The rate of the reaction is determined by the slowest step in its mechanism.
For, any mechanism the rate is determined by the concentration of the reactants.
Therefore, rate = k3[A2B][B]
Here, k3 is the rate constant, [A2B] is the concentration of A2B, and [B] is the concentration of B.
So, 3rd option is correct.
Please give good rating.
Question 5 (1 point) Saved The rate law for the mechanism below is A260) = 2A...
The rate law for the mechanism below is A2(g) ⇌ 2A(g) K1, fast A(g) + B(g) ⇌ AB(g) K2, fast AB(g) + A(g) → A2B(g) k3, slow Options: rate = K2k3[A]2[B] rate = K1K2k3 [A2][B] rate = K10.5k3 [A2]0.5[AB] rate = k3[A][AB] rate = K10.5K2k3 [A2]0.5[A] [B]
Predict the overall reaction from the following two-step mechanism: 2 A→A 2(fast) A2+B→A2B(slow) Express your answer as a chemical equation. Predict the rate law from the following two-step mechanism: 2A⇌ A2 (fast) A2+B→A2B (slow)
Derive the rate law for a uni-molecular reaction with the following mechanism A + M AM* (k1, k -1) AM* + M AM + M (k2, k -2) AM + A ? A2 + M (k3) when the pressure is low.
A chemical reaction, A+B → P, has the following mechanism: 2A< Ki>A, (fast to equilibrium) A+B&K, ™C (fast to equilibrium), A,+C-k>P+ 2A (slow) where Kį and K2 are the equilibrium constants for the first two reactions, respectively. k3 is the rate constant for the third reaction. (a) [5 points] Based on this mechanism, show that the rate of product (P) formation is: d[P] – k[A[B], where k is the rate constant of the overall reaction. Write k in terms of...
Consider the following mechanism. step 1 2A > B slow B+ C D fast step 2 overall: 2A+C D Determine the rate law for the overall reaction (where the overall rate constant is represented as k) rate=
PLEASE MAKE SURE C AND D ARE CORRECT! i have had two wrong answers so far. This is all one question, so I could not post the parts separately. Thanks. Consider the following reaction mechanism: Step 1 K 2A + B =D (fast equilibrium) Step 2 D+B kk E+F (slow) Step 3 F G (fast) a. Give the molecularity of each step. Step 1 O unimolecular bimolecular O termolecular Step 2 unimolecular Obimolecular termolecular Step 3 unimolecular O bimolecular termolecular...
The reaction 2A + B → C occurs by the following 2 step mechanism: A + B ------(k1)------> AB AB ------(-k1) ------> A+B AB + A ----- (k2)------> C Apply the steady-state approximation for the reaction intermediate concentration to obtain the overall rate law from this mechanism: a) k1 [A][B] b) k1k2[A][B]/((-k1) - k2[A]) c) k1k2[A]^2[B]/((-k1)+k2[A])
Below is a mechanism for the reaction A+B- P 5. 2A ? A+C rate constant kl AtC 2A rate constant k B+C Prate constant k2 In this mechanism, C is an intermediate. nd the steady state approximation if necessary, determine the rate law for the reaction. (B) Under what conditions does the rate law become first order in [AJ? (C) Under what conditions does the rate not depend on [B]?
Please answer all parts to this question! parts a through d with the explanation on question d! This is all one question so I could not post them separately, please also check your answers because the last person got it wrong! Thank you! WIll give thumbs up. Consider the following reaction mechanism: Step 1 K 2A + B =D (fast equilibrium) Step 2 D+B k? E+ F (slow) Step 3 F G (fast) a. Give the molecularity of each step....
Given the following proposed mechanism: A3 ⇌ A2 + A K1, fast step A + A3 → 2A2 k2, slow step The rate rate law is Question 5 options: a.)second order in A3 and first order in A2 b.)second order in A3 and zero order in A2 c.)first order in A3 and zero order in A2 d.)inverse order, i.e., -1, in A3 and second order in A2 e.)second order in A3 and inverse order, i.e., -1, in A2 f.)zero order...