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Assam ASSEB HS Chemistry study materials
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Assam ASSEB HS Chemistry practice questions

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Question 1Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

The number of atoms contained in one face-centred cubic unit cell of a metal is:

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

Explanation

In an fcc cell the eight corner atoms contribute 1/8 each, giving 1 atom, and the six face-centred atoms contribute 1/2 each, giving 3 atoms. The total is therefore 4 atoms per unit cell, the same count as in a ccp arrangement.

Question 2Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

The percentage of the total volume of a body-centred cubic unit cell that is occupied by the atoms is approximately:

  1. 68% Correct answer.
  2. 74%
  3. 52.4%
  4. 34%

Explanation

In a bcc cell the atoms touch along the body diagonal, so 4r = √3 a and r = √3a/4. Substituting this into the packing fraction (2 × 4/3 πr³)/a³ gives 0.68, that is about 68%.

Question 3Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

Colligative properties of a dilute solution depend mainly on which feature of the dissolved solute?

  1. the molar mass of the solute
  2. the number of solute particles Correct answer.
  3. the chemical nature of the solute
  4. the colour of the solute

Explanation

Colligative properties such as relative lowering of vapour pressure, boiling point elevation and osmotic pressure are governed by the number of solute particles present in a fixed amount of solvent. The identity, colour and molar mass of the particles do not enter the equations except through the particle count.

Question 4Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

For a metal that crystallises in a face-centred cubic lattice, the edge length a of the unit cell in terms of the atomic radius r is:

  1. 2√2 r Correct answer.
  2. √3 r
  3. 2r
  4. 4r/√3

Explanation

In an fcc lattice the atoms touch along the face diagonal, whose length is a√2. This diagonal spans four atomic radii, so a√2 = 4r and hence a = 4r/√2 = 2√2 r, about 2.83r.

Question 5Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

For a solution of a non-volatile solute in a volatile solvent, the relative lowering of vapour pressure (p° − p)/p° is equal to:

  1. the mass fraction of the solute
  2. the molality of the solution
  3. the mole fraction of the solute Correct answer.
  4. the mole fraction of the solvent

Explanation

Raoult's law for a non-volatile solute gives p = p° x_solvent, so (p° − p)/p° = 1 − x_solvent. Since the mole fractions sum to one, this equals x_solute, the mole fraction of the solute.

Question 6Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

Acetic acid dissolved in benzene dimerises almost completely through hydrogen bonding. For a solution in which association to dimers is complete, the van't Hoff factor i is:

  1. 0.5 Correct answer.
  2. 2
  3. 1
  4. 0

Explanation

The van't Hoff factor is the ratio of the number of particles actually present to the number of formula units dissolved. When two monomer units combine into one dimer, the particle count is halved, so i = 1/2 = 0.5, which is less than one, as expected for association.

Question 7Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

An aqueous solution is 0.1 molal in a non-electrolyte and the freezing point constant of water is 1.86 K kg mol⁻¹. The depression in freezing point is:

  1. 1.86 K
  2. 0.186 K Correct answer.
  3. 0.093 K
  4. 18.6 K

Explanation

For a non-electrolyte the depression is given by ΔTf = Kf × m. Substituting Kf = 1.86 K kg mol⁻¹ and m = 0.1 mol kg⁻¹ gives ΔTf = 0.186 K, so the solution freezes at −0.186 °C.

Question 8Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

Which defect in an ionic crystal causes its measured density to be lower than the value calculated from the ideal lattice?

  1. Schottky defect Correct answer.
  2. interstitial impurity defect
  3. Frenkel defect
  4. metal excess defect

Explanation

A Schottky defect consists of equal numbers of cations and anions missing from their lattice sites, so the unit cell loses mass without gaining volume. The calculated density therefore falls, whereas defects that only displace or add atoms leave the mass essentially unchanged.

Question 9Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

An ionic solid has its anions in a ccp arrangement and its cations occupying half of the octahedral voids. The formula of the solid is:

  1. A₄B
  2. A₂B
  3. AB₂ Correct answer.
  4. AB

Explanation

A ccp array of anions contributes 4 anions and produces 4 octahedral voids per unit cell. Half of these voids are occupied, giving 2 cations, so the cation to anion ratio is 2:4, that is 1:2, and the formula is AB₂.

Question 10Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

A mixture of ethanol and acetone shows a positive deviation from Raoult's law. The best explanation is that:

  1. unlike-molecule hydrogen bonding is stronger than in the pure liquids
  2. the solute is completely non-volatile
  3. unlike-molecule hydrogen bonding is weaker than in the pure liquids Correct answer.
  4. the two components do not interact at all

Explanation

In pure ethanol the molecules are held by strong hydrogen bonds, but dilution with acetone disrupts this network because the new unlike interactions are weaker. The molecules therefore escape more readily than Raoult's law predicts, giving a higher vapour pressure and a positive deviation.

Question 11Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

In a simple cubic unit cell, each atom is in contact with how many nearest neighbouring atoms?

  1. 6 Correct answer.
  2. 8
  3. 4
  4. 12

Explanation

A simple cubic lattice has atoms only at the eight corners, and each corner atom touches its nearest neighbours along the three cube axes. Along each axis there are two contacts, so the coordination number is 3 × 2 = 6.

Question 12Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

Which concentration term for an aqueous solution changes when the solution is warmed from 25 °C to 35 °C without any loss of solute or solvent?

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

Explanation

Molarity is defined as moles of solute per litre of solution, and the volume of the solution expands as temperature rises. Therefore its numerical value changes on warming, whereas molality, mole fraction and mass percentage are based on masses or mole ratios and remain essentially unchanged.

Question 13Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

A metal crystallises in a face-centred cubic lattice with edge length 400 pm. If its molar mass is 40 g mol⁻¹, its density is closest to (N_A = 6.022 × 10²³ mol⁻¹)

  1. 16.6 g cm⁻³
  2. 4.15 g cm⁻³ Correct answer.
  3. 2.07 g cm⁻³
  4. 8.30 g cm⁻³

Explanation

An fcc unit cell contains Z = 4 atoms, and 400 pm equals 4.00 × 10⁻⁸ cm, so the cell volume is (4.00 × 10⁻⁸)³ = 6.40 × 10⁻²³ cm³. Density = ZM/(N_A V) = (4 × 40)/(6.022 × 10²³ × 6.40 × 10⁻²³) ≈ 4.15 g cm⁻³.

Question 14Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

At a given temperature, pure benzene has a vapour pressure of 120 torr and pure toluene has a vapour pressure of 40 torr. An ideal liquid mixture has a benzene mole fraction of 0.25. What is the total vapour pressure of the mixture?

  1. 50 torr
  2. 60 torr Correct answer.
  3. 80 torr
  4. 40 torr

Explanation

For an ideal solution, Raoult's law gives p_total = x_benzene p°_benzene + x_toluene p°_toluene. With x_toluene = 0.75, p_total = (0.25 × 120) + (0.75 × 40) = 30 + 30 = 60 torr.

Question 15Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

In a close-packed arrangement containing N spheres, the number of tetrahedral voids is

  1. 4N
  2. 2N Correct answer.
  3. N
  4. N/2

Explanation

Each sphere in a close-packed structure is associated with two tetrahedral voids and one octahedral void. Therefore an array of N spheres contains 2N tetrahedral voids.

Question 16Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

For a body-centred cubic unit cell with edge length a, the atomic radius r is given by

  1. r = √3 a/2
  2. r = √2 a/4
  3. r = a/2
  4. r = √3 a/4 Correct answer.

Explanation

In a bcc cell the atoms touch along the body diagonal, whose length is √3 a. The body diagonal equals four atomic radii, so 4r = √3 a and r = √3 a/4.

Question 17Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

An aqueous solution containing 5.0 g of a non-electrolyte solute per litre exerts an osmotic pressure of 2.0 atm at 300 K. What is the molar mass of the solute? (R = 0.0821 L atm K⁻¹ mol⁻¹)

  1. 246.4 g mol⁻¹
  2. 30.8 g mol⁻¹
  3. 123.2 g mol⁻¹
  4. 61.6 g mol⁻¹ Correct answer.

Explanation

Osmotic pressure is given by π = CRT, so C = π/RT = 2.0/(0.0821 × 300) ≈ 0.0812 mol L⁻¹. Since one litre contains 5.0 g of solute, molar mass = 5.0/0.0812 ≈ 61.6 g mol⁻¹.

Question 18Solid State and Solutions (unit cell, packing, colligative properties, Raoult's law)

Assuming complete dissociation, what is the boiling point elevation of a 0.10 molal aqueous solution of K₃PO₄? (K_b for water = 0.512 K kg mol⁻¹)

  1. 0.051 K
  2. 0.205 K Correct answer.
  3. 0.154 K
  4. 0.102 K

Explanation

K₃PO₄ dissociates completely into four ions (3 K⁺ and 1 PO₄³⁻), so its van't Hoff factor is i = 4. Then ΔT_b = i K_b m = 4 × 0.512 × 0.10 ≈ 0.205 K.

Question 19Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

The SI unit of molar conductivity, Λm, is

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

Explanation

Molar conductivity is defined as Λm = κ/c, where κ is conductivity in S m⁻¹ and c is concentration in mol m⁻³. Dividing S m⁻¹ by mol m⁻³ gives S m² mol⁻¹, which is the SI unit. Values are often quoted in S cm² mol⁻¹, but that is not the SI unit.

Question 20Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

In a galvanic Daniell cell, the salt bridge or KCl bridge is used mainly to

  1. complete the circuit and keep the half-cells electrically neutral Correct answer.
  2. prevent any movement of ions between the anode and cathode compartments
  3. increase the standard electrode potential of the copper half-cell
  4. carry electrons from zinc to copper through the solution

Explanation

The salt bridge allows ions to migrate into or out of each half-cell so that charge does not build up, and it completes the internal circuit. Electrons travel through the external wire, not through the bridge. The bridge does not change the standard electrode potentials.

Question 21Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

Given Λ°m(CH3COONa) = 91.0, Λ°m(HCl) = 426.2 and Λ°m(NaCl) = 126.5 S cm² mol⁻¹, the limiting molar conductivity of acetic acid, Λ°m(CH3COOH), is

  1. 217.5 S cm² mol⁻¹
  2. 208.7 S cm² mol⁻¹
  3. 390.7 S cm² mol⁻¹ Correct answer.
  4. 643.7 S cm² mol⁻¹

Explanation

By Kohlrausch's law, Λ°m(CH3COOH) = Λ°m(CH3COO⁻) + Λ°m(H⁺). This equals Λ°m(CH3COONa) + Λ°m(HCl) − Λ°m(NaCl). Substitution gives 91.0 + 426.2 − 126.5 = 390.7 S cm² mol⁻¹.

Question 22Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

For the electrode reaction Zn²⁺(aq) + 2e⁻ ⇌ Zn(s), increasing [Zn²⁺] at 298 K will

  1. shift the reduction potential to a less negative value Correct answer.
  2. reverse the sign of the standard reduction potential
  3. make the reduction potential more negative
  4. leave the reduction potential unchanged because Zn is a solid

Explanation

The Nernst equation can be written E = E° + (0.0591/2) log[Zn²⁺] for this half-cell. Increasing [Zn²⁺] therefore raises E, making it less negative. The standard potential E° itself remains unchanged.

Question 23Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

Which statement correctly describes the variation of molar conductivity with concentration at 298 K?

  1. Λm increases for strong electrolytes and decreases for weak electrolytes as concentration is increased
  2. Λm is independent of concentration for both strong and weak electrolytes at all finite concentrations
  3. Λm decreases slowly with √c for strong electrolytes and rises sharply on dilution for weak ones Correct answer.
  4. Λm becomes zero at infinite dilution for weak electrolytes because ionization is complete

Explanation

Strong electrolytes are essentially fully ionized, and interionic attractions reduce Λm as concentration increases, giving an approximately linear decrease with √c. Weak electrolytes are only partly ionized at higher concentrations; dilution increases ionization, so Λm rises steeply.

Question 24Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

In the electrochemical rusting of iron in neutral aerated water, the cathodic reaction is

  1. 2H⁺(aq) + 2e⁻ → H2(g), the cathodic reaction in acid
  2. Fe(s) → Fe²⁺(aq) + 2e⁻, the anodic oxidation of iron
  3. Fe²⁺(aq) → Fe³⁺(aq) + e⁻, a secondary oxidation
  4. O2(g) + 2H2O(l) + 4e⁻ → 4OH⁻(aq) Correct answer.

Explanation

In neutral aerated water, oxygen dissolved in the water is reduced at cathodic regions: O2 + 2H2O + 4e⁻ → 4OH⁻. Iron is simultaneously oxidized at anodic regions to Fe²⁺. The hydroxide and Fe²⁺ then lead to hydrated iron(III) oxide rust.

Question 25Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

For the cell reaction Cu²⁺(aq) + Zn(s) → Cu(s) + Zn²⁺(aq), E°cell = 1.10 V. At standard conditions the reaction is

  1. non-spontaneous with ΔG° = +212 kJ mol⁻¹
  2. spontaneous with ΔG° = −106 kJ mol⁻¹
  3. at equilibrium with ΔG° = 0
  4. spontaneous with ΔG° = −212 kJ mol⁻¹ Correct answer.

Explanation

A positive E°cell means ΔG° = −nFE° is negative, so the reaction is spontaneous under standard conditions. Here n = 2 because two electrons are transferred. Thus ΔG° ≈ −2 × 96500 × 1.10 J mol⁻¹ = −212 kJ mol⁻¹.

Question 26Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

A current of 2.00 A is passed through molten Al2O3 for 1.50 hours. Using 1 F = 96500 C mol⁻¹, the mass of aluminium deposited is closest to

  1. 1.51 g
  2. 3.02 g
  3. 1.01 g Correct answer.
  4. 0.504 g

Explanation

The charge passed is Q = It = 2.00 × 1.50 × 3600 = 10800 C. This is 10800/96500 = 0.112 mol of electrons. Since Al³⁺ requires 3 electrons per aluminium atom, the amount of Al is 0.112/3 = 0.0373 mol, giving 0.0373 × 27 ≈ 1.01 g.

Question 27Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

The cell constant of a conductivity cell is

  1. the conductance of a solution of exactly 1 mol L⁻¹ concentration
  2. the area of the electrodes divided by their separation
  3. the resistance of the cell multiplied by the molar conductivity
  4. the distance between the electrodes divided by their area Correct answer.

Explanation

For a conductivity cell, the cell constant is l/A, where l is the distance between the electrodes and A is their area. Conductivity is then κ = cell constant / R. Its unit is therefore m⁻¹ or cm⁻¹.

Question 28Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

A hydrogen electrode is dipped into a solution of unknown pH and coupled with a standard hydrogen electrode at 298 K. The measured cell emf is 0.236 V with the unknown electrode negative. The pH of the solution is closest to

  1. 8.0
  2. 2.0
  3. 4.0 Correct answer.
  4. 6.0

Explanation

For a hydrogen electrode coupled to the standard hydrogen electrode at 298 K, Ecell = 0.0591 × pH when the unknown electrode is the negative electrode. Rearranging gives pH = 0.236/0.0591 ≈ 4.0. The sign confirms that the unknown solution is less acidic than 1 M H⁺.

Question 29Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

On progressive dilution of a solution of a strong electrolyte at constant temperature, the specific conductance (conductivity) of the solution

  1. increases, because dissociation increases
  2. decreases, as ions per unit volume fall Correct answer.
  3. first rises, then falls through a maximum
  4. remains fixed by the cell constant

Explanation

Specific conductance is the conductance of a unit volume of the solution, so it depends on the number of ions present per unit volume. On dilution the ion concentration falls, so conductivity decreases, even though molar conductivity increases as dissociation becomes more complete.

Question 30Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

For the electrode reaction Fe³⁺(aq) + e⁻ → Fe²⁺(aq) at 298 K, the Nernst equation for the reduction potential is

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

Explanation

For a reduction half-reaction, E = E° − (0.0591/n) log([reduced form]/[oxidised form]) at 298 K. Here Fe²⁺ is the reduced form, Fe³⁺ is the oxidised form and n = 1, so E = E° − 0.0591 log([Fe²⁺]/[Fe³⁺]).

Question 31Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

The EMF at 298 K of the concentration cell Cu(s)|Cu²⁺(0.010 M)||Cu²⁺(1.0 M)|Cu(s) is

  1. 0.118 V
  2. 0.000 V
  3. 0.030 V
  4. 0.059 V Correct answer.

Explanation

For a concentration cell, E = (0.0591/n) log(C_cathode/C_anode). With n = 2, C_cathode = 1.0 M and C_anode = 0.010 M, E = (0.0591/2) log(100) = 0.0591 V, so the cell gives about 0.059 V.

Question 32Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

For a cell reaction with n = 2 and E°cell = +0.46 V at 298 K, the standard Gibbs energy change ΔG° is closest to (F = 96500 C mol⁻¹)

  1. −177.6 kJ mol⁻¹
  2. +88.8 kJ mol⁻¹
  3. −44.4 kJ mol⁻¹
  4. −88.8 kJ mol⁻¹ Correct answer.

Explanation

The relation is ΔG° = −nFE°cell. Substituting n = 2, F = 96500 C mol⁻¹ and E° = 0.46 V gives ΔG° = −2 × 96500 × 0.46 = −88,780 J, that is about −88.8 kJ mol⁻¹. The negative sign shows the reaction is spontaneous under standard conditions.

Question 33Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

In the conductometric titration of a strong acid (HCl) with a strong base (NaOH), the conductivity of the solution

  1. falls to a minimum at equivalence and then rises Correct answer.
  2. stays unchanged until equivalence and then rises
  3. falls continuously even beyond equivalence
  4. rises steadily up to equivalence and then falls

Explanation

During the titration, highly mobile H⁺ ions are replaced by the less mobile Na⁺ ions, so the conductivity falls. After the equivalence point, excess OH⁻ ions, which have high mobility, are added and the conductivity rises again, giving a V-shaped curve.

Question 34Electrochemistry (conductance, Kohlrausch law, Nernst equation, cells, corrosion)

For the reaction Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s) with E°cell = 1.10 V at 298 K, the equilibrium constant K is closest to

  1. 1.6 × 10^74
  2. 1.6 × 10^18
  3. 3.2 × 10^37
  4. 1.6 × 10^37 Correct answer.

Explanation

At equilibrium, ΔG° = −nFE°cell = −RT ln K, so log K = nE°cell/0.0591 at 298 K. Substituting n = 2 and E°cell = 1.10 V gives log K = 2 × 1.10/0.0591 ≈ 37.2, hence K ≈ 1.6 × 10^37.

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

For the reaction A + B → products, the experimentally determined rate law is rate = k[A][B]^2. What is the overall order of the reaction?

  1. Third order Correct answer.
  2. First order
  3. Second order
  4. Zero order

Explanation

The overall order is the sum of the exponents in the experimental rate law. Here the exponent on A is 1 and on B is 2, so 1 + 2 = 3, making the reaction third order overall.

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

For a first-order reaction, the unit of the rate constant k is:

  1. L mol^-1 s^-1
  2. s^-1 Correct answer.
  3. mol L^-1 s^-1
  4. L^2 mol^-2 s^-1

Explanation

For a first-order reaction, rate = k[A], so k = rate/[A]. Substituting the units of rate (mol L^-1 s^-1) and concentration (mol L^-1) gives k in s^-1.

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

According to collision theory, which condition must be satisfied for a collision between reactant molecules to be effective in producing products?

  1. The molecules must remain in contact for a long time after every collision.
  2. The molecules must collide only when their masses and charges are equal.
  3. The molecules must collide in the presence of a catalyst at the container wall.
  4. The molecules must collide with sufficient energy and proper orientation. Correct answer.

Explanation

Collision theory states that only collisions with energy equal to or greater than the activation energy and with a favourable orientation can break and form bonds. A catalyst changes the reaction pathway, but it does not replace these two requirements.

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

A first-order reaction is 75% complete in 60 minutes. What is its half-life?

  1. 15 min
  2. 30 min Correct answer.
  3. 60 min
  4. 45 min

Explanation

If 75% has reacted, 25% remains, and 25% is (1/2)^2 of the initial amount. Therefore the 60 minutes correspond to two half-lives, so each half-life is 30 minutes.

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

For a reaction obeying the Arrhenius equation, a plot of ln k against 1/T gives a straight line. What does the slope of this line represent?

  1. -Ea/(RT)
  2. ln A
  3. Ea/R
  4. -Ea/R Correct answer.

Explanation

The Arrhenius equation can be written as ln k = ln A - Ea/(RT). Comparing this with y = mx + c shows that the slope of ln k versus 1/T is -Ea/R.

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

The reaction 2NO(g) + O2(g) → 2NO2(g) is proposed to proceed by a fast equilibrium 2NO ⇌ N2O2 followed by the slow step N2O2 + O2 → 2NO2. What rate law is predicted?

  1. rate = k[N2O2][O2]
  2. rate = k[NO]^2[O2] Correct answer.
  3. rate = k[NO][O2]
  4. rate = k[NO]^2

Explanation

The slow step controls the rate, so rate = k2[N2O2][O2]. From the fast equilibrium, [N2O2] = K[NO]^2; substituting this into the slow-step rate gives rate = k[NO]^2[O2].

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

For a zero-order reaction, which plot is a straight line with slope equal to -k?

  1. ln[A] versus t
  2. k versus t
  3. [A] versus t Correct answer.
  4. 1/[A] versus t

Explanation

The integrated rate equation for a zero-order reaction is [A] = [A]0 - kt. This has the form y = c + mx with y = [A], x = t, and slope -k.

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

The rate of a reaction doubles when the temperature is raised from 300 K to 310 K. Using the Arrhenius equation, what is the approximate activation energy? (R = 8.314 J K^-1 mol^-1, ln 2 = 0.693)

  1. 26.8 kJ mol^-1
  2. 107 kJ mol^-1
  3. 5.36 kJ mol^-1
  4. 53.6 kJ mol^-1 Correct answer.

Explanation

Using ln(k2/k1) = (Ea/R)(1/T1 - 1/T2), substitute ln 2 = 0.693 and the two temperatures. This gives Ea ≈ 0.693 × 8.314 / (1/300 - 1/310) ≈ 53.6 kJ mol^-1.

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

A first-order reaction has a half-life of 20 minutes. Approximately how many half-lives are needed for the reaction to be 99.9% complete?

  1. 5
  2. 3
  3. 10 Correct answer.
  4. 7

Explanation

For 99.9% completion, the remaining fraction is 0.001. Since each half-life reduces the amount by a factor of 1/2, solve (1/2)^n = 0.001; because 2^10 = 1024, about ten half-lives are needed.

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

Which statement correctly distinguishes molecularity from order of reaction?

  1. Molecularity is always equal to the overall order even for a complex reaction with several steps.
  2. Order is defined only for elementary reactions, whereas molecularity is defined for both elementary and complex reactions.
  3. Molecularity counts species in an elementary step; order comes from the experimental rate law. Correct answer.
  4. Molecularity can be fractional in some reactions, whereas the order must always be a whole number for every reaction.

Explanation

Molecularity is a theoretical count of the reacting species in a single elementary step and is always a small whole number. Order is an experimental quantity obtained by summing the exponents in the rate law and may be zero, fractional, or negative.

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

For the reaction 2N2O5(g) → 4NO2(g) + O2(g), the rate of the reaction is expressed in terms of the rates of change of concentration of the reactant and the products. Which expression correctly represents the rate of this reaction?

  1. −2d[N2O5]/dt = 4d[NO2]/dt = d[O2]/dt
  2. −(1/2)d[N2O5]/dt = (1/4)d[NO2]/dt = d[O2]/dt Correct answer.
  3. −d[N2O5]/dt = (1/4)d[NO2]/dt = (1/2)d[O2]/dt
  4. −d[N2O5]/dt = d[NO2]/dt = d[O2]/dt

Explanation

The rate of a reaction is obtained by dividing the rate of change of concentration of each species by its stoichiometric coefficient in the balanced equation. The reactant term carries a negative sign because N2O5 is consumed, while product terms are positive. Hence rate = −(1/2)d[N2O5]/dt = (1/4)d[NO2]/dt = d[O2]/dt.

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

A small amount of manganese dioxide is added to a flask in which hydrogen peroxide is decomposing. The rate of oxygen evolution increases sharply. The catalyst brings this about mainly by:

  1. providing an alternative pathway that has a lower activation energy Correct answer.
  2. increasing the enthalpy change of the decomposition reaction
  3. raising the average kinetic energy of the peroxide molecules
  4. shifting the equilibrium position further towards the products

Explanation

A catalyst offers a new reaction pathway whose activation energy is lower, so a larger fraction of collisions at the same temperature is effective and the rate rises. It is regenerated at the end and does not alter the enthalpy change or the position of equilibrium of the overall reaction.

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

A reaction obeys the rate law rate = k[A]^2. Which graph of a measured quantity against time gives a straight line whose slope is equal to k?

  1. [A]^2 against time
  2. 1/[A] against time Correct answer.
  3. ln[A] against time
  4. [A] against time

Explanation

For a second-order reaction the integrated rate equation is 1/[A] = 1/[A]0 + kt, which is of the form y = c + kx. A plot of 1/[A] against t is therefore linear and its slope equals the rate constant k.

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

In a kinetics experiment on the reaction X + Y → Z, doubling [X] while holding [Y] constant makes the initial rate four times larger, whereas doubling [Y] while holding [X] constant leaves the initial rate unchanged. The rate law consistent with these data is:

  1. rate = k[X]^2 Correct answer.
  2. rate = k[X][Y]^2
  3. rate = k[X]^2[Y]
  4. rate = k[X][Y]

Explanation

The rate depends on the square of [X] because doubling [X] multiplies the rate by 4, and it is independent of [Y] because changing [Y] has no effect, so the order in Y is zero. Combining these gives rate = k[X]^2[Y]^0 = k[X]^2.

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

In the Arrhenius equation k = A e^(−Ea/RT), the pre-exponential factor A is best described as:

  1. the fraction of collisions whose energy is greater than the activation energy
  2. the average kinetic energy of the reacting molecules at temperature T
  3. the frequency factor, related to the number of collisions with the proper orientation per unit time Correct answer.
  4. the rate of the reaction when each reactant has unit concentration

Explanation

The factor A is the frequency factor or pre-exponential factor; it reflects how often molecules collide and how often those collisions have a suitable orientation. The exponential term e^(−Ea/RT), not A, supplies the fraction of collisions with enough energy to react.

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

The rate constant of a reaction is 2.0 × 10^−3 s^−1 at 300 K and 3.2 × 10^−2 s^−1 at 320 K. Using R = 8.314 J K^−1 mol^−1 and ln 16 = 2.77, the activation energy is closest to:

  1. 111 kJ mol^−1 Correct answer.
  2. 166 kJ mol^−1
  3. 55 kJ mol^−1
  4. 222 kJ mol^−1

Explanation

Using ln(k2/k1) = (Ea/R)(1/T1 − 1/T2) with k2/k1 = 16 gives Ea = (8.314 × 2.77)/(1/300 − 1/320). The reciprocal difference is 2.08 × 10^−4 K^−1, so Ea ≈ 1.1 × 10^5 J mol^−1, that is about 111 kJ mol^−1.