Rate = k[acetone]1[I2]0[H+]1 rate = 0.001M/ 332.26s = 3.01 x 10-6 my question is what is the S.I unit of this answer? 3.01 x 10-6 = k[0.80M]1 [0.001M]0 [0.20]1 k= (3.01 x 10-6)/ (0.16) k = 1.88 x 10-5 also S.I unit of this answer with clear explanation and steps

Rate = k[acetone]1[I2]0[H+]1 rate = 0.001M/ 332.26s = 3.01 x 10-6 my question is what is the S.I unit of this answer? 3.01 x 10-6 = k[0.80M]1 [0.001M]0 [0.20]1 k= (3.01 x 10-6)/ (0.16) k = 1.88 x 10-5 also S.I unit of this answer with clear explanation and steps

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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Final Reactant Concentrations and Reaction Rates [I-] [BrO3-] [H+] Reaction Time Rate of Reaction 2.8 x 10-5/time Expt. # M M M sec. M/sec 1 0.00166 0.0066 0.0166 258 3.876 x 10^-6 2 0.0033 0.0666 0.0166 135 7.40 x 10^-6 3 0.00166 0.0133 0.0166 132 7.576 x 10^-6 4 0.00166 0.0066 0.033 78 1.282 x 10^-5 Using the general rate law equation below and your orders determined, calculate the rate constant for each experiment and then the average. Rate= k[I-]m[BrO3-]n[H+]p k______________ k________________ k________________ k ____________ Average value of k __________________ Units of k__________________ Rewrite the final rate law including values for m, n, p and the average value of k and units: _________________________________
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The data below were collected for the following reaction: CH3CI(g) + 3 Cl2(g) CCI4(g) + 3 HCI(g) Trial [CH3CI] / mol L-1 [Cl2] / mol L-1 Initial Rate / mol L-1 s-1 1 0.050 0.050 0.0142 2 0.100 0.050 0.0290 3 0.200 0.200 0.115 What is the rate constant?
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Final Reactant Concentrations and Reaction Rates [I-] [BrO3-] [H+] Reaction Time Rate of Reaction 2.8 x 10-5/time Expt. # M M M sec. M/sec 1 0.00166 0.0066 0.0166 258 1.08 x 10^-7 2 0.0033 0.0666 0.0166 135 2.07 x 10^-7 3 0.00166 0.0133 0.0166 132 2.12 x 10^-7 4 0.00166 0.0066 0.033 78 3.58 x 10^-7 1.Using the information above determine the rate law (m, n and p) for the reaction. Show all calculations/reasoning in your notebook. m_______________ n_________________ p ____________
13
I-(aq) + OCI-(aq) 10-(aq) + Cl-(aq) The above reaction was studied (at a particular temperature) and the following data were obtained: [1-10 [OCI-]0 Initial Rate (mol/L) (mol/L) (mol L-1s-1) 0.360 0.100 8.42\times 10-20.045 0.100 1.05\times 10-20.090 0.200 4.21\times 10-20.180 0.200 8.42\times 10-2 Which of the following expressions for the rate law are completely consistent with the above experimental data, d[Cl-]/dt = kl OCI-][-] d[10-]/dt = k[OCI-]3 k[OCI][I] = d[OCI-]/dt-d[-]/dt = k[-]2[OC]d[1]/dt = k[-][OC] 1 pts Tries 0/5 Calculate the rate constant (k). (Units required.)
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O KINETICS AND EQUILIBRIUM Using a second-order integrated rate law to find concentration... -1-1 At a certain temperature the rate of this reaction is second order in Cl,0, with a rate constant of 103.Ms: 2Cl,0; (g) 2Cl, (g) +50, (g) → Suppose a vessel contains Cl,0, at a concentration of 0.130M. Calculate the concentration of Cl,0, in the vessel 0.550 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits. OM x10 Check Explanation Terms of Use | Privacy Center Accessibility 2021 McGraw Hill LLC. All Rights Reserved. II
0 State what rate law you would expect from this mechanism (Rate = k[Acetone]"[H*]"[[₂]') • Compare the expected rate law from the mechanism to the rate law you got from the Method of Initial rates and the Method of Isolation. If they are different, suggest 1-2 valid, specific, experimental reasons why they might be different. (fast, equilibrium) CH3 ソース H3C ག་ CH3 + H* = H3C H3C H. + H (slow) CH3 H3C CH2 + ↳2 Т + HI