Need help filling this table please, also kindly use excel to do the graph that
ID: 1067833 • Letter: N
Question
Need help filling this table please, also kindly use excel to do the graph that is being asked below the table ( please slow your calculations, so i can understand and show it to teacher) Thanks
this was the lab experiment that was done
this were the result
Moles Determi mL Moles Moles [S202] Inisz0e j Elapsed Szoe2. remaining remaining time in s time in min S2082 nation Nazszos NazS20s consumed consumed remaining Prepare a graph of InIS2002 ]vs. time in min, determine the pseudo-first order rate constant and calculate the rate constant k.Explanation / Answer
The iodine-clock reaction is given by
2 I- (aq) + S2O82- (aq) -------> 2 SO42- (aq) + I2 (aq)
The thiosulfate reaction is used to estimate the concentration of I2 generated as per the equation
2 S2O32- (aq) + I2 (aq) -----> S4O62- (aq) + 2 I- (aq)
Moles S2O32-:moles I2 = 2:1
Moles I2:moles S2O82- = 1:1
Therefore, moles S2O32-:moles S2O82- = 2:1 ……(1)
The rate of the reaction is given by
R = k[I-][S2O82-] where k = rate constant for the reaction.
I- is used up to form I2 by thiosulfate. Hence in presence of thiosulfate, the entire I- concentration is converted to I2 quantitatively. Only when thiosulfate is completely used up do we get I2 in the system which reacts with starch to give the blue-brown end-point. Therefore,
[I-] [I2]0 where [I2]0 is the initial concentration of I2. Therefore,
R = k[I2]0[S2O82-] …..(2)
If [I2]0 is large and held constant throughout the experiment and only [S2O82-] is varied, then we can write the pseudo first order rate constant as
R = k’[S2O82-] ……(3)
where k’ = k[I2]0 is the pseudo first order rate constant.
We can obtain k’ by measuring [S2O82-] vs time and plotting the data according to the first order rate law:
ln [S2O82-] = ln [S2O82-]0 – k’t where [S2O8]0 is the initial concentration of S2O82-.
Slope = -k’
Now turn to the data; note that we add 3 mL of 0.15 M sodium persulfate to all the test tubes so that the moles of S2O82- initially present = (3 mL)*(1 L/1000 mL)*(0.15 mol/L) = 0.00045 mol.
The total volume of the solution in each case is 12.0 mL (you can add up the volume of all the reagents added to find the total volume present). Therefore, the initial molar concentration of S2O82- is
[S2O82-]0 = (0.00045 mol)/(12.0 mL*1 L/1000 mL) = 0.0375 M.
Now fill up the required table:
Determination
mL Na2S2O3
Moles Na2S2O3 consumed = (volume of Na2S2O3 in L)*(concentration of Na2S2O3 in mol/L)
Moles S2O82- consumed = 1/2*moles Na2S2O3 (see equation 1)
[S2O82-] remaining = [(moles S2O82- initially present) – (moles S2O82- consumed)]/volume of solution in L
ln [S2O82-] remaining
Elapsed time in sec
Elapsed time in min
1
3.0
(3.0 mL)*(1 L/1000 mL)*(0.2 mol/L) = 0.0006
0.0003
(0.00045 – 0.0003)mol/(12.0 mL*1L/1000 mL) = 0.0125
-4.382
6:30
0.1050
2
2.5
0.0005
0.00025
0.0167
-4.092
5:26
0.0877
3
2.0
0.0004
0.0002
0.0208
-3.873
4:46
0.0743
4
1.5
0.0003
0.00015
0.0250
-3.689
3:32
0.0553
5
1.0
0.0002
0.00010
0.0292
-3.533
2:16
0.0360
You haven’t mentioned the unit of time; I assumed the times recorded were in seconds. Now plot ln [S2O82-] vs t.
The data is plotted graphically above.
Slope = -12.218 = -k’
===> k’ = 12.218
Therefore, the pseudo first order rate constant is 12.218 min-1 (the unit will be the inverse of time since the left hand side is dimensionless) (ans).
Next, we need to find out [I]0. Move back to the first equation given
Moles I-:moles I2 = 2:1
Since the concentration of I- remains constant, we can calculate the molar concentration of I2 as below:
Moles of I- = (3 mL)*(1 L/1000 mL)*(1 mol/L) = 0.003 mol.
Volume of solution = 12.0 mL; therefore, [I2]0 = (0.003 mol)/(12 mL).(1 L/1000 mL) = 0.25 mol/L = 0.25 M.
Therefore, k’ = k[I2]0
====> 12.218 min-1 = k.(025 mol/L)
====> k = (12.218 min-1)/(0.25 mol/L) = 48.872 L mol-1min-1 (ans).
Determination
mL Na2S2O3
Moles Na2S2O3 consumed = (volume of Na2S2O3 in L)*(concentration of Na2S2O3 in mol/L)
Moles S2O82- consumed = 1/2*moles Na2S2O3 (see equation 1)
[S2O82-] remaining = [(moles S2O82- initially present) – (moles S2O82- consumed)]/volume of solution in L
ln [S2O82-] remaining
Elapsed time in sec
Elapsed time in min
1
3.0
(3.0 mL)*(1 L/1000 mL)*(0.2 mol/L) = 0.0006
0.0003
(0.00045 – 0.0003)mol/(12.0 mL*1L/1000 mL) = 0.0125
-4.382
6:30
0.1050
2
2.5
0.0005
0.00025
0.0167
-4.092
5:26
0.0877
3
2.0
0.0004
0.0002
0.0208
-3.873
4:46
0.0743
4
1.5
0.0003
0.00015
0.0250
-3.689
3:32
0.0553
5
1.0
0.0002
0.00010
0.0292
-3.533
2:16
0.0360
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