Reaction Rates - College Chemistry

For the reaction sequence $A \rightarrow B \rightarrow C$ , where both steps are first-order with rate constants $k_1 = 0.10 \text{ s}^{-1}$ and $k_2 = 0.050 \text{ s}^{-1}$ , which statement best describes the concentration of intermediate B over time?
Answer: $[B]$ increases to a maximum then decreases because it is both formed and consumed When you encounter consecutive first-order reactions, you're dealing with a classic kinetics scenario where an intermediate species is both formed and consumed simultaneously. The key insight is recognizing how the relative rate constants affect the intermediate's concentration profile. For the sequence $A \rightarrow B \rightarrow C$ , intermediate B is formed from A at rate $k_1[A]$ and consumed to form C at rate $k_2[B]$ . Initially, when $[A]$ is high and $[B]$ is zero, B forms faster than it's consumed. As A depletes and B accumulates, the formation rate decreases while the consumption rate increases. Eventually, B reaches a maximum concentration when its formation and consumption rates are equal, then continues to decrease as A becomes depleted. This creates the characteristic rise-and-fall profile described in choice B. Choice A is incorrect because first-order kinetics means the rate depends on concentration, not time, so B's formation rate decreases as A is consumed. Choice C describes a steady-state approximation that might apply when $k_2 >> k_1$ , but here $k_1 > k_2$ , so B accumulates significantly before declining. Choice D incorrectly focuses on the relative magnitudes of the rate constants - what matters is that $k_1 > k_2$ , meaning B forms faster initially than it's consumed, leading to accumulation rather than immediate decrease. Remember: in consecutive reactions, intermediates typically show a concentration maximum when the first step is faster than the second. This pattern appears frequently in reaction mechanisms and enzyme kinetics.
A first-order reaction has a half-life of 20.0 minutes. What fraction of the original reactant remains after 60.0 minutes?
Answer: $\frac{1}{8}$ When you encounter first-order kinetics problems, remember that the half-life remains constant throughout the reaction, making calculations straightforward using the half-life method. For first-order reactions, you can track the remaining reactant by counting how many half-lives have elapsed. With a half-life of 20.0 minutes, let's see what happens over 60.0 minutes: - After 0 minutes: 100% remains (starting amount) - After 20 minutes (1 half-life): 50% remains ( $\frac{1}{2}$ ) - After 40 minutes (2 half-lives): 25% remains ( $\frac{1}{4}$ ) - After 60 minutes (3 half-lives): 12.5% remains ( $\frac{1}{8}$ ) Since 60.0 ÷ 20.0 = 3 half-lives, the fraction remaining is $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$ , confirming answer B. Looking at the incorrect options: A) $\frac{1}{16}$ represents what would remain after 4 half-lives (80 minutes), suggesting you miscounted the time intervals. C) $\frac{1}{4}$ is what remains after only 2 half-lives (40 minutes), indicating you stopped your calculation one step too early. D) $\frac{1}{3}$ doesn't correspond to any whole number of half-lives and suggests confusion about the half-life concept entirely. Study tip: For first-order reactions, always use the formula: fraction remaining = $\left(\frac{1}{2}\right)^n$ where $n$ = total time ÷ half-life. This approach is faster and more reliable than using the integrated rate law for these straightforward problems.
A catalyst is added to a reaction system. Which of the following correctly describes the effect of the catalyst on the reaction?
Answer: The catalyst decreases the activation energy and increases the rate constant but does not affect the equilibrium position When you encounter questions about catalysts, focus on their fundamental role: they speed up reactions by providing alternative pathways with lower energy barriers, without being consumed or affecting where the reaction ultimately settles. A catalyst works by lowering the activation energy ( $E_a$ ) for both the forward and reverse reactions equally. This creates an alternative reaction pathway that requires less energy to proceed. According to the Arrhenius equation, $k = Ae^{-E_a/RT}$ , when activation energy decreases, the rate constant increases exponentially. Importantly, because catalysts lower the activation energy for both directions equally, they speed up both forward and reverse reactions proportionally, leaving the equilibrium position unchanged—they just help the system reach equilibrium faster. Option A incorrectly states that catalysts change the equilibrium position. Catalysts never alter equilibrium concentrations; they only affect how quickly equilibrium is reached. Option C contradicts the basic mechanism of catalysis by claiming activation energies increase. This is backwards—catalysts always decrease activation energy. Option D also incorrectly suggests that activation energy increases, then tries to compensate with an increased frequency factor. While the frequency factor can vary between pathways, the primary effect of catalysis is definitively the reduction of activation energy. Remember this key principle: catalysts are "speed helpers, not outcome changers." They make reactions faster by lowering energy barriers, but they never shift where the equilibrium lies. If you see answer choices suggesting catalysts change equilibrium positions or increase activation energies, eliminate them immediately.
At 25°C, a certain reaction has a rate constant of $2.5 \times 10^{-4} \text{ s}^{-1}$ . At 35°C, the rate constant is $7.5 \times 10^{-4} \text{ s}^{-1}$ . What is the activation energy for this reaction? (R = 8.314 J/mol·K)
Answer: $9.1 \times 10^4 \text{ J/mol}$ When you encounter a question about how reaction rates change with temperature, you're dealing with the Arrhenius equation, which relates rate constants to activation energy and temperature. To find activation energy, use the two-point form of the Arrhenius equation: $\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ First, convert temperatures to Kelvin: T₁ = 298 K (25°C) and T₂ = 308 K (35°C). Calculate the ratio of rate constants: $\ln\left(\frac{7.5 \times 10^{-4}}{2.5 \times 10^{-4}}\right) = \ln(3) = 1.099$ Calculate the temperature term: $\frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{308} - \frac{1}{298} = -1.09 \times 10^{-4} \text{ K}^{-1}$ Solve for activation energy: $1.099 = -\frac{E_a}{8.314} \times (-1.09 \times 10^{-4})$ $E_a = \frac{1.099}{8.314 \times 1.09 \times 10^{-4}} = 9.1 \times 10^4 \text{ J/mol}$ This confirms answer D is correct. Answer A ( $2.8 \times 10^4$ ) likely results from calculation errors in the natural logarithm or temperature conversion. Answer B ( $3.6 \times 10^4$ ) suggests using Celsius temperatures instead of Kelvin. Answer C ( $7.6 \times 10^4$ ) probably comes from sign errors in the temperature difference calculation. Study tip: Always convert Celsius to Kelvin in kinetics problems, and remember that higher temperatures give larger rate constants—if your calculation suggests otherwise, check your signs and temperature units.
For a reaction with rate law $\text{rate} = k[A]^2[B]$ , the initial rate is $2.0 \times 10^{-3} \text{ M/s}$ when $[A] = 0.20 \text{ M}$ and $[B] = 0.30 \text{ M}$ . What is the value of the rate constant k?
Answer: $0.17 \text{ M}^{-2}\text{s}^{-1}$ When you encounter rate law problems, you're working with the fundamental relationship between reaction rate, rate constant, and reactant concentrations. The rate law $\text{rate} = k[A]^2[B]$ tells you how the reaction rate depends on the concentrations of reactants A and B. To find the rate constant k, rearrange the rate law equation: $k = \frac{\text{rate}}{[A]^2[B]}$ . Substituting the given values: $k = \frac{2.0 \times 10^{-3}}{(0.20)^2(0.30)} = \frac{2.0 \times 10^{-3}}{(0.040)(0.30)} = \frac{2.0 \times 10^{-3}}{0.012} = 0.167 \text{ M}^{-2}\text{s}^{-1}$ This rounds to $0.17 \text{ M}^{-2}\text{s}^{-1}$ , confirming answer D is correct. Answer A ( $0.033 \text{ M}^{-2}\text{s}^{-1}$ ) results from incorrectly using $[A]^3$ instead of $[A]^2$ , creating an extra factor of 0.20 in the denominator. Answer B ( $0.067 \text{ M}^{-2}\text{s}^{-1}$ ) comes from mistakenly squaring both [A] and [B], giving $(0.20)^2(0.30)^2 = 0.0036$ in the denominator. Answer C ( $0.12 \text{ M}^{-2}\text{s}^{-1}$ ) occurs when you forget to square [A] and just use $[A][B] = (0.20)(0.30) = 0.060$ . Always pay careful attention to the exponents in rate laws—they directly affect your calculation. Double-check that you're raising each concentration to the correct power as specified in the rate law, and remember that the units of k depend on the overall reaction order.
The decomposition of nitrogen dioxide follows the reaction: $2NO_2(g) \rightarrow 2NO(g) + O_2(g)$ . In a 2.0 L container at 300°C, the concentration of $NO_2$ decreases from 0.80 M to 0.60 M over a 45-second period. What is the average rate of formation of $O_2$ during this time interval?
Answer: $2.2 \times 10^{-3} \text{ M/s}$ When you encounter reaction rate problems, remember that rates of reactants and products are related by their stoichiometric coefficients in the balanced equation. First, calculate the rate of $NO_2$ consumption. The concentration decreases from 0.80 M to 0.60 M over 45 seconds: $\text{Rate of }NO_2\text{ consumption} = \frac{0.80 - 0.60}{45} = \frac{0.20}{45} = 4.4 \times 10^{-3} \text{ M/s}$ Now use stoichiometry to find the $O_2$ formation rate. From the balanced equation $2NO_2 \rightarrow 2NO + O_2$ , you can see that 2 moles of $NO_2$ produce 1 mole of $O_2$ . This means $O_2$ forms at half the rate that $NO_2$ disappears: $\text{Rate of }O_2\text{ formation} = \frac{1}{2} \times 4.4 \times 10^{-3} = 2.2 \times 10^{-3} \text{ M/s}$ This confirms answer B is correct. Answer A ( $1.1 \times 10^{-3}$ M/s) incorrectly divides by 4 instead of 2, misunderstanding the stoichiometric relationship. Answer C ( $4.4 \times 10^{-3}$ M/s) gives the rate of $NO_2$ consumption without applying stoichiometry—this would be correct for $NO$ formation since it has the same coefficient as $NO_2$ . Answer D ( $8.9 \times 10^{-3}$ M/s) doubles the $NO_2$ rate, inverting the stoichiometric relationship. Remember: always check the stoichiometric coefficients when relating rates of different species. The rate of formation or consumption is inversely proportional to the coefficient in the balanced equation.
The reaction $2A \rightarrow B + C$ follows second-order kinetics in A. If the initial concentration of A is 0.50 M and the rate constant is $0.20 \text{ M}^{-1}\text{s}^{-1}$ , what is the half-life of this reaction?
Answer: $10.0 \text{ s}$ When you encounter a second-order kinetics problem, you need to use the integrated rate law specific to second-order reactions, which differs significantly from first-order reactions where half-life is constant. For a second-order reaction in one reactant, the integrated rate law is: $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$ At the half-life, $[A] = \frac{[A]_0}{2}$ , so we can substitute: $\frac{1}{[A]_0/2} = kt_{1/2} + \frac{1}{[A]_0}$ Simplifying: $\frac{2}{[A]_0} = kt_{1/2} + \frac{1}{[A]_0}$ Rearranging: $t_{1/2} = \frac{1}{k[A]_0}$ With $k = 0.20 \text{ M}^{-1}\text{s}^{-1}$ and $[A]_0 = 0.50 \text{ M}$ : $t_{1/2} = \frac{1}{(0.20)(0.50)} = \frac{1}{0.10} = 10.0 \text{ s}$ This confirms answer C is correct. Answer A (2.5 s) results from incorrectly using $t_{1/2} = \frac{1}{2k[A]_0}$ , adding an extra factor of 2. Answer B (5.0 s) comes from using $t_{1/2} = \frac{1}{2k}$ , ignoring the concentration dependence entirely. Answer D (20.0 s) suggests using $t_{1/2} = \frac{2}{k[A]_0}$ , incorrectly placing the factor of 2 in the numerator. Remember: Second-order half-life depends on initial concentration and gets longer as the reaction proceeds, unlike first-order reactions where half-life remains constant. Always use $t_{1/2} = \frac{1}{k[A]_0}$ for second-order kinetics.
The rate-determining step of a reaction mechanism is Step 2: $X + Y \rightarrow Z$ (slow), which follows Step 1: $A + B \rightleftharpoons X + C$ (fast equilibrium). If the concentration of A is tripled while all other concentrations remain constant, what happens to the overall reaction rate?
Answer: The rate increases by a factor of 3 because A appears once in the pre-equilibrium When you encounter reaction mechanisms with a fast pre-equilibrium followed by a slow rate-determining step, you need to derive the rate law by expressing concentrations of intermediates in terms of the original reactants. For this mechanism, the overall rate is determined by Step 2: $\text{rate} = k_2[X][Y]$ . However, X is an intermediate formed in the fast equilibrium Step 1. Since Step 1 reaches equilibrium quickly, you can write: $K_{eq} = \frac{[X][C]}{[A][B]}$ , which rearranges to $[X] = \frac{K_{eq}[A][B]}{[C]}$ . Substituting this into the rate expression: $\text{rate} = k_2 \cdot \frac{K_{eq}[A][B]}{[C]} \cdot [Y] = k_{obs}[A][B][Y]/[C]$ When [A] is tripled while other concentrations remain constant, the rate increases by a factor of 3 because A appears with a first-order dependence in the derived rate law. Answer A is wrong because even though A doesn't appear in the rate-determining step directly, it affects the concentration of intermediate X through the pre-equilibrium. Answer C incorrectly suggests the rate increases by a factor of 9 and mentions squaring the equilibrium constant, which has no basis in the mechanism. Answer D is incorrect because increasing A shifts the equilibrium forward (toward products X and C), not backward. Study tip: For mechanisms with fast pre-equilibria, always substitute the equilibrium expression for intermediates into the rate law of the slow step. The pre-equilibrium determines how reactant concentrations affect intermediate concentrations.
The decomposition of hydrogen peroxide follows first-order kinetics with a rate constant of $3.0 \times 10^{-4} \text{ s}^{-1}$ at 20°C. How long will it take for 75% of the original $H_2O_2$ to decompose?
Answer: $4.6 \times 10^3 \text{ s}$ When you encounter first-order kinetics problems, you're dealing with reactions where the rate depends on the concentration of one reactant raised to the first power. The key equation here is the integrated first-order rate law: $\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$ , where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time t, k is the rate constant, and t is time. If 75% of the original $H_2O_2$ decomposes, then 25% remains. This means $\frac{[A]_t}{[A]_0} = 0.25$ , so $\frac{[A]_0}{[A]_t} = 4.0$ . Substituting into the equation: $\ln(4.0) = (3.0 \times 10^{-4} \text{ s}^{-1}) \times t$ . Since $\ln(4.0) = 1.386$ , we get: $t = \frac{1.386}{3.0 \times 10^{-4}} = 4620 \text{ s} = 4.6 \times 10^3 \text{ s}$ . This matches answer C. Answer A ( $1.2 \times 10^3 \text{ s}$ ) represents only about 31% decomposition - you might get this if you confused the percentage remaining with percentage decomposed. Answer B ( $2.3 \times 10^3 \text{ s}$ ) corresponds to roughly 50% decomposition, which you'd calculate if you mistakenly used $\ln(2)$ instead of $\ln(4)$ . Answer D ( $6.9 \times 10^3 \text{ s}$ ) represents about 87% decomposition - this could result from calculation errors or using the wrong kinetic equation. Remember: for first-order reactions, always identify what fraction remains, not what's decomposed, then use the natural logarithm in your calculations.
A second-order reaction has a rate constant of $0.25 \text{ M}^{-1}\text{s}^{-1}$ at 25°C. If the initial concentration of the reactant is 0.80 M, what is the concentration after 10.0 seconds?
Answer: $0.27 \text{ M}$ When you encounter a second-order reaction kinetics problem, you need to use the integrated rate law that relates concentration to time. For a second-order reaction, the integrated rate law is: $\frac{1}{[A]} = \frac{1}{[A]_0} + kt$ , where $[A]$ is the final concentration, $[A]_0$ is the initial concentration, $k$ is the rate constant, and $t$ is time. Let's substitute the given values: $k = 0.25 \text{ M}^{-1}\text{s}^{-1}$ , $[A]_0 = 0.80 \text{ M}$ , and $t = 10.0 \text{ s}$ . $\frac{1}{[A]} = \frac{1}{0.80} + (0.25)(10.0) = 1.25 + 2.5 = 3.75 \text{ M}^{-1}$ Therefore: $[A] = \frac{1}{3.75} = 0.27 \text{ M}$ Answer A (0.20 M) likely comes from using the wrong integrated rate law or making a calculation error. Answer B (0.24 M) might result from incorrect manipulation of the equation or rounding errors. Answer D (0.40 M) represents exactly half the initial concentration, which would be correct for a first-order reaction at one half-life, but this is a second-order reaction where the math works differently. The key study tip here is to memorize the integrated rate laws for different reaction orders. Zero-order uses $[A] = [A]_0 - kt$ , first-order uses $\ln[A] = \ln[A]_0 - kt$ , and second-order uses $\frac{1}{[A]} = \frac{1}{[A]_0} + kt$ . The units of the rate constant always tell you the reaction order.
For the reaction $A + 2B \rightarrow C$ , the rate law is $\text{rate} = k[A][B]^2$ . If the concentration of A is doubled while the concentration of B is halved, how does the new reaction rate compare to the original rate?
Answer: The new rate is one-half the original rate When you encounter rate law problems, you're working with how concentration changes affect reaction speed. The rate law equation tells you exactly how each reactant's concentration influences the overall rate. Let's work through this systematically. The original rate is $\text{rate}_1 = k[A][B]^2$ . After the concentration changes, you have $[A]_{\text{new}} = 2[A]$ and $[B]_{\text{new}} = \frac{1}{2}[B]$ . Substituting into the rate law: $\text{rate}_2 = k(2[A])\left(\frac{1}{2}[B]\right)^2 = k(2[A])\left(\frac{1}{4}[B]^2\right) = \frac{2}{4}k[A][B]^2 = \frac{1}{2}k[A][B]^2$ Therefore, $\text{rate}_2 = \frac{1}{2}\text{rate}_1$ , making the new rate half the original rate. Looking at the wrong answers: (A) suggests one-fourth the original rate, which would occur if you forgot to account for doubling [A] and only considered the $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ effect from halving [B]. (C) claims the rates are equal, which ignores that the exponents in rate laws create non-linear relationships—equal and opposite percentage changes don't cancel out when exponents are involved. (D) suggests doubling the rate, which incorrectly focuses only on doubling [A] while ignoring the squared term's impact. Remember: in rate law calculations, pay careful attention to exponents. A concentration change gets raised to whatever power appears in the rate law, making the effects non-intuitive. Always substitute the new concentrations completely into the rate expression.
The reaction $A \rightarrow \text{products}$ shows the following kinetic data when the concentration of A is plotted in different ways. Which plot indicates that the reaction is second-order in A?
Answer: A linear plot of $\frac{1}{[A]}$ vs. time with a positive slope When you encounter kinetic data plots, you're being tested on integrated rate laws—mathematical relationships that describe how concentration changes over time for different reaction orders. For a second-order reaction in A, the integrated rate law is: $\frac{1}{[A]} = kt + \frac{1}{[A]_0}$ , where k is the rate constant and $[A]_0$ is the initial concentration. This equation has the form y = mx + b, meaning a plot of $\frac{1}{[A]}$ versus time will be linear with slope k (positive, since k is always positive). This confirms that choice C is correct. Let's examine why the other options are wrong. Choice A describes a zero-order reaction, where $[A] = -kt + [A]_0$ —concentration decreases linearly with time. Choice B represents a first-order reaction, where $\ln[A] = -kt + \ln[A]_0$ —the natural log of concentration decreases linearly with time. Choice D has no basis in kinetics; $[A]^2$ versus time doesn't correspond to any standard integrated rate law. Each reaction order has its own characteristic linear plot: zero-order uses $[A]$ vs. time, first-order uses $\ln[A]$ vs. time, and second-order uses $\frac{1}{[A]}$ vs. time. The key pattern to remember is that the mathematical manipulation needed to linearize the data reveals the reaction order—and second-order kinetics always requires plotting the reciprocal of concentration.
A reaction has an activation energy of 85 kJ/mol. By what factor does the rate constant increase when the temperature is raised from 25°C to 45°C? (R = 8.314 J/mol·K)
Answer: $7.3$ This question tests your understanding of the Arrhenius equation, which describes how reaction rates depend on temperature. When you see activation energy and temperature changes together, you're dealing with the relationship between kinetic energy and the fraction of molecules that can overcome the energy barrier to react. To find how the rate constant changes, you'll use the Arrhenius equation in its ratio form: $\frac{k_2}{k_1} = e^{\frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})}$ First, convert temperatures to Kelvin: T₁ = 25°C + 273 = 298 K, and T₂ = 45°C + 273 = 318 K. Also convert activation energy to consistent units: 85 kJ/mol = 85,000 J/mol. Now substitute: $\frac{k_2}{k_1} = e^{\frac{85,000}{8.314}(\frac{1}{298} - \frac{1}{318})}$ Calculate the temperature term: $\frac{1}{298} - \frac{1}{318} = 0.003356 - 0.003145 = 0.000211$ Then: $\frac{k_2}{k_1} = e^{10,225 \times 0.000211} = e^{2.16} = 7.3$ Choice C (7.3) is correct. Choice A (2.1) likely results from calculation errors or using incorrect temperature units. Choice B (4.7) might come from forgetting to convert activation energy to J/mol or making arithmetic mistakes. Choice D (12.2) could result from sign errors in the temperature difference or using degrees Celsius instead of Kelvin. Remember: always convert to Kelvin for temperature-dependent rate calculations, and ensure your energy units match R's units. The Arrhenius equation shows that even modest temperature increases can dramatically accelerate reactions.
The concentration of reactant X decreases from 1.20 M to 0.30 M over 150 seconds in a reaction that follows first-order kinetics. What is the rate constant for this reaction?
Answer: $9.2 \times 10^{-3} \text{ s}^{-1}$ When you encounter first-order kinetics problems, you're working with the integrated rate law: $\ln[A] = \ln[A_0] - kt$ , where [A] is the final concentration, [A₀] is the initial concentration, k is the rate constant, and t is time. To find the rate constant, rearrange this equation to solve for k: $k = \frac{\ln[A_0] - \ln[A]}{t} = \frac{\ln\left(\frac{[A_0]}{[A]}\right)}{t}$ Substituting your values: $k = \frac{\ln\left(\frac{1.20}{0.30}\right)}{150\text{ s}} = \frac{\ln(4.0)}{150\text{ s}} = \frac{1.386}{150\text{ s}} = 9.2 \times 10^{-3}\text{ s}^{-1}$ This confirms answer B is correct. Answer A ( $6.0 \times 10^{-3}\text{ s}^{-1}$ ) likely results from using the wrong mathematical approach, perhaps trying to use a zero-order or second-order rate law instead of the first-order integrated equation. Answer C ( $1.4 \times 10^{-2}\text{ s}^{-1}$ ) could come from calculation errors in the natural logarithm or incorrectly handling the concentration ratio. Answer D ( $2.1 \times 10^{-2}\text{ s}^{-1}$ ) might result from using $\ln(1.20) - \ln(0.30)$ incorrectly or making arithmetic mistakes in the division. Remember: For first-order kinetics, always use the natural logarithm form of the integrated rate law. Double-check that you're taking $\ln(\text{initial}/\text{final})$ and dividing by time. The units for a first-order rate constant are always $\text{s}^{-1}$ (or $\text{time}^{-1}$ ).
For a zero-order reaction, the concentration of reactant A decreases from 0.60 M to 0.20 M in 40 seconds. What is the rate constant for this reaction?
Answer: $0.010 \text{ M/s}$ When you encounter zero-order reaction problems, remember that these reactions have a constant rate that's independent of reactant concentration. The rate law is simply: rate = k, where k is the rate constant with units of M/s. For zero-order reactions, concentration changes linearly with time according to: $[A] = [A]_0 - kt$ , where $[A]_0$ is initial concentration and $[A]$ is concentration at time t. You can rearrange this to solve for k: $k = \frac{[A]_0 - [A]}{t}$ . Plugging in your values: $k = \frac{0.60 \text{ M} - 0.20 \text{ M}}{40 \text{ s}} = \frac{0.40 \text{ M}}{40 \text{ s}} = 0.010 \text{ M/s}$ This confirms answer B is correct. Answer A (0.005 M/s) would result if you incorrectly divided the concentration change by twice the actual time. Answer C (0.020 M/s) occurs if you mistakenly used 20 seconds instead of 40 seconds in your calculation. Answer D (0.025 M/s) would arise from using 16 seconds instead of 40, perhaps from a calculation error. The key study tip: Zero-order kinetics always involves linear concentration changes over time, and the rate constant equals the slope of the concentration vs. time plot. Always double-check that your rate constant has the correct units (M/s for zero-order) and verify your time units match throughout the calculation.
The energy diagram for a catalyzed vs. uncatalyzed reaction is shown. Based on this diagram, which statement is correct about the effect of the catalyst?
Answer: The catalyst decreases the activation energy for both forward and reverse reactions by the same amount A catalyst provides an alternative pathway with lower activation energy for both forward and reverse directions. The diagram shows that both Ea(forward) and Ea(reverse) are reduced by the same amount, maintaining the same ΔH (difference between reactant and product energies). Choice A incorrectly describes opposite effects. Choice C is wrong because catalysts don't change ΔH. Choice D misinterprets the pathway complexity. Choice E contradicts the principle that catalysts affect both directions equally.
The reaction $N_2O_5 \rightarrow N_2O_4 + \frac{1}{2}O_2$ follows first-order kinetics. If 80% of the original $N_2O_5$ remains after 100 seconds, what is the rate constant for this reaction?
Answer: $2.2 \times 10^{-3} \text{ s}^{-1}$ When you encounter a first-order kinetics problem, you need to use the integrated rate law: $\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$ , where $[A]_t$ is the concentration at time t, $[A]_0$ is the initial concentration, k is the rate constant, and t is time. Since 80% of the original $N_2O_5$ remains after 100 seconds, we have $\frac{[N_2O_5]_t}{[N_2O_5]_0} = 0.80$ . Substituting into the integrated rate law: $\ln(0.80) = -k(100 \text{ s})$ $-0.223 = -k(100 \text{ s})$ $k = \frac{0.223}{100} = 2.23 \times 10^{-3} \text{ s}^{-1}$ This matches answer choice B: $2.2 \times 10^{-3} \text{ s}^{-1}$ . Answer A ( $1.4 \times 10^{-3} \text{ s}^{-1}$ ) results from incorrectly using $\ln(0.20)$ instead of $\ln(0.80)$ - this would be the calculation if 20% remained rather than 80%. Answer C ( $3.1 \times 10^{-3} \text{ s}^{-1}$ ) likely comes from using the wrong logarithm or making an arithmetic error. Answer D ( $4.6 \times 10^{-3} \text{ s}^{-1}$ ) appears to result from using $\ln(2)$ instead of $\ln(0.80)$ , possibly confusing this with a half-life calculation. Remember: for first-order kinetics problems, always identify what fraction of the original substance remains, then use the natural logarithm of that fraction in the integrated rate law. Don't confuse "percent remaining" with "percent decomposed."
A proposed mechanism for the reaction $2NO + O_2 \rightarrow 2NO_2$ is: Step 1: $NO + NO \rightleftharpoons N_2O_2$ (fast equilibrium); Step 2: $N_2O_2 + O_2 \rightarrow 2NO_2$ (slow). What experimental observation would most strongly support this mechanism?
Answer: The reaction rate is proportional to $[NO]^2[O_2]$ over a wide range of concentrations When you encounter reaction mechanism questions, you need to derive the rate law from the proposed steps and see which experimental observation matches your prediction. For this mechanism, Step 1 is a fast equilibrium, so you can write: $K_{eq} = \frac{[N_2O_2]}{[NO]^2}$ , which gives $[N_2O_2] = K_{eq}[NO]^2$ . Since Step 2 is the slow step, it determines the overall rate: $\text{rate} = k_2[N_2O_2][O_2]$ . Substituting the equilibrium expression: $\text{rate} = k_2K_{eq}[NO]^2[O_2]$ . This predicts the rate should be proportional to $[NO]^2[O_2]$ , making choice B correct. Choice A is wrong because your derived rate law shows the reaction should depend on $[O_2]$ concentration, not be independent of it. Choice C is incorrect because most reactions speed up with temperature increases due to higher kinetic energy—a decrease would be unusual and isn't predicted by this mechanism. Choice D is wrong because your rate law predicts second-order dependence on $[NO]$ , not fractional order dependence. Study tip: For mechanism problems, always identify the rate-determining step (usually the slow one), write its rate law, then substitute equilibrium expressions for any intermediates from fast pre-equilibrium steps. The final rate law you derive should match experimental observations that support the mechanism.
The reaction $2NO(g) + Cl_2(g) \rightarrow 2NOCl(g)$ proceeds via the mechanism: Step 1: $NO + Cl_2 \rightleftharpoons NOCl_2$ (fast equilibrium); Step 2: $NOCl_2 + NO \rightarrow 2NOCl$ (slow). What is the predicted rate law based on this mechanism?
Answer: $\text{rate} = k[NO]^2[Cl_2]$ When you encounter a reaction mechanism problem, you need to derive the rate law from the slowest step while expressing any intermediates in terms of the original reactants. The rate law is determined by the slow step (Step 2): $\text{rate} = k[NOCl_2][NO]$ . However, $NOCl_2$ is an intermediate that doesn't appear in the overall reaction, so you must express it in terms of the original reactants using the fast equilibrium from Step 1. For the fast equilibrium $NO + Cl_2 \rightleftharpoons NOCl_2$ , the equilibrium constant is: $K_{eq} = \frac{[NOCl_2]}{[NO][Cl_2]}$ Solving for the intermediate: $[NOCl_2] = K_{eq}[NO][Cl_2]$ Substituting this into the rate law from Step 2: $\text{rate} = k[NOCl_2][NO] = k \cdot K_{eq}[NO][Cl_2] \cdot [NO] = k'[NO]^2[Cl_2]$ where $k' = k \cdot K_{eq}$ . Choice A ( $\text{rate} = k[NO][Cl_2]$ ) incorrectly assumes the reaction is elementary and ignores the mechanism. Choice C ( $\text{rate} = k[NOCl_2][NO]$ ) fails to eliminate the intermediate, which isn't acceptable since intermediates don't appear in final rate laws. Choice D ( $\text{rate} = k[NO][Cl_2]^2$ ) has the wrong stoichiometry - there's only one $Cl_2$ molecule involved in forming the intermediate. Study tip: Always identify the rate-determining step first, then use pre-equilibrium approximations to eliminate any intermediates by expressing them in terms of the original reactants.
For the elementary reaction $2A + B \rightarrow C$ , which of the following statements about the rate law is correct?
Answer: The rate law is $\text{rate} = k[A]^2[B]$ because the exponents match the stoichiometric coefficients When you encounter questions about rate laws for elementary reactions, remember that elementary reactions are unique because their rate laws can be written directly from the balanced chemical equation—unlike overall reactions that require experimental determination. For the elementary reaction $2A + B \rightarrow C$ , the rate law is $\text{rate} = k[A]^2[B]$ . In elementary reactions, the exponents in the rate law always equal the stoichiometric coefficients from the balanced equation. Since two molecules of A must collide with one molecule of B simultaneously for this reaction to occur, the rate depends on $[A]^2$ and $[B]^1$ . This makes answer B correct. Answer A incorrectly suggests that each reactant appears with an exponent of 1, ignoring the stoichiometric coefficient of 2 for species A. This represents a fundamental misunderstanding of how elementary reactions work at the molecular level. Answer C incorrectly squares all reactants regardless of their stoichiometric coefficients. This reflects confusion about the relationship between molecularity and rate law exponents—only the actual stoichiometric coefficients determine the exponents. Answer D would be correct for overall reactions (which often involve multiple elementary steps), but it's wrong here because elementary reactions are special cases where we can determine the rate law directly from stoichiometry. Key takeaway: For elementary reactions only, rate law exponents always equal stoichiometric coefficients. This is because elementary reactions represent the actual molecular collision event, unlike complex reactions that proceed through multiple steps.

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