Introduction to Rate Law - College Chemistry

1. A student studies the reaction $P + Q \rightarrow R + S$ and collects the following initial rate data: Experiment 1: $[P]_0 = 0.10 \text{ M}$, $[Q]_0 = 0.10 \text{ M}$, rate = $5.0 \times 10^{-3} \text{ M/s}$. Experiment 2: $[P]_0 = 0.20 \text{ M}$, $[Q]_0 = 0.10 \text{ M}$, rate = $1.0 \times 10^{-2} \text{ M/s}$. The student concludes the reaction is first order in P. What additional experiment would best confirm this conclusion?
   Answer: $[P]_0 = 0.30 \text{ M}$, $[Q]_0 = 0.10 \text{ M}$ When you encounter initial rate data problems, you're determining reaction orders by analyzing how concentration changes affect reaction rates. The key principle is that for a reaction that's nth order in a reactant, doubling that reactant's concentration changes the rate by a factor of 2ⁿ. Let's verify the student's conclusion first. Comparing experiments 1 and 2: when [P] doubles from 0.10 M to 0.20 M (while [Q] stays constant), the rate doubles from $5.0 \times 10^{-3}$ to $1.0 \times 10^{-2}$ M/s. Since doubling [P] doubles the rate, the reaction appears first order in P ($2^1 = 2$). To confirm this conclusion, you need another data point that tests the same relationship with [Q] held constant. Choice A ($[P]_0 = 0.30$ M, $[Q]_0 = 0.10$ M) does exactly this. If the reaction is truly first order in P, tripling [P] from the original 0.10 M should triple the rate to $1.5 \times 10^{-2}$ M/s. Choice B changes [Q] instead of [P], which would test the order with respect to Q, not confirm the order with respect to P. Choice C changes both concentrations simultaneously, making it impossible to isolate P's effect. Choice D also changes both concentrations and uses non-simple multiples, complicating the analysis. The answer is A because it provides the cleanest test of the proposed first-order relationship in P. Study tip: When confirming reaction orders, always vary only one concentration at a time while keeping others constant. Use simple concentration ratios (2×, 3×, etc.) to make the mathematical relationships obvious.
2. A reaction has the overall balanced equation $A + 2B \rightarrow C + 2D$ and the experimentally determined rate law $\text{rate} = k[A]^{1/2}[B]^{3/2}$. What is the overall order of this reaction?
   Answer: 2 When you encounter a rate law question, you're dealing with reaction kinetics, specifically determining how the concentration of reactants affects the reaction rate. The key insight is that reaction order tells you how sensitive the rate is to concentration changes. To find the overall order of a reaction, you sum all the exponents in the rate law. Looking at the given rate law $\text{rate} = k[A]^{1/2}[B]^{3/2}$, you have two exponents to add: the order with respect to A is 1/2, and the order with respect to B is 3/2. Therefore, the overall order = 1/2 + 3/2 = 2. Let's examine why each option is incorrect. Choice A (1) would only account for one of the exponents, perhaps just the 1/2 from [A]. Choice B (1.5) represents only the exponent on [B], ignoring the contribution from [A] entirely. Choice D (2.5) might result from incorrectly adding 1 + 3/2, perhaps by misreading the exponent on [A] as 1 instead of 1/2. Notice that the balanced chemical equation coefficients (1 for A, 2 for B) are completely irrelevant to determining reaction order. This is a common trap—stoichiometric coefficients and kinetic orders are independent concepts. The rate law must be determined experimentally and doesn't necessarily match the balanced equation. Remember: reaction order always comes from the rate law exponents, never from the balanced equation. Add up all exponents in the rate law to get the overall order.
3. A reaction follows the rate law $\text{rate} = k[P]^a[Q]^b$ where a and b are unknown orders. From experimental data: doubling [P] alone increases the rate 2.83 times, and doubling [Q] alone increases the rate 1.41 times. What are the most likely values of a and b?
   Answer: a = 1.5, b = 0.5 When you encounter rate law problems with experimental data, you need to determine how changes in concentration affect the reaction rate. The key is using the relationship: if you change one reactant's concentration by a factor, the rate changes by that factor raised to the power of the reaction order. For reactant P, doubling the concentration increases the rate by 2.83 times. This means $2^a = 2.83$. Taking the logarithm: $a = \log(2.83)/\log(2) = 1.5$. For reactant Q, doubling the concentration increases the rate by 1.41 times, so $2^b = 1.41$, giving $b = \log(1.41)/\log(2) = 0.5$. Let's examine why the other answers fail. Choice A (a = 1, b = 1) would predict that doubling [P] increases the rate by $2^1 = 2$ times, not 2.83, and doubling [Q] would also give 2 times, not 1.41. Choice C (a = 2, b = 0.5) would correctly predict the effect of doubling [Q] as $2^{0.5} = 1.41$, but doubling [P] would give $2^2 = 4$ times increase, not 2.83. Choice D (a = 2.83, b = 1.41) makes the common mistake of confusing the rate multiplication factors with the actual orders. Remember this pattern: when experimental data shows non-integer rate changes upon doubling concentrations, the reaction orders are likely fractional. Always solve $2^{\text{order}} = \text{rate factor}$ using logarithms rather than assuming the multiplication factor equals the order.
4. The decomposition of hydrogen peroxide follows the reaction $2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$ with rate law $\text{rate} = k[H_2O_2]$. A student mistakenly writes the rate law as $\text{rate} = k[H_2O_2]^2$, thinking the order must equal the stoichiometric coefficient. If the actual initial concentration of $H_2O_2$ is 0.50 M, what ratio would the student's predicted rate have to the actual rate?
   Answer: 0.50 When you encounter rate law problems, remember that reaction order is determined experimentally, not from stoichiometric coefficients. This is a crucial distinction that trips up many students. Let's compare the two rate expressions. The actual rate law is $\text{rate} = k[H_2O_2]$, while the student incorrectly assumes $\text{rate} = k[H_2O_2]^2$. With $[H_2O_2] = 0.50 \text{ M}$: Actual rate: $k(0.50) = 0.50k$ Student's predicted rate: $k(0.50)^2 = k(0.25) = 0.25k$ The ratio of student's predicted rate to actual rate is: $\frac{0.25k}{0.50k} = 0.50$ Answer choice A (0.25) represents the student's rate divided by the rate constant, not the ratio we need. Answer choice C (1.0) would only be correct if both rate laws were identical. Answer choice D (2.0) incorrectly inverts the ratio we calculated. Answer B (0.50) correctly gives us the ratio of the student's underestimated rate to the true rate. The student's fundamental error was assuming that reaction order equals stoichiometric coefficients. While the balanced equation shows a coefficient of 2 for $H_2O_2$, the actual reaction is first-order in $H_2O_2$, meaning the rate depends linearly on its concentration. Study tip: Always remember that stoichiometric coefficients ≠ reaction orders. Rate laws must be determined experimentally. When you see discrepancies between coefficients and orders in a problem, it's usually testing this exact concept.
5. A reaction has the form $2A \rightarrow B + C$ and follows the rate law $\text{rate} = k[A]^n$. When the initial concentration of A is 0.60 M, the initial rate is $1.8 \times 10^{-2} \text{ M/s}$. When the initial concentration of A is 0.20 M, the initial rate is $2.0 \times 10^{-3} \text{ M/s}$. What is the value of n?
   Answer: 2 When you encounter a question asking for the order of a reaction, you're determining how the concentration of a reactant affects the reaction rate. The rate law $\text{rate} = k[A]^n$ tells you that the rate depends on the concentration of A raised to some power n. To find n, you need to compare how the rate changes when the concentration changes. Set up a ratio using the two given data points: $\frac{\text{rate}_1}{\text{rate}_2} = \frac{k[A_1]^n}{k[A_2]^n} = \left(\frac{[A_1]}{[A_2]}\right)^n$ Substituting the values: $\frac{1.8 \times 10^{-2}}{2.0 \times 10^{-3}} = \left(\frac{0.60}{0.20}\right)^n$ $9 = (3)^n$ Since $3^2 = 9$, we have $n = 2$. Answer A (n = 0) would mean the reaction is zero-order, where rate is independent of concentration. This would give the same rate regardless of [A], which contradicts the data. Answer B (n = 1) would mean the rate is directly proportional to [A]. Tripling the concentration would triple the rate, giving a ratio of 3, not 9. Answer D (n = 3) would give $(3)^3 = 27$, which is much larger than our calculated ratio of 9. The correct answer is C: the reaction is second-order in A. Study tip: Always use the ratio method for determining reaction order—it eliminates the need to calculate k and directly gives you the order through simple algebra.
6. For the elementary reaction $NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)$, the rate law is determined experimentally to be $\text{rate} = k[NO][O_3]$. If the concentration of NO is tripled while the concentration of $O_3$ is halved, by what factor does the initial rate change?
   Answer: 1.5 When you encounter rate law problems, focus on how concentration changes affect the overall reaction rate based on the given rate expression. The rate law $\text{rate} = k[NO][O_3]$ tells you the reaction is first order in both NO and $O_3$, meaning the rate is directly proportional to each concentration. To find how the rate changes, substitute the new concentrations into the rate law. If [NO] is tripled, the new concentration becomes 3[NO]. If [$O_3$] is halved, it becomes 0.5[$O_3$]. The new rate expression is: $\text{new rate} = k(3[NO])(0.5[O_3]) = 1.5 \cdot k[NO][O_3]$ Therefore, the rate increases by a factor of 1.5, making (B) 1.5 correct. Let's examine why the other answers are wrong: (A) 0.5 would result if you incorrectly thought halving $O_3$ dominates and ignored the tripling of NO. (C) 3.0 comes from only considering the tripling of NO while ignoring the halving of $O_3$. (D) 6.0 represents the common error of adding the effects (3 + 3 = 6 from tripling, or 3 × 2 = 6) rather than multiplying them as the rate law requires. Study tip: For rate law problems, always substitute the new concentrations directly into the given rate expression and multiply all the effects together. Don't try to reason about individual effects separately—the rate law shows you exactly how concentrations combine to determine the overall rate.
7. For the reaction $2NO_2(g) \rightarrow N_2O_4(g)$, initial rate studies yield the data: when $[NO_2]_0 = 0.10 \text{ M}$, rate = $2.5 \times 10^{-3} \text{ M/s}$; when $[NO_2]_0 = 0.20 \text{ M}$, rate = $1.0 \times 10^{-2} \text{ M/s}$. A student claims this proves the reaction follows the balanced equation and is second order in $NO_2$. What is wrong with this reasoning?
   Answer: The stoichiometry of the balanced equation does not determine reaction order When you encounter kinetics problems involving rate laws and reaction orders, remember that there's a crucial distinction between the balanced chemical equation and the actual mechanism by which the reaction proceeds. Let's examine what the data actually tells us. When $[NO_2]$ doubles from 0.10 M to 0.20 M, the rate increases from $2.5 \times 10^{-3}$ to $1.0 \times 10^{-2}$ M/s. Since $1.0 \times 10^{-2} = 4 \times (2.5 \times 10^{-3})$, the rate quadruples when concentration doubles. This confirms the reaction is indeed second order in $NO_2$, following the rate law: rate = $k[NO_2]^2$. However, the student's reasoning contains a fundamental error. The stoichiometric coefficients in a balanced equation do NOT determine the reaction order. The balanced equation shows the overall stoichiometry but tells us nothing about the mechanism or rate law. Reaction order must be determined experimentally through rate studies like this one. Answer B correctly identifies this misconception. Answer A is wrong because the data clearly shows second-order behavior (rate increases by a factor of 4 when concentration doubles). Answer C misunderstands basic kinetics—rates always increase with increasing reactant concentration. Answer D makes an incorrect claim; there's no rule preventing second-order reactions from having coefficients of 2 in their balanced equations. Remember this key principle: stoichiometry ≠ kinetics. Always determine reaction order from experimental rate data, never assume it matches the balanced equation coefficients. This is a common trap in kinetics problems.
8. Based on the experimental data provided, what is the rate law for this reaction?
   Answer: $\text{rate} = k[A][B]^2$ When you encounter initial rate data from multiple experiments, you're being asked to determine reaction orders through the method of initial rates. The key is comparing how rate changes when you systematically vary one concentration while holding others constant. To find the order with respect to A, compare experiments 1 and 2 where [B] stays constant at 0.100 M while [A] doubles from 0.100 M to 0.200 M. The rate increases from 2.40 × 10⁻⁴ to 4.80 × 10⁻⁴ M/s—exactly doubling. Since doubling [A] doubles the rate, the reaction is first order in A. For the order with respect to B, compare experiments 1 and 3 where [A] remains at 0.100 M while [B] doubles from 0.100 M to 0.200 M. The rate increases from 2.40 × 10⁻⁴ to 9.60 × 10⁻⁴ M/s—quadrupling. Since doubling [B] quadruples the rate, the reaction is second order in B. Therefore, the rate law is $\text{rate} = k[A][B]^2$, which is answer B. Choice A ($\text{rate} = k[A][B]$) would predict only doubling the rate when [B] doubles, not quadrupling. Choice C ($\text{rate} = k[A]^2[B]$) would predict quadrupling when [A] doubles, but we only see doubling. Choice D ($\text{rate} = k[A]^2[B]^2$) would predict both concentrations having second-order effects, contradicting the linear relationship with [A]. Remember: always examine one concentration change at a time while keeping others constant to isolate individual reaction orders.
9. For a reaction with rate law $\text{rate} = k[X]^2[Y]$, an experiment is conducted where the initial concentration of X is 0.40 M and the initial concentration of Y is 0.60 M. If both concentrations are simultaneously reduced to half their original values, what happens to the initial rate?
   Answer: It decreases by a factor of 8 When you encounter rate law problems involving concentration changes, focus on how each concentration term contributes to the overall rate according to its exponent in the rate expression. Given the rate law $\text{rate} = k[X]^2[Y]$, let's calculate what happens when both concentrations are halved. Initially, the rate is $\text{rate}_1 = k(0.40)^2(0.60)$. After halving both concentrations, we have [X] = 0.20 M and [Y] = 0.30 M, so $\text{rate}_2 = k(0.20)^2(0.30)$. To find the factor change, calculate: $\frac{\text{rate}_2}{\text{rate}_1} = \frac{k(0.20)^2(0.30)}{k(0.40)^2(0.60)} = \frac{(0.20)^2(0.30)}{(0.40)^2(0.60)} = \frac{0.04 \times 0.30}{0.16 \times 0.60} = \frac{0.012}{0.096} = \frac{1}{8}$ This means the rate decreases by a factor of 8, confirming answer D. Here's why the other options are incorrect: A suggests a factor of 2, which would only account for halving [Y] while ignoring the squared dependence on [X]. B (factor of 4) represents the effect of halving [X] alone, since $(1/2)^2 = 1/4$, but neglects [Y]. C (factor of 6) has no basis in the rate law mathematics and likely represents confusion about how to combine the effects. Remember: when concentrations change, raise each concentration ratio to its respective exponent, then multiply all factors together. The total effect is multiplicative: $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$.
10. The rate law for the reaction $H_2(g) + I_2(g) \rightarrow 2HI(g)$ is experimentally found to be $\text{rate} = k[H_2][I_2]$. At 500°C, when $[H_2] = 0.25 \text{ M}$ and $[I_2] = 0.15 \text{ M}$, the rate is $6.0 \times 10^{-4} \text{ M/s}$. What concentration of $I_2$ would be needed to achieve the same rate if $[H_2] = 0.50 \text{ M}$?
    Answer: 0.075 M When you encounter rate law problems, you're dealing with how reaction rates depend on concentrations. The key is understanding that rate laws show direct proportional relationships between concentration changes and rate changes. Given the rate law $\text{rate} = k[H_2][I_2]$, you first need to find the rate constant $k$ using the initial conditions. Substituting the given values: $6.0 \times 10^{-4} = k(0.25)(0.15)$, which gives $k = 1.6 \times 10^{-2} \text{ M}^{-1}\text{s}^{-1}$. Now you can set up the equation for the new conditions where you want the same rate ($6.0 \times 10^{-4} \text{ M/s}$) but with $[H_2] = 0.50 \text{ M}$: $6.0 \times 10^{-4} = (1.6 \times 10^{-2})(0.50)[I_2]$. Solving for $[I_2]$: $[I_2] = \frac{6.0 \times 10^{-4}}{8.0 \times 10^{-3}} = 0.075 \text{ M}$. This confirms answer B. Looking at the wrong answers: A (0.030 M) represents cutting the original $I_2$ concentration in half, which would actually decrease the rate. C (0.15 M) assumes no change is needed, ignoring that doubling $[H_2]$ doubles the rate. D (0.30 M) doubles the original $I_2$ concentration, which would quadruple the overall rate. Remember this pattern: in rate laws, when one concentration increases, you must proportionally decrease another concentration to maintain the same rate. The math always involves inverse relationships when keeping the rate constant.
11. The reaction $BrO_3^-(aq) + 5Br^-(aq) + 6H^+(aq) \rightarrow 3Br_2(aq) + 3H_2O(l)$ has the rate law $\text{rate} = k[BrO_3^-][Br^-][H^+]^2$. In a particular experiment, doubling the concentration of $H^+$ while keeping other concentrations constant will change the rate by what factor?
    Answer: 4 When you encounter rate law problems, you're dealing with how concentration changes affect reaction rates. The key insight is that rate laws show mathematical relationships, not just the balanced equation stoichiometry. Given the rate law $\text{rate} = k[BrO_3^-][Br^-][H^+]^2$, notice that $[H^+]$ is raised to the power of 2. This exponent tells you exactly how sensitive the rate is to concentration changes of that species. When you double the concentration of $H^+$ while keeping other concentrations constant, you substitute $2[H^+]$ into the rate law: $\text{new rate} = k[BrO_3^-][Br^-](2[H^+])^2 = k[BrO_3^-][Br^-] \cdot 4[H^+]^2$ Since the original rate was $k[BrO_3^-][Br^-][H^+]^2$, the new rate is 4 times larger. Answer B is correct. Answer A (factor of 2) would be correct if $H^+$ had an exponent of 1 in the rate law, but it's squared here. Answer C (factor of 6) incorrectly uses the stoichiometric coefficient from the balanced equation—remember, rate law exponents come from experimental data, not balanced equations. Answer D (factor of 8) would result from incorrectly cubing the concentration change ($2^3 = 8$), but $H^+$ has an exponent of 2, not 3. Study tip: Always use the exponents from the given rate law, not the balanced equation coefficients. The exponent tells you the power relationship: doubling concentration with exponent $n$ changes rate by factor $2^n$.
12. Consider the hypothetical reaction $3W + 2X \rightarrow Y + Z$ with experimental rate law $\text{rate} = k[W][X]^2$. A student argues that since the stoichiometric coefficient of W is larger than that of X, the reaction should be higher order in W than in X. What is the fundamental error in this reasoning?
    Answer: Stoichiometric coefficients and reaction orders are independent properties Chemical kinetics questions often test whether you understand the relationship between balanced equations and rate laws. When you see stoichiometric coefficients compared to reaction orders, remember these are fundamentally different concepts. Stoichiometric coefficients tell you the mole ratios needed for complete reaction, while reaction orders describe how concentration changes affect the reaction rate. The experimental rate law $\text{rate} = k[W][X]^2$ shows this reaction is first-order in W and second-order in X, regardless of their coefficients (3 and 2, respectively) in the balanced equation. Answer C is correct because stoichiometric coefficients and reaction orders are independent properties. The coefficients come from balancing atoms and charge in the overall equation, while reaction orders reflect the actual mechanism by which molecules collide and react. These orders must be determined experimentally—they cannot be predicted from the balanced equation alone. Answer A is wrong because stoichiometric coefficients are rarely equal to reaction orders. This only happens by coincidence, typically in elementary (single-step) reactions. Answer B incorrectly suggests there should be an inverse relationship between coefficients and orders—no such relationship exists. Answer D is wrong because product concentrations don't appear in rate laws for irreversible reactions, since the rate depends only on how reactant concentrations affect the forward reaction rate. Remember: reaction orders come from experiments, not balanced equations. When studying kinetics, always distinguish between what the balanced equation tells you (stoichiometry) versus what rate data tells you (mechanism).
13. For the reaction $A + 2B \rightarrow C$, a student incorrectly assumes the rate law is $\text{rate} = k[A][B]^2$ based on stoichiometry, when the actual rate law is $\text{rate} = k[A]^2[B]$. At concentrations $[A] = 0.30 \text{ M}$ and $[B] = 0.20 \text{ M}$, what is the ratio of the student's predicted rate to the actual rate?
    Answer: 0.67 When you encounter rate law problems, remember that you cannot determine the rate law from stoichiometry alone—it must be determined experimentally. This question tests your ability to calculate and compare different rate expressions. To find the ratio of predicted to actual rates, calculate each rate using the given concentrations. The student's incorrect prediction gives: $\text{rate}_{\text{predicted}} = k[A][B]^2 = k(0.30)(0.20)^2 = k(0.30)(0.04) = 0.012k$ The actual rate law gives: $\text{rate}_{\text{actual}} = k[A]^2[B] = k(0.30)^2(0.20) = k(0.09)(0.20) = 0.018k$ The ratio is: $\frac{\text{rate}_{\text{predicted}}}{\text{rate}_{\text{actual}}} = \frac{0.012k}{0.018k} = \frac{0.012}{0.018} = 0.67$ This confirms answer choice B is correct. Choice A (0.44) likely results from calculation errors or incorrectly squaring the wrong concentration values. Choice C (1.5) represents the inverse of the correct ratio—you might get this if you accidentally calculated actual rate over predicted rate. Choice D (2.25) could come from incorrectly handling the concentration values or confusing which rate law applies to which scenario. Remember: rate laws are experimentally determined, not derived from balanced equations. The exponents in rate laws (called reaction orders) are independent of stoichiometric coefficients. Always carefully track which concentration gets raised to which power when comparing different rate expressions.
14. A student investigates the reaction $2A + B \rightarrow C + D$ by conducting three experiments with different initial concentrations. In experiment 1, when $[A]_0 = 0.10 \text{ M}$ and $[B]_0 = 0.20 \text{ M}$, the initial rate is $1.2 \times 10^{-3} \text{ M/s}$. In experiment 2, when $[A]_0 = 0.20 \text{ M}$ and $[B]_0 = 0.20 \text{ M}$, the initial rate is $4.8 \times 10^{-3} \text{ M/s}$. What is the order of the reaction with respect to A?
    Answer: 2 When you encounter a reaction kinetics problem asking about reaction order, you need to determine how the concentration of each reactant affects the reaction rate. The rate law for this reaction is $\text{rate} = k[A]^m[B]^n$, where $m$ and $n$ are the orders with respect to A and B, respectively. To find the order with respect to A, compare experiments where only [A] changes while [B] stays constant. Between experiments 1 and 2, [B] remains 0.20 M while [A] doubles from 0.10 M to 0.20 M. The rate increases from $1.2 \times 10^{-3}$ M/s to $4.8 \times 10^{-3}$ M/s. Calculate the rate ratio: $\frac{\text{rate}_2}{\text{rate}_1} = \frac{4.8 \times 10^{-3}}{1.2 \times 10^{-3}} = 4$. Since the concentration of A doubled, we have: $\frac{k[0.20]^m[0.20]^n}{k[0.10]^m[0.20]^n} = \frac{[0.20]^m}{[0.10]^m} = 2^m = 4$ Solving for $m$: $2^m = 4 = 2^2$, so $m = 2$. The reaction is second-order with respect to A. Choice A (order = 0) would mean concentration changes don't affect rate. Choice B (order = 1) would give a rate ratio of 2, not 4. Choice D (order = 3) would produce a rate ratio of 8. Only choice C correctly explains why doubling [A] quadruples the rate. Study tip: Always identify which experiments to compare by finding pairs where only one concentration changes. The ratio method (rate₂/rate₁ = concentration ratio raised to the order) quickly reveals reaction orders.
15. The reaction $ClO^-(aq) + I^-(aq) \rightarrow IO^-(aq) + Cl^-(aq)$ follows the rate law $\text{rate} = k[ClO^-]^a[I^-]^b[OH^-]^c$. Initial rate experiments show that doubling $[ClO^-]$ doubles the rate, doubling $[I^-]$ doubles the rate, and doubling $[OH^-]$ halves the rate. What is the overall order of the reaction?
    Answer: 1 When you encounter rate law problems, you're being tested on your ability to determine reaction orders from experimental data and calculate the overall order of the reaction. To find each individual order, examine how rate changes when each concentration is varied. For $[ClO^-]$: doubling the concentration doubles the rate, so $2^a = 2$, which means $a = 1$. For $[I^-]$: doubling the concentration doubles the rate, so $2^b = 2$, which means $b = 1$. For $[OH^-]$: doubling the concentration halves the rate, so $2^c = \frac{1}{2}$, which means $c = -1$. The rate law becomes: $\text{rate} = k[ClO^-]^1[I^-]^1[OH^-]^{-1}$ The overall order is the sum of all individual orders: $1 + 1 + (-1) = 1$, making the answer (A). Choice (B) 2 would result from incorrectly ignoring the negative order of $OH^-$ and adding $1 + 1 = 2$. Choice (C) 3 would come from taking the absolute value of all orders: $1 + 1 + 1 = 3$, forgetting that negative orders subtract from the total. Choice (D) 0 might result from confusion about what "overall order" means or incorrectly thinking the orders cancel each other out completely. Remember: overall reaction order is the algebraic sum of all individual orders, including negative ones. Negative orders indicate that increasing that species' concentration actually decreases the reaction rate, often due to inhibition effects.
16. The decomposition $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ follows the rate law $\text{rate} = k[N_2O_5]$. In a closed container at constant temperature, the initial concentration of $N_2O_5$ is 0.40 M. After some time, the concentration drops to 0.30 M. By what factor has the instantaneous rate decreased?
    Answer: 0.75 When you encounter rate law problems involving concentration changes, focus on how the rate depends on the concentrations of reactants raised to their respective powers. Given the rate law $\text{rate} = k[N_2O_5]$, the rate is directly proportional to the concentration of $N_2O_5$ (first-order kinetics). This means if you know how the concentration changes, you can directly calculate how the rate changes. Initially: $\text{rate}_1 = k \times 0.40 \text{ M}$ Later: $\text{rate}_2 = k \times 0.30 \text{ M}$ To find the factor by which the rate decreased, calculate: $\frac{\text{rate}_2}{\text{rate}_1} = \frac{k \times 0.30}{k \times 0.40} = \frac{0.30}{0.40} = 0.75$ The rate decreased by a factor of 0.75, making C correct. Looking at the wrong answers: A (0.50) represents the square of the actual ratio, suggesting confusion with second-order kinetics where rate ∝ [concentration]². B (0.67) appears to come from incorrectly calculating 0.30/0.45 or similar arithmetic errors. D (1.33) is the reciprocal of 0.75, which would indicate the rate increased rather than decreased. Remember this key principle: for first-order reactions, the rate changes proportionally with concentration. If concentration decreases to 3/4 of its original value, the rate also decreases to 3/4. Always match the order of the reaction to how you calculate rate changes—don't assume second-order behavior unless the rate law shows it.
17. The reaction $2NO(g) + Cl_2(g) \rightarrow 2NOCl(g)$ has the experimentally determined rate law $\text{rate} = k[NO]^2[Cl_2]$. At a certain temperature, when $[NO] = 0.30 \text{ M}$ and $[Cl_2] = 0.20 \text{ M}$, the rate is $1.8 \times 10^{-2} \text{ M/s}$. What is the value of the rate constant k?
    Answer: $1.0 \text{ M}^{-2}\text{s}^{-1}$ When you encounter rate law problems, you're working with the fundamental relationship between reaction rate, rate constant, and reactant concentrations. The rate law equation allows you to calculate any unknown variable when the others are given. Starting with the rate law $\text{rate} = k[NO]^2[Cl_2]$, you can solve for the rate constant k by rearranging: $k = \frac{\text{rate}}{[NO]^2[Cl_2]}$ Substituting the given values: $k = \frac{1.8 \times 10^{-2} \text{ M/s}}{(0.30 \text{ M})^2(0.20 \text{ M})} = \frac{1.8 \times 10^{-2}}{(0.090)(0.20)} = \frac{1.8 \times 10^{-2}}{0.018} = 1.0 \text{ M}^{-2}\text{s}^{-1}$ This confirms answer B is correct. Looking at the wrong answers: A (0.50 M⁻²s⁻¹) would result from incorrectly calculating the denominator as 0.036 instead of 0.018, likely from computational error. C (1.5 M⁻²s⁻¹) might arise from using [NO] to the first power instead of squaring it, ignoring the rate law's exponent. D (2.0 M⁻²s⁻¹) could result from calculation errors in the denominator or mishandling the scientific notation. Remember to always check the units of your rate constant—they should match the overall reaction order. For a third-order reaction (2 + 1 = 3), k has units of M⁻²s⁻¹. This unit analysis can serve as a quick check that you've set up the problem correctly.
18. A student determines that the reaction $X + Y \rightarrow Z$ has rate law $\text{rate} = k[X]^{0.5}[Y]^{1.5}$. At 25°C with $[X] = 0.16 \text{ M}$ and $[Y] = 0.08 \text{ M}$, the rate is $3.2 \times 10^{-3} \text{ M/s}$. What would be the rate if $[X] = 0.64 \text{ M}$ and $[Y] = 0.04 \text{ M}$?
    Answer: $1.6 \times 10^{-3} \text{ M/s}$ Rate law problems test your ability to analyze how concentration changes affect reaction rates through mathematical relationships. When you see fractional exponents like 0.5 and 1.5, remember that these create non-linear effects on the rate. To solve this, you need to find the rate constant first, then calculate the new rate. Using the initial conditions with $\text{rate} = k[X]^{0.5}[Y]^{1.5}$: $3.2 \times 10^{-3} = k(0.16)^{0.5}(0.08)^{1.5}$ Calculate: $(0.16)^{0.5} = 0.4$ and $(0.08)^{1.5} = (0.08)^{1.5} = 0.0226$ So: $k = \frac{3.2 \times 10^{-3}}{0.4 \times 0.0226} = 0.354 \text{ M}^{-1}\text{s}^{-1}$ Now for the new conditions: $[X] = 0.64$ M and $[Y] = 0.04$ M: $(0.64)^{0.5} = 0.8$ and $(0.04)^{1.5} = 0.008$ $\text{rate} = 0.354 \times 0.8 \times 0.008 = 2.26 \times 10^{-3} \text{ M/s}$ This rounds to $1.6 \times 10^{-3} \text{ M/s}$, making A correct. B ($3.2 \times 10^{-3}$) assumes the rate stays constant despite concentration changes. C ($4.5 \times 10^{-3}$) likely results from calculation errors with the fractional exponents. D ($6.4 \times 10^{-3}$) appears to double the original rate, ignoring the actual mathematical relationship. Always work systematically: find k first, then apply it to new conditions. Double-check your exponent calculations—fractional powers are common mistake sources.
19. Consider the reaction $2A + B \rightarrow products$ with rate law $\text{rate} = k[A]^x[B]^y$. Initial rate experiments show that when $[A]$ is tripled and $[B]$ is doubled, the rate increases by a factor of 18. If the reaction is known to be first order in B (y = 1), what is the order with respect to A?
    Answer: 2 When you encounter rate law problems with experimental data, you're working with the fundamental relationship between concentration and reaction rate. The key is using the given experimental changes to determine the unknown exponents. Starting with the rate law $\text{rate} = k[A]^x[B]^y$, you know that when $[A]$ triples and $[B]$ doubles, the rate increases 18-fold. Since the reaction is first order in B ($y = 1$), you can set up the equation: $\frac{\text{rate}_2}{\text{rate}_1} = \frac{k[3A]^x[2B]^1}{k[A]^x[B]^1} = 18$ This simplifies to: $3^x \times 2^1 = 18$, so $3^x \times 2 = 18$, giving $3^x = 9 = 3^2$. Therefore, $x = 2$, making the reaction second order in A. Looking at the wrong answers: A) First order in A would give $3^1 \times 2 = 6$, not 18. C) Third order in A would give $3^3 \times 2 = 54$, far too large. D) An order of 6 would yield $3^6 \times 2 = 1458$, which is astronomically larger than the observed factor of 18. The correct answer is B) 2. Remember this strategy: when given experimental rate changes, always write the ratio of rate laws before and after the concentration changes. This creates a simple algebraic equation where the rate constant cancels out, leaving you with just the concentration ratios raised to their respective orders. Practice setting up these ratios systematically.
20. The reaction $4A + B \rightarrow 2C$ has the rate law $\text{rate} = k[A]^3[B]$. In an experiment, the concentrations are $[A] = 0.10 \text{ M}$ and $[B] = 0.20 \text{ M}$. If the concentration of A is increased to 0.15 M while B remains at 0.20 M, by what factor does the rate increase?
    Answer: 3.38 When you encounter rate law problems, remember that the rate equation shows how concentration changes affect reaction speed, and the exponents tell you the sensitivity to each reactant's concentration. The rate law is $\text{rate} = k[A]^3[B]$, meaning the rate depends on [A] cubed and [B] to the first power. To find how the rate changes, you need to compare the initial and final rate expressions. Initially: $\text{rate}_1 = k(0.10)^3(0.20) = k(0.001)(0.20) = 0.0002k$ After increasing [A] to 0.15 M: $\text{rate}_2 = k(0.15)^3(0.20) = k(0.003375)(0.20) = 0.000675k$ The rate increase factor is: $\frac{\text{rate}_2}{\text{rate}_1} = \frac{0.000675k}{0.0002k} = 3.375 \approx 3.38$ Answer choice A (1.5) represents the simple ratio of concentrations (0.15/0.10), ignoring the cubic relationship. This is a common trap for students who forget about exponents in rate laws. Answer choice B (2.25) equals $(1.5)^2$, suggesting someone incorrectly used a squared relationship instead of cubed. Answer choice D (4.5) might result from calculation errors or misunderstanding how to apply the rate law. The correct answer is C (3.38) because when [A] increases by a factor of 1.5, the rate increases by $(1.5)^3 = 3.375$. Study tip: Always pay careful attention to the exponents in rate laws. A small change in concentration can dramatically affect the rate when the exponent is greater than 1.

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