Please answer Data for (A) versus time is given below for the reaction:A - productsTime, s [A), M0.001.0025.00.7875.00.47150.0 0.22Fit this data into the integrated rate laws of first and second-orderreactions.38. What is the linear correlation coefficient (R) for each linearregression?R (First order) = (Select I Fit this data into the integrated rate laws of first and second-orderreactions.38, What is the linear correlation coefficient (R) for each linearregression?R(First order) = (Select )R (Second order).[ Select]39. What is the order of the reaction?| Select )40. What is the value of the rate constant?[ Select)

Data for (A) versus time is given below for the reaction: A + products Time, s [A), M 0.00 1.00 25.0 0.78 75.0 0.47 150.0 0.22 Fit this data into the integrated rate laws of first and second-order reactions. 38. What is the linear correlation coefficient (R) for each linear regression? R(First order) = [Select| !!

Data for (A) versus time is given below for the reaction: A + products Time, s [A), M 0.00 1.00 25.0 0.78 75.0 0.47 150.0 0.22 Fit this data into the integrated rate laws of first and second-order reactions. 38. What is the linear correlation coefficient (R) for each linear regression? R(First order) = [Select| !!

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter18: Chemical Kinetics

Section: Chapter Questions

Problem 75AP

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The color change accompanying the reaction of phenolphthalein with strong base is illustrated below. The change in concentration of the dye can be followed by spectrophotometry (Section 4.9), and some data collected by that approach are given below. The initial concentrations were [phenolphthalein] = 0.0050 mol/L and [OH] = 0.61 mol/L. (Data are taken from review materials for kinetics at chemed.chem.purdue.edu.) (For more details on this reaction see L Nicholson, Journal of Chemical Education, Vol. 66, p. 725, 1989.) (a) Plot the data above as [phenolphthalein] versus time, and determine the average rate from t = 0 to t = 15 seconds and from t = 100 seconds to t = 125 seconds. Does the rate change? If so, why? (b) Use a graphical method to determine the order of the reaction with respect to phenolphthalein. Write the rate law, and determine the rate constant. (c) What is the half-life for the reaction?
Instantaneous rates for the reaction of hydroxide ion with Cv+ can be determined from the slope of the curve in Figure 11.3 at various concentrations. They are (1) At 4.0 105 mol/L, rate = 12.3 107 mol L1 s1 (2) At 3.0 105 mol/L, rate = 9.25 107 mol L1 s1 (3) At 2.0 105 mol/L, rate = 6.16 107 mol L1 s1 (4) At 1.5 105 mol/L, rate = 4.60 107 mol L1 s1 (5) At 1.0 105 mol/L, rate = 3.09 107 mol L1 s1 (a) What is the relationship between the rates in (1) and (3)? Between (2) and (4)? Between (3) and (5)? (b) What is the relationship between the concentrations in each of these cases? (c) Is the rate of the reaction proportional to the concentration of Cv+? Explain your answer.
Part B: Integrated rate law expressions for zero- and half-order reactionsDerive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form A  products where k is the rate constant. Also, deduce the graphical linear relations, units of rate constant k, and expression for half-life, t1/2 , for all cases. Assume that at t = 0 only A is present with an initial concentration of [A]0. Fill in the table below by following the example given for first-order reaction1.
Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.340 0.210 0.0204 2 0.340 0.420 0.0204 3 0.680 0.210 0.0816 ?=
Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.360 0.290 0.0144 2 0.360 0.580 0.0144 3 0.720 0.290 0.0576 k=
For the gas phase decomposition of cyclobutane,(CH2)4----->2 C2H4the rate constant in s-1 has been determined at several temperatures. When ln k is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -3.15×104 K and a y-intercept of 35.9.The value of the rate constant for the gas phase decomposition of cyclobutane at 711 K is _____ s-1.(Enter your answer to one significant figure.)
Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form A ® products where k is the rate constant. Also, deduce the graphical linear relations, units of rate constant k, and expression for half-life, t1/2 , for all cases. Assume that at t = 0 only A is present with an initial concentration of [A]0. Fill in the table below by following the example given for first-order reaction1. Show all your work. order Integrated Rate Law Graphical Linear Relation Unit of k Half-life 1 ln[A][A]0=−?? ln[A] vs t ⇒ slope = −k , y-intercept = ln[A]0 1? ?12⁄=ln2? 1/2 3/2 3 1Atkins’ Physical Chemistry, Atkins and de Paula, 10th Ed., 2014, OUP, pp. 827-828.
The kinetic data shown below were observed for the reaction: BF3 (g) + NH3 (g) ---------> F3B·NH3 (g) Trial # [BF3] (mol/L) [NH3] (mol/L) Rate (M/s) 1 0.250 0.250 0.2130 2 0.250 0.125 0.1065 3 0.200 0.100 0.0682 4 0.350 0.100 0.1193 5 0.175 0.100 0.0596 What is the rate constant for this experiment?
The half-life (t1/2) of a first-order reaction is 0.950s What is the rate constant? Solve with complete solution.
For the gas phase decomposition of cyclobutane,(CH2)4 -----> 2 C2H4the rate constant has been determined at several temperatures. When ln k in s-1 is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -3.15×104 K and a y-intercept of 35.9.The activation energy for the gas phase decomposition of cyclobutane is _____ kJ/mol.
A first-order reaction A → products is to be carried out in a CMFR.The reaction rate constant was determined to be 1.0 h−1. What is the residence time required for 90 percent conversion of the reactantprovided there are no changes in temperature and density of thesystem?
For enzymatic reaction, a mechanism was proposed by Michaelis and Menten as follows: E + S == ES k1 and k1’ ES -> E and P k2. Use steady-state assumption, derive an expression for the reaction rate. Where E is the concentration of enzyme, S substrate, ES complex of E and S, E = E0 – ES. (If you have difficulty in doing it, please consult lecture note) Assume Km = 0.038 mol.L-1 at 25 oC, when the substrate concentration is 0.156 Mol.L-1, the rate of the reaction is 1.21 m mol/L.s. The maximum rate of conversion reaction is reached at high substrate concentrations. Calculate the maximum rate of this enzyme-catalyzed reaction.