What is the rate law for the uncatalyzed reaction?

What is the rate law for the uncatalyzed reaction?

Metals often form several cations with different charges. Cerium, for example, forms and ions, and thallium forms and ions. Cerium and thallium ions react as follows:  

2Ce^4+(aq) + Tl^+(aq) ---> 2Ce^3+(aq)+Tl^3+(aq)  

This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the addition of Mn^2+ according to the following mechanism:  

Ce^4+(aq)+Mn^2+(aq) ---> Ce^3+(aq)+Mn^3+(aq)  

Ce^4+(aq)+Mn^3+(aq) ---> Ce^3+ (aq) + Mn^4+ (aq)  

Mn^4+(aq) + Tl^+(aq) ---> Mn^2+(aq) + Tl^3+(aq)  

If the uncatalyzed reaction occurs in a single elementary step, why is it a slow reaction?  

1)The probability of an effective three-particle collision is low.  

2) The transition state is low in energy.  

3) The reaction requires the collision of three particles with the correct energy and orientation.  

4) All reactions that occur in one step are slow.  

The catalyzed reaction is first order in [Ce^4+] and first order in [Mn^2+]. Which of the steps in the catalyzed mechanism is rate determining?  

Based on their available oxidation states, rank the following metals on their ability to catalyze this and other oxidation-reduction reactions.  

Rank from best to worst catalyst.  

Vanadium, titanium, manganese

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.