### Reaction Rate EquationsThis section provides different rate equations for chemical reactions, each representing various dependencies on the concentration of reactants.**A)** Rate = \( k_1[A][B][C] \)- This equation suggests a rate that is dependent on the concentrations of reactants A, B, and C, with each appearing linearly in the rate expression.**B)** Rate = \( k_2[C] \)- Here, the rate is only dependent on the concentration of reactant C, indicating a simpler reaction mechanism involving C.**C)** Rate = \( k_1[A]^2[B] \)- This equation indicates that the reaction rate depends on the concentration of A squared and B, implying a second-order dependence on A.**D)** Rate = \( \frac{k_1[A][B]}{k_2[C]} \)- The rate is given by a ratio, suggesting a dependence on the concentrations of A and B divided by C. This could reflect a situation where C negatively impacts the rate.**E)** Rate = \( k[A]^2[B]^2 \)- This indicates a reaction where the rate depends on the square of the concentrations of both A and B, suggesting a more complex interaction.Explore the rate laws to understand the order of reactions and how each reactant influences the reaction pace. Engage with additional resources for a deeper dive into reaction kinetics by tapping below. **Multistep Reaction Analysis and Rate Law Determination****Problem Statement:**Consider the multistep reaction. What is the best rate law for the overall reaction?**Reaction Steps:**1. **First Step (Slow):** \[ 2A + B \xrightarrow{k_1} C + D \] - This step is characterized as slow.2. **Second Step (Fast Reversible):** \[ B + C \xleftrightarrow{k_2} 2E \] - This step is fast and reversible.**Explanation:**In this multistep reaction, the slow step is the rate-determining step. The rate law for the overall reaction is typically determined by this rate-determining step. The reaction proceeds via an intermediate, C, which is formed and consumed in subsequent steps.1. **Rate of the Slow Step:** - The rate can be expressed as: \[ \text{Rate} = k_1 [A]^2 [B] \] Here, \( k_1 \) is the rate constant for the slow step, and \( [A] \) and \( [B] \) represent the concentrations of A and B, respectively.2. **Fast Step Equilibrium:** - The fast equilibrium implies that concentrations can be expressed in terms of each other using the equilibrium constant if needed.Efforts to derive the overall reaction rate must focus on substituting for any intermediates based on the fast equilibrium. However, the dominant concentration terms in the initial slow step predominantly dictate the overall rate law.This multistep mechanism suggests a reaction heavily dependent on the concentrations of A and B relevant to the first, slow step.

Answer:The Law of mass action states that rate of reaction is directly proportional to all the…

Answered: Consider the multistep reaction. What…

Consider the multistep reaction. What is the best rate law for the overall reaction? 2A K₁ + B → B + C 11 5 C + D (slow) 2E (fast)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A reaction was experimentally determined to follow the rate law, rate = k[X]2 where k = 0.860/M.s. If the starting concentration of X = 0.800 M, how many seconds will it take for the concentration of X to become 0.400 M? O (3) 0.465 sec O (4) 1.45 sec O (1) 0.805 sec O (2) 1.16 sec A Moving to the next question prevents changes to this answer. # T3 7 To D *3 E C 54 R F % 5 V T G 6 MacBook Pro B Y & N 7 H U N * 00 8 J 1 ( 9 K M O L P Question 12 of 37 > = است 11
What is the rate law for the following mechanism: NO (g) + Br2 (g) 2 NOB12 (g) (fast equilibrium) (rate of forward reaction: k1, rate of reverse reaction: k.1) NOB12 (g) + NO (g) → 2 NOB1 (g) (slow, rate: k2) Rate= k2 [NOB12[NO] O A. Rate= k1 [NO][Br2] B. Rate= (k1/k-1) [NO][Br2] Rate= k2 (k1k.1) [NOJ²[Br2] D. Rate= k2 [NO]?[Br2]
The reaction 2 NO(g) + O2(g) --> 2 NO2(g) proceeds through the following mechanism: 2 NO(g) --> N2O2(g)N2O2(g) + O2(g) --> 2 NO2(g) (a) The second step of this mechanism is rate-determining (slow). What is the rate law for this reaction? Rate = k [NO] [O2] Rate = k [NO]2 [O2] Rate = k [NO] [O2]2 Rate = k [NO]1/2 [O2] Rate = k [NO] [O2]1/2 Rate = k [NO]2 Rate = k [NO]2 [O2]1/2 (b) What would the rate law be if the first step of this mechanism were rate-determining? Rate = k [NO] [O2] Rate = k [NO]2 [O2] Rate = k [NO] [O2]2 Rate = k [NO]1/2 [O2] Rate = k [NO] [O2]1/2 Rate = k [NO]2 Rate = k [NO]2 [O2]1/2 Rate = k [NO]
8 OKONETICS AND EQUILIBRIUM Deducing a rate law from the change in concentration over time A chemistry graduate student is studying the rate of this reaction: 2N₂O, (g) → 2N₂O, (g) + O₂(g) He fills a reaction vessel with N₂O, and measures its concentration as the reaction proceeds: time (minutes) 0 10. [N₂O,] Explanation 0.300M 0.121 M 20. 0.0762 M 30. 0.0555 M 40. 0.0436M Use this data to answer the following questions. Write the rate law for this reaction. Calculate the value of the rate constant k Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol. Check rate & D www-awa.aleks.com k= D.D X
Determine the rate law and calculate the rate constant for this reaction: 2 NO (g) + Br₂(1)→2 NOBr (g) Initial [Br₂]/M Expt. 1 2 3 Initial [NO]/M 0.10 0.20 0.20 0.10 0.10 0.30 Initial rate/M.s-¹ 1.30 x 10-³ 5.20 x 10-³ 1.56 x 10-²
Consider the reaction and its rate law. 3A+3B⟶productsrate=?[A][B]2 What is the order with respect to A?
Consider the reaction and its rate law. 4A + 3B What is the order with respect to A? order: products What is the order with respect to B? order: What is the overall reaction order? overall order: rate = k[A]²[B]²
What is the rate constant of the reaction?
Questions 2A 2B and 2C please
The reaction 2 NO(g) + O2(g) --> 2 NO2(g) proceeds through the following mechanism: 2 NO(g) --> N2O2(g)N2O2(g) + O2(g) --> 2 NO2(g) (a) The second step of this mechanism is rate-determining (slow). What is the rate law for this reaction? Rate = k [NO] [O2] Rate = k [NO]2 [O2] Rate = k [NO] [O2]2 Rate = k [NO]1/2 [O2] Rate = k [NO] [O2]1/2 Rate = k [NO]2 Rate = k [NO]2 [O2]1/2 (b) What would the rate law be if the first step of this mechanism were rate-determining? Rate = k [NO] [O2] Rate = k [NO]2 [O2] Rate = k [NO] [O2]2 Rate = k [NO]1/2 [O2] Rate = k [NO] [O2]1/2 Rate = k [NO]2 Rate = k [NO]2 [O2]1/2 Rate = k [NO]
Consider the gas phase reaction 2 NO + H₂ → N₂O + H₂O The experimental rate law is found to be R = [NO]²[H₂] Which of the mechanisms shown below is consistent with the information about this reaction given in this question? Step 1: NO + H₂ → N + H₂O (slow) Step2: N + NO → N₂O (fast) Step 1: NO + NO 2 NO₂ + N (slow) (fast) Step2: NO₂ + H₂ Step 3: N + NO → NO + H₂O N₂O (fast) Step 1: NO + NO>>> N₂O₂ Step2: N₂O₂ + H2 →→ N₂O + H₂O 2H (fast) Step 1: H₂ Step2: H+2 NO → Step 3: H+ OH → N₂O + OH (slow) H₂O (fast) (fast) (slow)
Write the reaction rate equation and identify the order of the reaction: Chemical Equation Type of Order Reaction Rate H2(g) + I2(g) => 2HI(g) NO(g) + O3(g) => NO2(g) + O2(g) 4OH(G) + H2S(g) => SO2(g) + 2H2O(aq) + H2(g)