What is the rate law for the following mechanism: A-B+C=D (fast equilibrium; rate of forward reaction: k1, rate of reverse reaction: k.1) A+D -E (slow, rate: k2) E +G - H (fast, rate: k3) Rate= k2 (k1 k.j) [Aj°{B][C] OA Rate= k2 (k1 k.1)[A][D] OB. oc Rate= k1 [A][B][C] Rate= k2 [AJ°{B][C] OD. O E. Rate= k2 (k1/k.1) [A][B][C]

What is the rate law for the following mechanism: A-B+C=D (fast equilibrium; rate of forward reaction: k1, rate of reverse reaction: k.1) A+D -E (slow, rate: k2) E +G - H (fast, rate: k3) Rate= k2 (k1 k.j) [Aj°{B][C] OA Rate= k2 (k1 k.1)[A][D] OB. oc Rate= k1 [A][B][C] Rate= k2 [AJ°{B][C] OD. O E. Rate= k2 (k1/k.1) [A][B][C]

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

Enzymes facilitate reactions in living systems. The figure below presents an energy diagram for the reaction X®Y. The solid line in the energy diagram represents changes in energy as the reactant is converted to product under standard conditions. The dashed line shows changes observed when the same reaction takes place in the presence of a dedicated enzyme. Choose the sentence that best describes this graph. а X reaction pathway O a DG° = 0 Ob. DG° = 1 O. DG° =-1 Od. DG° = -5 free energy →
See image below.
3.3333 p O3+0 406 kJ 392 kJ O2+ 02 Reaction Progress Using the above reaction coordinate diagram, what is the activation energy for the formation of the prodcuts (assuming the reaction is proceeding in the forward direction)? And is the forward reaction represented an endothermic or an exothermic reaction? 406 kJ, exothermic 392 kJ, endothermic 392 kJ, exothermic 406 kJ, endothermic 14 kJ, exothermic 3.3333 p QUESTION 29 acer Energy kJ 000
a) The following are typical reactions of alkyl halides. In each case, draw the structure of the main organic product and classify the reaction mechanism as SN1, SN2, E1 or E2. NH2 (1) CH,CH,CH,Br Br CH,0° (ii) CH,CH,C-CH3 ČH3 heat Br H,0 (iii) CH;CH,C-CH3 heat CH, ČH3 (iv) CH;CH,CH,CH,Br CH3
9:54 AM Tue Nov 21 For the following reaction mechanism, what is the catalyst? NO(g)+O3(g) → NO2(g) + O2(g) A) O(g) O(g) +NO2(g) O2(g) + NO(g) B) O₂(g) C) O3(g) Question 20 of 20 D) NO(g) E) NO₂(g) O3(g) O2(g) + O(g) Tap here or pull up for additional resources 36% Submit
The following mechanism has been suggested for the reaction: H:O2 + 2H* + 21-→ I2 + 2H:0 H:O2 +I-+ HOI + OH- OH-+H* + H;O HOI + H* +I-1+H:O Slow Fast Fast Identify the rate law that is consistent with this mechanism. Multiple Choice Rate - k(H2021) Rate - k(HOIIOH] Rate - K[HOITH'I] Rate - K(H202?H?
Which of the following statements is/are correct? (1) Catalysts are substances that increase the rate of reaction but are not consumed in the reaction. (2) Enzymes are catalysts. (3) Raising the temperature will have no effect on the rate of a reaction. OA. 2 only OB. 2 and 3 O C. 1 and 2 O D. 1, 2, and 3 OE. 3 only 81°F Sunny Esc A @ 2 W S F3 #m 3 F4 E D Use the References to access important values if needed for this c $ 4 F5 R 6 % соо 5 F F6 a T DELL F7 1.0 G F8 & 7 7 CE H F9 2 0000 8
8. (15 marks) A reaction is written as A(g) At 298 K, k₁ =0.21 s.1. k.₁=5×10 Pa¹·s¹. B(g) + C(g). k-1 As the temperature increased from 298 K to 310 K, the values of k₁ and k., are doubled (denoted as k2 and K.2). (1) Calculate the rate constant k at 298K; (2) Calculate the activation energy of both forward and reverse reactions; (3) Calculate AH and AU for the reaction at 298K; (4) Calculate the reaction time at the whole pressure reaching to 152 kPa. The initial pressure of A(g) is 100 kPa and the reaction temperature is 298 K.
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]
2A(g) + 3/2B(g) Experiment e b) D d) k= 12345 Initial rate (mol/L.s) →→ C(g) 0.25*10-4 2.25x10-4 0.50×10-4 1.00×10-4 0.75*10-4 Initial [A] (mol/L) 5x10-2 15x10-2 5x10-2 10×10-2 5x10-2 unit: Overall reaction order is Rate = K[A][B] Initial [B] (mol/L) Reaction order with respect to A is Reaction order with respect to B is 5x10-2 5x10-2 10x10-2 5x10-2 15x10-2 O " O Calculate rate constant (k) based on data from experiment 3:
Energy diagrams for two reactions are shown. Energy (kJ/mol) 1502 100- 50 Reaction progress Reaction A What is the heat of reaction for Reaction A? AH rxn A = What is the activation energy for Reaction A? kJ mol kJ Energy (kJ/mol) 150. 100- 50- Reaction progress Reaction B What is the heat of reaction for Reaction B? AHrxn B = What is the activation energy for Reaction B? Hint kJ mol kJ
Q27. The gas phase reaction between SiH2 and H20 was studied, and the following mechanism was proposed. M is a buffer gas which takes away energy as heat; you can treat it as a regular reactant. 1) SİH2 + H20 F SİH2-H2O* K 2) SİH2-H2O* + M SİH3OH Assuming that the first step is a pre-equilibrium step and the second step is rate-limiting, derive the rate law for this mechanism.