4. You have two reactions that you wish to carry out. Each reaction only involves a single reactant compound. To do this you make two solutions each having an initial concentration of 0.20 M in the reactant. The first reaction is first order and proceeds with a rate constant of 2.0 s-1. The second reaction is second order in the reactant and proceeds with a rate constant of 4.0 M-1s-1. Calculate the initial rate for each reaction. Why does the rate constant with the larger numerical value lead to a slower rate?

4. You have two reactions that you wish to carry out. Each reaction only involves a single reactant compound. To do this you make two solutions each having an initial concentration of 0.20 M in the reactant. The first reaction is first order and proceeds with a rate constant of 2.0 s-1. The second reaction is second order in the reactant and proceeds with a rate constant of 4.0 M-1s-1. Calculate the initial rate for each reaction. Why does the rate constant with the larger numerical value lead to a slower rate?

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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4. The reaction 2NO (g) + 2H2 (g) - N2 (g) + 2H2O (g) is found experimentally to be second order in NO (g) and first-order in H2 (g). a. Write the rate law for the reaction. b. What is the overall order of the reaction? c. What are the units for the rate constant k? d. If [NO] is doubled (while keeping [H2] constant), by what factor will the reaction rate increase? e. If [H2] is doubled (while keeping [NO] constant), by what factor will the reaction rate increase?
Order and rate law of a reaction The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an elementary step. For example, the rate law for step 1 is rate = k[NO2]? The exponent "2" is used because the reaction involves two NO2 molecules. The rate law for step 2 is rate = k[NO3]'[C0]! = k[NO3][CO] because the reaction involves only one molecule of each reactant the exponents are omitted. Analyzing a new reaction Consider the following elementary steps that make up the mechanism of a certain reaction: 1. 2A→B+ C 2. B +D→E+C
can you help w/ c?
d. The mechanism of the reaction is suggested as two steps: NO2+ F2 → NO₂F + F (fast) NO₂+ F→ NO₂F (slow) Is your rate law consistent with this mechanism? Explain your answer, making any changes necessary to create consistency.
What is the rate law for the following reaction? OH O Rate = k [CSH15 OH H2 SO4 ORate = k CSH15 OH [H20] ORate = k[CsH14 ORate = k[C3H15 OH
The rate constant for the second order reaction 2 NO2 2 NO + O2 is 0.31 M-1s-1. If a 4.33 M sample of NO2reacts for 31.7 seconds, what concentration of NO2 will remain? Report your answer to two decimal places.
2ClO2+2OH- —> ClO-3+ClO-2+H2O R=k[ClO2]2[OH-] What is the rate constant for the reaction given the rate law and data?
Order and rate law of a reaction The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an elementary step. For example, the rate law for step 1 is rate = k[NO₂)² The exponent "2" is used because the reaction involves two NO₂ molecules. The rate law for step 2 is because the reaction involves only one molecule of each reactant the exponents are omitted. Analyzing a new reaction Consider the following elementary steps that make up the mechanism of a certain reaction: 1.2A B+C 2. B+D+E+C Part A What is the overall reaction? Express your answer as a chemical equation. ► View Available Hint(s) for Part A for Pardo for Part redo foPart A refor Part A keyboard shortcuts for Part A help for Part A A chemical reaction does not occur for this question. Submit Part B Which species is a reaction…
Consider the following set of reactions and the related kinetics data. The reaction conditions and initial reactant concentrations were the same for all three reactions. Select the one statement that is FALSE. Reaction Rate Law Rate constant, k (M-1s-1) CH3F + SCN¯ → CH3SCN + F¯ Rate = k [CH3F] [SCN¯] k = 8.3 x 10-14 CH3Cl + SCN¯ → CH3SCN + Cl¯ Rate = k [CH3Cl] [SCN¯] k = 2.5 x 10-4 CH3I + SCN¯ → CH3SCN + I¯ Rate = k [CH3I] [SCN¯] k = 1.5 x 10-2 OPTIONS: A) Increasing the temperature by 20ºC will increase all three reaction rates by the exact same amount. B) Doubling the concentration of SCN¯ will double the reaction rate. C) It is possible that each reaction occurs in one elementary step. D) Each of these reactions is second order overall and first order with respect to SCN¯. E) The reaction with CH3I as the reactant will produce CH3SCN the fastest initially.
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The half-life (t1/2) of a first-order reaction is 1.00 s. What is the value of the rate constant? A. 6.93 s–1 B. 0.144 s–1 C. 0.0693 s–1 D. 0.693 s–1 E. 3.01 s–1
2. Imagine you run a reaction whose free energy of activation (AG*) is 18 kcal/mol. If it were a first order reaction, its’ half life (the time for half the starting material to go away) or half the product to form (assuming a 100% yield) would by In2/k where k is the rate constant. How long will it take for the reaction to be done if you run the reaction in an ice-salt bath at room temperature (20 °C) if the rate constant were 1 x 104/sec?