All questions
Question 1
An endothermic reaction has an activation energy of +150 kJ mol−1. The enthalpy change for the reaction is +50 kJ mol−1. What is the activation energy for the reverse reaction?
- (-150 \text{ kJ mol}^{-1}\
- (-100 \text{ kJ mol}^{-1}\
- (+100 \text{ kJ mol}^{-1}\ (correct answer)
- (+200 \text{ kJ mol}^{-1}\
Explanation: For any reaction, the relationship between the forward activation energy (Ea,fwd), reverse activation energy (Ea,rev), and enthalpy change (ΔH) is ΔH=Ea,fwd−Ea,rev. Rearranging gives Ea,rev=Ea,fwd−ΔH. Substituting the given values: Ea,rev=(+150 kJ mol−1)−(+50 kJ mol−1)=+100 kJ mol−1. Activation energies are always positive values. Question 2
Hydrogen peroxide decomposes in the presence of a solid manganese(IV) oxide catalyst: 2H2O2(aq)→2H2O(l)+O2(g). To achieve the fastest initial rate of decomposition with 1.0 g of the catalyst, the manganese(IV) oxide should be used as:
- a single 1.0 g pellet to ensure it is not washed away by the effervescence.
- a fine powder to maximize the number of active sites available for reaction. (correct answer)
- dissolved in the aqueous solution to create a homogeneous catalytic system.
- a 1.0 g pellet heated to a high temperature before being added to the solution.
Explanation: This reaction involves a heterogeneous catalyst (solid catalyst, aqueous reactant). The rate of such a reaction is dependent on the surface area of the catalyst. Using the catalyst as a fine powder maximizes its surface area, which in turn maximizes the number of active sites available to the reactant molecules, leading to the fastest rate. Manganese(IV) oxide is insoluble (C). Pre-heating the catalyst (D) would briefly warm the solution, but the effect of surface area is far more significant and persistent.
Question 3
An iron catalyst is used in the Haber process: N2(g)+3H2(g)⇌2NH3(g). What is the primary effect of this catalyst on the reaction system?
- It increases the rates of both the forward and reverse reactions. (correct answer)
- It shifts the equilibrium position to the right to favor product formation.
- It increases the percentage yield of ammonia at equilibrium.
- It decreases the activation energy for the forward reaction but not the reverse reaction.
Explanation: A catalyst speeds up the rate at which equilibrium is reached but does not affect the position of the equilibrium itself. It achieves this by lowering the activation energy for both the forward and reverse reactions by an equal amount. Therefore, it increases the rates of both reactions equally. It does not change the percentage yield at equilibrium (A, B).
Question 4
The reaction CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g) is monitored in an open beaker. Which set of measurements is suitable for determining the rate of this reaction?
I. Change in mass of the beaker and contents over time.
II. Volume of gas collected in a gas syringe over time.
III. Change in pH of the solution over time.
- I and II only
- I and III only
- II and III only
- I, II and III (correct answer)
Explanation: All three methods can be used to monitor the rate. I: The mass of the system will decrease as CO2 gas escapes from the open beaker. II: The volume of CO2 gas produced can be collected and measured over time. III: As the hydrochloric acid (HCl) is consumed, the concentration of H⁺ ions decreases, so the pH of the solution will increase over time. This change in pH can be monitored with a pH meter. Question 5
A student plans an experiment to determine how the concentration of an acid affects the rate of its reaction with a metal. To ensure a fair test, which variable is the most crucial to keep constant across all experimental trials?
- The total time taken for the reaction to complete.
- The total volume of gas produced by the end of the reaction.
- The method used for measuring the volume of gas.
- The initial temperature of the acid solution. (correct answer)
Explanation: The student is investigating the effect of concentration (the independent variable) on the rate (the dependent variable). To isolate the effect of concentration, all other factors that could influence the reaction rate must be controlled (kept constant). Temperature has a significant effect on reaction rate, so it is crucial to keep the initial temperature of the acid constant for each trial. The time taken (A) and total volume of gas (B) are dependent variables that will be measured. The measurement method (C) should be consistent but is part of the procedure, not a physical variable affecting the rate itself.
Question 6
For the reaction A(aq)+B(aq)→C(aq), the concentration of product C is plotted against time. The slope of the resulting curve is observed to decrease as the reaction proceeds. What is the best explanation for this observation?
- The frequency of effective collisions between reactant particles A and B decreases. (correct answer)
- The temperature of the system decreases as the reaction is endothermic.
- The activation energy of the reaction increases as the concentration of product C increases.
- The catalyst for the reaction is being consumed, making it less effective over time.
Explanation: The slope of a concentration vs. time graph represents the rate of reaction. The rate decreases over time because the concentrations of the reactants (A and B) are decreasing as they are converted into product C. A lower concentration of reactants leads to a lower frequency of collisions, and therefore a lower frequency of effective collisions, slowing the rate. The activation energy is constant (A). Catalysts are not consumed (D). While temperature might change (B), the primary and universal reason for the rate decrease is reactant depletion.
Question 7
The rate of a chemical reaction is found to decrease significantly when the reactants are changed from a powder to large lumps, keeping total mass and other conditions constant. What is the primary reason for this change in rate?
- The total surface area of the reactant is decreased. (correct answer)
- The concentration of the solid reactant is lower in the lumps.
- The activation energy is higher for the lumps than for the powder.
- The lumps are less pure than the powdered form of the reactant.
Explanation: For a given mass of a solid, grinding it into a powder dramatically increases its total surface area. For reactions involving a solid, collisions can only occur on the surface. By using large lumps instead of a powder, the available surface area for reaction is significantly decreased. This reduces the frequency of collisions between reactant particles, leading to a much slower reaction rate. The chemical nature of the substance, and thus its activation energy (A), does not change with particle size.
Question 8
For the gaseous equilibrium 2NO(g)+O2(g)⇌2NO2(g), what is the effect of doubling the pressure by reducing the volume at a constant temperature?
- The rate of the forward reaction increases because the activation energy is decreased.
- The rate of the forward reaction increases because the frequency of molecular collisions increases. (correct answer)
- The rate of the forward reaction decreases because the average distance between molecules increases.
- The rate of the forward reaction is unchanged because the kinetic energy of the molecules is constant.
Explanation: Increasing the pressure by reducing the volume increases the concentration of the gaseous reactants. This leads to a greater frequency of collisions between reactant molecules per unit time, thus increasing the rate of the forward reaction. Pressure does not affect the activation energy (A). Increasing pressure decreases the average distance between molecules (C). While the average kinetic energy is constant at constant temperature, the collision frequency is not, so the rate changes (D).
Question 9
Two experiments are conducted. In Experiment 1, 1.0 g of magnesium powder reacts with 100 cm³ of 1.0 mol dm⁻³ HCl. In Experiment 2, 1.0 g of magnesium powder reacts with 100 cm³ of 2.0 mol dm⁻³ HCl. In both experiments, magnesium is the limiting reactant. How will the initial rate of reaction and the total volume of H₂ produced in Experiment 2 compare to Experiment 1?
- Rate is faster; total volume of H₂ is greater.
- Rate is faster; total volume of H₂ is the same. (correct answer)
- Rate is the same; total volume of H₂ is greater.
- Rate is the same; total volume of H₂ is the same.
Explanation: The initial rate of reaction depends on the concentration of the reactants. Since Experiment 2 uses a higher concentration of HCl (2.0 mol dm⁻³ vs 1.0 mol dm⁻³), its initial rate will be faster. The total volume of product is determined by the limiting reactant. Since magnesium is stated to be the limiting reactant in both cases, and the mass of magnesium is the same (1.0 g), the total amount (moles) of H₂ produced will be identical in both experiments.
Question 10
Consider the reaction: 2KMnO4(aq)+5H2C2O4(aq)+3H2SO4(aq)→K2SO4(aq)+2MnSO4(aq)+10CO2(g)+8H2O(l). Which change would NOT cause an increase in the initial rate of this reaction?
- Increasing the concentration of H2C2O4(aq).
- Increasing the pressure above the reaction mixture. (correct answer)
- Warming the reaction mixture from 20 °C to 30 °C.
- Adding a small amount of MnSO4(aq), a known autocatalyst for this reaction.
Explanation: The rate of reaction is affected by the concentration of reactants, temperature, and catalysts. Increasing reactant concentration (A), increasing temperature (C), and adding a catalyst (D) will all increase the rate. However, pressure only has a significant effect on the rate of reactions involving gases as reactants. In this reaction, all reactants are in the aqueous phase. Although a gas is produced, increasing the external pressure will not affect the rate at which the aqueous reactants collide and react. Therefore, increasing pressure will not increase the initial rate.
Question 11
A piece of magnesium ribbon is added to an excess of hydrochloric acid. The volume of hydrogen gas produced is measured over time. Which statement best describes the change in the rate of reaction as time proceeds?
- The rate is constant throughout the reaction because the acid is in excess.
- The rate is fastest at the beginning and gradually decreases until the reaction stops. (correct answer)
- The rate increases over time as the reaction is exothermic, which heats the mixture.
- The rate is zero at the start and increases to a maximum before decreasing to zero.
Explanation: The rate of reaction is highest at the beginning (t=0) when the concentration of reactants is at its maximum. As the reaction proceeds, the magnesium is consumed (its surface area decreases) and the concentration of HCl decreases, leading to fewer frequent and effective collisions. Therefore, the rate gradually slows down and becomes zero when the limiting reactant (magnesium) is fully consumed. While the reaction is exothermic (C), the effect of decreasing reactant concentration is the dominant factor determining the shape of the rate curve.
Question 12
The reaction 2NO(g)+Cl2(g)→2NOCl(g) proceeds via a proposed mechanism where the rate-determining step is NOCl2+NO→2NOCl with rate constant k₂. The fast pre-equilibrium is NO+Cl2⇌NOCl2 with equilibrium constant K₁. Experimental data shows that doubling [NO] increases the rate by a factor of 4, while doubling [Cl₂] doubles the rate. What can be concluded about this mechanism?
- The mechanism is consistent but requires modification of the equilibrium constant expression for the pre-equilibrium step
- The mechanism is inconsistent because the predicted rate law would show first-order dependence on [NO]
- The mechanism is inconsistent because the fast pre-equilibrium assumption cannot account for second-order NO dependence
- The mechanism is consistent with the experimental data and predicts a rate law of rate = k₂K₁[NO]²[Cl₂] (correct answer)
Explanation: When you encounter reaction mechanisms with pre-equilibria, you need to derive the rate law from the proposed steps and compare it to experimental observations. The key is understanding how fast pre-equilibrium steps affect the overall rate expression.
For this mechanism, start with the rate-determining step: rate = k₂[NOCl₂][NO]. Since NOCl₂ is an intermediate, you must express its concentration using the pre-equilibrium. For the fast equilibrium NO+Cl2⇌NOCl2, we have K1=[NO][Cl2][NOCl2], so [NOCl2]=K1[NO][Cl2].
Substituting this into the rate law gives: rate = k₂K₁[NO][Cl₂][NO] = k₂K₁[NO]²[Cl₂]. This predicts second-order dependence on [NO] and first-order on [Cl₂], which perfectly matches the experimental data: doubling [NO] increases rate by 4× (2² = 4), while doubling [Cl₂] doubles the rate (2¹ = 2).
Option A is wrong because the equilibrium constant expression is correctly written as given. Option B incorrectly claims the mechanism predicts first-order NO dependence—it actually predicts second-order. Option C suggests the pre-equilibrium assumption fails, but it successfully explains the second-order NO dependence through the combination of NO appearing in both the pre-equilibrium and rate-determining step. Option D correctly identifies that the mechanism is consistent and properly states the derived rate law.
Remember: when analyzing mechanisms, always derive the complete rate law by substituting equilibrium expressions for intermediates, then compare orders of dependence to experimental data. Question 13
The half-life of a second-order reaction with initial concentration 0.60 mol L⁻¹ is 240 seconds. What will be the concentration of the reactant after 480 seconds?
- 0.15 mol L⁻¹
- 0.20 mol L⁻¹ (correct answer)
- 0.24 mol L⁻¹
- 0.30 mol L⁻¹
Explanation: For a second-order reaction, the half-life depends on concentration: t₁/₂ = 1/(k[A₀]). Given t₁/₂ = 240 s and [A₀] = 0.60 mol L⁻¹: 240 = 1/(k × 0.60), so k = 1/(240 × 0.60) = 0.00694 L mol⁻¹ s⁻¹. For second-order kinetics: 1/[A] = 1/[A₀] + kt. After 480 s: 1/[A] = 1/0.60 + 0.00694(480) = 1.67 + 3.33 = 5.00. Therefore [A] = 1/5.00 = 0.20 mol L⁻¹. Choice A incorrectly applies first-order kinetics (0.60/4 after 2 half-lives). Choice C uses incorrect rate constant calculation. Choice D assumes linear decrease.
Question 14
Two reactions, X and Y, have the same activation energy but different pre-exponential factors (A). At 298 K, reaction X has a rate constant that is 5 times larger than reaction Y. If both reactions are heated to 350 K, what will be the ratio of their rate constants (kₓ/kᵧ) at this new temperature?
- The ratio will be greater than 5 because the faster reaction benefits more from temperature increase
- The ratio will be less than 5 because temperature increases favor the slower reaction more significantly
- The ratio will remain exactly 5 because both reactions have the same activation energy (correct answer)
- The ratio cannot be determined without knowing the specific activation energy values
Explanation: Using the Arrhenius equation: k = Ae^(-Ea/RT). Since both reactions have the same Ea but different A values: kₓ = Aₓe^(-Ea/RT) and kᵧ = Aᵧe^(-Ea/RT). The ratio kₓ/kᵧ = Aₓ/Aᵧ at any temperature, since the exponential terms cancel out. At 298 K, kₓ/kᵧ = 5, which means Aₓ/Aᵧ = 5. At 350 K, the ratio will still be Aₓ/Aᵧ = 5. Choice A incorrectly assumes the faster reaction gains more benefit. Choice B incorrectly assumes the slower reaction catches up. Choice D incorrectly suggests we need Ea values when the ratio depends only on the pre-exponential factors.
Question 15
The rate of formation of product P in the reaction 2A+B→3P+2Q is measured as 4.2×10−3 mol L⁻¹ s⁻¹ at a particular instant. What is the rate of consumption of reactant A at this same instant?
- 2.8×10−3 mol L⁻¹ s⁻¹ (correct answer)
- 6.3×10−3 mol L⁻¹ s⁻¹
- 1.4×10−3 mol L⁻¹ s⁻¹
- 8.4×10−3 mol L⁻¹ s⁻¹
Explanation: Using stoichiometric relationships for reaction rates: −21dtd[A]=−dtd[B]=31dtd[P]=21dtd[Q]. Given that dtd[P]=4.2×10−3 mol L⁻¹ s⁻¹, we can find: −21dtd[A]=31(4.2×10−3). Solving: dtd[A]=−2×34.2×10−3=−2.8×10−3 mol L⁻¹ s⁻¹. The rate of consumption is the magnitude: 2.8×10−3 mol L⁻¹ s⁻¹. Choice B uses incorrect stoichiometry (3:2 instead of 3:2). Choice C forgets the factor of 2 from stoichiometry. Choice D incorrectly multiplies by 2 instead of dividing. Question 16
A reaction mechanism consists of two elementary steps: Step 1 (fast equilibrium): A+B⇌C with Keq=2.5; Step 2 (slow): C+D→E+F. If the rate law for step 2 is rate = k₂[C][D], what is the overall rate law for the reaction A+B+D→E+F?
- rate = k₂K_{eq}[A][B][D] (correct answer)
- rate = k₂[A][B][D]/K_{eq}
- rate = k₂K_{eq}[C][D]
- rate = k₂[C][D]
Explanation: Since step 1 is a fast equilibrium, we can use the equilibrium approximation: K_{eq} = [C]/([A][B]) = 2.5, so [C] = 2.5[A][B]. The rate-determining step is step 2: rate = k₂[C][D]. Substituting the expression for [C]: rate = k₂(2.5[A][B])[D] = 2.5k₂[A][B][D] = k₂K_{eq}[A][B][D]. Choice B incorrectly inverts the equilibrium constant. Choice C fails to substitute for [C] in terms of reactants. Choice D doesn't account for the fast pre-equilibrium.
Question 17
How does the addition of a suitable catalyst affect the Maxwell-Boltzmann energy distribution curve and the activation energy, Ea, of a reaction?
- The curve flattens and shifts to the right, and Ea decreases.
- The curve is unchanged, but Ea decreases. (correct answer)
- The curve's peak increases in height and shifts left, and Ea is unchanged.
- The curve is unchanged, but Ea increases.
Explanation: A catalyst provides an alternative reaction pathway with a lower activation energy (Ea). It does not affect the kinetic energy distribution of the molecules themselves. Therefore, the Maxwell-Boltzmann curve, which represents this distribution, remains unchanged. An inhibitor might increase Ea (D), and a change in temperature would change the curve (A, C). Question 18
Which statement correctly describes the function of a catalyst in a chemical reaction?
- It increases the average kinetic energy of reactant particles, making collisions more effective.
- It is consumed during the reaction while providing an alternative, faster reaction pathway.
- It provides an alternative reaction mechanism which has a lower activation energy. (correct answer)
- It increases the magnitude of the enthalpy change (ΔH) of the reaction, causing it to proceed faster.
Explanation: A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. It does not change the kinetic energy of particles (A, effect of temperature), it is not consumed in the overall reaction (B), and it does not alter the overall enthalpy change (ΔH) of the reaction (D). Question 19
The rate of reaction between a solid reactant and an aqueous reactant is investigated. Which combination of changes will cause the largest increase in the reaction rate?
- Decreasing the concentration of the aqueous reactant and grinding the solid into a powder.
- Increasing the temperature and using a single large piece of the solid reactant.
- Increasing the temperature and grinding the solid reactant into a powder. (correct answer)
- Decreasing the temperature and increasing the concentration of the aqueous reactant.
Explanation: To achieve the largest increase in rate, factors that increase rate should be combined. Increasing the temperature increases the kinetic energy of particles and the frequency of collisions. Grinding a solid reactant into a powder increases its surface area, which also increases the frequency of collisions. Therefore, combining these two changes (C) will have the most significant positive effect on the rate. The other options combine a factor that increases rate with one that decreases it.
Question 20
Some substances, known as inhibitors, slow down chemical reactions. Which is the most plausible mechanism for an inhibitor's action?
- It decreases the concentration of reactants by forming a stable compound with them.
- It provides an alternative reaction pathway with a higher activation energy. (correct answer)
- It lowers the temperature of the system by absorbing energy from the surroundings.
- It reverses the enthalpy change of the reaction, making it less favorable.
Explanation: An inhibitor acts in the opposite way to a catalyst. While a catalyst provides a pathway with a lower activation energy, an inhibitor works by blocking the normal reaction mechanism or providing an alternative pathway that has a higher activation energy, thus slowing the reaction. Reacting with a reactant (A) is a competing reaction, not inhibition of the main pathway. Inhibitors do not change the thermodynamics of the system (C, D).