101 Chem101b Answered: Step 1X + Y XY (slow) XB Please answer in format required XG Consider the proposed mechanis x+->A app.101edu.coI Appsb bartlebyB Brainly.com - For st. = Google Docs M gmailO eagle advO school emailS canvasm Monthly | My ClassesE Reading listemail101 Chem101generalQuestion 36 of 44SubmitConsider the proposed mechanism for the decomposition reaction 2 H2O2(aq) → 2H:O(I) + O2(g) in the presence of l-(aq). What is the rate law for this reaction?A) rate = k[H2O2]Step 1 (slow)B) rate =k[H:O:]?H2O2(aq) + l-(aq)H2O(1) + 10-(aq)C) rate = k[H2O:][I]Step 2 (fast)D) rate = k[H2O:][IO-]H2O2(aq) + 10-(aq) → H20(1) + Oz(g) + l(aq)E) rate = k[H2O2]°[I¬]/[H±O]4:47 PMP Type here to search82°F AQI 517/21/2021

Consider the proposed mechanism for the decomposition reaction 2 H:O:(aq) → 2 H:O(I) + Oz(g) in the presence of I-(aq). What is the rate law for this reaction? A) rate = k[H2O:] Step 1 (slow) B) rate = k[H2O:]² H2O2(aq) + I(aq) → H2O(1) + 10-(aq) C) rate = k[H2O:][I] Step 2 (fast) D) rate = k[H2O:][IO¯] H2O2(aq) + 10-(aq) → H2O(1) + O2(g) + I-(aq) E) rate = k[H:O2]°[I¬/[[H±O]

Consider the proposed mechanism for the decomposition reaction 2 H:O:(aq) → 2 H:O(I) + Oz(g) in the presence of I-(aq). What is the rate law for this reaction? A) rate = k[H2O:] Step 1 (slow) B) rate = k[H2O:]² H2O2(aq) + I(aq) → H2O(1) + 10-(aq) C) rate = k[H2O:][I] Step 2 (fast) D) rate = k[H2O:][IO¯] H2O2(aq) + 10-(aq) → H2O(1) + O2(g) + I-(aq) E) rate = k[H:O2]°[I¬/[[H±O]

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...

See similar textbooks

Similar questions

The rate law for the reaction: H2(g) + 2NO(g) ⟶ N2 O(g) + H2 O(g)has been determined to be rate = k[NO]2[H2]. What are the orders with respect to each reactant, and what is the overall order of the reaction?
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]
The rearrangement of ammonium cyanate to urea in aqueous solution at 50 °C NHẠNCO(aq)–→(NH2)½CO(aq) is second order in NH NCO with a rate constant of 1.10x10² M' min! If the initial concentration of NH,NCO is 0.945 M, the concentration of NH NCO will be 0.160 M after minutes have passed.
4. The rate law for the reaction H₂ (g) + I₂ (g) → 2 HI (g) was determined to be rate = k [H₂] [1₂], which led to a simple single-step bimolecular mechanism where an H₂ molecule simply collided with an I₂ molecule. In the 1960s, though, different teams found spectroscopic evidence for this mechanism: (1) (2) (3) I₂ (g) →21 (g) H₂(g) + I (g) → H₂I (g) H₂I (g) + I (g) → 2 HI (g) (fast) (fast) (slow) Show that this mechanism is consistent with the rate law: rate = k [H₂] [1₂]. To do this, assign individual rate constants for ALL forward and reverse reactions in the mechanism (I'm guessing k₁, k-1, k2, k-2 and k3). Then show that the overall rate law from combining individual rate laws for the mechanism steps yields rate = (some combination of k's) [H₂] [1₂], and thus k= (some combination of k's). Remember that there should be no reaction intermediates in the rate law, only reactants. \
5. You studied the chemical reaction, 2NO2(g) → 2NO(g) + O2(g), at 25°C by monitoring the concentration of NO2(g) as a function of time and constructed the following graph. y= 0.5408x + 125.0 R2=0.998 Time (s) What is the rate constant for this reaction at 25°C? Include the proper units. (W/T) [(3)ONI/T
Consider the mechanism. 2 A = B В +С — D Step 1: equilibrium Step 2: slow Overall: 2A +С — D Determine the rate law for the overall reaction, where the overall rate constant is represented as k. rate =
The reaction described by H, (g) + L,(g) 2 HI(g) has an experimentally determined rate law of rate = k [H2] [I2] Some proposed mechanisms for this reaction are: Mechanism A k1 (1) H,(g) + L,(g) → 2 HI(g) (one-step reaction) Mechanism B k1 =2 I(g) k_1 (1) L(g) ; (fast, equilibrium) k2 (2) H, (g) + 2 I(g) 2 HI(g) (slow) Mechanism C k1 = 2 I(g) (fast, equilibrium) (1) k2 (2) I(g) + H, (g) HI(g) + H(g) (slow) (3) H(g) + I(g) → HI(g) (fast) Question Sourc
Consider the following reaction: A + B + C → D The rate law for this reaction is as follows: [A][C]* Rate = k [B]L/2 Suppose the rate of the reaction at certain initial concentrations of A, B, and C is 1.17x10-2 mol · L-ls¬1. Part A What is the rate of the reaction if the concentrations of A and C are doubled and the concentration of B is tripled? Templates Symbols uado redo reset keyboard shortcuts help Rate = mol · L-!s-1
The observed rate law for the reaction 20₂(g) →30₂(g) is Rate = k[O3]²/[0₂]. Is it improbable that the reaction occurs in one step: O₂(g) + O₂(g) → O₂(g) + O₂(g) + O₂(8) ? Why or why not?
Be sure to answer all parts. For the decomposition of gaseous dinitrogen pentaoxide, 2 N2O5(g) → 4 NO2(g) + O2(g) the rate constant is k = 2.8 × 10¯³ s−1 at 60°C. The initial concentration of N2O5 is 2.31 mol/L. (a) What is [N2O5] after 5.00 min? mol L (b) What fraction of the N2O5 has decomposed after 5.00 min? X
For the reaction: IO3 (aq) + 5 I (aq) + 6 H* (aq) + 3I2 (aq) + 3 H2O (l) the reaction rate of I° (aq) is 2.6 x 10-3 M/s Calculate the rate at which (ac) is formed.
Suppose the formation of nitrogen dioxide proceeds by the following mechanism: elementary reaction 2NO (g) → N₂O₂ (g) step 1 2 N₂O₂(g) + O₂ (g) Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction. 2NO₂ (g) Suppose also k₁ « k₂. That is, the first step is much slower than the second. << Note: your answer should not contain the concentrations of any intermediates. rate constant 0 k₁ k₂ rate = k 1-0 010 X S