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BSEB Intermediate Chemistry study materials
BSEB Intermediate Chemistry study resources

BSEB Intermediate Chemistry practice questions

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Question 1Solutions

At 298 K, the vapour pressure of pure benzene is 75.0 mm Hg and that of pure toluene is 22.5 mm Hg. A mixture contains 0.40 mole fraction of benzene. Assuming ideal behaviour, what is the total vapour pressure of the solution?

  1. 30.0 mm Hg
  2. 97.5 mm Hg
  3. 52.5 mm Hg
  4. 43.5 mm Hg Correct answer.

Explanation

For an ideal binary solution the total pressure is the sum of partial pressures, each given by Raoult's law. Benzene contributes 0.40 × 75.0 = 30.0 mm Hg and toluene contributes 0.60 × 22.5 = 13.5 mm Hg. Adding them gives 43.5 mm Hg.

Question 2Solutions

Which statement correctly describes a colligative property of a dilute solution?

  1. It depends on the nature of both the solute and the solvent particles.
  2. It depends on the molar mass of the solute dissolved in the solvent.
  3. It depends on the number of solute particles but not on their chemical identity. Correct answer.
  4. It depends on the total mass of solute present in the solution.

Explanation

Colligative properties such as osmotic pressure and boiling point elevation are governed by the count of dissolved particles. The identity and mass of the particles do not enter the equations, which is why equal molalities of different non-electrolytes give equal effects.

Question 3Solutions

Which pair of liquids is most likely to form an ideal solution when mixed in all proportions?

  1. Acetone and chloroform
  2. Ethanol and water
  3. Benzene and toluene Correct answer.
  4. Water and ethanol

Explanation

Benzene and toluene are structurally similar non-polar liquids with comparable intermolecular forces, so their A–B interactions resemble A–A and B–B interactions. This near-equality of forces is the condition for ideal behaviour.

Question 4Solutions

The osmotic pressure of a solution is measured at 300 K and found to be 2.46 atm. If the concentration of the solute is 0.100 mol per litre, what is the value of the gas constant R obtained from these data?

  1. 0.246 L atm K⁻¹ mol⁻¹
  2. 0.0082 L atm K⁻¹ mol⁻¹
  3. 0.0820 L atm K⁻¹ mol⁻¹ Correct answer.
  4. 8.20 L atm K⁻¹ mol⁻¹

Explanation

Osmotic pressure is given by π = CRT, so R = π/(CT). Substituting gives 2.46/(0.100 × 300) = 2.46/30 = 0.0820 L atm K⁻¹ mol⁻¹, matching the known gas constant.

Question 5Solutions

A solution of a non-volatile solute in a solvent shows a vapour pressure lower than that of the pure solvent. Which expression correctly represents the relative lowering of vapour pressure?

  1. (p⁰ − p)/p⁰ = mole fraction of solvent
  2. (p⁰ − p)/p⁰ = mole fraction of solute Correct answer.
  3. (p⁰ − p)/p = mole fraction of solute
  4. (p − p⁰)/p⁰ = mole fraction of solute

Explanation

Raoult's law for a non-volatile solute gives p = x₁p⁰, where x₁ is the mole fraction of solvent. Rearranging gives (p⁰ − p)/p⁰ = 1 − x₁ = x₂, the mole fraction of solute.

Question 6Solutions

The boiling point elevation constant of water is 0.52 K kg mol⁻¹. What is the boiling point of an aqueous solution containing 18.0 g of glucose (molar mass 180 g mol⁻¹) in 500 g of water?

  1. 100.104 °C Correct answer.
  2. 100.208 °C
  3. 100.052 °C
  4. 100.026 °C

Explanation

Moles of glucose = 18.0/180 = 0.100 mol, so molality = 0.100/0.500 = 0.200 mol kg⁻¹. Elevation ΔTb = 0.52 × 0.200 = 0.104 K, giving a boiling point of 100.104 °C.

Question 7Solutions

Two solutions are prepared in the same solvent: solution A contains 0.1 mol of NaCl per kg of solvent and solution B contains 0.1 mol of glucose per kg of solvent. Assuming complete dissociation, how do their freezing points compare?

  1. Solution B freezes at a lower temperature than solution A.
  2. Solution A freezes above 0 °C while B freezes below 0 °C.
  3. Both solutions freeze at the same temperature.
  4. Solution A freezes at a lower temperature than solution B. Correct answer.

Explanation

Freezing point depression depends on the total number of particles. NaCl dissociates into two ions per formula unit, giving an effective particle count near 0.2 mol kg⁻¹, while glucose remains one particle. The greater particle count in A produces the larger depression.

Question 8Solutions

Which statement about Henry's law is correct?

  1. The solubility of a gas in a liquid is directly proportional to its partial pressure above the liquid. Correct answer.
  2. The solubility of a gas is independent of the pressure above the liquid.
  3. The solubility of a gas in a liquid is inversely proportional to its partial pressure above the liquid.
  4. The solubility of a gas depends only on the temperature of the liquid.

Explanation

Henry's law states that at constant temperature the mole fraction of dissolved gas is proportional to its partial pressure in the vapour above the solution, with the Henry's law constant as the proportionality factor. This explains why carbonated drinks fizz when the pressure is released.

Question 9Solutions

A solution is described as showing positive deviation from Raoult's law. Which explanation is consistent with this behaviour?

  1. A–B interactions are weaker than A–A and B–B interactions, so escaping tendency increases. Correct answer.
  2. The solute is non-volatile and its vapour pressure is negligible.
  3. A–B interactions are stronger than A–A and B–B interactions, so escaping tendency decreases.
  4. The solution forms a maximum boiling azeotrope with a higher vapour pressure than predicted.

Explanation

When unlike molecules attract each other less than like molecules, molecules escape more readily than in the pure liquids. The measured vapour pressure then exceeds the Raoult's law prediction, which is positive deviation.

Question 10Solutions

In the van't Hoff equation for osmotic pressure of an electrolyte solution, what does the factor i represent?

  1. The ratio of vapour pressure of solution to that of pure solvent
  2. The ratio of observed colligative effect to that expected for a non-electrolyte of the same molality Correct answer.
  3. The ratio of solvent mass to solute mass in the solution
  4. The degree of hydration of the ions in solution

Explanation

The van't Hoff factor i accounts for dissociation or association by comparing the measured colligative effect with the value expected if the solute did not dissociate. For complete dissociation of NaCl, i approaches 2.

Question 11Solutions

The vapour pressure of pure liquid A is 400 mm Hg and that of pure liquid B is 600 mm Hg. An equimolar mixture of A and B is found to have a total vapour pressure of 560 mm Hg. What does this indicate about the mixture?

  1. It behaves as an ideal solution.
  2. It forms a maximum boiling azeotrope.
  3. It shows positive deviation from Raoult's law.
  4. It shows negative deviation from Raoult's law. Correct answer.

Explanation

For an ideal equimolar mixture the expected pressure is 0.5 × 400 + 0.5 × 600 = 500 mm Hg. The observed 560 mm Hg would be higher, but since the question states 560 is greater than 500 this is positive deviation. However, the keyed answer reflects that the observed value here is lower than the ideal value only if the ideal value exceeds it; recomputing gives 500, so the correct interpretation is positive deviation.

Question 12Solutions

Which aqueous solution will have the highest osmotic pressure at 298 K, assuming complete dissociation where applicable?

  1. 0.10 M NaCl
  2. 0.10 M urea
  3. 0.10 M CaCl₂ Correct answer.
  4. 0.10 M glucose

Explanation

Osmotic pressure is proportional to the total concentration of particles. CaCl₂ gives three ions per formula unit, so its effective particle concentration is about 0.30 M, higher than NaCl (0.20 M) and the non-electrolytes (0.10 M).

Question 13Solutions

For a binary solution of two volatile liquids, which condition must hold at the azeotropic composition?

  1. The mole fraction of each component equals 0.5.
  2. The vapour phase has the same composition as the liquid phase. Correct answer.
  3. The solution obeys Raoult's law over the whole composition range.
  4. The total vapour pressure equals the sum of the pure component pressures.

Explanation

At an azeotrope the liquid and vapour compositions are identical, so the mixture distils without change in composition. This occurs because the two components have equal escaping tendencies at that particular composition.

Question 14Solutions

A solute associates into dimers in solution. If the van't Hoff factor is 0.60, what fraction of the solute has associated into dimers?

  1. 0.20
  2. 0.80 Correct answer.
  3. 0.60
  4. 0.40

Explanation

For association of two molecules into a dimer, i = 1 − α/2 where α is the fraction associated. Setting 0.60 = 1 − α/2 gives α/2 = 0.40, so α = 0.80, meaning 80 percent of the solute exists as dimers.

Question 15Electrochemistry

In the Daniell cell, the overall reaction is Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Which electrode is the anode and what is its polarity?

  1. Copper electrode, and it is the negative terminal
  2. Zinc electrode, and it is the negative terminal Correct answer.
  3. Copper electrode, and it is the positive terminal
  4. Zinc electrode, and it is the positive terminal

Explanation

Oxidation always occurs at the anode, and in this cell zinc metal is oxidised to Zn²⁺, releasing electrons. The electrode where electrons are released is the negative terminal, so the zinc electrode is the negative anode. Copper is reduced and acts as the positive cathode.

Question 16Electrochemistry

Which statement correctly describes the sign convention for standard electrode potential of the Zn²⁺/Zn couple?

  1. It is zero, because the hydrogen electrode is the reference
  2. It is positive, because zinc lies above hydrogen in the activity series
  3. It is negative, because zinc is oxidised when coupled with the standard hydrogen electrode Correct answer.
  4. It is negative, because zinc is reduced in preference to hydrogen

Explanation

Standard electrode potentials are quoted for the reduction half-reaction. For Zn²⁺ + 2e⁻ → Zn the value is about −0.76 V, meaning zinc has a weaker tendency than hydrogen to be reduced and is therefore oxidised when paired with the SHE.

Question 17Electrochemistry

What is the SI unit of molar conductivity, and how is it related to conductivity?

  1. S m² mol⁻¹, obtained by dividing conductivity by concentration in mol m⁻³ Correct answer.
  2. S m⁻¹, obtained by multiplying conductivity by concentration
  3. Ω m, obtained by taking the reciprocal of conductivity
  4. S mol m⁻³, obtained by dividing concentration by conductivity

Explanation

Molar conductivity Λm is defined as κ/c, where κ is conductivity in S m⁻¹ and c is concentration in mol m⁻³. The resulting unit is S m² mol⁻¹. It measures the conducting power of one mole of electrolyte in solution.

Question 18Electrochemistry

Which quantity is conventionally assigned the value zero at all temperatures to define the electrode potential scale?

  1. The standard calomel electrode potential
  2. The standard hydrogen electrode potential Correct answer.
  3. The standard copper electrode potential
  4. The standard zinc electrode potential

Explanation

By international convention the standard hydrogen electrode, H⁺(aq, 1 M)|H₂(g, 1 bar), is assigned E° = 0.00 V. All other standard electrode potentials are measured relative to it. This makes it the reference point of the electrochemical series.

Question 19Electrochemistry

For the reaction 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), how many electrons are transferred per formula unit as written?

  1. 3
  2. 2
  3. 6 Correct answer.
  4. 5

Explanation

Each aluminium atom loses three electrons and there are two aluminium atoms, giving six electrons lost. Each copper ion gains two electrons and there are three such ions, also giving six electrons gained. The balanced electron count is therefore six.

Question 20Electrochemistry

The standard electrode potentials are E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V. What is the standard cell potential of the Daniell cell?

  1. +1.10 V Correct answer.
  2. +0.42 V
  3. −0.42 V
  4. −1.10 V

Explanation

For a galvanic cell, E°cell = E°cathode − E°anode. Copper has the higher reduction potential, so it is the cathode: +0.34 − (−0.76) = +1.10 V. A positive value confirms the reaction is spontaneous under standard conditions.

Question 21Electrochemistry

For the half-cell reaction Fe³⁺(aq) + e⁻ → Fe²⁺(aq), what is the Nernst equation at 298 K?

  1. E = E° + 0.0591 log([Fe²⁺]/[Fe³⁺])
  2. E = E° − 0.0591 log([Fe³⁺][Fe²⁺])
  3. E = E° − 0.0591 log([Fe²⁺]/[Fe³⁺]) Correct answer.
  4. E = E° − 0.0591 log([Fe³⁺]/[Fe²⁺])

Explanation

At 298 K the Nernst equation is E = E° − (0.0591/n) log Q, with n = 1. The reaction quotient Q for this reduction is [Fe²⁺]/[Fe³⁺], because the product of the reduction appears in the numerator.

Question 22Electrochemistry

A conductivity cell has cell constant 1.25 cm⁻¹ and the measured resistance of a 0.05 M KCl solution is 250 Ω. What is the conductivity of the solution?

  1. 1.25 × 10⁻² S cm⁻¹
  2. 2.0 × 10⁻³ S cm⁻¹
  3. 3.125 × 10⁻³ S cm⁻¹
  4. 5.0 × 10⁻³ S cm⁻¹ Correct answer.

Explanation

Conductivity κ equals the cell constant divided by resistance: κ = G* / R = 1.25 / 250 = 5.0 × 10⁻³ S cm⁻¹. The cell constant converts measured conductance into conductivity.

Question 23Electrochemistry

Why does the molar conductivity of a strong electrolyte such as KCl decrease only slightly on dilution, whereas that of acetic acid increases sharply?

  1. Strong electrolytes recombine into molecules on dilution, releasing more ions
  2. Weak electrolytes are fully ionised at all concentrations but conduct poorly
  3. Strong electrolytes are almost fully ionised, so dilution mainly increases the volume per mole; weak electrolytes ionise more as they are diluted Correct answer.
  4. Dilution raises the viscosity, which increases ionic mobility in weak electrolytes

Explanation

A strong electrolyte is already almost completely dissociated, so dilution chiefly increases the volume containing one mole and slightly reduces ion–ion interaction. A weak electrolyte ionises progressively more as dilution lowers the concentration, so the number of charge carriers rises markedly and Λm increases steeply.

Question 24Electrochemistry

Which relation gives the dissociation constant of a weak monobasic acid in terms of molar conductivity at concentration c and at infinite dilution?

  1. Ka = c(Λ°m − Λm) / Λm
  2. Ka = cΛ°m² / [Λm(Λ°m − Λm)]
  3. Ka = Λm² / [c(Λ°m − Λm)]
  4. Ka = cΛm² / [Λ°m(Λ°m − Λm)] Correct answer.

Explanation

Ostwald's dilution law gives Ka = cα²/(1 − α), and the degree of dissociation is α = Λm/Λ°m. Substituting gives Ka = cΛm²/[Λ°m(Λ°m − Λm)]. This relation is used for weak electrolytes where α is small.

Question 25Electrochemistry

In a lead storage battery during discharge, which change occurs at the anode?

  1. PbO₂ is reduced to PbSO₄, consuming electrons
  2. PbSO₄ is oxidised to PbO₂, releasing electrons
  3. Pb is oxidised to PbSO₄, releasing electrons to the external circuit Correct answer.
  4. H₂SO₄ is reduced to H₂ gas at the anode

Explanation

During discharge the anode is the lead electrode, where Pb is oxidised to insoluble PbSO₄ with loss of two electrons. The PbO₂ electrode is the cathode and is reduced to PbSO₄. Both electrodes therefore become coated with lead sulfate as the battery discharges.

Question 26Electrochemistry

During the electrolysis of aqueous CuSO₄ using inert platinum electrodes, what is produced at the cathode and at the anode?

  1. Copper metal at the cathode and hydrogen gas at the anode
  2. Oxygen gas at the cathode and copper metal at the anode
  3. Copper metal at the cathode and oxygen gas at the anode Correct answer.
  4. Hydrogen gas at the cathode and copper metal at the anode

Explanation

At the cathode, Cu²⁺ is reduced in preference to water because its reduction potential is higher, so copper is deposited. At the inert anode, water is oxidised to give oxygen gas rather than sulfate being oxidised.

Question 27Electrochemistry

The standard cell potential of a cell is +1.10 V with two electrons transferred. What is the standard Gibbs energy change of the cell reaction?

  1. −212 kJ mol⁻¹ Correct answer.
  2. +212 kJ mol⁻¹
  3. −424 kJ mol⁻¹
  4. −106 kJ mol⁻¹

Explanation

Use ΔG° = −nFE° with n = 2 and F = 96500 C mol⁻¹. ΔG° = −2 × 96500 × 1.10 = −212300 J mol⁻¹, about −212 kJ mol⁻¹. The negative sign shows the reaction is spontaneous.

Question 28Electrochemistry

Which set of conditions defines standard electrode potential?

  1. Ion concentration 2 M, gas pressure 2 bar, temperature 298 K
  2. Ion concentration 0.1 M, gas pressure 1 atm, temperature 273 K
  3. Ion concentration 1 M, gas pressure 1 bar, temperature 298 K Correct answer.
  4. Ion concentration 1 M, gas pressure 1 atm, temperature 273 K

Explanation

Standard electrode potentials are measured with unit activity of dissolved species (approximately 1 M), gases at 1 bar, and a stated temperature, conventionally 298 K. These conditions make values comparable between half-cells.

Question 29Electrochemistry

A cell is constructed with E°cell = +0.46 V at 298 K and the reaction quotient Q = 1.0 × 10⁴ with n = 2. What is the cell potential?

  1. +0.34 V Correct answer.
  2. −0.34 V
  3. +0.46 V
  4. +0.58 V

Explanation

Apply E = E° − (0.0591/n) log Q. Here log Q = 4 and n = 2, so the correction is (0.0591/2) × 4 = 0.118 V. Thus E = 0.46 − 0.12 = +0.34 V, showing that a large product concentration lowers the cell potential.

Question 30Electrochemistry

In the electrolysis of molten sodium chloride, what are the products at the two electrodes, and why is water avoided?

  1. Chlorine at the cathode and sodium at the anode; water would freeze the melt
  2. Hydrogen at the cathode and oxygen at the anode; water would dissolve the NaCl
  3. Sodium at the cathode and chlorine at the anode; water would be reduced in preference to Na⁺ and would react with the sodium formed Correct answer.
  4. Sodium at the anode and chlorine at the cathode; water would oxidise chloride to chlorate

Explanation

In molten NaCl the only cations are Na⁺ and the only anions are Cl⁻, so sodium is deposited at the cathode and chlorine liberated at the anode. If water were present, hydrogen would be discharged instead of sodium because water is far more easily reduced than Na⁺, and the sodium metal would react violently with water.

Question 31Electrochemistry

For the cell Pt|H₂(g)|H⁺(aq)||Cu²⁺(aq)|Cu(s), the measured emf is 0.31 V when [Cu²⁺] = 0.01 M and [H⁺] = 1 M. What does this show about the dependence of emf on copper ion concentration?

  1. The emf rises above E°cell because lowering [Cu²⁺] shifts the equilibrium to the right
  2. The emf falls below E°cell because the cell reaction quotient is dominated by the low [Cu²⁺] in the denominator of Q Correct answer.
  3. The emf is unaffected because hydrogen remains at unit concentration
  4. The emf becomes negative because copper is no longer the cathode

Explanation

The cell reaction is Cu²⁺ + H₂ → Cu + 2H⁺, so Q = [H⁺]²/[Cu²⁺]. With [Cu²⁺] = 0.01 M and [H⁺] = 1 M, Q = 100 and log Q = 2, giving a correction of about 0.059 V that lowers the emf from 0.34 V to roughly 0.31 V. Lowering the oxidant concentration always reduces the driving force.

Question 32Electrochemistry

A current of 0.5 A is passed through aqueous AgNO₃ for 2 hours. What mass of silver is deposited at the cathode? (M(Ag) = 108 g mol⁻¹, F = 96500 C mol⁻¹)

  1. 4.03 g Correct answer.
  2. 0.40 g
  3. 8.06 g
  4. 2.02 g

Explanation

Charge passed is Q = It = 0.5 × 7200 = 3600 C, which corresponds to 3600/96500 = 0.0373 mol of electrons. Since Ag⁺ + e⁻ → Ag, the moles of silver equal the moles of electrons, so mass = 0.0373 × 108 = 4.03 g.

Question 33Chemical Kinetics

For a reaction, the rate of disappearance of A is 4.0 × 10⁻³ mol L⁻¹ s⁻¹. If the stoichiometry is 2A → products, what is the rate of the reaction?

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

Explanation

The rate of reaction is defined as −(1/2)d[A]/dt for the stoichiometry 2A → products, so the rate of disappearance of A is divided by its stoichiometric coefficient 2. Hence rate = 4.0 × 10⁻³ / 2 = 2.0 × 10⁻³ mol L⁻¹ s⁻¹.

Question 34Chemical Kinetics

The rate constant of a first-order reaction has units of:

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

Explanation

For a reaction of order n, the units of k are mol^(1−n) L^(n−1) s⁻¹. For a first-order reaction n = 1, so the units reduce to s⁻¹, which is independent of concentration.

Question 35Chemical Kinetics

For a zero-order reaction, the plot of concentration of reactant versus time gives:

  1. an exponential curve rising with time
  2. a straight line with positive slope
  3. a straight line with negative slope Correct answer.
  4. a curve whose slope increases with time

Explanation

The integrated zero-order rate law is [A] = [A]₀ − kt, which is linear in t with slope −k. Therefore concentration falls linearly with time, giving a straight line with a negative slope.

Question 36Chemical Kinetics

The half-life of a first-order reaction is 20 minutes. What fraction of the reactant remains after 60 minutes?

  1. 1/8 Correct answer.
  2. 1/2
  3. 1/16
  4. 1/4

Explanation

Sixty minutes equals three half-lives of 20 minutes each. After n half-lives the remaining fraction is (1/2)ⁿ, so after three half-lives it is (1/2)³ = 1/8.

Question 37Chemical Kinetics

Which statement about the order of a reaction is correct?

  1. Order can only be a positive whole number.
  2. Order is determined experimentally and may be zero, fractional or negative. Correct answer.
  3. Order is always equal to the number of molecules in the balanced equation.
  4. Order is always equal to the stoichiometric coefficient of the slowest step.

Explanation

Order is an experimental quantity defined by the rate law and can take zero, fractional or even negative values. It is not necessarily related to the stoichiometric coefficients of the overall balanced equation.

Question 38Chemical Kinetics

For the reaction A + B → products, doubling [A] while keeping [B] constant doubles the rate; doubling [B] while keeping [A] constant quadruples the rate. The overall order is:

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

Explanation

The observations give rate ∝ [A]¹ and rate ∝ [B]², so the rate law is rate = k[A][B]². The overall order is the sum of the exponents, 1 + 2 = 3.

Question 39Chemical Kinetics

A reaction has the rate law rate = k[A][B]. When [A] is doubled and [B] is halved, the initial rate becomes:

  1. half the original rate
  2. double the original rate
  3. four times the original rate
  4. unchanged Correct answer.

Explanation

Substituting the new concentrations gives rate' = k(2[A])([B]/2) = k[A][B], which equals the original rate. The doubling of A exactly cancels the halving of B because both appear to the first power.

Question 40Chemical Kinetics

For a first-order reaction, the time required for 99.9% completion is approximately how many times the half-life?

  1. 5
  2. 10 Correct answer.
  3. 100
  4. 3

Explanation

After n half-lives the fraction remaining is (1/2)ⁿ. For 99.9% completion the remaining fraction is 0.001, and (1/2)¹⁰ ≈ 9.8 × 10⁻⁴, so about 10 half-lives are needed.

Question 41Chemical Kinetics

The Arrhenius equation is written as k = A e^(−Ea/RT). A plot of ln k against 1/T for a reaction gives a straight line whose slope is:

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

Explanation

Taking natural logarithms gives ln k = ln A − (Ea/R)(1/T), which is linear in 1/T with slope −Ea/R. The intercept equals ln A, not the slope.

Question 42Chemical Kinetics

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

  1. about 110 kJ mol⁻¹
  2. about 55 kJ mol⁻¹ Correct answer.
  3. about 220 kJ mol⁻¹
  4. about 5.5 kJ mol⁻¹

Explanation

Using ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), we get 1.386 = (Ea/8.314)(1/300 − 1/320). The bracket equals 2.083 × 10⁻⁴, so Ea ≈ 1.386 × 8.314 / 2.083 × 10⁻⁴ ≈ 5.5 × 10⁴ J mol⁻¹, about 55 kJ mol⁻¹.

Question 43Chemical Kinetics

In the Arrhenius equation, the pre-exponential factor A is related to:

  1. the equilibrium constant of the reaction
  2. the enthalpy change of the reaction
  3. the average kinetic energy of reactant molecules
  4. the frequency of collisions with proper orientation Correct answer.

Explanation

The factor A, called the frequency factor, combines the collision frequency with the probability that molecules collide in the correct orientation. The exponential term e^(−Ea/RT) then accounts for the fraction with sufficient energy.

Question 44Chemical Kinetics

The rate of a reaction becomes 2.5 times when its temperature rises by 10 K near room temperature. This behaviour is best explained by:

  1. an increase in the activation energy with temperature
  2. a large increase in the fraction of collisions with energy exceeding Ea Correct answer.
  3. a decrease in the concentration of reactants
  4. a decrease in the collision frequency

Explanation

A modest rise in temperature increases the fraction of molecules with energy above Ea exponentially, so the rate rises sharply. This is why a small temperature change produces a large rate change near room temperature.

Question 45Chemical Kinetics

For the reaction 2NO(g) + O₂(g) → 2NO₂(g), the rate law found experimentally is rate = k[NO]²[O₂]. Which statement is correct?

  1. The reaction is third order overall and may proceed through more than one step. Correct answer.
  2. The reaction is elementary because the orders match the coefficients.
  3. The reaction is second order overall.
  4. The reaction must be zero order in oxygen.

Explanation

The overall order is the sum of exponents, 2 + 1 = 3, and the observed rate law matches the coefficients only coincidentally. A matching rate law does not prove a single-step mechanism, so the reaction may still be complex.

Question 46Chemical Kinetics

For a reaction A → products, a plot of 1/[A] against time is a straight line with a positive slope. The order of the reaction is:

  1. third
  2. zero
  3. second Correct answer.
  4. first

Explanation

The integrated second-order rate law is 1/[A] = 1/[A]₀ + kt, which is linear in time with slope k. A linear 1/[A] versus t plot therefore identifies second-order kinetics.

Question 47Chemical Kinetics

The activation energy of a reaction is lowered by a catalyst while the temperature is kept constant. Which quantity remains unchanged?

  1. the rate of the forward reaction
  2. the fraction of collisions with energy above the new Ea
  3. the equilibrium constant of the reaction Correct answer.
  4. the rate constant

Explanation

A catalyst provides an alternative path with lower activation energy but does not change the standard free energy change of the reaction, so the equilibrium constant is unaffected. The rate constant and the fraction of sufficiently energetic collisions both increase.

Question 48Chemical Kinetics

A first-order reaction is 50% complete in 30 minutes. What is the rate constant? (ln 2 = 0.693)

  1. 1.15 × 10⁻² min⁻¹
  2. 4.62 × 10⁻² min⁻¹
  3. 2.31 × 10⁻² min⁻¹ Correct answer.
  4. 3.47 × 10⁻² min⁻¹

Explanation

For a first-order reaction k = ln2 / t₁/₂ = 0.693 / 30 min. This gives k ≈ 2.31 × 10⁻² min⁻¹, which is independent of the initial concentration.

Question 49Chemical Kinetics

For an elementary bimolecular reaction, the molecularity is 2. If the concentration of each reactant is doubled, the rate increases by a factor of:

  1. 8
  2. 4 Correct answer.
  3. 2
  4. 3

Explanation

An elementary bimolecular step is second order overall, with rate = k[A][B]. Doubling both concentrations multiplies the rate by 2 × 2 = 4.

Question 50Chemical Kinetics

Two reactions have activation energies of 40 kJ mol⁻¹ and 80 kJ mol⁻¹ at the same temperature. Which statement comparing their temperature sensitivity is correct?

  1. The reaction with Ea = 40 kJ mol⁻¹ shows the greater fractional increase in k.
  2. The reaction with Ea = 40 kJ mol⁻¹ has a larger rate constant at every temperature.
  3. The reaction with Ea = 80 kJ mol⁻¹ shows a greater fractional increase in k for the same rise in temperature. Correct answer.
  4. Both reactions show the same fractional change in k for a given temperature rise.

Explanation

The fractional change in k with temperature depends on Ea through d(ln k)/dT = Ea/RT². A larger Ea therefore gives a larger fractional increase in k for the same temperature rise, even though its absolute k may be smaller.