Answer to Question 3? THE AMOUNT OF KI USED IN RUN ONE IS 10.0 ML AND THE CONCEN

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Answer to Question 3?

THE AMOUNT OF KI USED IN RUN ONE IS 10.0 ML AND THE CONCENRATION IS 0.250 M.

INTRODUCTION The purpose of this experiment is to study the kinetics of the reaction between 10dide, a peroxydisulfate, S208-, ions: tics of the reaction between iodide, F, and S2O32-(aq) + 2 F(aq) I2(S) + 2 SO42-(aq) (1) The rate of this reaction is equal to the rate of disappearance of S20:2, -A[S2O8]/At, which equan of formation of iodine, A[I]/At. The rate law expression has the form: A[S,0; . A[I,1 = k[5,0;]' rate = - At At (2) The experiment involves determining the rate of the reaction with differing Concentrations of S2O8 and I present by measuring the rate of formation of I2. From the data you will calculate: (1) "x", the order of the reaction with respect to S2032- (2) "y", the order of the reaction with respect to IF (3) "K", the specific rate constant for the reaction Iodine, I2, is produced continuously by the reaction being studied: S2082-(aq) + 2 F(aq) I2(s) + 2 SO42-(aq) Che rate of formation of I2 (which equals the rate of the reaction) is determined by measuring the time equired, At, for a measured amount of iodine, A[12], to be formed. This is carried out by adding a measured nannt undium thiosulfate, Na2S203, which reacts extremely rapidly an (3)

Explanation / Answer

Q3)

Data required:

               Rate (Ms^-1)              [S2O82-]                                                [I-]

Run1 1.0*10^-11                      0.250M *(10ml/50ml)=0.05M           0.250M *(10ml/50ml)=0.05M

Run2 1.0*10^-11                      0.250M*(20ml/50ml)=0.1M              0.250M *(10ml/50ml)=0.05M

Run3 1.0*10^-11                       0.250M *(10ml/50ml)=0.05M          0.250M*(20ml/50ml)= 0.1M                        

Using equation,

Rate=k[S2O82-]^x [I-]^y   x=order of rxn with respect to S2O8 ,y=order of rxn with respect to I-

Putting the above data in the equation,

1.0*10^-11=(0.05)^x (0.05)^y .......(1)

3.0*10^-11=(0.1)^x (0.05)^y ..........(2)

3.0*10^-11=(0.05)^x (0.1)^y ..........(2)

Eqn(1)/eqn(2) gives,

1/3=(0.5)^x

or, 0.33=(0.5)^x

log 0.33=x log 0.5

x=log 0.33/log 0.5=-0.477/-0.301=1.6

x=1.6

Similarly,

eqn2/eqn3 gives

1=2^x (0.5)^y

putting x=1.6

(0.5)^y =0.33

y log 0.5=log 0.33

y=log0.33/log0.5=1.6

x=1.6,y=1.6

approximation x=1.0,y=1.0

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