Data for (A) versus time is given below for the reaction: A+ products Time, s JA 1. 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)- 1 Select) R(Second order) - Select] 39. What is the order of the reaction? (Select] 40. What is the value of the rate constant? I Select

Data for (A) versus time is given below for the reaction: A+ products Time, s JA 1. 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)- 1 Select) R(Second order) - Select] 39. What is the order of the reaction? (Select] 40. What is the value of the rate constant? I Select

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter8: Sampling, Standardization, And Calibration

Section: Chapter Questions

Problem 8.19QAP

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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?
The following experimental data were obtained for the reaction of \'I14* and NOf in acidic solution. NH/(aq) + NO2-(aq) — N;(g) + 2 H,O(f) INH/I (mol L1) [NO21 (mol L-1, Rate = A[NJ/At (mol L-1 s’) 0.0092 0.098 3.33 X IO"7 0.0092 0.049 1.66 X 10‘7 0.0488 0.196 3.51 X 10"6 0.0249 0.196 1.80 X 10-6 Determine the rate law for this reaction and calculate the rate constant.
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=
Using the data in the table, calculate the rate constant of this reaction. A+B⟶C+D Trial [?] (?) [?] (?)[) Rate (M/s) 1 0.370 0.390 0.0234 2 0.370 0.936 0.135 3 0.592 0.390 0.0374 ?= ? units= ?
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.)
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?
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.
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.
Consider the base hydrolysis of an ester.RCOOR' + OH− → RCOO− + R'OH (where R and R' represent organic groups).The kinetics of this reaction, for a particular ester, were studied using the isolation method. When [OH−] << [RCOOR'] a pseudo-rate law can be posed:-d[OH-] / dt = k'' [OH-]mWith initial concentrations: [OH−]0 = 0.0015 M [RCOOR']0 = 0.700 M a plot of ln[OH-] versus t gave a straight line with slope = -0.0176 s-1 and y-intercept = -6.50.What is the order "m"?m = ? What is the corresponding pseudo rate constant k''? (include appropriate units)k'' = ?
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.
Fill in the table with the correct responses. Reaction Order Integrated Rate Law Linear Plot Rate constant (k) from the linear plot Half life (t1/2) zero [a] [b] [c] [d] first [e] [f] [g] [h] second [i] [j] [k] [l] Answer bank: 1/At = kt + 1/Ao ln A vs time negative of the slope (-m) log A vs time At = -kt + Ao At/k Ao/2k y-intercept A vs time slope (m) At = kt + Ao 1/A vs t lnAt = -kt + lnAo ln2/k 1/kAo