Answer to question 3? ONLY NEED TO USE RUN 1-3 FOR CALCULATION ** INFORMATION IS

Question

Answer to question 3? ONLY NEED TO USE RUN 1-3 FOR CALCULATION ** INFORMATION IS NOT MISSING **

IN ALL RUNS CONCENTRATION IS THE SAME

Concentration of KI is 0.250 M

Concentration of (NH4)2SO8 is 0.250 M

Concentration of KNO3 is 0.250 M

Concentration of Na2S203 is 0.200M

IN RUN 1 THE AMOUNT OF KI IS 10.0 ML AND THE CONCENTRATION IS 0.250 M

KI IS 10.0 ML IS RUN 1

10.0 ML OF KI IS ADDED IN RUN 1

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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