USE ALL THIS INFORMATION TO SOLVE QUESTION 1 ONLY - BELOW:- Consider the reaction A + 2B → C. To determine the rate law, a common experiment used: the method of initial rates. The experiment monitors concentrations of each reactant as close to the beginning of the reaction as possible. An advantage of this method is the fact that the initial concentration of each reactant is precisely known at time zero. Step 1: Pick any two datasets where only one reactant concentration changes.Which pairs of datasets in Table 13.3 involve only one reactant concentration change? There may be more than one answer.a. 1 and 2-Your answerc. 2 and 3 Your answerExplanationIn data sets 1 and 2, [A] doubles while [B] does not change. In data sets 1 and 3, both [A] and [B] double. In sets 2 and 3, [A] is constant, while [B] doubles.After identifying the dataset pairs, pick any pair to work with in Step 2. We’ll select experiments 1 and 2.Step 2: Write out the complete rate law for one experiment and set this as a ratio to the rate law expression for the other experiment. Use actual concentrations and initial rates from the data table for each experiment number. Your ratio will resemble Equation 13.8. initial rate 1 initial rate 2=�[�]�[�]��[�]�[�]� initial rate 2 initial rate 1=k[A]X[B]Yk[A]X[B]Y What is the rate law expression for experiment 1? a. 0.002 M/s = k - Your answerIn data set 1, [A] = 0.0100 M, [B] = 0.0100 M, and the initial rate = 0.002 M/s [/math]. What is the rate law expression for experiment 2? b. 0.002 Ms = k(0.0200 M) (0.0100 M) Your answerIn data set 2, [A] = 0.0200 M, [B] = 0.0100 M, and the initial rate = 0.002 Ms.Step 3: You now see why we keep one reactant concentration the same while varying the other! By doing so, everything cancels, leaving only one variable to solve for: the reaction order x. What is the net equation once like terms have been canceled? d. 1.0 = 0.500 Your answer Step 4: Solve for x, which gives us the reaction order with respect to [A]. 1.0=0.500, log(1.0) = x log(0.500), 0 = x log0.500 ∴ x=0 Therefore, the reaction is zeroth order with respect to [A]. In other words, the reaction rate is independent of [A]. This correlates with the data table. We see that when changing [A] while holding [B] constant, the initial rate does not change.Step 5: Repeat steps 2 through 4, this time using the other pair of datasets, which we determined to be experiments 2 and 3.What is the rate law expression for experiment 3? c. 0.004 Ms = k(0.0200 M) (0.0200 M) Your answerIn data set 3, [A] = 0.0200 M, [B] = 0.0200 M, and the initial rate = 0.004 Ms. What is the value of y? Because x = 0, our equation simplifies to: 0.50 = 1.000 × 0.500y = 0.500y.0.50 = 0.500, log 0.50 = y log 0.500, y = 1.0 Hence, the reaction is first order with respect to [B].Step 6: Once all reaction orders are known, it’s time to solve for the last missing parameter of the rate law: the rate constant k. This is the easiest part of the process. Simply choose any dataset and plug in all values into the dataset’s expression for the rate law. Be sure to use the solved values for the reaction ordersSolve for the rate constant k. Do not enter units as part of your answer.0.21. What is the rate law for the example reaction?Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.a. Rate = 0.2 Ms [A][B]b. Rate = 0.2 Ms [B] 2. Consider the following reaction:2NO(g) + Br(g) → 2NOBr(g)What is the rate constant k?[Image description: Expt. 1: M = 0.0010 [NO2]o; M = 0.0020[Br2]o; Initial rate (Ms-1) = 2.4e-5Expt. 2: M = 0.0010[NO2]o; M = 0.0030[Br2]o; Initial rate (Ms-1) = 3.6e-5Expt. 3: M = 0.0020[NO2]o; M = 0.0020[Br2]o; Initial rate (Ms-1) = 9.6e-5SELECT ONE a. 6000 M-2s-1b. 12,000 M-2s-1 3. In a zeroth-order reaction, it takes 342 s for 75% of a hypothetical reactant to decompose. Determine the half-life t in units of seconds. Do not enter units with your numerical answer.WRONG ANSWERS:- 342 & 171

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USE ALL THIS INFORMATION TO SOLVE QUESTION 1 ONLY - BELOW:- Consider the reaction A + 2B → C.  To determine the rate law, a common experiment used: the method of initial rates. The experiment monitors concentrations of each reactant as close to the beginning of the reaction as possible. An advantage of this method is the fact that the initial concentration of each reactant is precisely known at time zero. Step 1: Pick any two datasets where only one reactant concentration changes.Which pairs of datasets in Table 13.3 involve only one reactant concentration change? There may be more than one answer.a. 1 and 2-Your answerc. 2 and 3 Your answerExplanationIn data sets 1 and 2, ​[A] doubles while ​[B] does not change. In data sets 1 and 3, both ​[A] and ​[B] double. In sets 2 and 3, ​[A] is constant, while ​[B] doubles.After identifying the dataset pairs, pick any pair to work with in Step 2. We’ll select experiments 1 and 2.Step 2: Write out the complete rate law for one experiment and set this as a ratio to the rate law expression for the other experiment. Use actual concentrations and initial rates from the data table for each experiment number. Your ratio will resemble Equation 13.8.  initial rate 1 initial rate 2=�[�]�[�]��[�]�[�]� initial rate 2 initial rate 1​=k[A]X[B]Yk[A]X[B]Y​ What is the rate law expression for experiment 1? a. 0.002 M/s = k - Your answerIn data set 1, ​[A] = 0.0100 M, ​[B] = 0.0100 M, and the initial rate = 0.002 M/s [/math].  What is the rate law expression for experiment 2? b. 0.002 Ms = k(0.0200 M)  (0.0100 M)  Your answerIn data set 2, ​[A] = 0.0200 M, ​[B] = 0.0100 M, and the initial rate = 0.002 Ms.Step 3: You now see why we keep one reactant concentration the same while varying the other! By doing so, everything cancels, leaving only one variable to solve for: the reaction order x. What is the net equation once like terms have been canceled? d. 1.0 = 0.500 Your answer   Step 4: Solve for x, which gives us the reaction order with respect to [A]. 1.0=0.500, log(1.0) = x log(0.500), 0 = x log0.500 ∴ x=0   Therefore, the reaction is zeroth order with respect to [A]. In other words, the reaction rate is independent of [A]. This correlates with the data table. We see that when changing [A] while holding [B] constant, the initial rate does not change.Step 5: Repeat steps 2 through 4, this time using the other pair of datasets, which we determined to be experiments 2 and 3.What is the rate law expression for experiment 3?  c. 0.004 Ms = k(0.0200 M)  (0.0200 M)  Your answerIn data set 3, ​[A] = 0.0200 M, ​[B] = 0.0200 M, and the initial rate = 0.004 Ms. What is the value of y? Because x = 0, our equation simplifies to: 0.50 = 1.000 × 0.500y = 0.500y.0.50 = 0.500, log 0.50 = y log 0.500, y = 1.0 Hence, the reaction is first order with respect to [B].Step 6: Once all reaction orders are known, it’s time to solve for the last missing parameter of the rate law: the rate constant k. This is the easiest part of the process. Simply choose any dataset and plug in all values into the dataset’s expression for the rate law. Be sure to use the solved values for the reaction ordersSolve for the rate constant k. Do not enter units as part of your answer.0.21. What is the rate law for the example reaction?Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.a. Rate = 0.2 Ms ​[A]​[B]b. Rate = 0.2 Ms ​[B] 2. Consider the following reaction:2NO(g) + Br(g) → 2NOBr(g)What is the rate constant k?​[Image description: Expt. 1:  M = 0.0010 [NO2]o;  M = 0.0020[Br2]o; Initial rate (Ms-1) = 2.4e-5Expt. 2:  M = 0.0010[NO2]o;  M = 0.0030[Br2]o; Initial rate (Ms-1) = 3.6e-5Expt. 3:  M = 0.0020[NO2]o;  M = 0.0020[Br2]o; Initial rate (Ms-1) = 9.6e-5SELECT ONE a. 6000 M-2s-1b. 12,000 M-2s-1 3. In a zeroth-order reaction, it takes 342 s for 75% of a hypothetical reactant to decompose. Determine the half-life t in units of seconds. Do not enter units with your numerical answer.WRONG ANSWERS:- 342 & 171

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