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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.The first-order rate constant, k1, for the decomposition of ampicillin at pH 5.8 and 35?C is k1 = 2 × 10-7 sec-1. The solubility of ampicillin is 1.1 g/100 mL. If it is desired to prepare a suspension of the drug containing 2.5 g/100 mL, calculate the zero-order rate constant, k0, and the shelf-life, that is, the time in days required for the drug to decompose to 90% of its original concentration (at 35?C) in solution. Note: 100 mL = 1 deciliter = 1 dL.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
- 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=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 1.084Using 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 ?=
- what is the activation energy? rate = r = k [I-]^1 [S2O82-]^1 Mixtures: Temperatures (celsius): Initial Rates: 1 21.9 6.45 * 10-4 2 38.9 3.54 * 10-4 3 2.4 4.35 * 10-4The equilibrium NH3(aq) + H2O(l) ↔NH4+(aq) + OH−(aq) at 25 °C is subjected to a temperature jump which slightly increases the concentration of NH4+(aq) and OH−(aq). The measured relaxation time is 7.61 ns. The equilibrium constant for the system is 1.78 × 10−5 at 25 °C, and the equilibrium concentration of NH3(aq) is 0.15 mol dm−3. (a) Calculate the rate constant for the forward step. kf = _____________. Just value in 3 sig. fig., normal or exponential format, e.g. type in 1.16E6 meaning 1.16 x 106, must use capital E here. Choose a unit in the next question, must be in one of those.The rate constant for the fi rst-order decomposition of N2O5 in the reaction 2 N2 O5(g) → 4 NO2(g) + O2(g) with v = kr[N2O5] is kr = 3.38 x 10-5 s-1 at 25 oC. What is the ha lf- life of N2O5? What w ill be the total pressure, init ial ly 78.4 kPa for the pure N2O5 vapour, (a) 5.0 s, (b) 5.0 min after init iation of the reaction?
- 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.Given: 2NO2 --> 2NO + O2 [NO2]o (M) Rate (M/s) 0.01 7.1 x 10-5 0.02 28.1 x 10-5 a) Give the rate law and overall order. b) What is [NO2] at 125 s if [NO2]o = 0.015 M?Use this data to determine the value of the rate constant. Expt 1 [Cl2] [Br2] Rate [ClBr]/min 1 0.011 0.025 0.0028 2 0.033 0.025 0.0252 3 0.033 0.100 0.0252 4 0.011 0.050 0.0028 Question 3 options: 0.056 / min 23 / min 30 / min 1.1 / min 925 / min 102 / min 2.5 / min 5.1 / min 463 /min 44.8 / min 10.2 / min
