This the DRA of my Chem lab experiment: Chemical Kinetics of the Iodine Clock Re

Question

This the DRA of my Chem lab experiment: Chemical Kinetics of the Iodine Clock Reaction. I have been fumbling though most of these questions not sure if I’m doing them right. The data in questions one & three is what I collected in lab, all the other calculations I’m not super confident in.
Dat Reduction and Analysis Experiment 12 Chemical Kinetics of the lodine Clock Reaction Name Lab Partner Section TA Name Score _ Graph of In(rate) vs. Inang,1) with regression line Graph of ln(rate) vs. In(11-1) with regression line Data sheet stapled to back of plots Calculate the initial molar concentration of lodi le, r, in mole&Lfor; the reaction in test tube 1 . 1. General Equation molur concentew tionmolesLI-)/Liks of 2 moles of L-1.2xlo- Liters of Solvtion-.oIL O.OIL 2. In the following table, give the initial concentration (in moles/L) for 1,H2O2.S03, and Mo for each reaction. Test Tube (H20,) s,o,21 N/A N/A 012 |«O24mal eog mel 2x10-4| 3 N/A 4 5 mul roul 6 olb mol 8 N/A 153

Explanation / Answer

Lets see it with an example:

N2 + 3H2 --------->2NH3

here by the time1 mol N2 is reacted , 3 mol H2 also has to be reacted

so rate of consumption of hydrogen is 3 times that of nitrogen

mathematically it can be represented as:

3[N2]/t = [H2]/t

if you rearrange it

[N2]/t = [H2]/t x 1/3

It is what you call divide by stoichiometric ratio,whatever way you understand

Now lets look at the other rate equation

rate = k[S2O32-]m[I2]n

till now we don't know the order of reaction (complete as well as individual rate orders)

the reaction is as follows

first, slow reaction, iodine is produced:

H2O2 + 2I + 2H+ I2 + 2H2O

In the second, fast reaction, iodine is reconverted to 2 iodide ions by the thiosulfate reaction:

2S2O32 + I2 S4O62 + 2I

to find individual order of reaction

[reactant(1)]/t = rate = k[reactant(1)]m[reactant(2)]n

now change say double the reactant concentration and see if rate has become four times or not

rate = k[reactant(1)]m[reactant(2)]n

if m = 2;

= [2 x reactant(1)]2[reactant(2)]n = 4[reactant(1)][reactant(2)]n

if it remains same , then the reaction is first order

similarly if m=3 ,the rate will change by cube = 23 = 8

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