Hw1-G2-RxnDyn - Exercise Set 1 Reaction Dynamics CHM 2046...

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Unformatted text preview: Exercise Set 1 Reaction Dynamics CHM 2046 1. Why are some collisions between molecules of reactants not effective? 2. 'Describe with the aid of a diagram a possible transition state for the single step reaction between CH3Br and OH"1 to give methyl alcohol, CH30H, and Br“ 3. In order to get charcoal' to burn, it must first be heated. Why? Would you consider this reaction exothermic or endothermic? Explain. 15 H for this reaction positive or negative? 4. The single-step reaction, “0201(9) + “0(9) =;i= “02(9) i ONC1(g) is reversible; Eact (forward) is 28.9 KJ and Eact (reverse) is 41.8 KJ . Draw a potential energy diagram for the reaction, and indicate both energies of activation and the position of the transition state. Indicate 13H and deduce its magnitude and sign. Is this an endothermic or exothermic reaction? 5. The following statements are sometimes encountered with reference to catalysis, but they are not stated as carefully as they might be. What modifications should be made in each of them? a) A catalyst is a substance which speeds up a chemical reactin but does not take part in the reaction. b) The function of a catalyst is to lower the activation energy for a chemical reaction. 6. How will the activation energy for the oxidation of sulfur dioxide 2 $02 + 02 ——-da— 2 $03 be changed (if at all) when a platinum catalyst is present? How will the activation energy for the reverse reaction be affected? How will AE (~AH) for the oxidation be affected for the reverse reaction? 7. Indentify the reaction.intermediates, catalysts, and overall stoichiometric reaction for each of the mechanisms given below. a) Hzoz+ I‘1 -—=- H20 + 10‘1 1 10— + H202 “'4’ H20 + 02 + 1—1 + -———>— + b) “0(9) 03(9) “02(9) 02(9) 03(9) "‘*’ 02(9) + 0(9) “02(9) + 0(9) “"' “0(9) + 02(9) Nielsen/7745-H 8. 10. 11. A particuiar reaction is found to have the foiiowing overaii rate law: Rate = k [A] 2 [3] Which terms in this rate law are changed by each of the foiiowing changes: a. the concentration of A is doubied; b. a cataiyst is added; c. the concentration of A is increased by a factor of 2, whereas the concentration of B is decreased by a factor of 4; d. temperature is increased. For the hypotheticai equation A + 23 -——-4’- D + E the rate equation is A Bi (1 [D] What wiii happen to the reaction rate if (a) [A] is doubied, (b) [B] is doubied, (c) more catalyst is added, (d) D is removed as it is formed to keep [0] at a smaii value, (e) [A] and [B] are both doubied, and (f) the temperature is increased? rate = k The reaction + --———45— 2 IC1( + 2 HC1(g) 9) “2(9) 12(9) at temperatures above 200°C is first order in H2 and first order in 101. Suggest a mechanism of two steps with the first step rate determining to account for the rate equation. Write the overaii rate equation for each of the foliowing reactions, based on the experimental findings that (a) is first order, (b) is second order, and (c) is third order: ———-> a. 2 N02C1 2 N02 + C12 b. 2 NOCi ————»- 2 N0 + (:12 c. 2 no + 012 ——> 2 N001 The overaii rate Taw for the reaction N0 + 03 --- N02 + 02 is of the erm rate = k [N0] [03] Which of the foTTowing mechanisms is consistent with this rate Taw? Expiain. a. N0 ——4—- N + O siow N + 03 ——45- N02 + 0 fast 0 + O -4>- 02 fast b. NO + 03 -—A- N02 + 03 c. 03 -—=>- 02 + O sTow O + N0 —-‘%- N02 fast The reaction, N02(0) + COég) -—fi-602(g) + NO(g), appears to have the mechanism (at Tow tempera ure), + "—9 + S1OW N03 + co ——aa- N02 + 002 fast Expiain why the overaTT reaction rate equation is zero-order with respect to C0 . The thermaT decomposition of ozone of oxygen, 2 03(9) 2(9) has a rate Taw given by, rate = k [03] 2 ‘ [02] Show that the foTTowing mechanism is consistent with this overaiT rate Taw: k 03 --l=' 02 + 0 fast 0 + 03 ——44—A7— 2 02 siow 15. 16. 17. For the mechanism postulated in (14) above, the proposed energies of activation are as follows: 4KJ; and 2nd step, 24 KG. —285 KJ. lst step forward, 100 KJ; lst step reverse, The value of 15H for the overall reaction is Draw a reaction energy profile for this reaction. for the reaction: 2 N0ég% + 02(9) -—d5— 2 N02(g) , the overall rate law 15 experimentally foun o be- Rate = k [no] 2 [CS Is the rate law consistent with the one~step mechanism 2 NO + 02 -—-a-v 2 N02 ? Why is this one—step mechanism not very likely to be correct? Show that the rate law is consistent with the following mechanism: k1 2 NO :;:::Er N202 fast k N202 + 02 __EZ_,_ 2 N02 slow Why does the rate of a reaction increase as the temperature increases by a factor much larger than the increase in the number of molecular collisions? ...
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This note was uploaded on 11/16/2011 for the course CHM 2046 taught by Professor Chamusco during the Summer '10 term at Santa Fe College.

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Hw1-G2-RxnDyn - Exercise Set 1 Reaction Dynamics CHM 2046...

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