Mixture I- (M) S208-2 (M) Initial Rate (M/s) 1 0.08 0.04 6.45 * 10 -4 2 0.04 0.04 3.54 * 10 -4 3 0.08 0.02 4.35 * 10 -4 PART B Mixture Temperature Rate Constant (k) 1 0.003390979993 4 0.003206155819
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determine the rate constant k for mixtures 1, 4, 5
rate = r = k[I-]^1 [S2O82-]^1
Mixture: Time (s):
1 61
4 28
5 85
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- 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.Determine the average rate of change of BB from ?=0 st=0 s to ?=272 s.t=272 s. A⟶2BA⟶2B Time (s) Concentration of A (M) 0 0.7300.730 136136 0.4450.445 272272 0.1600.160 rateB= __________M/sDetermine the missing initial rate: 2N2O5 → 4NO2 + O2 [N2O5] Initial rate (Ms-1) 0.093 4.84x10-4 0.084 4.37x10-4 0.224 ??
- Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [?] (?)[A] (M) [?] (?)[B] (M) Rate (M/s) 1 0.360 0.330 0.0164 2 0.360 0.660 0.0164 3 0.720 0.330 0.0656 ?=? Units =?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 ?=A group of students compiled the data shown in data table 1 below. What is the units for the average rate constant value (k)? Exp. # [IO3-]0 (M) [I-]0 (M) [H+]0 (M) Time (s) 1 0.005 0.05 2 x 10-5 22.12 2 0.010 0.05 2 x 10-5 86.84 3 0.005 0.10 2 x 10-5 5.35 4 0.005 0.05 4 x 10-5 2.65 Question 8 options: M-2 s-1 M-3 s-2 M2 s-2 M-4 s-2 M3 s3 M-4 s-3 M-2 s-2 M2 s2 M s M s-1 M-3 s-1 M-1 s-1 M-3 s-3
- 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=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.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
- 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.)Table 2: Molarity of H2O2 and KI and Reaction Rate Trial H2O2 Concentration, M KI Concentration, M Reaction Rate(Reciprocal Slope) 1 0.29 M 0.40 M 14.08 2 0.29 M 0.20M 25 3 0.023 M 0.40 M 20 Order with respect to H2O2: -0.138 Order with respect to KI: -0.828 Rate constant value Trial 1: 5.84 Trial 2: 5.55 Trial 3: 5.56 Average: 5.6 -Please help me with this part, please 1. Should the rate constant (k) be the same for all three trials in this experiment? Explain your answer. 2.Write the complete rate law for this reaction.The following results were found after completion of Part C in the Experimental procedure: 0.057 M I- and 0.065 M H2O2 were used Run Catalyst Calculated Rate of Reaction (M/s) 1 none 0.053 2 10.00mL of 0.54 M FeCl2 0.446 Assuming the Rate Law = k[I-][H2O2] 2 Calculate the value of k for run 2. Give your answer to the nearest whole number.