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Methane (CH4) is a greenhouse gas emitted by industry, agriculture, and waste systems. It is the second most prevalent greenhouse gas (after carbon dioxide). Methane plays an important role in climate change because it absorbs infrared radiation more efficiently than carbon dioxide. Methane is broken down in the atmosphere by ozone (O3), making its atmospheric Iifetime shorter than that of carbon dioxide.
A research group studied the
Initial Rate versus Initial Concentrations
[CH4] | [O3] | Initial Rate (M/ s) |
0.010 | 0.010 | |
0.020 | 0.010 | |
0.020 | 0.020 |
Rate Constant versus Temperature
Temperature (K) | K (M-1.s-1) |
260 | |
265 | |
270 | |
275 | 0.000114 |
280 | 0.000187 |
285 | 0.000303 |
290 | 0.000483 |
a. Use the data in the first table to determine the order of the reaction with respect to each reactant.
b. Use the data in the second table to determine the activation barrier and pre-exponential factor for the reaction.
c. Atmospheric concentrations of methane and ozone can vary depending on the location and altitude. Calculate the rate of the reaction at 273 K for a methane concentration of 1.8 ppm (by volume) and an ozone concentration of 5.0 ppm (by volume). Note that 1 ppm of CH4 by volume means 1 L CH4 /106 L air. Assume STP (standard temperature and pressure) so that 1 mol gas occupies 22.4 L.
d. What is the half-life of methane in the atmosphere in years at 323 K? (Assume that [CH4] = [O3] = [A]0 =
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Chapter 14 Solutions
Chemistry: Structure and Properties (2nd Edition)
- 1) The isoamyl acetate report requires eight paragraphs - four for comparison of isoamyl alcohol and isoamyl acetate (one paragraph each devoted to MS, HNMR, CNMR and IR) and four for comparison of acetic acid and isoamyl acetate ((one paragraph each devoted to MS, HNMR, CNMR and IR. 2) For MS, the differing masses of molecular ions are a popular starting point. Including a unique fragmentation is important, too. 3) For HNMR, CNMR and IR state the peaks that are different and what makes them different (usually the presence or absence of certain groups). See if you can find two differences (in each set of IR, HNMR and CNMR spectra) due to the presence or absence of a functional group. Include peak locations. Alternatively, you can state a shift of a peak due to a change near a given functional group. Including peak locations for shifted peaks, as well as what these peaks are due to. Ideally, your focus should be on not just identifying the differences but explaining them in terms of…arrow_forward№3 Fill in the below boxes. HN 1. LAH 2. H3O+ NH2arrow_forwardFor the photochemical halogenation reaction below, draw both propagation steps and include the mechanism arrows for each step. H CH ot CH3 CI-CI MM hv of CH H-CI CH3 2nd attempt See Periodic Table See Hint Draw only radical electrons; do not add lone pair electrons. Note that arrows cannot meet in "space," and must end at either bonds or at atoms. 1 i Add the missing curved arrow notation to this propagation step. 20 H ن S F P H CI Br 品arrow_forward
- The radical below can be stabilized by resonance. 4th attempt Draw the resulting resonance structure. DOCEarrow_forwardUse curved arrows to generate a second resonance form for the allylic radical formed from 2-methyl-2-pentene. 1 Draw the curved arrows that would generate a second resonance form for this radical. D 2 H S F A Бг Iarrow_forwardDraw the resulting product(s) from the coupling of the given radicals. Inlcude all applicable electrons and non-zero formal charges. H.C öö- CH3 2nd attempt +1 : 招 H₂C CH CH₂ See Periodic Table See H H C S F P Br CH₂ Iarrow_forward
- Please, help me out with the calculation, step by step on how to find what's blank with the given information.arrow_forwardPredict the following products. Then show the mechanism. H₂N NH2arrow_forwardBF3, Boron Trifluoride, known to contain three covalent boron-fluorine bonds. suggest and illustrate all of the processes as well as their energetical consequences for the formation of BF3 from its elements.arrow_forward
- Draw the mechanism of the reaction.arrow_forward9. Draw all of the possible Monochlorination Products that would Result From the Free Radical Chlormation OF 23,4-TRIMethyl Pentane b. Calculate the To Yield For the major • Product given the Following Relative Restritus For 1° 2° and 30 Hydrogens toward Free Radical Chloration 5.0: 38 : 1 30 2° 1° C. what would be the major product in the Free Radical brominator Of the Same Molecule. Explain your Reasoning.arrow_forwardWhat is the complete reaction mechanism for the chlorination of Ethane, C2H6?arrow_forward
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