34. A key step in the formation of sulfuric acid from dissolvedSO, in acid precipitation is the oxidation of hydrogensulfite ion by hydrogen peroxide:HSO5 (aq) + H2O2 (aq) -The mechanism involves peroxymonosulfurous acid,SO,0OH:HSO, (aq) + H20(e)kiHSO3 (аg) + H02(aq) — SO20OH (аg)+ H:0(€)k-1k2SO20OH¯(aq) + H3O* (aq)→ HSO, (aq) + H3O* (aq)By making a steady-state approximation for the reactiveintermediate concentration, [SO2OOH¯(aq)], express therate of formation of HSO, (aq) in terms of the concentra-tions of HSO3 (aq), H,O2(aq), and H;O*(aq).

Given that, the mechanism of formation of sulfuric acid from the dissolved SO2 in acid precipitation…

34. A key step in the formation of sulfuric acid from dissolved SO, in acid precipitation is the oxidation of hydrogen sulfite ion by hydrogen peroxide: HSO3 (aq) + H2O2(aq) → HSO, (aq) + H2O(e) The mechanism involves peroxymonosulfurous acid, SO,OOH: ki HSO5 (aq) + H2O2(aq) = k-1 SO2OOH¯(aq) + H2O(€) k2 SO20OH (aq) + H:0" (аg) HSO, (aq) + H30* (aq) By making a steady-state approximation for the reactive intermediate concentration, [SO,OOH¯(aq)], express the rate of formation of HSO, (aq) in terms of the concentra- tions of HSO, (aq), H¿O2(aq), and H3O*(aq).

34. A key step in the formation of sulfuric acid from dissolved SO, in acid precipitation is the oxidation of hydrogen sulfite ion by hydrogen peroxide: HSO3 (aq) + H2O2(aq) → HSO, (aq) + H2O(e) The mechanism involves peroxymonosulfurous acid, SO,OOH: ki HSO5 (aq) + H2O2(aq) = k-1 SO2OOH¯(aq) + H2O(€) k2 SO20OH (aq) + H:0" (аg) HSO, (aq) + H30* (aq) By making a steady-state approximation for the reactive intermediate concentration, [SO,OOH¯(aq)], express the rate of formation of HSO, (aq) in terms of the concentra- tions of HSO, (aq), H¿O2(aq), and H3O*(aq).

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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(3) 13. For the following questions, use the following diagrams. Uncatalyzed Reaction Catalyzed Reaction 400 400 300 300 reactants reactants 200 200 100 100 products products Reaction coordinate Reaction coordinate (1) (a) What is the value of the forward activation energy of the uncatalyzed reaction? (1) (b) What is the value of the reverse activation energy of the catalyzed reaction? (1) (c) What is the overall enthalpy (AH) of the forward reaction? Potential energy (kJ) Potential energy (kJ)
Reaction CaCO3(s) Ca²+ (aq) + CO3²- (aq) AgCl(s) Ag+ (aq) + Cl(aq) AgBr(s) = Ag*(aq) + Br(aq) Agl(s) Ag+ (aq) + I¯(aq) Initial Concentrations ← [Ca²+] = 1.0 x 10-⁹ M [CO32] = 1.0 x 10⁹ M Ksp = [Ca²+][CO3²-1 X = 3.8 x 10-9 Initial Conditions Q = [1.0 × 10-⁹][1.0 × 10-⁹] = 1.0 x 10-18 Equilibrium Conditions [Ca2+] = 1.0 x 10⁹ M [CO32] 1.0 x 10.⁹ M Mass of CaCO3(s) = 0.0 g 42 of 7 What is the value of the reaction quotient, Q, which is calculated by substituting the actual initial values of [Ca²+] and [CO32] into the equilibrium expression?
Q:4-1 Consider the following reaction: 2 NO(g) + 2 H2(8) → N;(g) + 2 H,O(g) (a) The rate law for this reaction is first order in H, and second order in NO. Write the rate law. (b) If the rate constant for this reaction at 1000 K is 6.0 × 10ª M¯²s¯!, what is the re- action rate when [NO] = 0.035 M and [H,] = 0.015 M? (c) What is the reaction rate at 1000 K when the concentration of NO is increased to 0.10 M, while the concentration of H, is 0.010 M? (d) What is the reaction rate at 1000 K if [NO] is decreased to 0.010 M and [H,] is increased to 0.030 M? %3D Q:4-2 The iodide ion reacts with hypochlorite ion (the ac- tive ingredient in chlorine bleaches) in the following way: OCI- + I- –→ OI¯ + Cl¯. This rapid reaction gives the following rate data: [OCI¯] (M) [I-] (M) Initial Rate (M/s) 1.5 X 10-3 1.5 X 10¬3 1.36 X 10¬4 3.0 × 10–3 1.5 X 10–3 2.72 X 10¬4 1.5 × 10-3 3.0 × 10-3 2.72 X 10¬4 (a) Write the rate law for this reaction. (b) Calculate the rate constant with proper units. (c)…
Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl2F2, catalyze the decomposition of ozone in the atmosphere. One simplifiedmechanism for the decomposition is:O3 ⎯⎯⎯⎯⎯⎯⎯→ sunlight O2 + OO3 + Cl ⟶ O2 + ClOClO + O ⟶ Cl + O2(a) Explain why chlorine atoms are catalysts in the gas-phase transformation: 2O3 ⟶ 3O2(b) Nitric oxide is also involved in the decomposition of ozone by the mechanism:O3 ⎯⎯⎯⎯⎯⎯⎯→ sunlight O2 + OO3 + NO ⟶ NO2 + O2NO2 + O ⟶ NO + O2Is NO a catalyst for the decomposition? Explain your answer.
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3a
3. The following reaction: A' (a) + B (ag + C (an)→ G* (ae) +2 H ) AH = -250 kJ There are two proposed mechanism: Mechanism 1 A' () + B aa) AB' a) + D (a) Step 1 AB" (a) (fast) + E (a) + Hm → G' sae) + H ) + D² (n) Step 2 (slow) Step 3 E (agi+ C (a) (fast) Mechanism 2: Step 1 AB' (a) A' (an) + B (ag) AB' (a) + C ja) (fast) Step 2 + G' (a) + 2 H() (slow) The reactions was determined to have the following rate law: r= k[A'][B] a) Which of the two proposed mechanisms is valid? Explain. b) List any reaction intermediates: catalysts: c) What is a rate determining step? d) If the activation energy for the overall forward reaction is 150 kJ, sketch the potential energy diagram for the above reaction; showing the overall reaction, the reaction mechanism, labelling the AHu, AHy, Eapu, Eag, reaction intermediates, and activated complex. rvs fwd drvs/
The overall reaction of hydroxide radical with hydrogen gas occurs as follows: OH (g) + H2(g) – H20(g) + H'(g) Part A This reaction plays a significant role in the chemistry of the earth's atmosphere by eventually converting OH to HO, through the subsequent reaction of H with O2. HO2 is involved in the destruction of stratospheric ozone (J. Phys. Chem. A, 2006, 110. 6978). Determine the activation energy for the reaction of OH with H, reaction. O 16.4 kJ mol The rate constants for the reaction of OH with H2 were found to be 1.83x10' L mol s'at 107. °C and 5.68×10° L mol's at 37. C. -1 O -85.9 kJ mol -1 O -16.4 kJ mol O 54.8 kJ mol O 65.9 kJ mol
Ethyl acetate reacts with sodium hydroxide when the two are mixed. The reaction, called ester hydrolysis, yields products ethyl alcohol and sodium acetate. The equation is CH3CO₂C₂H5(aq) + NaOH(aq) --> C₂H5OH(aq) + CH3CO₂Na(aq) The following data were obtained: Initial Concentration of CH3CO₂C₂Hs (mol L-¹) Initial Concentration of NaOH (mol L-¹) 0.005 0.010 0.005 0.010 0.020 0.020 The rate law for this reaction is: O Rate-k[CH3CO₂C₂H5]²[NaOH] O Rate-k[CH3CO₂C₂H5] + [NaOH] O Rate-k[CH3CO₂C₂H5]²[NaOH)² O Rate-k[CH3CO₂C₂H5] [NaOH] Initial Rate of Formation of C₂H5OH (mol L-1¹) 5.40 x 10-6 2.16 x 10-5 1.08 X 105
6. To observe the effect of temperature on reaction rate, the reaction between MnO, ag) and C204(ag) in acidic media is performed according to the given redox reaction: 2 MnO4 (aq) + 16 H*(aq) + 5 C204²(aq) → 10 CO2(g) + 2 Mn²"(aq) + 8 H2O(1) At various temperatures, the rate constants were determined experimentally as follows: k (M s') 5.03x105 T (°C) 20 35 3.68×104 60 6.71×103 90 0.119 (a) Determine Ea for this reaction by a graphical method (see the graph given below). (b) Determine E. using Arrhenius equation using tabulated data. (c) Calculate the value of the rate constant, k at 100°C. -2 -4 兰 -6 -8 -10 -12 0.0025 0.0027 0.0029 0.0031 0.0033 0,0035 1/T (K)
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