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BSEH SS Chemistry study materials
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BSEH SS Chemistry practice questions

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Question 1Solid State and Solutions

In a face-centred cubic (fcc) unit cell, how many atoms are effectively present per unit cell?

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

Explanation

An fcc cell has atoms at all eight corners and at the centres of all six faces. Corner contributions total 8 × 1/8 = 1, and face contributions total 6 × 1/2 = 3, giving 4 atoms per unit cell.

Question 2Solid State and Solutions

For an ideal dilute solution, the magnitude of a colligative property is determined mainly by which factor?

  1. The nature of solute-solvent interactions
  2. The colour of the solute
  3. The number of solute particles present Correct answer.
  4. The molar mass of the solvent

Explanation

Colligative properties depend on the number of solute particles in an ideal dilute solution, not on their chemical identity. Therefore equal molalities of different non-electrolytes produce similar effects.

Question 3Solid State and Solutions

A Schottky defect in an ionic crystal is formed when:

  1. an ion leaves its lattice site and occupies an interstitial site
  2. an extra electron is trapped at an anion vacancy
  3. a cation is replaced by an ion of higher charge
  4. a pair of oppositely charged ions is missing from the lattice Correct answer.

Explanation

A Schottky defect is a stoichiometric vacancy pair in which one cation and one anion are missing from their normal lattice sites. This maintains electrical neutrality and lowers the crystal density.

Question 4Solid State and Solutions

The packing efficiency of atoms in a body-centred cubic (bcc) unit cell is approximately:

  1. 52%
  2. 90%
  3. 74%
  4. 68% Correct answer.

Explanation

In a bcc cell, atoms touch along the body diagonal, giving a relation between atomic radius and edge length that leads to 68% space filling. The remaining 32% is empty space.

Question 5Solid State and Solutions

For a dilute aqueous solution of K2SO4 that dissociates completely, the van't Hoff factor i is:

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

Explanation

K2SO4 dissociates into two K+ ions and one SO4^2- ion, giving three particles per formula unit. Hence i = 3 for complete dissociation.

Question 6Solid State and Solutions

The vapour pressure of a pure solvent is 100 mmHg. On dissolving a non-volatile, non-electrolyte solute, the vapour pressure of the solution becomes 90 mmHg. The mole fraction of the solute is:

  1. 0.10 Correct answer.
  2. 0.11
  3. 0.20
  4. 0.90

Explanation

Raoult's law gives relative lowering (p° − p)/p° = x_solute. Here (100 − 90)/100 = 0.10, so the solute mole fraction is 0.10.

Question 7Solid State and Solutions

Doping silicon with phosphorus produces:

  1. an insulating material because phosphorus disrupts the lattice completely
  2. an n-type semiconductor because phosphorus donates an extra electron Correct answer.
  3. a p-type semiconductor because phosphorus has fewer valence electrons
  4. an intrinsic semiconductor because phosphorus is isoelectronic with silicon

Explanation

Silicon has four valence electrons, while phosphorus has five. The extra electron is weakly bound and becomes mobile, making the doped crystal n-type.

Question 8Solid State and Solutions

The freezing point depression constant Kf of water is 1.86 K kg mol^-1. What is the freezing point of a 0.5 molal aqueous solution of a non-electrolyte?

  1. -0.93 °C Correct answer.
  2. -1.86 °C
  3. +0.93 °C
  4. -3.72 °C

Explanation

ΔTf = Kf × molality = 1.86 × 0.5 = 0.93 K. Freezing point depression lowers the freezing point, so the solution freezes at −0.93 °C.

Question 9Solid State and Solutions

A cubic unit cell has edge length 400 pm and contains 4 atoms of molar mass 60 g mol^-1. Using N_A = 6.0 × 10^23 mol^-1, the density is closest to:

  1. 0.63 g cm^-3
  2. 6.3 g cm^-3 Correct answer.
  3. 2.5 g cm^-3
  4. 12.5 g cm^-3

Explanation

Density = ZM/(N_A a^3). With Z = 4, M = 60 g mol^-1, and a = 4.0 × 10^-8 cm, the volume is 6.4 × 10^-23 cm^3 and the density is about 6.3 g cm^-3.

Question 10Solid State and Solutions

When benzoic acid dimerises completely in benzene, its observed molar mass becomes:

  1. half its normal molar mass
  2. twice its normal molar mass Correct answer.
  3. equal to its normal molar mass
  4. one-fourth its normal molar mass

Explanation

Complete dimerisation halves the number of solute particles, so the van't Hoff factor is 0.5. Since observed molar mass is inversely proportional to particle number, it becomes twice the normal value.

Question 11Solid State and Solutions

In the caesium chloride (CsCl) structure, the coordination number of each Cs+ ion is:

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

Explanation

CsCl adopts a simple cubic chloride lattice with Cs+ at the body centre, so each Cs+ is surrounded by eight Cl- ions at the corners of a cube. The structure is not body-centred cubic in the metallic sense, but the 8:8 coordination is the key point.

Question 12Solid State and Solutions

Which concentration term is independent of temperature?

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

Explanation

Molality is defined as moles of solute per kilogram of solvent, and both mass terms are unaffected by thermal expansion. Molarity, normality and formality are volume-based and therefore change with temperature.

Question 13Solid State and Solutions

The number of atoms per unit cell in a simple cubic lattice is:

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

Explanation

A simple cubic cell has atoms only at its eight corners, and each corner atom is shared by eight adjacent cells. Thus 8 × 1/8 = 1 atom per unit cell.

Question 14Solid State and Solutions

A metal crystallises in a face-centred cubic lattice with edge length 400 pm. The radius of the metal atom is approximately:

  1. 200 pm
  2. 100 pm
  3. 141 pm Correct answer.
  4. 173 pm

Explanation

In an fcc cell, atoms touch along the face diagonal, so 4r = a√2 and r = a/(2√2). Substituting a = 400 pm gives r ≈ 141 pm.

Question 15Solid State and Solutions

Henry's law constant for a gas in water is 1.0 × 10^5 Pa at 298 K. If the partial pressure of the gas is 0.20 × 10^5 Pa, the mole fraction of dissolved gas is:

  1. 0.020
  2. 0.80
  3. 5.0
  4. 0.20 Correct answer.

Explanation

Henry's law states p = KH x, so x = p/KH. Substituting gives 0.20 × 10^5 / 1.0 × 10^5 = 0.20.

Question 16Solid State and Solutions

For a binary solution to behave ideally, which condition must hold?

  1. ΔHmix < 0 and ΔVmix > 0
  2. ΔHmix = 0 and ΔVmix > 0
  3. ΔHmix = 0 and ΔVmix = 0 Correct answer.
  4. ΔHmix > 0 and ΔVmix = 0

Explanation

An ideal solution forms with no enthalpy change and no volume change on mixing, so ΔHmix = 0 and ΔVmix = 0. It also obeys Raoult's law over the whole composition range.

Question 17Solid State and Solutions

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

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

Explanation

For glucose, i = 1, so π = CRT = 0.10 × 0.0821 × 300. This gives π ≈ 2.46 atm.

Question 18Solid State and Solutions

When a small amount of CdCl2 is added to AgCl, the predominant defect introduced is:

  1. Schottky defect
  2. Cation vacancies Correct answer.
  3. Frenkel defect
  4. Anion vacancies

Explanation

Cd2+ substitutes for Ag+ in the AgCl lattice, and each Cd2+ carries one extra positive charge. To maintain electrical neutrality, one Ag+ vacancy is created for every Cd2+ incorporated, giving cation vacancies.

Question 19Solid State and Solutions

A solute associates into trimers in solution. If the degree of association is 0.60, the van't Hoff factor is closest to:

  1. 1.60
  2. 0.60 Correct answer.
  3. 0.40
  4. 1.40

Explanation

For association into n particles, i = 1 + α(1/n − 1). With n = 3 and α = 0.60, i = 1 + 0.60(1/3 − 1) = 0.60.

Question 20Solid State and Solutions

In a hexagonal close-packed (hcp) structure containing 6 atoms per unit cell, the numbers of tetrahedral and octahedral voids per unit cell are, respectively:

  1. 12 and 6 Correct answer.
  2. 6 and 12
  3. 18 and 6
  4. 12 and 3

Explanation

In any close-packed structure, the number of tetrahedral voids is twice the number of atoms and the number of octahedral voids equals the number of atoms. For 6 atoms per hcp cell, these are 12 tetrahedral and 6 octahedral voids.

Question 21Electrochemistry

In a galvanic cell such as the Daniell cell, what is the primary function of the salt bridge?

  1. It carries electrons between the two electrode compartments.
  2. It provides the cell with its standard electrode potential.
  3. It prevents ions from leaving either half-cell solution.
  4. It completes the circuit and maintains electrical neutrality. Correct answer.

Explanation

The salt bridge contains an electrolyte that allows ions to migrate between the half-cells. This completes the internal circuit and prevents the build-up of charge that would otherwise stop the cell reaction. Electrons travel through the external wire, not through the salt bridge.

Question 22Electrochemistry

By convention, the standard electrode potential of the hydrogen electrode is taken as zero when which set of conditions is satisfied?

  1. H2 at 1 bar, H+ at 1 M, and temperature at 298 K Correct answer.
  2. H2 at 0.5 bar, H+ at 2 M, and temperature at 298 K
  3. H2 at 2 bar, H+ at 0.5 M, and temperature at 273 K
  4. H2 at 1 bar, H+ at 0.1 M, and temperature at 373 K

Explanation

The standard hydrogen electrode is assigned E° = 0.00 V by convention. Its standard state uses hydrogen gas at 1 bar, hydrogen ion concentration of 1 M, and a temperature of 298 K. Changing these conditions changes the electrode potential, so the zero value no longer applies.

Question 23Electrochemistry

In a zinc-copper Daniell cell, oxidation takes place at the anode. Which statement correctly describes the anode?

  1. It is the copper electrode and it loses mass as Cu2+ is reduced.
  2. It is the copper electrode and it gains mass as Cu is oxidised.
  3. It is the zinc electrode and it gains mass as Zn2+ is reduced.
  4. It is the zinc electrode and it loses mass as Zn is oxidised. Correct answer.

Explanation

In a Daniell cell the zinc electrode is the anode because zinc is oxidised to Zn2+, releasing electrons. The zinc electrode therefore loses mass as the reaction proceeds. Copper ions are reduced at the copper cathode, which gains mass.

Question 24Electrochemistry

For the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s), what happens to the cell EMF when the concentration of Zn2+ is increased while the Cu2+ concentration is held constant at 298 K?

  1. The EMF remains unchanged because Zn2+ is not involved in the cell reaction.
  2. The EMF becomes zero because the cell reaches equilibrium.
  3. The EMF increases because the reaction quotient decreases.
  4. The EMF decreases because the reaction quotient increases. Correct answer.

Explanation

The Nernst equation for this cell is E = E° − (0.059/2) log([Zn2+]/[Cu2+]). Increasing [Zn2+] raises the reaction quotient Q, so the logarithmic term becomes more positive and E decreases. The cell EMF falls but remains positive until equilibrium is reached.

Question 25Electrochemistry

Why does the molar conductivity of a weak electrolyte increase sharply on dilution, while that of a strong electrolyte increases only slightly?

  1. It raises the viscosity, allowing ions to move more quickly.
  2. It raises the number of ions per unit volume in the solution.
  3. It lowers the temperature and therefore raises ionic mobility.
  4. It increases dissociation of the weak electrolyte and reduces ion pairing. Correct answer.

Explanation

Molar conductivity depends on the number of ions produced per mole of electrolyte and on their mobility. For a weak electrolyte, dilution increases the degree of dissociation, so the number of ions rises sharply. For a strong electrolyte, dilution mainly reduces ion–ion attractions, giving only a small increase.

Question 26Electrochemistry

Using Kohlrausch's law, the limiting molar conductivity of acetic acid, Λ°(CH3COOH), can be calculated from the limiting molar conductivities of which substances?

  1. CH3COONa, HCl, and NaCl Correct answer.
  2. CH3COONa, NaOH, and CH3OH
  3. CH3OH, HCl, and NaOH
  4. CH3COOH, CH3COONa, and NaCl only

Explanation

Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities. For acetic acid, Λ°(CH3COOH) = Λ°(CH3COONa) + Λ°(HCl) − Λ°(NaCl), because the sodium and chloride terms cancel. This allows a weak electrolyte's limiting value to be found from strong electrolytes.

Question 27Electrochemistry

A current of 9.65 A is passed through aqueous copper(II) sulphate for 100 s. What mass of copper is deposited at the cathode? (Molar mass of Cu = 63.5 g mol−1, F = 96500 C mol−1)

  1. 1.27 g
  2. 3.175 g
  3. 0.3175 g Correct answer.
  4. 0.635 g

Explanation

The charge passed is Q = It = 9.65 A × 100 s = 965 C. This corresponds to 965/96500 = 0.0100 mol of electrons. Since Cu2+ + 2e− → Cu, the moles of copper are 0.0050, giving a mass of 0.0050 × 63.5 = 0.3175 g.

Question 28Electrochemistry

During the electrolysis of concentrated aqueous sodium chloride using inert electrodes, which products are formed at the cathode and anode respectively?

  1. Hydrogen at the cathode and oxygen at the anode
  2. Sodium at the cathode and oxygen at the anode
  3. Chlorine at the cathode and hydrogen at the anode
  4. Hydrogen at the cathode and chlorine at the anode Correct answer.

Explanation

In concentrated aqueous NaCl, water is reduced at the cathode in preference to Na+ because the reduction potential of water is less negative. This produces hydrogen gas and hydroxide ions. Chloride ions are oxidised at the anode to chlorine gas, while sodium ions remain in solution.

Question 29Electrochemistry

For a cell reaction with n = 2 and standard cell potential E° = 1.10 V at 298 K, what is the standard Gibbs free energy change, ΔG°, and does the reaction proceed spontaneously under standard conditions? (F = 96500 C mol−1)

  1. −212.3 kJ mol−1, and the reaction is spontaneous Correct answer.
  2. +212.3 kJ mol−1, and the reaction is spontaneous
  3. −106.15 kJ mol−1, and the reaction is non-spontaneous
  4. −212.3 kJ mol−1, and the reaction is non-spontaneous

Explanation

The standard Gibbs free energy change is ΔG° = −nFE° = −2 × 96500 C mol−1 × 1.10 V = −212300 J mol−1, or −212.3 kJ mol−1. A negative ΔG° means the reaction is spontaneous under standard conditions. The value is not divided by two because n is already included.

Question 30Electrochemistry

In the electrochemical mechanism of rusting of iron, which statement correctly describes the role of oxygen and water?

  1. Water acts as the oxidising agent and oxygen is reduced to hydroxide.
  2. Oxygen is reduced at the anode while iron is oxidised at the cathode.
  3. Oxygen acts as the oxidising agent and water is the conducting medium. Correct answer.
  4. Oxygen acts as the reducing agent and water is oxidised to hydrogen gas.

Explanation

Rusting is an electrochemical process in which iron is oxidised at anodic sites: Fe → Fe2+ + 2e−. Oxygen is reduced at cathodic sites in the presence of water: O2 + 2H2O + 4e− → 4OH−. Water provides the medium that allows ions to move, so both oxygen and water are required.

Question 31Electrochemistry

A galvanic cell generates electricity because it is based on which type of change?

  1. A spontaneous redox reaction that releases electrical energy Correct answer.
  2. A spontaneous acid–base neutralisation that releases heat
  3. A non-spontaneous precipitation reaction driven by light
  4. A non-spontaneous redox reaction driven by an external current

Explanation

A galvanic cell couples two spontaneous half-reactions in separate compartments. The electrons released by oxidation travel through an external circuit to the reduction half-cell, so chemical energy is converted into electrical energy.

Question 32Electrochemistry

The charge carried by one mole of electrons, called one faraday, is closest to which value?

  1. 8.31 × 10^4 C
  2. 1.60 × 10^-19 C
  3. 9.65 × 10^4 C Correct answer.
  4. 6.02 × 10^23 C

Explanation

One faraday is the product of the Avogadro number and the elementary charge on one electron. This gives approximately 6.02 × 10^23 × 1.60 × 10^-19 C, which is about 9.65 × 10^4 C per mole of electrons.

Question 33Electrochemistry

In the electrolytic extraction of aluminium, the cathode reaction is Al3+ + 3e− → Al. How many faradays are required to deposit 1.0 mol of aluminium?

  1. 0.5 F
  2. 3.0 F Correct answer.
  3. 1.0 F
  4. 6.0 F

Explanation

The half-equation shows that three moles of electrons are consumed for every mole of aluminium formed. Since one faraday is one mole of electrons, 1.0 mol of Al requires 3.0 F.

Question 34Electrochemistry

A hydrogen electrode is dipped in a solution of pH 3 at 298 K, and hydrogen gas at 1 bar is passed through it. Taking E = −0.059 × pH for this electrode, what is its potential relative to the standard hydrogen electrode?

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

Explanation

For the hydrogen electrode at 298 K, the potential falls by 0.059 V for each unit increase in pH. Substituting pH = 3 gives E = −0.059 × 3 = −0.177 V relative to the standard hydrogen electrode.

Question 35Electrochemistry

Given E°(Zn2+/Zn) = −0.76 V, E°(Cu2+/Cu) = +0.34 V and E°(Ag+/Ag) = +0.80 V, which reaction is spontaneous under standard conditions?

  1. 2Ag(s) + Cu2+(aq) → 2Ag+(aq) + Cu(s)
  2. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Correct answer.
  3. Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s)
  4. Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)

Explanation

A reaction is spontaneous when the standard cell potential is positive. Zinc has the lowest reduction potential, so it is oxidised while copper(II) ions are reduced; the cell potential is +0.34 − (−0.76) = +1.10 V.

Question 36Electrochemistry

Aqueous copper(II) sulfate is electrolysed between inert platinum electrodes. Which half-reaction occurs at the anode?

  1. Cu2+(aq) + 2e− → Cu(s)
  2. 2H+(aq) + 2e− → H2(g)
  3. 2H2O(l) → O2(g) + 4H+(aq) + 4e− Correct answer.
  4. SO4^2−(aq) → SO2(g) + O2(g) + 2e−

Explanation

At the anode, oxidation occurs, and in aqueous sulfate solution water is oxidised in preference to sulfate ions. This produces oxygen gas and hydrogen ions while releasing electrons into the external circuit.

Question 37Electrochemistry

For a cell reaction at 298 K with n = 2, the standard cell potential is 0.295 V. Using E° = (0.059 V/n) log Kc, what is the value of Kc?

  1. 1 × 10^20
  2. 1 × 10^5
  3. 1 × 10^10 Correct answer.
  4. 1 × 10^8

Explanation

Rearranging the relation gives log Kc = nE°/0.059 = (2 × 0.295)/0.059 = 10. Therefore Kc = 10^10, showing that the reaction is strongly product-favoured under standard conditions.

Question 38Electrochemistry

In a hydrogen–oxygen fuel cell operating with an alkaline electrolyte, which pair of electrode reactions is correct?

  1. Anode: H2 → 2H+ + 2e−; Cathode: O2 + 4H+ + 4e− → 2H2O
  2. Anode: O2 + 2H2O + 4e− → 4OH−; Cathode: H2 + 2OH− → 2H2O + 2e−
  3. Anode: 2H2 + O2 → 2H2O; Cathode: no electrode reaction occurs
  4. Anode: H2 + 2OH− → 2H2O + 2e−; Cathode: O2 + 2H2O + 4e− → 4OH− Correct answer.

Explanation

In alkaline solution, hydrogen is oxidised at the anode by hydroxide ions, forming water and releasing electrons. At the cathode, oxygen is reduced with water to produce hydroxide ions, and the overall cell reaction is the formation of water.

Question 39Chemical Kinetics

For the reaction A(aq) + B(aq) → C(aq), the average rate measured over a time interval is expressed in which unit?

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

Explanation

Rate of reaction is the change in concentration of a reactant or product divided by the time interval over which the change occurs. Concentration is measured in mol L⁻¹ and time in seconds, so the unit of reaction rate is mol L⁻¹ s⁻¹.

Question 40Chemical Kinetics

Zinc reacts with dilute hydrochloric acid to give hydrogen. Which change increases the initial rate without changing the amounts of zinc and acid used?

  1. Cooling the reaction mixture from 25 °C to 5 °C
  2. Carrying out the reaction in a wider vessel
  3. Using the same mass of zinc as a powder Correct answer.
  4. Adding a large volume of water to the acid

Explanation

The rate depends on the surface area of the solid reactant that is exposed to the acid. The same mass of zinc powder exposes far more surface than granules, so more collisions occur per second and the initial rate increases. Cooling or diluting the acid lowers the rate, and a wider vessel does not change the concentrations.

Question 41Chemical Kinetics

For the elementary reaction NO2(g) + CO(g) → NO(g) + CO2(g), which statement about molecularity and order is correct?

  1. The molecularity is one because only one product is formed
  2. The reaction is second order in NO2 and zero order in CO
  3. The order must be found by experiment and always differs from molecularity
  4. The molecularity is two and the order is also two Correct answer.

Explanation

In an elementary step the reaction takes place in a single collision, so the exponents of the rate law equal the stoichiometric coefficients of the reactants. Two species collide in this step, so the molecularity is two and the rate law is rate = k[NO2][CO], making the reaction second order overall.

Question 42Chemical Kinetics

For the reaction A + B → products, the initial rates in three experiments are: (1) [A] = 0.10 mol L⁻¹, [B] = 0.10 mol L⁻¹, rate = 2.0 × 10⁻³ mol L⁻¹ s⁻¹; (2) [A] = 0.20 mol L⁻¹, [B] = 0.10 mol L⁻¹, rate = 4.0 × 10⁻³ mol L⁻¹ s⁻¹; (3) [A] = 0.10 mol L⁻¹, [B] = 0.20 mol L⁻¹, rate = 8.0 × 10⁻³ mol L⁻¹ s⁻¹. What are the orders with respect to A and B, and the overall order?

  1. Second order in A, second order in B, overall fourth order
  2. First order in A, first order in B, overall second order
  3. Second order in A, first order in B, overall third order
  4. First order in A, second order in B, overall third order Correct answer.

Explanation

Doubling [A] while [B] is held constant doubles the rate, so the reaction is first order with respect to A. Doubling [B] while [A] is constant quadruples the rate, so it is second order with respect to B. The overall order is the sum, 1 + 2 = 3.

Question 43Chemical Kinetics

A first-order reaction has a rate constant k = 6.93 × 10⁻³ s⁻¹. What is the half-life of the reaction?

  1. 100 s Correct answer.
  2. 693 s
  3. 10 s
  4. 50 s

Explanation

For a first-order reaction the half-life is given by t1/2 = 0.693/k and is independent of the initial concentration. Substituting k = 6.93 × 10⁻³ s⁻¹ gives t1/2 = 0.693/(6.93 × 10⁻³) ≈ 100 s.

Question 44Chemical Kinetics

A zero-order reaction has a rate constant k = 2.0 × 10⁻² mol L⁻¹ s⁻¹ and an initial concentration of A equal to 0.10 mol L⁻¹. What is the half-life?

  1. 50 s
  2. 0.20 s
  3. 2.5 s Correct answer.
  4. 5.0 s

Explanation

For a zero-order reaction the concentration falls linearly with time, and the rate stays constant until the reactant is used up. The half-life is t1/2 = [A]0/2k = 0.10/(2 × 2.0 × 10⁻²) = 2.5 s.

Question 45Chemical Kinetics

A catalyst increases the rate of a reversible gaseous reaction. Which statement about the action of the catalyst is correct?

  1. It lowers the activation energy of the forward reaction only
  2. It increases the equilibrium constant by speeding up the forward reaction
  3. It provides a different pathway of lower activation energy Correct answer.
  4. It raises the activation energy so that more collisions become effective

Explanation

A catalyst offers an alternative reaction pathway with a lower activation energy, so a larger fraction of collisions at a given temperature is effective. It lowers the activation energy of the forward and reverse reactions by the same amount, so the enthalpy change and the equilibrium constant remain unchanged.

Question 46Chemical Kinetics

The oxidation of nitric oxide, 2NO(g) + O2(g) → 2NO2(g), is thought to proceed by a fast pre-equilibrium 2NO ⇌ N2O2 followed by the slow step N2O2 + O2 → 2NO2. Which rate law is consistent with this mechanism?

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

Explanation

The slow step controls the overall rate, so rate = k2[N2O2][O2]. The fast pre-equilibrium gives [N2O2] = (k1/k−1)[NO]², and substituting this into the rate expression gives rate = k[NO]²[O2], a third-order overall reaction.

Question 47Chemical Kinetics

For a reaction, a graph of ln k against 1/T is a straight line with slope −1.2 × 10⁴ K. Using R = 8.314 J K⁻¹ mol⁻¹, what is the activation energy?

  1. about 10 kJ mol⁻¹
  2. about 8.3 kJ mol⁻¹
  3. about 1.0 kJ mol⁻¹
  4. about 100 kJ mol⁻¹ Correct answer.

Explanation

The Arrhenius equation in logarithmic form gives a straight line of slope −Ea/R when ln k is plotted against 1/T. Therefore Ea = −slope × R = 1.2 × 10⁴ K × 8.314 J K⁻¹ mol⁻¹ ≈ 1.0 × 10⁵ J mol⁻¹, which is about 100 kJ mol⁻¹.

Question 48Chemical Kinetics

The hydrolysis of an ester in water containing a large excess of water follows pseudo first-order kinetics. What is the reason for this description?

  1. The ester concentration stays constant while water is slowly consumed
  2. Water acts as a catalyst, so its concentration is left out of the rate law
  3. The reaction is zero order with respect to the ester
  4. Water is in large excess, so its concentration stays almost constant Correct answer.

Explanation

The rate law for the hydrolysis is rate = k[ester][H2O]. In dilute aqueous solution water is present in large excess, so its concentration changes very little during the reaction and can be treated as constant. The observed rate law therefore becomes rate = k′[ester], with k′ = k[H2O], which is why the reaction is called pseudo first order.

Question 49Chemical Kinetics

For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), which expression gives the rate of reaction correctly in terms of concentration changes?

  1. −Δ[N2O5]/Δt = +Δ[NO2]/Δt = +Δ[O2]/Δt
  2. −(1/2) Δ[N2O5]/Δt = +(1/2) Δ[NO2]/Δt = +Δ[O2]/Δt
  3. −(1/2) Δ[N2O5]/Δt = +(1/4) Δ[NO2]/Δt = +(1/2) Δ[O2]/Δt
  4. −(1/2) Δ[N2O5]/Δt = +(1/4) Δ[NO2]/Δt = +Δ[O2]/Δt Correct answer.

Explanation

The rate of reaction is defined by dividing the change in concentration of each species by its stoichiometric coefficient and by the time interval, with a negative sign for reactants and a positive sign for products. For 2 N2O5 → 4 NO2 + O2, this gives −(1/2)Δ[N2O5]/Δt = +(1/4)Δ[NO2]/Δt = +Δ[O2]/Δt. Only this expression preserves the 2:4:1 stoichiometric relationship among the three species.

Question 50Chemical Kinetics

The rate constant of a reaction has units L mol⁻¹ s⁻¹. What is the overall order of the reaction?

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

Explanation

For an overall order n, the units of the rate constant are (mol L⁻¹)^(1−n) s⁻¹, which can also be written as L^(n−1) mol^(1−n) s⁻¹. The given units L mol⁻¹ s⁻¹ correspond to n = 2. Therefore the reaction is second order overall.