Understand Rate of Chemical Change - IB Chemistry

1. 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?
   Answer: 0.20 mol L⁻¹ 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.
2. Enzymes are proteins that act as highly specific biological catalysts. How does an enzyme such as catalase increase the rate of decomposition of hydrogen peroxide?
   Answer: By providing an active site that lowers the activation energy of the decomposition. Enzymes, like all catalysts, increase reaction rates by providing an alternative reaction pathway with a lower activation energy. They do this via their specific three-dimensional structure, which includes an 'active site' where the reactant (substrate) binds and is converted to product. Enzymes do not change the temperature of the system (A), nor do they alter the overall thermodynamics ($\Delta H$) of the reaction (D). While they bind the substrate (B), the fundamental mechanism is the lowering of the activation energy.
3. An iron catalyst is used in the Haber process: $\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g})$. What is the primary effect of this catalyst on the reaction system?
   Answer: It increases the rates of both the forward and reverse reactions. 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).
4. 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?
   Answer: The initial temperature of the acid solution. 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.
5. Which statement correctly describes the function of a catalyst in a chemical reaction?
   Answer: It provides an alternative reaction mechanism which has a lower activation energy. 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 ($\Delta H$) of the reaction (D).
6. 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?
   Answer: Increasing the temperature and grinding the solid reactant into a powder. 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.
7. Some substances, known as inhibitors, slow down chemical reactions. Which is the most plausible mechanism for an inhibitor's action?
   Answer: It provides an alternative reaction pathway with a higher activation energy. 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).
8. Consider the reaction: $2\text{KMnO}_4(\text{aq}) + 5\text{H}_2\text{C}_2\text{O}_4(\text{aq}) + 3\text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{K}_2\text{SO}_4(\text{aq}) + 2\text{MnSO}_4(\text{aq}) + 10\text{CO}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l})$. Which change would NOT cause an increase in the initial rate of this reaction?
   Answer: Increasing the pressure above the reaction mixture. 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.
9. The rate of formation of product P in the reaction $2A + B \rightarrow 3P + 2Q$ is measured as $4.2 \times 10^{-3}$ mol L⁻¹ s⁻¹ at a particular instant. What is the rate of consumption of reactant A at this same instant?
   Answer: $2.8 \times 10^{-3}$ mol L⁻¹ s⁻¹ Using stoichiometric relationships for reaction rates: $-\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3}\frac{d[P]}{dt} = \frac{1}{2}\frac{d[Q]}{dt}$. Given that $\frac{d[P]}{dt} = 4.2 \times 10^{-3}$ mol L⁻¹ s⁻¹, we can find: $-\frac{1}{2}\frac{d[A]}{dt} = \frac{1}{3}(4.2 \times 10^{-3})$. Solving: $\frac{d[A]}{dt} = -2 \times \frac{4.2 \times 10^{-3}}{3} = -2.8 \times 10^{-3}$ mol L⁻¹ s⁻¹. The rate of consumption is the magnitude: $2.8 \times 10^{-3}$ 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.
10. Hydrogen peroxide decomposes in the presence of a solid manganese(IV) oxide catalyst: $2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})$. To achieve the fastest initial rate of decomposition with 1.0 g of the catalyst, the manganese(IV) oxide should be used as:
    Answer: a fine powder to maximize the number of active sites available for reaction. 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.
11. 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?
    Answer: The total surface area of the reactant is decreased. 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.
12. The reaction $\text{CaCO}_3(\text{s}) + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O(l)} + \text{CO}_2(\text{g})$ 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.
    Answer: I, II and III All three methods can be used to monitor the rate. I: The mass of the system will decrease as $\text{CO}_2$ gas escapes from the open beaker. II: The volume of $\text{CO}_2$ 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.
13. An endothermic reaction has an activation energy of $+150 \text{ kJ mol}^{-1}$. The enthalpy change for the reaction is $+50 \text{ kJ mol}^{-1}$. What is the activation energy for the reverse reaction?
    Answer: +100 \text{ kJ mol}^{-1}\ For any reaction, the relationship between the forward activation energy (\(E_{a,fwd}), reverse activation energy ($E_{a,rev}$), and enthalpy change ($\Delta H$) is $\Delta H = E_{a,fwd} - E_{a,rev}$. Rearranging gives $E_{a,rev} = E_{a,fwd} - \Delta H$. Substituting the given values: $E_{a,rev} = (+150 \text{ kJ mol}^{-1}) - (+50 \text{ kJ mol}^{-1}) = +100 \text{ kJ mol}^{-1}$. Activation energies are always positive values.
14. For the gaseous equilibrium $2\text{NO(g)} + \text{O}_2(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})$, what is the effect of doubling the pressure by reducing the volume at a constant temperature?
    Answer: The rate of the forward reaction increases because the frequency of molecular collisions increases. 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).
15. Which statement best explains why most molecular collisions do not result in a chemical reaction?
    Answer: The colliding particles must possess both sufficient kinetic energy and the correct reactive orientation. According to collision theory, for a reaction to occur, colliding particles must meet two conditions simultaneously: they must have kinetic energy equal to or greater than the activation energy, and they must collide with the correct geometric orientation. Choices A and B only state one of the necessary conditions, making them incomplete. Choice D describes a factor that determines the overall rate, not why an individual collision is unsuccessful.
16. The reaction $2NO(g) + Cl_2(g) \rightarrow 2NOCl(g)$ proceeds via a proposed mechanism where the rate-determining step is $NOCl_2 + NO \rightarrow 2NOCl$ with rate constant k₂. The fast pre-equilibrium is $NO + Cl_2 \rightleftharpoons NOCl_2$ 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?
    Answer: The mechanism is consistent with the experimental data and predicts a rate law of rate = k₂K₁[NO]²[Cl₂] 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 $NO + Cl_2 \rightleftharpoons NOCl_2$, we have $K_1 = \frac{[NOCl_2]}{[NO][Cl_2]}$, so $[NOCl_2] = K_1[NO][Cl_2]$. 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.
17. When the temperature of a gaseous reaction system is increased, what is the principal reason for the significant increase in reaction rate?
    Answer: The fraction of collisions possessing energy greater than the activation energy increases significantly. Increasing the temperature increases the average kinetic energy of the molecules. This causes the Maxwell-Boltzmann distribution curve to flatten and shift to the right, meaning a significantly larger proportion of molecules will have kinetic energy equal to or greater than the activation energy (Ea). While collision frequency also increases slightly, the primary effect on the rate is this increase in the fraction of effective collisions. Temperature does not lower the activation energy (B), and the average kinetic energy increases, not decreases (A, D).
18. 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?
    Answer: The ratio will remain exactly 5 because both reactions have the same activation energy Using the Arrhenius equation: k = $Ae^{-Ea/RT}$. Since both reactions have the same Ea but different A values: kₓ = Aₓ$e^{-Ea/RT}$ and kᵧ = Aᵧ$e^{-Ea/RT}$. 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.
19. A reaction mechanism consists of two elementary steps: Step 1 (fast equilibrium): $A + B \rightleftharpoons C$ with $K_{eq} = 2.5$; Step 2 (slow): $C + D \rightarrow E + F$. If the rate law for step 2 is rate = k₂[C][D], what is the overall rate law for the reaction $A + B + D \rightarrow E + F$?
    Answer: rate = k₂$K_{eq}$[A][B][D] 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] = k₂$K_{eq}$[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.
20. For the reaction $\text{A(aq)} + \text{B(aq)} \rightarrow \text{C(aq)}$, 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?
    Answer: The frequency of effective collisions between reactant particles A and B decreases. 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.

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