II.7- Saponification, initial rates and half-lifeAn esterification reaction is studied at 350 K under conditions where the reaction is complete:ROH + R'COOH→ R'COOR + H20For the same initial alcohol concentration, and variable acid concentrations, we obtain the followingcurves of evolution of the acid concentration versus time:[R'COOH] (mol L'')-0.8-0.6-0.4--0.2406080100120140160180200The rate v of this reaction follows the law v = k[R°COOH]ª [ROH]F.1) Explain how to graphically determine an instantaneous velocity (rate) at time t.20

A esterification reaction is a reaction between an alcohol and a carboxylic acid that results in the…

Answered: II.7- Saponification, initial rates and…

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...

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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?
Concentration-Time Relationships (See Section 15.4 and Examples 15.5-15.7.) 15. The rate equation for the hydrolysis of sucrose to fruc- tose and glucose CRH2O1(aq) + H¿O(€) → 2 C,H12O6(aq) is -A[sucrose]/At = k[C¡2H2O1]. After 27 minutes %3D at 27 °C, the sucrose concentration decreased from 0.0146 M to 0.0132 M. Find the rate constant, k.
Consider the reaction data. A products T (K) | k (s-1) 325 0.383 825 0.611 What two points should be plotted to graphically determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values. X1 Yi = X2 = Y2 = Determine the rise, run, and slope of the line formed by these points. rise: run: slope: What is the activation energy of this reaction? Ea J/mol ? II
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'' = ?
For the gas phase decomposition of bromoethane, CH;CH,Br >C,H4 + HBr -1 the rate constant in s has been determined at several temperatures. When In k is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of-2.63x104 K and a y-intercept of 29.6. The value of the rate constant for the gas phase decomposition of bromoethane at 703 K is -1 (Enter your answer to one significant figure.)
For the gas phase isomerization of cis-1,2-diphenylethene, cis-CGH5CH=CHC6Hstrans-C6H5CH=CHCGH5 the rate constant in s has been determined at several temperatures. When In k is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -2.15x10* K and a y-intercept of 29.4. The value of the rate constant for the gas phase isomerization of cis-1,2-diphenylethene at 591 K is sl. (Enter your answer to one significant figure.)
The following initial rate data are for the ozonization of pentene in carbon tetrachloride solution at 25 °C: C5 H10+ Os C5H1003 Experiment [C5H100, M 0.116 0.116 0.231 0.231 2 4 Rate = Complete the rate law for this reaction in the box below. Use the form k[A] [B]", where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n. k= [Oslo. M 0.0170 0.0340 0.0170 0.0340 M¹.8-1 Initial Rate, M-s 192 385 383 766
Consider the reaction data. A-products T (K)& (s- 325 0.376 875 0.719 What two points should be plotted to graphically determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values. yi = X2 = Determine the rise, run, and slope of the line formed by these points. rise: run: slope: What is the activation energy of this reaction? J/mol Ea =
he rate constant of a chemical reaction increased from 0.100 s−1s−1 to 3.30 s−1s−1 upon raising the temperature from 25.0 ∘C∘C to 43.0 ∘C∘C . Calculate the value of (1T2−1T1)(1�2−1�1) where T1�1 is the initial temperature and T2�2 is the final temperature. Express your answer numerically. View Available Hint(s)for Part ACalculate the value of ln(k2k1) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.Express your answer numerically.View Available Hint(s)for Part What is the activation energy of the reaction?Express your answer numerically in kilojoules per mole.View Available Hint(s)
Chlorine dioxide, Cl02, is a reddish-yellow gas that is soluble in water. In basic solution it gives Cl03- and CIO2 ions. 2C1O:(aq) + 20н (ад) —> СIО3 (aд) + CIO, (aq) + H;0(1) To obtain the rate law for this reaction, the following experiments were run and, for each, the initial rate of reaction of ClO2 was determined. Initial Concentration Initial Concentration Initial Rate of Cl02 (mol/L) of OH (mol/L) (mol(L's) Exp. 1 0.21 1.6 x 10-2 0.162 Exp. 2 7.0 x 10-2 1.6 x 10-2 1.80 x 10-2 Exp. 3 7.0 x 10-2 4.8 x 10-2 5.39 х 10-2 a Obtain the rate law. (Use k for the rate constant.) Rate law =
The starting concentration (time zero) of crystal violet is 1 x 10-4 M. Using the proper integrated rate law and the calculated value for k, calculate the concentration of crystal violet after 240 seconds. Look at the equation of the trend line for your lost linear graph and the beer-lambert law. Equation from graph from NaOH 0.0660M is y= -0.0105x-0.6277
The gas phase decomposition of hydrogen peroxide at 400 °C H,O2(gH,0(g) +¾ O2(g) is second order in H2O2 with a rate constant of 0.650 M- s-1 If the initial concentration of H,O, is 0.218 M, the concentration of H,0, will be |Mafter 58.3 seconds have passed.

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