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Homework answers / question archive / Part A What is the rate law for the uncatalyzed reaction? View Available Hint(s) O rate = k[Ce4+][TH+12 O rate = k[Ti'] O rate = f[Ce!!] O rate = k04|TL!! rate = k1C4?TI rate = k Ce412TH 12 Submit Part B If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction? Check all that apply
Part A What is the rate law for the uncatalyzed reaction? View Available Hint(s) O rate = k[Ce4+][TH+12 O rate = k[Ti'] O rate = f[Ce!!] O rate = k04|TL!! rate = k1C4?TI rate = k Ce412TH 12 Submit Part B
If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction? Check all that apply. View Available Hints) All reactions that occur in one step are slow. The transition state is low in energy. The reaction requires the collision of three particles with the correct energy and orientation The probability of an effective three-particle collision is low. Submit
Part The catalyzed reaction is first order in Ce" and first order in (Mna+] Which of the steps in the catalyzed mechanism is rate determining View Available Hint(s) Ce+ (aq) + Mn?' (aq) --Ce (aq) + Mn (24) Cem (aq) + Mu? (aq)-Cett (ng) + Mn(n) Mn" (aq) + 'T' (aq) →Mn? (14) + Ti (0) Submit
Part A) we cannot determine the rate law without a chemical equation. Incomplete question.
Part B) If the uncatalyzed reaction occurs in a single elementary step it is slow because the reaction requires the collision of three particles with the correct energy and orientation and the probability of an effective three-particle collision is low.
Part C) the slowest step is the rate-determining step. In the above question, the information is not complete so not able to answer the question correctly.
