Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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- The isomerization of cyclopropane, C3H6, is believed to occur by the mechanism shown in the equations above. Here C3H6* is an excited cyclopropane molecule. At low pressure, Step 1 is much slower than Step 2. Derive the rate law for this mechanism at low pressure?arrow_forwardA proposed mechanism for the decomposition of hydrogen peroxide consists of three elementary steps: (1) H₂O2(g) →>> 2OH(g) (2) H₂O2(g) + OH(g) → H₂O(g) + HO₂(g) (3) HO₂(g) + OH(g) → H₂O(g) + O₂(g) If the rate law for the reaction is first order in H₂O2, which step in the mechanism is the rate-determining step? O Step 1 O None of the steps O Step 3 O Impossible to determine Steps 1 and 2 O All of the steps O Step 2 Hint: The slowest step in a mechanism determines the rate law. From each elementary reaction we can write a rate-law expression. Submit Questionarrow_forward4. The rate law for the reaction H₂ (g) + I₂ (g) → 2 HI (g) was determined to be rate = k [H₂] [1₂], which led to a simple single-step bimolecular mechanism where an H₂ molecule simply collided with an I₂ molecule. In the 1960s, though, different teams found spectroscopic evidence for this mechanism: (1) (2) (3) I₂ (g) →21 (g) H₂(g) + I (g) → H₂I (g) H₂I (g) + I (g) → 2 HI (g) (fast) (fast) (slow) Show that this mechanism is consistent with the rate law: rate = k [H₂] [1₂]. To do this, assign individual rate constants for ALL forward and reverse reactions in the mechanism (I'm guessing k₁, k-1, k2, k-2 and k3). Then show that the overall rate law from combining individual rate laws for the mechanism steps yields rate = (some combination of k's) [H₂] [1₂], and thus k= (some combination of k's). Remember that there should be no reaction intermediates in the rate law, only reactants. \arrow_forward
- Consider the following reaction: (a) The rate law for this reaction is first order in NO₂(g) and first order in O3(g). What is the rate law for this reaction? Rate = k [NO₂(g)] [03(g)] Rate = k [NO₂(g)]² [03(g)] Rate = k [NO₂(g)] [03(g)]² O Rate = k [NO₂(g)]² [03(g)]² Rate = k [NO₂(g)] [03(g)]³ Rate = k [NO₂(g)]4 [03(g)] (b) If the rate constant for this reaction at a certain temperature is 97900, what is the reaction rate when [NO₂(g)] = 0.587 M and [03(9)] = 1.40 M? Rate = 2 NO₂(g) + 03(9) → N₂O5(9) + O₂(g) M/s. Rate = (c) What is the reaction rate when the concentration of NO₂(g) is doubled, to 1.17 M while the concentration of O3(g) is 1.40 M? M/Sarrow_forwardThe rate law for the reaction NO2(g) + O2(g) is given by rate = - k[NO₂] NO(g) + 03 (g) If the following is the mechanism for the reaction, which of the following statements correctly describe the reaction? Check all that apply. NO2(g) NO(g) + O(g) - O(g) + O2(g) 03 (g) The reaction is 2nd order overall. The first step is the slow step. Doubling NO, would quadruple the rate. Cutting O, in half would decrease the rate by a factor of 2. The molecularity of the first step is 1. Both steps are termolecular. None of the abovearrow_forwardUnder certain conditions the rate of this reaction zero order in dinitrogen monoxide with a rate constant of 0.0064 M · s: 2N,0 (g) – 2N, (g) +0, (g) Suppose a 3.0 L flask is charged under these conditions with 400. mmol of dinitrogen monoxide. How much is left 5.0 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.arrow_forward
- The following reaction is observed to be second order with respect to NO and first order with respect to Br2. Br2 (g) + 2NO (g) --> 2NOBr (g)arrow_forwardThe reaction described by H, (g) + L,(g) 2 HI(g) has an experimentally determined rate law of rate = k [H2] [I2] Some proposed mechanisms for this reaction are: Mechanism A k1 (1) H,(g) + L,(g) → 2 HI(g) (one-step reaction) Mechanism B k1 =2 I(g) k_1 (1) L(g) ; (fast, equilibrium) k2 (2) H, (g) + 2 I(g) 2 HI(g) (slow) Mechanism C k1 = 2 I(g) (fast, equilibrium) (1) k2 (2) I(g) + H, (g) HI(g) + H(g) (slow) (3) H(g) + I(g) → HI(g) (fast) Question Sourcarrow_forwardDuring kinetic studies, the following reaction was determined to be second order in respect to NO and zeroth order in respect to O2. What is the rate law for the reaction? 2 NO (g) + O2(g) -----> 2NO2 (g)arrow_forward
- Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of 0.0039 M²s¯¹: 2N₂O(g) 2 N₂(g) + O₂(g) Suppose a 4.0 L flask is charged under these conditions with 400. mmol of dinitrogen monoxide. How much is left 6.0 s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. 0 00 X 3arrow_forwardThe reaction 2 NO(g) + O2(g) --> 2 NO2(g) proceeds through the following mechanism: 2 NO(g) --> N2O2(g)N2O2(g) + O2(g) --> 2 NO2(g) (a) The second step of this mechanism is rate-determining (slow). What is the rate law for this reaction? Rate = k [NO] [O2] Rate = k [NO]2 [O2] Rate = k [NO] [O2]2 Rate = k [NO]1/2 [O2] Rate = k [NO] [O2]1/2 Rate = k [NO]2 Rate = k [NO]2 [O2]1/2 (b) What would the rate law be if the first step of this mechanism were rate-determining? Rate = k [NO] [O2] Rate = k [NO]2 [O2] Rate = k [NO] [O2]2 Rate = k [NO]1/2 [O2] Rate = k [NO] [O2]1/2 Rate = k [NO]2 Rate = k [NO]2 [O2]1/2 Rate = k [NO]arrow_forwardSuppose the formation of nitrogen dioxide proceeds by the following mechanism: elementary reaction 2NO (g) → N₂O₂ (g) step 1 2 N₂O₂(g) + O₂ (g) Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction. 2NO₂ (g) Suppose also k₁ « k₂. That is, the first step is much slower than the second. << Note: your answer should not contain the concentrations of any intermediates. rate constant 0 k₁ k₂ rate = k 1-0 010 X Sarrow_forward
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