °C. Given Kf for water = 1.86 K kg mol⁻¹, the degree of dissociation of HA is

  1. 5%
  2. 10% Correct answer.
  3. 90%
  4. 20%

Explanation

The observed depression is 0.2046 K, so i = ΔTf/(Kf × m) = 0.2046/(1.86 × 0.10) = 1.10. For a monobasic acid HA ⇌ H⁺ + A⁻, i = 1 + α, hence α = 0.10, that is 10% dissociation at this concentration.

Question 10Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

Liquid mixtures of ethanol and acetone show a positive deviation from Raoult's law. This behaviour is best explained by the fact that

  1. ethanol–acetone attractions are weaker than the attractions in the pure liquids Correct answer.
  2. the mixture forms a maximum-boiling azeotrope over the whole composition range
  3. the enthalpy of mixing is large and negative
  4. ethanol–acetone attractions are stronger than the attractions in the pure liquids

Explanation

Positive deviation arises when the new solute-solvent interactions are weaker than the interactions in the pure components, so molecules escape into the vapour phase more readily and the vapour pressure exceeds the Raoult's law prediction. Such systems also show a positive enthalpy of mixing and can form minimum-boiling azeotropes.

Question 11Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

According to Henry's law, the solubility of a sparingly soluble gas in a liquid at a fixed temperature is directly proportional to which quantity?

  1. The partial pressure of that gas in the vapour phase above the liquid Correct answer.
  2. The total pressure exerted by all gases present above the liquid
  3. The temperature of the gas–liquid system
  4. The volume of the liquid solvent taken in the container

Explanation

Henry's law states that p = KH·x, where x is the mole fraction of the dissolved gas and KH is the Henry's law constant at that temperature. At constant temperature, the mole fraction of dissolved gas therefore rises in direct proportion to the partial pressure of the gas above the solution. It is the partial pressure of the individual gas, not the total pressure of the gas mixture, that controls solubility.

Question 12Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

A student warms a dilute aqueous solution of sodium chloride from 25 °C to 60 °C in a sealed container. Which concentration term changes because of this temperature change alone?

  1. Molality
  2. Mass percentage of solute
  3. Molarity Correct answer.
  4. Mole fraction of solute

Explanation

Molarity is defined as moles of solute per litre of solution, and the volume of a solution expands on heating, so molarity falls when temperature rises. Molality, mole fraction and mass percentage are all based on masses or mole ratios, which are unaffected by thermal expansion. Hence only molarity is temperature dependent.

Question 13Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

At a given temperature, the vapour pressure of a pure solvent (molar mass 100 g mol⁻¹) is 100 mm Hg. When 5.0 g of a non-volatile, non-electrolyte solute is dissolved in 98 g of this solvent, the vapour pressure of the solution becomes 98 mm Hg. The molar mass of the solute is:

  1. 250 g mol⁻¹ Correct answer.
  2. 200 g mol⁻¹
  3. 500 g mol⁻¹
  4. 100 g mol⁻¹

Explanation

Moles of solvent = 98/100 = 0.98 mol. Since p = p°·x₁, the solvent mole fraction is 98/100 = 0.98, so the solute mole fraction is 0.02 and moles of solute = 0.02 mol. The molar mass is therefore 5.0 g ÷ 0.02 mol = 250 g mol⁻¹.

Question 14Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

Aluminium sulphate dissociates in water as Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻. If it is 60% dissociated in a dilute aqueous solution, the van't Hoff factor i is:

  1. 3.4 Correct answer.
  2. 2.4
  3. 1.0
  4. 5.0

Explanation

The salt produces five ions per formula unit, so n = 5. The van't Hoff factor for partial dissociation is i = 1 + (n − 1)α = 1 + 4(0.60) = 3.4. A value of 5.0 would apply only for complete dissociation.

Question 15Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

At the same temperature, which aqueous solution is isotonic with a 0.10 M glucose solution?

  1. 0.10 M NaCl
  2. 0.20 M urea
  3. 0.05 M NaCl Correct answer.
  4. 0.05 M glucose

Explanation

Osmotic pressure is π = iCRT, so isotonic solutions must have equal values of iC at the same temperature. Glucose has i = 1, giving iC = 0.10; NaCl has i = 2, so C must be 0.05 M to give iC = 0.10. Thus 0.05 M NaCl is isotonic with 0.10 M glucose.

Question 16Solutions (concentration terms, colligative properties, Raoult's law, van't Hoff factor)

An aqueous sodium chloride solution freezes at a temperature 0.372 K below the freezing point of pure water. Taking Kf for water as 1.86 K kg mol⁻¹ and assuming complete dissociation of NaCl, the molality of the solution is:

  1. 0.20 mol kg⁻¹
  2. 0.40 mol kg⁻¹
  3. 0.10 mol kg⁻¹ Correct answer.
  4. 0.050 mol kg⁻¹

Explanation

For an electrolyte, ΔTf = i·Kf·m. NaCl gives two ions per formula unit, so i = 2. Rearranging gives m = 0.372/(2 × 1.86) = 0.372/3.72 = 0.10 mol kg⁻¹.

Question 17Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

The SI unit of conductivity (κ) of an electrolyte solution is

  1. Ω m
  2. S m² mol⁻¹
  3. S m⁻¹ Correct answer.
  4. mol L⁻¹

Explanation

Conductivity is the conductance of a unit volume of the solution, so it is expressed in siemens per metre, S m⁻¹. Molar conductivity, which refers to the conductance due to one mole of electrolyte, is the quantity that carries the unit S m² mol⁻¹.

Question 18Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

The electrolyte used in a lead-acid storage battery is

  1. aqueous potassium hydroxide
  2. concentrated hydrochloric acid
  3. molten sodium chloride
  4. dilute sulphuric acid Correct answer.

Explanation

In the lead-acid accumulator the plates are lead and lead dioxide dipped in dilute sulphuric acid of about 38% strength when fully charged. During discharge both plates are converted to lead(II) sulphate while the acid is diluted further.

Question 19Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

Rusting of iron requires the simultaneous presence of

  1. oxygen alone
  2. moisture alone
  3. nitrogen and carbon dioxide
  4. oxygen and moisture Correct answer.

Explanation

Rusting is an electrochemical corrosion in which iron is oxidised at anodic sites while dissolved oxygen is reduced at cathodic sites. Water acts as the medium for ion movement and oxygen is the electron acceptor, so both must be present; the product is hydrated iron(III) oxide.

Question 20Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

For the cell Zn(s)|Zn²⁺(0.10 M)||Cu²⁺(0.010 M)|Cu(s) at 298 K, E°cell = 1.10 V. The emf of the cell is

  1. 1.13 V
  2. 1.07 V Correct answer.
  3. 1.10 V
  4. 1.04 V

Explanation

Apply the Nernst equation E = E° − (0.0591/n) log([Zn²⁺]/[Cu²⁺]) with n = 2. The reaction quotient is 0.10/0.010 = 10, so E = 1.10 − (0.0591/2)(1) = 1.10 − 0.0296 ≈ 1.07 V.

Question 21Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

A current of 2.0 A is passed for 30 minutes through aqueous copper(II) sulphate between platinum electrodes. The mass of copper deposited at the cathode is (Cu = 63.5 g mol⁻¹, F = 96500 C mol⁻¹)

  1. 2.37 g
  2. 1.18 g Correct answer.
  3. 0.59 g
  4. 4.74 g

Explanation

Charge passed Q = It = 2.0 × 1800 = 3600 C, which corresponds to 3600/96500 = 0.0373 mol of electrons. Since Cu²⁺ + 2e⁻ → Cu, the copper deposited is 0.0373/2 = 0.01865 mol, giving 0.01865 × 63.5 ≈ 1.18 g.

Question 22Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

At 298 K, Λ°(NaCl) = 126.4, Λ°(HCl) = 426.2 and Λ°(CH₃COONa) = 91.0 S cm² mol⁻¹. The limiting molar conductivity of acetic acid, Λ°(CH₃COOH), is

  1. 461.6 S cm² mol⁻¹
  2. 643.6 S cm² mol⁻¹
  3. 217.4 S cm² mol⁻¹
  4. 390.8 S cm² mol⁻¹ Correct answer.

Explanation

By Kohlrausch's law, Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl), because Na⁺ and Cl⁻ cancel as spectator ions. Substituting gives 91.0 + 426.2 − 126.4 = 390.8 S cm² mol⁻¹.

Question 23Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

The molar conductivity of 0.010 M acetic acid is 16.0 S cm² mol⁻¹ and its limiting molar conductivity is 400 S cm² mol⁻¹ at 298 K. The degree of dissociation and the pH of the solution are

  1. α = 0.04, pH = 2.4
  2. α = 0.04, pH = 3.4 Correct answer.
  3. α = 0.40, pH = 3.4
  4. α = 0.04, pH = 4.0

Explanation

The degree of dissociation is α = Λm/Λ°m = 16.0/400 = 0.040. Then [H⁺] = cα = 0.010 × 0.040 = 4.0 × 10⁻⁴ M, so pH = −log(4.0 × 10⁻⁴) ≈ 3.4.

Question 24Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

An underground iron pipeline is connected to a block of magnesium metal to slow corrosion. This protection works because

  1. magnesium is oxidised in preference to iron and acts as a sacrificial anode Correct answer.
  2. magnesium acts as the cathode and is reduced while iron dissolves
  3. magnesium raises the potential of iron above its standard value
  4. magnesium blocks oxygen from reaching the iron surface

Explanation

Magnesium has a much more negative standard reduction potential (−2.37 V) than iron (−0.44 V), so it is oxidised preferentially and becomes the sacrificial anode. The pipeline is forced to behave as the cathode, where reduction of oxygen occurs instead of iron dissolution.

Question 25Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

Aqueous copper(II) sulphate is electrolysed between platinum electrodes. How does the pH of the solution change and why?

  1. pH rises because H₂ is evolved at the cathode
  2. pH stays unchanged because Cu²⁺ is replaced by H⁺
  3. pH falls because sulphate ions are oxidised to SO₃
  4. pH falls because water is oxidised at the anode, giving O₂ and H⁺ Correct answer.

Explanation

With inert platinum electrodes, Cu²⁺ is reduced at the cathode while water is oxidised at the anode: 2H₂O → O₂ + 4H⁺ + 4e⁻. The hydrogen ions released make the solution progressively acidic, so the pH falls as electrolysis proceeds.

Question 26Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

For the reaction Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), E°cell = 0.46 V at 298 K. Using log K = nE°cell/0.0591, the equilibrium constant is closest to

  1. 1.8 × 10¹⁵
  2. 3.7 × 10¹⁵ Correct answer.
  3. 1.3 × 10³¹
  4. 6.0 × 10⁷

Explanation

Here n = 2, so log K = (2 × 0.46)/0.0591 = 0.92/0.0591 ≈ 15.57. Therefore K = 10^15.57 ≈ 3.7 × 10¹⁵, a very large value consistent with the positive cell potential.

Question 27Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

In a galvanic cell such as Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s), a salt bridge of saturated KCl in agar joins the two half-cells. Its main function is to

  1. complete the circuit and maintain electrical neutrality of the two half-cell solutions Correct answer.
  2. provide a path for electrons to travel from the cathode back to the anode
  3. prevent the reduction of Cu²⁺ ions at the cathode surface
  4. increase the standard electrode potential of the copper half-cell

Explanation

The salt bridge contains an inert electrolyte such as saturated KCl. Its ions migrate into the two half-cells, completing the internal circuit and neutralising the excess charge that builds up as oxidation and reduction proceed. It also keeps the liquid junction potential between the two solutions as small as possible.

Question 28Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

A hydrogen–oxygen fuel cell uses aqueous KOH as the electrolyte. The overall cell reaction is

  1. 2H₂O(l) → 2H₂(g) + O₂(g)
  2. 2H₂(g) + O₂(g) → 2H₂O(l) Correct answer.
  3. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
  4. H₂(g) + O₂(g) → H₂O₂(l)

Explanation

In the H₂–O₂ fuel cell, hydrogen is oxidised at the anode and oxygen is reduced at the cathode, so the overall change is 2H₂ + O₂ → 2H₂O. The only product is water, which is why the cell is regarded as non-polluting. It converts chemical energy directly into electrical energy with high efficiency.

Question 29Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

For a cell reaction involving two electrons, E°cell = +0.295 V at 298 K. The standard Gibbs energy change of the reaction is

  1. −28.5 kJ mol⁻¹
  2. −56.9 kJ mol⁻¹ Correct answer.
  3. −569 kJ mol⁻¹
  4. +56.9 kJ mol⁻¹

Explanation

The relation between cell emf and Gibbs energy is ΔG° = −nFE°cell. Substituting n = 2, F = 96,500 C mol⁻¹ and E°cell = 0.295 V gives ΔG° = −2 × 96,500 × 0.295 = −56,935 J mol⁻¹, that is about −56.9 kJ mol⁻¹. The negative sign confirms that the cell reaction is spontaneous under standard conditions.

Question 30Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

A conductivity cell has a cell constant of 0.50 cm⁻¹. When it is filled with 0.10 M KCl solution, its resistance at 298 K is 100 Ω. The conductivity of the solution is

  1. 2.0 × 10⁻³ S cm⁻¹
  2. 5.0 × 10⁻⁵ S cm⁻¹
  3. 5.0 × 10⁻³ S cm⁻¹ Correct answer.
  4. 5.0 × 10⁻¹ S cm⁻¹

Explanation

Conductivity is calculated as κ = cell constant / R, so κ = 0.50 cm⁻¹ / 100 Ω = 5.0 × 10⁻³ S cm⁻¹. The cell constant l/A converts the measured resistance into conductivity, and a larger cell constant or a smaller resistance gives a larger κ. Conductivity depends on concentration and temperature, not on the amount of solution taken.

Question 31Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

Concentrated aqueous sodium chloride (brine) is electrolysed using inert platinum electrodes. The products liberated at the cathode and the anode respectively are

  1. Na and Cl₂
  2. H₂ and Cl₂ Correct answer.
  3. H₂ and O₂
  4. Cl₂ and H₂

Explanation

In concentrated brine, chloride ion is preferentially oxidised at the anode to give chlorine, while water is reduced at the cathode in preference to Na⁺, giving hydrogen and hydroxide ion. The overall result is Cl₂ at the anode, H₂ at the cathode and NaOH left in solution. With dilute NaCl, or with a mercury cathode, the products differ.

Question 32Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

Which statement about the variation of molar conductivity (Λm) with concentration at 298 K is correct?

  1. Λm of a strong electrolyte decreases steeply on dilution while that of a weak electrolyte remains almost constant.
  2. Λm of a strong electrolyte rises steeply on dilution because its degree of dissociation increases greatly.
  3. For strong electrolytes Λm rises only slightly and linearly with √c, while for weak electrolytes it rises steeply on dilution. Correct answer.
  4. For both strong and weak electrolytes, Λm is independent of concentration at 298 K.

Explanation

For strong electrolytes, dilution only weakens ion–ion interactions, so Λm increases slightly and follows Kohlrausch's relation Λm = Λ°m − A√c. For weak electrolytes, dilution increases the degree of dissociation, so Λm rises steeply and reaches Λ°m only at infinite dilution.

Question 33Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

For the concentration cell Ni(s)|Ni²⁺(0.010 M)||Ni²⁺(1.0 M)|Ni(s) at 298 K, the emf is

  1. 0.000 V
  2. +0.0591 V Correct answer.
  3. −0.0591 V
  4. +0.0296 V

Explanation

In a concentration cell both electrodes are of the same material, so E°cell = 0 and the emf is given by E = (0.0591/n) log(c_cathode/c_anode). The dilute compartment must be the anode, so E = (0.0591/2) log(1.0/0.010) = 0.0591 V. The driving force is the tendency of Ni²⁺ ions to move from the concentrated to the dilute half-cell.

Question 34Electrochemistry (Nernst equation, conductance, electrolysis, batteries, corrosion)

An underground iron pipe passes from well-aerated sandy soil into poorly aerated waterlogged clay. Which statement explains the corrosion observed?

  1. Both regions corrode equally because oxygen concentration has no effect on iron corrosion.
  2. The poorly aerated clay region becomes anodic and corrodes, while the aerated sandy region acts as cathode. Correct answer.
  3. The clay region becomes cathodic and is therefore protected, while hydrogen gas is evolved there.
  4. The sandy region becomes anodic because higher oxygen concentration favours iron dissolution.

Explanation

In differential aeration the oxygen-poor region has the lower reduction potential, so it becomes the anode and iron dissolves there as Fe²⁺. The oxygen-rich sandy region is cathodic, where O₂ is reduced to OH⁻, and the two regions form a small galvanic cell along the pipe. This is why corrosion is often severe where a buried pipe passes from aerated to waterlogged soil.

Question 35Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

For a reaction in aqueous solution, the rate is expressed in mol L⁻¹ s⁻¹. What are the units of the rate constant for a reaction that is first order overall?

  1. s⁻¹ Correct answer.
  2. L mol⁻¹ s⁻¹
  3. mol L⁻¹ s⁻¹
  4. L² mol⁻² s⁻¹

Explanation

For a reaction of overall order n, the units of the rate constant are mol^(1−n) L^(n−1) s⁻¹. When n = 1, this simplifies to mol⁰ L⁰ s⁻¹, which is s⁻¹. This follows directly from the rate law rate = k[A] for a first-order reaction.

Question 36Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

A first-order reaction has a half-life of 20 minutes. If the initial concentration of the reactant is 0.80 mol L⁻¹, what is its concentration after 40 minutes?

  1. 0.60 mol L⁻¹
  2. 0.40 mol L⁻¹
  3. 0.20 mol L⁻¹ Correct answer.
  4. 0.10 mol L⁻¹

Explanation

For a first-order reaction, each half-life reduces the reactant concentration to one half of its value at the start of that interval. The interval of 40 minutes is exactly two half-lives, so the concentration halves twice: 0.80 → 0.40 → 0.20 mol L⁻¹.

Question 37Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

For the reaction with rate law rate = k[NO]²[O₂], what are the order with respect to NO and the overall order, respectively?

  1. 2 and 3 Correct answer.
  2. 1 and 3
  3. 2 and 2
  4. 3 and 2

Explanation

The order with respect to a reactant is the exponent of its concentration in the experimental rate law. Here NO has exponent 2, so the reaction is second order in NO. The overall order is the sum of the exponents, 2 + 1 = 3, so the correct pair is 2 and 3.

Question 38Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

A zero-order reaction has rate constant k = 2.0 × 10⁻² mol L⁻¹ s⁻¹ and initial concentration [A]₀ = 0.40 mol L⁻¹. What is the half-life of the reaction?

  1. 5 s
  2. 40 s
  3. 20 s
  4. 10 s Correct answer.

Explanation

For a zero-order reaction, the half-life is given by t₁/₂ = [A]₀/(2k). Substituting the values gives 0.40/(2 × 0.020) = 0.40/0.040 = 10 s. The half-life of a zero-order reaction therefore depends on the initial concentration.

Question 39Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

A first-order reaction has rate constant k = 1.386 × 10⁻² s⁻¹. How long will it take for the reactant concentration to fall to 25% of its initial value?

  1. 25 s
  2. 100 s Correct answer.
  3. 50 s
  4. 200 s

Explanation

For a first-order reaction, t₁/₂ = 0.693/k = 0.693/(1.386 × 10⁻²) = 50 s. Falling to 25% means halving twice, so the required time is two half-lives, 2 × 50 s = 100 s. The integrated form t = (2.303/k) log(100/25) gives the same result.

Question 40Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

According to the Arrhenius equation, why does the rate constant of most reactions increase when the temperature is raised?

  1. The order of the reaction always increases with temperature.
  2. The concentration of the reactants increases with temperature.
  3. The activation energy decreases as temperature rises.
  4. A larger fraction of collisions has energy equal to or greater than the activation energy. Correct answer.

Explanation

The Arrhenius equation k = A e^(−Ea/RT) shows that raising T makes the exponent less negative and increases k. In molecular terms, a larger fraction of molecules then possesses energy at least equal to the activation energy, so more collisions are effective. The activation energy itself is treated as essentially constant over the usual temperature range.

Question 41Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

The hydrolysis of ethyl acetate in water is described as a pseudo-first-order reaction because

  1. the reaction is catalysed by an acid.
  2. the concentration of ethyl acetate remains constant.
  3. the rate is independent of temperature.
  4. water is present in large excess, so its concentration remains almost constant. Correct answer.

Explanation

In dilute aqueous solution, water is the solvent and is present in large excess, so its concentration changes by only a negligible amount during the reaction. The rate law therefore appears to depend only on the concentration of ethyl acetate, giving pseudo-first-order behaviour. The true rate law still contains the water concentration.

Question 42Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

The initial rate of the reaction A + B → C was measured at different initial concentrations: Experiment 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻³ M s⁻¹ Experiment 2: [A] = 0.20 M, [B] = 0.10 M, rate = 8.0 × 10⁻³ M s⁻¹ Experiment 3: [A] = 0.10 M, [B] = 0.20 M, rate = 4.0 × 10⁻³ M s⁻¹ What is the rate law for the reaction?

  1. rate = k[A]²[B]²
  2. rate = k[A][B]²
  3. rate = k[A][B]
  4. rate = k[A]²[B] Correct answer.

Explanation

Comparing experiments 1 and 2, doubling [A] while keeping [B] constant increases the rate from 2.0 × 10⁻³ to 8.0 × 10⁻³, a factor of 4, so the order in A is 2. Comparing experiments 1 and 3, doubling [B] while keeping [A] constant doubles the rate, so the order in B is 1. The rate law is therefore rate = k[A]²[B].

Question 43Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

For a reaction obeying the Arrhenius equation, a graph of ln k against 1/T is a straight line. What does the slope of this line equal?

  1. Ea/(2.303R)
  2. −Ea/(2.303R)
  3. −Ea/R Correct answer.
  4. Ea/R

Explanation

Taking natural logarithms of k = A e^(−Ea/RT) gives ln k = ln A − Ea/(RT). A plot of ln k against 1/T therefore has slope −Ea/R and intercept ln A. The factor 2.303 appears only when log₁₀ k is plotted instead of ln k.

Question 44Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

A catalyst increases the rate of a reaction. Which statement correctly describes the main reason in collision theory terms?

  1. It provides an alternative path with lower activation energy, so a larger fraction of collisions is effective. Correct answer.
  2. It increases the activation energy so that only fast molecules react.
  3. It changes the order of the reaction from first to zero.
  4. It increases the frequency of collisions by raising the temperature.

Explanation

A catalyst offers a different reaction pathway whose activation energy is lower than that of the uncatalysed path. With a lower energy barrier, a larger fraction of collisions has enough energy to be effective, so the rate increases. The catalyst is not consumed and does not usually alter the overall order of the reaction.

Question 45Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

For an elementary reaction, what does the term molecularity represent?

  1. The sum of the exponents in the experimental rate law
  2. The number of intermediates formed before the slow step
  3. The total number of product molecules in the overall equation
  4. The number of reactant particles colliding in the elementary step Correct answer.

Explanation

Molecularity is defined only for an elementary step and equals the number of reactant particles that collide in that step. It is a theoretical count and is not obtained from the experimental rate law; order is the experimental counterpart.

Question 46Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

For a first-order reaction, a graph of ln[A] against time is a straight line. What does the slope of this graph equal?

  1. +k
  2. ln[A]₀
  3. −k Correct answer.
  4. −k/2.303

Explanation

The integrated rate equation is ln[A] = ln[A]₀ − kt, so a plot of ln[A] against t has slope −k and intercept ln[A]₀. The factor 2.303 appears only when log₁₀[A] is plotted instead of ln[A].

Question 47Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

For the reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g), the rate of disappearance of N₂O₅ is 2.0 × 10⁻⁴ mol L⁻¹ s⁻¹. What is the rate of formation of NO₂ under the same conditions?

  1. 2.0 × 10⁻⁴ mol L⁻¹ s⁻¹
  2. 4.0 × 10⁻⁴ mol L⁻¹ s⁻¹ Correct answer.
  3. 8.0 × 10⁻⁴ mol L⁻¹ s⁻¹
  4. 1.0 × 10⁻⁴ mol L⁻¹ s⁻¹

Explanation

From the stoichiometry, −(1/2)d[N₂O₅]/dt = +(1/4)d[NO₂]/dt. The reaction rate is (1/2)(2.0 × 10⁻⁴) = 1.0 × 10⁻⁴ mol L⁻¹ s⁻¹, so d[NO₂]/dt = 4 × 1.0 × 10⁻⁴ = 4.0 × 10⁻⁴ mol L⁻¹ s⁻¹.

Question 48Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

A zero-order reaction has k = 5.0 × 10⁻³ mol L⁻¹ s⁻¹ and [A]₀ = 0.50 mol L⁻¹. How long will it take for [A] to fall to 0.20 mol L⁻¹?

  1. 60 s Correct answer.
  2. 140 s
  3. 100 s
  4. 40 s

Explanation

For a zero-order reaction, [A] = [A]₀ − kt, so t = ([A]₀ − [A])/k. Substituting gives (0.50 − 0.20)/(5.0 × 10⁻³) = 60 s.

Question 49Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

In collision theory, the steric factor P in the rate expression is included mainly to account for

  1. the fraction of collisions with energy greater than the activation energy
  2. the change in activation energy caused by a catalyst
  3. the increase in collision frequency when temperature rises
  4. the requirement that colliding molecules have a suitable orientation Correct answer.

Explanation

The exponential term exp(−Ea/RT) gives the fraction of collisions with enough energy, while P corrects for orientation and other probability factors. For simple spherical species P is close to 1, but complex molecules usually need a suitable orientation.

Question 50Chemical Kinetics (rate laws, order, integrated rate equations, Arrhenius, collision theory)

The rate constant of a reaction becomes four times larger when the temperature rises from 300 K to 320 K. What is the activation energy? (R = 8.314 J K⁻¹ mol⁻¹; ln 4 = 1.386)

  1. 28 kJ mol⁻¹
  2. 13 kJ mol⁻¹
  3. 110 kJ mol⁻¹
  4. 55 kJ mol⁻¹ Correct answer.

Explanation

Using ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), ln 4 = Ea/8.314 (1/300 − 1/320). The temperature term is 2.083 × 10⁻⁴ K⁻¹, giving Ea ≈ 5.53 × 10⁴ J mol⁻¹ ≈ 55 kJ mol⁻¹.