MSJ2e_Ch13_ISM2_June26_p593

MSJ2e_Ch13_ISM2_June26_p593 - Chapter 13 Chemical Kinetics...

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Unformatted text preview: Chapter 13: Chemical Kinetics: Rates of Reactions (b) E a,reverse = E a,forward – ∆ H = (95 kJ mol –1 ) – (–43 kJ mol –1 ) = 138 kJ mol –1 95 kJ/mol 138 kJ/mol E reactants products Reaction Progress 43. kJ/mol (c) E a,reverse = E a,forward – ∆ H = (55 kJ mol –1 ) – (15 kJ mol –1 ) = 40. kJ mol –1 55 kJ/mol 40. kJ/mol E reactants products Reaction Progress 15 kJ/mol 67. Assuming everything about the forward reactions in Question 67 is identical except the activation energy, the reaction with the largest E a,forward will be the slowest, and that with the smallest E a,forward will be the fastest. (a) Of the three, the smallest E a,forward is 65 kJ mol –1 . That is reaction (b). (b) Of the three, the largest E a,forward is 85 kJ mol –1 . That is reaction (c). 68. Assuming everything about the forward reactions in Question 68 is identical except the activation energy, the reaction with the largest E a,forward will be the slowest, and that with the smallest E a,forward will be the fastest. (a) Of the three, the smallest E a,forward is 55 kJ mol –1 . That is reaction (c). (b) Of the three, the largest E a,forward is 175 kJ mol –1 . That is reaction (a). 69. Assuming everything about the reverse reactions in Question 69 is identical except the activation energy, the reaction with the largest E a,reverse will be the slowest, and that with the smallest E a,reverse will be the fastest. (a) Of the three, the smallest E a,reverse is 15 kJ mol –1 . That is the reverse of reaction (c). (b) Of the three, the largest E a,reverse is 220 kJ mol –1 . That is the reverse of reaction (a). 54 Chapter 13: Chemical Kinetics: Rates of Reactions 70. Assuming everything about the forward reactions in Question 70 is identical except the activation energy, the reaction with the largest E a,reverse will be the slowest, and that with the smallest E a,reverse will be the fastest. (a) Of the three, the smallest E a,reverse is 40 kJ mol –1 . That is the reverse of reaction (c). (b) Of the three, the largest E a,reverse is 138 kJ mol –1 . That is the reverse of reaction (b). 71 . Define the problem : Given the chemical equations of elementary reactions write their rate laws. Develop a plan : As described in Section 13.6, the stoichiometric coefficient of a reactant in an elementary reaction is the reaction order for that reactant. Execute the plan : (a) NO + NO 3 products rate = k[NO][NO 3 ] (b) O + O 3 products rate = k[O][O 3 ] (c) (CH 3 ) 3 CBr products rate = [(CH 3 ) 3 CBr] (d) 2 HI products rate = k[HI] 2 72. Define the problem : Given the chemical equations of elementary reactions write their rate laws. Develop a plan : As described in Section 13.6, the stoichiometric coefficient of a reactant in an elementary reaction is the reaction order for that reactant....
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This note was uploaded on 09/10/2008 for the course CHE 131 taught by Professor Kerber during the Spring '08 term at SUNY Stony Brook.

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MSJ2e_Ch13_ISM2_June26_p593 - Chapter 13 Chemical Kinetics...

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