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the data in the table below were obtained for the reaction
(see image)
what is the magnitude of the rate constant for the reaaction
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- The label on a bottle of 3% (by volume) hydrogen peroxide, H2O2, purchased at a grocery store, states that the solution should be stored in a cool, dark place. H2O2decomposes slowly over time, and the rate of decomposition increases with an increase in temperature and in the presence of light. However, the rate of decomposition increases dramatically if a small amount of powdered MnO- is added to the solution. The decomposition products are H2O and O2. MnO2 is not consumed in the reaction. Write the equation for the decomposition of H2O2. What role does MnO2 play? In the chemistry lab, a student substituted a chunk of MnO2 for the powdered compound. The reaction rate was not appreciably increased. WTiat is one possible explanation for this observation? Is MnO2 part of the stoichiometry of the decomposition of H2O2?Iodomethane (CH3I) is a commonly used reagent in organic chemistry. When used properly, this reagent allows chemists to introduce methyl groups in many different useful applications. The chemical does pose a risk as a carcinogen, possibly owing to iodomethanes ability to react with portions of the DNA strand (if they were to come in contact). Consider the following hypothetical initial rates data: [DNA]0 ( mol/L) [CH3I]0 ( mol/L) Initial Rate (mol/Ls) 0.100 0.100 3.20 104 0.100 0.200 6.40 104 0.200 0.200 1.28 103 Which of the following could be a possible mechanism to explain the initial rate data? MechanismIDNA+CH3IDNACH3++IMechanismIICH3ICH3++ISlowDNA+CH3+DNACH3+Fast8. Consider the reaction: Trial # 1 2 3 4 5 6 2NH (g) + 302(g) + 2CH(g) → 2 HCN(g) + 6H₂O(g) [CH₂] (M) 0.37 0.45 0.45 0.26 0.45 0.17 [NH)] (M) 0.035 0.035 0.054 0.054 0.054 0.079 c. Write the rate law. b. Determine the overall reaction order. [0₂] (M) 0.12 0.27 0.12 0.27 a. Determine the order with respect to each of the reactants. 0.27 0.32 d. Solve for the value of the rate constant with units. Rate M/hr 2.3 x 10² 1.6 x 102 5.5 x 10² 3.7 x 102 3.7 x 10²
- The following reaction was analyzed and data was collected. 3 A (g) + 2 B4 (g) + C2 (g) – D (g) Initial Concentration (mol L) (Ba] 0.100 Trial Initial Rate (mol Ls4) 9.072 x 109 3.629 x 10 2.449 X 107 3.135 х 10% [A] [C:] 1 0.100 0.400 2. 0.400 0.100 0.400 0.100 0.300 0.400 4 0.100 0.600 1.60 0.600 0.75 0.100 a) Determine the rate law and rate constant for the reaction. Be sure to show all of your work. b) Determine the rate for the fifth reaction.A 20-mm cube of copper metal is placed in 250 mL of 12 M nitric acid at 25°C and the reaction below occurs: Cu(s) + 4H+ (aq) + 2NO3(aq) -> Cu²+ (aq) + 2NO2(g) + 2H₂O(1) At a particular instant in time, nitrogen dioxide is being produced at the rate of 4.4 x 10-4 M/min. At this same instant, what is the rate at which hydrogen ions are being consumed in M/min? Do not write unit in answer. Report your answer with five places past the decimal point. Type your answer...Peroxynitric acid (HOONO2) is an unstable molecule that decomposes to nitric acid and oxygen: 2HOONO2(aq) → 2HNO3(aq) + O2(g)When the concentration of peroxynitic acid is graphed against time, the resulting plot is curved, but if the logarithm of this concentration is plotted, we instead get a straight line. Based on this, which statement is true? a) This decay is a second order in peroxynitric acid. b) The slope of the straight-line graph is the rate constant. c) One needs the concentration of peroxynitric acid to calculate its half-life. d) The rate law appears to be of the form -Δ[HOONO2]/Δt = k[HOONO2].
- Consider the following balanced equation: Cl 2(g) + 2 Br (aq) → Br 2(aq) + 2 CI (aq) [CI 2] [Br ] Rate (M/s) 1.0.509 0.382 0.01544 2.0.613 0.382 0.02239 3.0.485 0.436 0.01600 What is the order with respect to chlorine gas? O -1 O 1/2 O 1 O 3/2 O 2 O 3 O 4 cannot be determinedDetermine the rate law and rate constant for the following reaction. Don't forget the units. Some of the numbers are blurry. (The concentrations for H* are x 10-5 and the rates are x 10-7 with the last value being x10-8) 10, (aq) + 81(aq) + 6 H(aq) → 31 (aq) + 2 H₂O(aq) Trial [103] Initial Rate (M/s) 1 0.0050 2.70 x 10 2 0.0100 5.40 x 10 3 0.0050 6.07 x 10 4 0.0050 6.75 x 10 [I]. 0.030 0.030 0.045 0.030 [H*]. 2.0 x 10 2.0 x 10 2.0 x 10 1.0 x 10Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2][I–][H+]rate I 0.100 M 5.00 × 10–4 M 1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec a. The rate law for this reaction is a. rate = k[H2O2][I–][H+] b. rate = k[H2O2]2[I–]2[H+]2 c. rate = k[I–][H+] d. rate = k[H2O2][H+] e. rate = k[H2O2][I–]
- Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M 1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec The rate law for this reaction is a. rate = k[H2O2]2[I–]2[H+]2b. rate = k[I–]c. None of thesed. rate = k[I–][H+]e. rate = k[H2O2][I–][H+]f. rate = k[H2O2][H+]Consider the following reaction: 2MnO4–(aq) + 10Br–(aq) + 16H+(aq) --> 2Mn2+(aq) + 5Br2(aq) + 8H2O Under a certain reaction condition, if the rate -D[MnO4-]/Dt = 2.4 x 10-4 mol/(L.s), what is the relative rate D[Br2]/Dt in mol/(L.s)? (A) 1.2 x 10–3 (B) 6.0 x 10–4 (C) 2.4 x 10–4 (D) 9.6 x 10–5Consider the following data concerning the equation: H2O2 + 3I– + 2H+ → I3– + 2H2O [H2O2] [I–] [H+] rate I 0.100 M 5.00 × 10–4 M 1.00 × 10–2 M 0.137 M/sec II. 0.100 M 1.00 × 10–3 M 1.00 × 10–2 M 0.268 M/sec III. 0.200 M 1.00 × 10–3 M 1.00 × 10–2 M 0.542 M/sec IV. 0.400 M 1.00 × 10–3 M 2.00 × 10–2 M 1.084 M/sec The rate law for this reaction is a. rate = k[H2O2][I–][H+] b. rate = k[H2O2][H+] c. rate = k[H2O2]2[I–]2[H+]2 d. None of these e. rate = k[H2O2][I–] f. rate = k[I–][H+]