2cl02(aq) + 20Haq cloz(aq) + Cl02 (aq) + H20(1) Experiment [CIO2-], M [ОН-1, М Initial Rate (M/s) 0.050 0.040 0.025 2 0.030 0.040 0.0028 3 0.030 1.000 0.0083
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Determine the rate law
![2cl02(aq) + 20Hag
→ cl03(ag) + Cl0= (aq)
+ H20(1)
Experiment
[CIO2-], M
[ОН-], М
Initial Rate (M/s)
1
0.050
0.040
0.025
0.030
0.040
0.0028
3
0.030
1.000
0.0083](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd48a7d5b-5164-4c1c-b884-866073cb729c%2Fa1313a61-08cf-48ea-acd4-5b753d7043ef%2Fu7op4o1d_processed.png&w=3840&q=75)
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- In Experiment 4- Chemical Kinetics “The Rate of Chemical Reaction”, the following data were obtained: Rxn Mixture time(s) temperature (˚C) 1 190 23.0 2 378 23.0 3 97 23.0 4 205 23.0 5 44 23.0 6 90 35.0 Concentrations and portions for reaction mixture number one: H2O 0.050 M Buffer 0.050 M KI (mL) 0.1% Starch (mL) 0.0450 MNa2S2O3(mL) 0.8525 M H202 (mL) 75.0mL 30.0mL 25.0 5.00 5.00 10.00 Use the data above to answer the following questions where applicable. (a) Write the formulas of the reactants, there are three, in the reaction being studied.Signs + or - or subscripts may be entered all on the same line; S2O32- can be entered as S2O3^2-, the symbol, ^, is not required if the…Explain why the reverse reaction ES ← E + P can be neglected. Plot the dependence of the reaction rate, V, on the molarity of the substrate, [S], in the solution.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?
- A general reaction written as A + 2B ® C + 2D is studied and yields the following data: [A]0 [B]0 Initial D[C]/Dt 0.150 M 0.150 M 8.00 ´ 10-3 mol/L·s 0.150 M 0.300 M 1.60 ´ 10-2 mol/L·s 0.300 M 0.150 M 3.20 ´ 10-2 mol/L·s Determine the initial rate of consumption ([deltaB]/delta t) for the first trial:Given the following data, determine the rate law and calculate K Experiment [NO] (M) [Cl2] (M) Rate (M/s) 1 0.0300 0.0100 3.4 x 10-4 2 0.0150 0.0100 8.5 x 10-5 3 0.0150 0.0400 3.4 x 10-4 the units on K are M-2s-1. You should enter the answer without units to 2 sig figs.The binding of Mg2+ to glycolate anion, (HO-CH2-COO-) proceeds by thisequilibrium:Mg2+ + HO-CH2-COO- ->Mg2+(HO-CH2-COO-) a) Write the expression for Keq for this reversible reaction. b) If you begin with 5 mM Mg2+ and 0.1 mM HO-CH2-COO-, and with a value of Keq = 103 M-1, calculate the concentrations of Mg2+ , HO-CH2-COO-, and Mg2+(HO-CH2-COO-) at equilibrium.
- Use this data to determine the value of the rate constant. Ex # [A] (M) [B] (M) Rate (M/hr) 1 .240 .120 2.00 2 .120 .120 .500 3 .240 .0600 1.00 Question 4 options: 289 per hour 579 per hour 8.3 per hour 0.0144 per hour 69.4 per hour .00346 per hour 0.060 per hour 0.0173 per hour 17 per hour 0.12 per hour5. (Show or describe your work) Suppose the reaction A <__> B has an equilibrium constant of Keq = [B] eq/[A] eq. The forward reaction proceeds to the right by an uncatalyzed pathway with a rate = kof[A] AND by a proton- catalyzed pathway with a rate = kHf[H+][A]. Thus the overall rate law for the forward reaction involves the sum of these two terms. ? =(??? +???[?+])[?]=??[?] What is the corresponding rate law for the backward reactions?Combination # [IO3-]0 [H+]0 [I-]0 average dt(sec) Initial rate, M s-1 kR (including units) 1 0.010 0.00002 0.10 18.1 0.00055 2 0.020 0.00002 0.10 9.1 0.0022 3 0.010 0.00002 0.20 5.1 0.0020 4 0.010 0.00002 0.10 5.1 0.0020 1. Determine the orders x,y and z 2. Calculatethe rate constant, kR and average kR for combimnations 1-4 3. Write the experimenally determined rate law Rate =kR[IO3-]x [I-]y [H+]z
- Consider the reaction Mn+ + ne- ⇌ M. If ΔG*c = ΔG* + (1-α)nFE and η = E – Eeq, prove that ic = io exp[-(1-α)nFηa)/RT)].Give the values of the following: Slope - [Select from 1.35x10^4, -1.09x10^4, 1.09x10^4, -1.35x10^4] y-intercept - [Select from -27.8, 27.8, 37.4, -37.4] Pearson coefficient (R) - [Select from -1.000, -0.962, -0.982] activation energy of this reaction in kJ/mol = [Select from 1.13x10^5, 90.6, 112.51] collision frequency factor A= [Select from 1.82x10^16, 1.54x10^14, 1.15x10^12]A + 3B + 2C --> D + 2E Determine the rate law and rate constant using the experimental data below [A] (M) [B] (M) [C] (M) Rate (M/sec) Exp. 1 0.100 5.00 x 10-4 1.00 x 10-2 0.137 Exp. 2 0.100 1.00 x 10-3 1.00 x 10-2 0.268 Exp. 3 0.200 1.00 x 10-3 1.00 x 10-2 0.542 Exp. 4 0.400 1.00 x 10-3 2.00 x 10-2 1.084