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CHSE Odisha Elective Chemistry study materials
CHSE Odisha Elective Chemistry study resources

CHSE Odisha Elective Chemistry practice questions

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Question 1Solid State and Solutions

How many atoms are present per unit cell in a face-centred cubic (fcc) crystal lattice?

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

Explanation

In a face-centred cubic unit cell, each of the eight corner atoms contributes one-eighth and each of the six face-centred atoms contributes one-half to the cell. The total is 8(1/8)+6(1/2)=4 atoms per unit cell.

Question 2Solid State and Solutions

Which concentration unit remains unchanged when the temperature of a solution is changed?

  1. Formality
  2. Molarity
  3. Normality
  4. Molality Correct answer.

Explanation

Molality is defined using the mass of solvent, so it does not depend on the volume of the solution. Molarity, normality and formality are volume-based and therefore change when temperature alters the solution volume.

Question 3Solid State and Solutions

In a sodium chloride crystal, each Na+ ion is surrounded by how many Cl- ions?

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

Explanation

Sodium chloride has the rock-salt structure, in which each Na+ is octahedrally surrounded by six Cl- ions and each Cl- by six Na+ ions. The coordination number is therefore 6:6.

Question 4Solid State and Solutions

For a metal crystallising in a body-centred cubic lattice, if a is the unit cell edge and r is the atomic radius, the correct relation is

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

Explanation

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

Question 5Solid State and Solutions

For a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the

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

Explanation

For a solution containing a non-volatile solute, Raoult's law gives (p°−p)/p° = x_solute, the mole fraction of solute. This quantity is the relative lowering of vapour pressure.

Question 6Solid State and Solutions

A metal crystallises in a body-centred cubic unit cell with edge length 4.0×10^-8 cm and molar mass 56 g mol^-1. Using N_A = 6.022×10^23 mol^-1, its density is approximately

  1. 1.45 g cm^-3
  2. 11.6 g cm^-3
  3. 5.81 g cm^-3
  4. 2.91 g cm^-3 Correct answer.

Explanation

For a bcc unit cell, Z = 2. The density is d = ZM/(N_A a^3). Substituting gives d = (2×56)/(6.022×10^23×(4.0×10^-8)^3) ≈ 2.91 g cm^-3.

Question 7Solid State and Solutions

Which 0.1 mol kg^-1 aqueous solution is expected to have the highest boiling point, assuming complete dissociation?

  1. 0.1 m CaCl2 Correct answer.
  2. 0.1 m NaCl
  3. 0.1 m glucose
  4. 0.1 m urea

Explanation

Boiling-point elevation is proportional to the total number of solute particles in solution. CaCl2 gives three ions per formula unit, NaCl gives two, and glucose and urea give one each, so 0.1 m CaCl2 has the largest effective particle concentration.

Question 8Solid State and Solutions

X-rays of wavelength 154 pm are reflected from a crystal plane at a glancing angle of 30°. For first-order reflection, the interplanar spacing is

  1. 154 pm Correct answer.
  2. 308 pm
  3. 77 pm
  4. 267 pm

Explanation

Bragg's law is nλ = 2d sinθ. For first-order reflection n = 1 and sin30° = 0.5, so d = 154/(2×0.5) pm = 154 pm.

Question 9Solid State and Solutions

A 0.10 M aqueous solution of a weak monobasic acid shows a van't Hoff factor of 1.05. Its degree of dissociation is

  1. 0.95
  2. 0.05 Correct answer.
  3. 0.10
  4. 0.50

Explanation

For a weak monobasic acid HA, dissociation produces two particles from one, so i = 1 + α. Given i = 1.05, α = i − 1 = 0.05.

Question 10Solid State and Solutions

A binary liquid mixture showing positive deviation from Raoult's law is most likely to form

  1. a solution obeying Henry's law over the entire range
  2. a minimum-boiling azeotrope Correct answer.
  3. a maximum-boiling azeotrope
  4. an ideal solution

Explanation

Positive deviation means the components escape more readily than in an ideal solution, raising vapour pressure and lowering boiling point. Such a mixture forms a minimum-boiling azeotrope at a composition where the liquid and vapour have the same composition.

Question 11Solid State and Solutions

In the classification of crystalline solids, the number of Bravais lattices possible in three dimensions is

  1. 32
  2. 14 Correct answer.
  3. 230
  4. 7

Explanation

The seven crystal systems combine with the centring patterns (primitive, body-centred, face-centred, base-centred) that are compatible with lattice symmetry to give exactly 14 distinct Bravais lattices. The larger numbers 32 and 230 refer to crystallographic point groups and space groups respectively, not to lattices.

Question 12Solid State and Solutions

Which point defect is produced when equal numbers of cations and anions are missing from their normal sites in an ionic crystal?

  1. Metal excess defect due to anionic vacancies
  2. Frenkel defect
  3. Interstitial defect caused by a foreign atom
  4. Schottky defect Correct answer.

Explanation

A Schottky defect consists of a cation vacancy together with an anion vacancy created in equal numbers, so electrical neutrality is preserved. Because whole lattice sites are left empty, the mass per unit volume falls and the density of the crystal decreases.

Question 13Solid State and Solutions

Which colligative property is most suitable for determining the molar mass of macromolecules such as proteins?

  1. Depression of freezing point
  2. Osmotic pressure Correct answer.
  3. Relative lowering of vapour pressure
  4. Elevation of boiling point

Explanation

Osmotic pressure is measurable at ordinary temperature and is appreciable even in quite dilute solutions, so it gives a reliable molar mass for substances of very large molar mass. Boiling-point elevation and freezing-point depression would be too small to measure accurately for such solutes.

Question 14Solid State and Solutions

The packing fraction, that is the fraction of the total volume occupied by spheres, in a body-centred cubic unit cell is closest to

  1. 52.4 %
  2. 68 % Correct answer.
  3. 90.7 %
  4. 74 %

Explanation

In a bcc cell the central sphere touches the eight corner spheres along the body diagonal, giving 4r = √3 a. Substituting this relation into the volume expressions yields a packing fraction of √3π/8, which is about 68 %.

Question 15Solid State and Solutions

In a cubic close-packed array built from N atoms, the number of octahedral voids present is

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

Explanation

Each atom in a close-packed structure is associated with one octahedral void and two tetrahedral voids, so N atoms generate N octahedral voids. In an fcc cell this appears as four octahedral voids, one at the body centre and one on each edge, for four atoms.

Question 16Solid State and Solutions

X-rays of wavelength 0.500 Å are incident on a crystal whose (100) planes are separated by 2.00 Å. The highest order of Bragg reflection that can be observed from these planes is

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

Explanation

Bragg's law is nλ = 2d sin θ, and since sin θ cannot exceed 1 the maximum order satisfies n ≤ 2d/λ. Substituting gives n ≤ (2 × 2.00)/0.500 = 8, so the eighth order is the highest possible before sin θ exceeds unity.

Question 17Solid State and Solutions

At the same temperature, the Henry's law constants of two gases in water are KH(A) = 4.0 × 10^4 bar and KH(B) = 1.0 × 10^5 bar. At equal partial pressures, which statement is correct?

  1. A is more soluble in water than B Correct answer.
  2. B is more soluble in water than A
  3. A and B dissolve to the same extent in water
  4. The solubilities cannot be compared without the molar masses

Explanation

Henry's law gives p = KH x, so at a fixed partial pressure the mole fraction dissolved is inversely proportional to KH. Gas A has the smaller Henry's law constant and therefore the larger solubility in water.

Question 18Solid State and Solutions

An aqueous glucose solution (molar mass of glucose = 180 g mol^-1) contains 10 % glucose by mass. Its molality is

  1. 1.11 mol kg^-1
  2. 0.62 mol kg^-1 Correct answer.
  3. 0.56 mol kg^-1
  4. 0.11 mol kg^-1

Explanation

A 100 g sample contains 10 g of glucose, which is 10/180 = 0.0556 mol, dissolved in 90 g (0.090 kg) of water. Dividing the amount of solute by the mass of solvent in kilograms gives 0.0556/0.090 = 0.62 mol kg^-1.

Question 19Solid State and Solutions

A 0.10 mol kg^-1 aqueous solution of a solute freezes at −0.372 °C. Taking Kf for water as 1.86 K kg mol^-1, the van't Hoff factor of the solute is

  1. 2.0 Correct answer.
  2. 1.0
  3. 3.0
  4. 0.5

Explanation

Freezing-point depression follows ΔTf = i Kf m, so i = 0.372/(1.86 × 0.10) = 2.0. A value of 2 shows that each formula unit produces two particles in solution, as expected for a solute such as NaCl that dissociates completely.

Question 20Solid State and Solutions

In an ionic solid AB with the rock-salt arrangement, the radius ratio r(A+)/r(B−) is 0.52. The coordination number of the cation A+ is

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

Explanation

A radius ratio of 0.52 falls in the range 0.414 to 0.732, which is the stability field of octahedral holes. The cation therefore sits in an octahedral hole and is surrounded by six anions, consistent with the 6 : 6 coordination of the rock-salt structure.

Question 21Electrochemistry

For the Daniell cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) at 298 K, the standard cell potential is (E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V):

  1. 1.10 V Correct answer.
  2. 0.42 V
  3. −1.10 V
  4. 0.76 V

Explanation

The standard cell potential equals the cathode reduction potential minus the anode reduction potential: E°cell = 0.34 V − (−0.76 V) = 1.10 V. The positive sign confirms that the cell reaction is spontaneous when all species are in their standard states.

Question 22Electrochemistry

The molar conductivity of an electrolyte solution, Λm, is expressed in the unit:

  1. S cm² mol⁻¹ Correct answer.
  2. S cm⁻¹
  3. S cm⁻¹ mol⁻¹
  4. Ω cm² mol⁻¹

Explanation

Molar conductivity is defined as Λm = κ/c, where κ is conductivity in S cm⁻¹ and c is concentration in mol cm⁻³. Dividing S cm⁻¹ by mol cm⁻³ gives S cm² mol⁻¹, the standard unit used for Λm.

Question 23Electrochemistry

During the electrolysis of concentrated aqueous sodium chloride (brine) using inert electrodes, the gas liberated at the anode is:

  1. Hydrogen
  2. Chlorine Correct answer.
  3. Oxygen
  4. Carbon dioxide

Explanation

In strong brine the chloride concentration is high, so chloride ions are oxidised at the anode: 2Cl⁻ → Cl₂ + 2e⁻. Water can also be oxidised to oxygen, but the high chloride concentration and the overpotential for oxygen evolution make chlorine the actual anode product.

Question 24Electrochemistry

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

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

Explanation

The Nernst equation gives E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]) at 298 K. Substituting, E = 1.10 − 0.0295 log(0.10/0.010) = 1.10 − 0.0295 = 1.07 V.

Question 25Electrochemistry

A current of 0.50 A is passed through aqueous AgNO₃ solution for 1930 s. The mass of silver deposited at the cathode is (M(Ag) = 108 g mol⁻¹, F = 96500 C mol⁻¹):

  1. 0.54 g
  2. 10.8 g
  3. 1.08 g Correct answer.
  4. 2.16 g

Explanation

Charge passed = I × t = 0.50 A × 1930 s = 965 C, which equals 965/96500 = 0.010 mol of electrons. Since Ag⁺ + e⁻ → Ag, 0.010 mol of silver is deposited, giving 0.010 × 108 = 1.08 g.

Question 26Electrochemistry

For CH₃COOH, the limiting molar conductivities of HCl, CH₃COONa and NaCl are 426, 91 and 126 S cm² mol⁻¹ respectively. The limiting molar conductivity of acetic acid is:

  1. 391 S cm² mol⁻¹ Correct answer.
  2. 209 S cm² mol⁻¹
  3. 265 S cm² mol⁻¹
  4. 643 S cm² mol⁻¹

Explanation

Kohlrausch's law allows the required value to be built from strong electrolytes: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl). The Na⁺ and Cl⁻ contributions cancel, so Λ° = 91 + 426 − 126 = 391 S cm² mol⁻¹.

Question 27Electrochemistry

Given E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V and E°(Ag⁺/Ag) = +0.80 V, which pair of substances reacts spontaneously when mixed in aqueous medium?

  1. Ag⁺ and Cu Correct answer.
  2. Cu²⁺ and Ag
  3. Zn²⁺ and Ag
  4. Zn²⁺ and Cu

Explanation

A reaction is spontaneous when E°cell is positive, which happens if the ion being reduced has a higher reduction potential than the metal being oxidised. E°(Ag⁺/Ag) = +0.80 V exceeds E°(Cu²⁺/Cu) = +0.34 V, so Ag⁺ oxidises copper with E°cell = +0.46 V.

Question 28Electrochemistry

When an aqueous solution of a strong electrolyte such as KCl is progressively diluted at constant temperature:

  1. conductivity stays constant while molar conductivity increases
  2. conductivity increases while molar conductivity decreases
  3. both conductivity and molar conductivity decrease
  4. conductivity decreases while molar conductivity increases Correct answer.

Explanation

Dilution lowers the number of ions per unit volume, so the conductivity κ of the solution falls. Molar conductivity Λm = κ/c rises because the conducting ions produced by one mole of electrolyte are spread through a larger volume and face less ionic interference.

Question 29Electrochemistry

For the Daniell cell at 298 K, E°cell = 1.10 V. The equilibrium constant of Zn(s) + Cu²⁺(aq) ⇌ Zn²⁺(aq) + Cu(s) is closest to:

  1. 10⁷⁴
  2. 10³⁷ Correct answer.
  3. 10⁵⁵
  4. 10¹⁸

Explanation

At equilibrium Ecell = 0, so the Nernst equation gives log K = nE°/0.0591. With n = 2 and E° = 1.10 V, log K = 2.20/0.0591 ≈ 37, so K is of the order of 10³⁷.

Question 30Electrochemistry

For the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) at 298 K, E°cell = 1.56 V. The standard Gibbs energy change of the reaction is:

  1. −150.5 kJ mol⁻¹
  2. −602 kJ mol⁻¹
  3. −301 kJ mol⁻¹ Correct answer.
  4. +301 kJ mol⁻¹

Explanation

The relation ΔG° = −nFE° links cell potential to free energy. Here n = 2 and F = 96500 C mol⁻¹, so ΔG° = −2 × 96500 × 1.56 = −301080 J mol⁻¹, that is about −301 kJ mol⁻¹. A positive E°cell always gives a negative ΔG°.

Question 31Electrochemistry

The standard hydrogen electrode is constructed using platinized platinum and hydrogen gas. Which set of conditions defines its standard state at 298 K?

  1. [H⁺] = 0.1 M and H₂ at 0.1 bar
  2. [H⁺] = 0.1 M and H₂ at 1 bar
  3. [H⁺] = 1 M and H₂ at 1 bar Correct answer.
  4. [H⁺] = 1 M and O₂ at 1 bar

Explanation

By convention, the standard hydrogen electrode has H⁺ activity equal to 1 (approximately 1 M) and H₂ gas at 1 bar pressure, with platinized platinum as the inert conductor. Its potential is assigned E° = 0.00 V at all temperatures, providing the reference against which other electrode potentials are measured.

Question 32Electrochemistry

In a galvanic cell such as Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s), a salt bridge is used. Its main function is to:

  1. prevent any ion movement between the two solutions
  2. complete the circuit and maintain electrical neutrality Correct answer.
  3. supply electrons to the cathode during cell operation
  4. permit electrons to travel between the two solutions

Explanation

The salt bridge contains a concentrated inert electrolyte such as KCl in agar. It completes the electrical circuit by allowing ions to migrate between the half-cells and maintains electrical neutrality in each compartment as reaction proceeds; electrons travel through the external wire, not through the bridge.

Question 33Electrochemistry

The potential of a hydrogen electrode (H₂ at 1 bar) immersed in an aqueous solution of pH = 3.0 at 298 K is:

  1. +0.059 V
  2. −0.059 V
  3. −0.177 V Correct answer.
  4. +0.177 V

Explanation

For a hydrogen electrode at 1 bar, the Nernst equation gives E = E° − 0.0591 log(1/[H⁺]) = −0.0591 pH at 298 K. Substituting pH = 3.0 gives E = −0.0591 × 3 = −0.177 V.

Question 34Electrochemistry

A current is passed through molten Al₂O₃ until 0.10 mol of aluminium metal is deposited at the cathode. The quantity of charge required is (F = 96500 C mol⁻¹):

  1. 19300 C
  2. 9650 C
  3. 28950 C Correct answer.
  4. 38600 C

Explanation

The cathode reaction is Al³⁺ + 3e⁻ → Al, so one mole of aluminium requires three moles of electrons, i.e. 3 F. For 0.10 mol Al, charge = 0.10 × 3 × 96500 C = 28950 C.

Question 35Electrochemistry

During the discharge of a lead storage battery containing aqueous H₂SO₄, the half-reaction at the anode is:

  1. PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O
  2. Pb → Pb²⁺ + 2e⁻
  3. PbSO₄ + 2e⁻ → Pb + SO₄²⁻
  4. Pb + SO₄²⁻ → PbSO₄ + 2e⁻ Correct answer.

Explanation

At the anode of a lead storage battery, lead metal is oxidized to PbSO₄ in the presence of sulfuric acid: Pb + SO₄²⁻ → PbSO₄ + 2e⁻. The cathode reaction involves reduction of PbO₂, and both electrodes become coated with PbSO₄ during discharge.

Question 36Electrochemistry

The resistance of a 0.10 M KCl solution measured in a conductivity cell is 500 Ω. If the cell constant is 0.50 cm⁻¹, the conductivity of the solution is:

  1. 1.0 × 10⁻³ S cm⁻¹ Correct answer.
  2. 5.0 × 10⁻³ S cm⁻¹
  3. 1.0 × 10⁻⁴ S cm⁻¹
  4. 1.0 × 10⁻² S cm⁻¹

Explanation

Conductivity is obtained from the measured resistance and the cell constant: κ = cell constant / R. Here κ = 0.50 cm⁻¹ / 500 Ω = 1.0 × 10⁻³ S cm⁻¹.

Question 37Electrochemistry

The EMF at 298 K of the concentration cell Ag(s)|Ag⁺(0.001 M)||Ag⁺(0.10 M)|Ag(s) is:

  1. 0.000 V
  2. 0.236 V
  3. 0.118 V Correct answer.
  4. 0.059 V

Explanation

For a concentration cell with identical silver electrodes, the electrode in the more concentrated solution is the cathode. EMF = (0.0591/n) log([Ag⁺]cathode/[Ag⁺]anode), with n = 1, so E = 0.0591 log(0.10/0.001) = 0.0591 × 2 = 0.118 V.

Question 38Electrochemistry

For copper in aqueous solution, E°(Cu²⁺/Cu⁺) = +0.15 V and E°(Cu⁺/Cu) = +0.52 V. Which statement about the disproportionation of Cu⁺ is correct?

  1. Cu⁺ disproportionates spontaneously with E°cell = +0.37 V. Correct answer.
  2. Cu⁺ disproportionates spontaneously with E°cell = +0.67 V.
  3. Cu⁺ is stable in aqueous solution because E°(Cu⁺/Cu) is more positive than E°(Cu²⁺/Cu⁺).
  4. Cu⁺ does not disproportionate because the calculated E°cell is negative.

Explanation

Disproportionation 2Cu⁺ → Cu²⁺ + Cu has Cu⁺/Cu as the cathode (E° = +0.52 V) and Cu²⁺/Cu⁺ as the anode (E° = +0.15 V). E°cell = 0.52 − 0.15 = +0.37 V, and the positive value means Cu⁺ is thermodynamically unstable with respect to disproportionation in aqueous solution.

Question 39Chemical Kinetics

For a reaction A → products, the rate of reaction is correctly expressed as

  1. −d[A]/dt Correct answer.
  2. d[A]/dt
  3. [A]/t
  4. −dt/d[A]

Explanation

Rate is defined as the decrease in concentration of a reactant per unit time. Because [A] decreases as the reaction proceeds, d[A]/dt is negative, so the rate is written with a minus sign as −d[A]/dt. The expression [A]/t is a concentration divided by time, not the instantaneous change in concentration.

Question 40Chemical Kinetics

The sum of the powers to which reactant concentrations are raised in the experimentally determined rate law is called the

  1. molecularity of the reaction
  2. order of the reaction Correct answer.
  3. activation energy of the reaction
  4. rate constant of the reaction

Explanation

The order of a reaction is obtained from the rate law as the sum of the exponents of the concentration terms. It is an experimental quantity and may differ from molecularity, which is the number of species colliding in an elementary step. For example, rate = k[A][B]^2 is third order overall.

Question 41Chemical Kinetics

The SI unit of the rate constant for a first-order reaction is

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

Explanation

For a first-order reaction, rate = k[A], where rate has units mol L⁻¹ s⁻¹ and [A] has units mol L⁻¹. Dividing the units of rate by the units of concentration leaves s⁻¹ for k. This unit is characteristic of first-order processes.

Question 42Chemical Kinetics

A first-order reaction has a half-life of 20 minutes. The time required for the reaction to be 75% complete is

  1. 20 minutes
  2. 40 minutes Correct answer.
  3. 10 minutes
  4. 30 minutes

Explanation

For a first-order reaction, the half-life is constant and 75% completion corresponds to two half-lives. After one half-life 50% remains, and after a second half-life 25% remains, so 75% has reacted. Thus the required time is 2 × 20 = 40 minutes.

Question 43Chemical Kinetics

The Arrhenius equation is k = A e^(−Ea/RT). Which statement correctly describes the effect of increasing temperature on k?

  1. Increasing T increases k because the exponential factor e^(−Ea/RT) becomes larger. Correct answer.
  2. Increasing T has no effect on k when the frequency factor A is constant.
  3. Increasing T increases k because the gas constant R becomes smaller.
  4. Increasing T decreases k because the activation energy Ea becomes larger.

Explanation

In the Arrhenius equation, raising T makes the exponent −Ea/RT less negative, so e^(−Ea/RT) increases and k increases. This reflects the larger fraction of molecules that possess energy equal to or greater than the activation energy. The activation energy and the gas constant are not changed by a simple temperature increase.

Question 44Chemical Kinetics

For a zero-order reaction in A, which plot against time t gives a straight line with slope −k?

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

Explanation

The integrated rate law for a zero-order reaction is [A] = [A]0 − kt, which is linear in [A] versus t with slope −k. A plot of ln[A] versus t is linear for a first-order reaction, and 1/[A] versus t is linear for a second-order reaction. Hence the zero-order dependence is identified by the direct concentration plot.

Question 45Chemical Kinetics

For the reaction 2NO + O2 → 2NO2, the rate law is rate = k[NO]^2[O2]. If [NO] is doubled while [O2] is kept constant, the rate becomes

  1. eight times the original rate
  2. twice the original rate
  3. one-half of the original rate
  4. four times the original rate Correct answer.

Explanation

The rate law shows that the reaction is second order with respect to NO, so doubling [NO] multiplies the rate by 2^2 = 4 when [O2] is unchanged. The concentration of O2 does not change in this comparison. Hence the new rate is four times the original rate.

Question 46Chemical Kinetics

The hydrolysis of ethyl acetate in aqueous acid is regarded as a pseudo first-order reaction because

  1. the reaction proceeds only in the presence of an acid catalyst
  2. the reaction is independent of the concentration of ethyl acetate
  3. the rate constant has units of mol L⁻¹ s⁻¹
  4. water is in large excess, so its concentration is nearly constant Correct answer.

Explanation

In dilute aqueous solution, water is the solvent and is present in large excess, so its concentration changes very little during the hydrolysis. The rate law therefore appears to depend only on the concentration of ethyl acetate, giving pseudo first-order behaviour. The true rate law still involves water, but its nearly constant concentration is absorbed into the rate constant.

Question 47Chemical Kinetics

A first-order reaction has a rate constant of 0.0693 min⁻¹. The time required for 99% completion is approximately

  1. 10.0 min
  2. 66.5 min Correct answer.
  3. 33.2 min
  4. 99.0 min

Explanation

For a first-order reaction, t = (2.303/k) log([A]0/[A]). For 99% completion, [A]/[A]0 = 0.01, so t = (2.303/0.0693) × log(100) = (2.303/0.0693) × 2 ≈ 66.5 min. This is about ten half-lives, consistent with the very small fraction of reactant remaining.

Question 48Chemical Kinetics

A proposed mechanism is: 2NO ⇌ N2O2 (fast equilibrium); N2O2 + O2 → 2NO2 (slow). The rate law predicted by this mechanism is

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

Explanation

The slow step governs the rate, so rate = k2[N2O2][O2]. From the fast equilibrium, [N2O2] = K[NO]^2, and substituting this into the slow-step rate gives rate = k2K[NO]^2[O2], or rate = k[NO]^2[O2]. The intermediate N2O2 must be replaced by reactants in the final rate law.

Question 49Chemical Kinetics

For a reaction R → products, the instantaneous rate of disappearance of R at time t is represented by which expression?

  1. d[R]/dt
  2. Δ[R]/Δt
  3. −d[R]/dt Correct answer.
  4. −Δ[R]/Δt

Explanation

Instantaneous rate is the derivative of concentration with respect to time evaluated at a particular instant. Because R is a reactant, its concentration falls with time, so a negative sign is used to make the rate positive. Expressions containing Δ refer to an average rate over a finite time interval, not to the instantaneous value.

Question 50Chemical Kinetics

A catalyst increases the rate of a chemical reaction because it

  1. raises the enthalpy change of the reaction
  2. provides an alternative path of lower activation energy Correct answer.
  3. changes the equilibrium constant of the reaction
  4. increases the activation energy of the forward reaction

Explanation

A catalyst offers a different reaction pathway whose transition state lies at lower energy, so the activation energy is reduced. A smaller activation energy increases the fraction of molecules having sufficient energy to react, increasing the rate constant and hence the rate. The catalyst does not alter the enthalpy change or the equilibrium constant, since these depend on the initial and final states.