Show that if it can be assumed that there is a pre-equilibrium involving step(1), the rate of formation of ethylene oxide is v = k,K[CH,CICH,OH][OH ],where K is the equilibrium constant for the first step and k, is the rate constantfor the second step.E17E.2(b) 'The mechanism for the reaction between 2-chloroethanol,CH,CICH,OH, and hydroxide ions in aqueous solution to form ethyleneoxide, (CH,CH,)0, is thought to consist of the steps(1) CH,CICH,OH+OH 2CH,CICH,O¯+H,O(2) CH,CICH,O¯ –→(CH,CH,)O+CI

Answered: Show that if it can be assumed that…

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter11: Chemical Kinetics: Rates Of Reactions

Section: Chapter Questions

Problem 83QRT

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E17C.1(b) The equilibrium constant for the binding of a drug molecule to a protein was measured as 200. In a separate experiment, the rate constant for the binding process, which is second order overall, was found to be 1.5 x 10 dm mol s. What is the rate constant for the first-order dissociation of the drug molecule from the protein- drug complex?
The first step in the heterogeneoushydrogenation of ethylene is adsorption of theethylene molecule on a metal surface. One proposed explanation for the “sticking” of ethylene to a metalsurface is the interaction of the electrons in the C—C π bond with vacant orbitals on the metal surface. (a) If thisnotion is correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bindcompared to ethylene? (b) Based on its Lewis structure,would you expect ammonia to adsorb to a metal surfaceusing a similar explanation as for ethylene?
20B.2(b) The rate constant for the first-order decomposition of a compound A in the reaction 2 A→P is krr=3.56×10−7 s−1 at 25 °C. What is the half-life of A? What will be the pressure, initially 33.0 kPa after (i) 50 s, (ii) 20 min after initiation of the reaction?
Suppose the formation of tert-butanol proceeds by the following mechanism: step elementary reaction rate constant CBr(aq) (CH), С"(аg) + Br (ag) c* k 1 2 (CH3),C"(aq) + OH (aq) → (CH,),COH(aq) k2 3 3 Suppose also k, «k,. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction: Write the experimentally- observable rate law for the overall chemical reaction. rate = k U Note: your answer should not contain the concentrations of any intermediates. Express the rate constant k for the overall chemical reaction in terms of k1, k2, and (if necessary) the rate constants k.1 and k-2 for 0 k = the reverse of the two elementary reactions in the mechanism.
E17B.6(a) The reaction A + B → P is found to be first order in both A and B. The reaction was carried out in a solution that was initially O.080 mol dm-3 in A and 0.060 mol dm-3 in B. After 1.0 h the concentration of B had fallen to 0.030 mol dm-3. (i) Calculate the rate constant. (ii) What are the half-lives of the reactants?
3 - Consider the gas phase decomposition of acetic acid at 1200 K. It proceeds via two parallel reactions: CH:COOH → CH+ + CO: ki = 3.74 s CH:COOH – CH:co + H2O k2 = 4.65 s'! (a) What is the maximum percentage yield of the ketene CH:CO at this temperature? (b) What is the concentration of the ketene when the extent of reaction when is 0.5 (50%)? (c) To favor the second reaction, should you increase or decrease temperature?
When the rate of the reversible reaction A + BSC is studied under a certain set of conditions, it is found that the rate of the forward reaction is kA]. What can be concluded about the rate law for the reverse reaction under these conditions? (A) Rate = k-[C] %3D (B) Rate = k [B] (C) The rate law of the reverse reaction cannot be determined from the information given. (D) An crror must have been made, since if the reaction is reversible, the forward rate law must be Rate = k{A][B]. %3D
The following mechanism is proposed: C6H5COOH ⇌ C6H5CO2H2+ (fast) C6H5CO2H2+ ⇌ C6H5CO+ + H2O (fast) C6H5CO+ + HX + H2O ⟶ C6H6NH2 + N2 + CO2 (slow) The overall enthalpy change is -237.6 kJ. Sketch a reaction-energy diagram consistent with the above mechanism and the enthalpy change. Label the enthalpy change and each transition state. Explain the appearance of your graph.
Derive the integrated form of a third-order rate law v = kr[A]2[B] in which the stoichiometry is 2 A + B → P and the reactants are initially present in (a) their stoichiometric proportions ([B]0 = 12[A]0); (b) with B present initially in twice that amount ([B]0 = [A]0). Express your rate law in terms of [A]0, [B]0, and x, where [A] = [A]0 − 2x.
Conventional equilibrium considerations do not apply when a reaction is driven by light absorption and the steady-state concentration of products and reactants might differ significantly from equilibrium values. For instance, suppose the reaction A → B is driven by light absorption, and that its rate is Ia, but that the reverse reaction B → A is bimolecular and second order with a rate kr[B]2. What is the stationary state concentration of B? Why does this ‘photostationary state’ differ from the equilibrium state?
E17B.1(b) In a study of the enzyme-catalysed oxidation of ethanol, the molar concentration of ethanol decreased in a first-order reaction from 220 mmol dm3 to 56.0 mmol dm in 1.22 x 10 s. What is the rate constant of the reaction?
Show that the initial rate law predicted by the reaction mechanism 6-12a-c, with the first step rate-limiting, is rate = 2k,[Cl,]. Assume that the Cl - produced in step (3) can be neglected initially.

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