RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
1A) You mix together in the proper manner the
following:
10.0 mL of .0100 M Potassium Iodide
10.0 mL of .00100 M Sodium thiosulfate
10.0 mL of water
10.0 mL of .0400 M Potassium Bromate
10.0 mL of .100 M HCl
The time to turn blue is... 53 sec
1B) You also mix together in the proper manner the following:
10.0 mL of .0100 M Potassium Iodide
10.0 mL of .00100 M Sodium thiosulfate
10.0 mL of .0400 M Potassium Bromate
20.0 mL of .100 M HCl
The time to turn blue is... 14 sec
Calculate:
The Experimental value of exponent c..._______________ (a)
2) If the exponents a, b, and c have the values:
a = 1; b = 0; c = 2
CALCULATE:
Rate constant k for data in 1A...______________ (b)
1. As we can see in the data given , the concentration of all the reactants are kept constant in second experiment except HCl.
The concentration of HCl was made double which reduces the time approx four times to the first experiment. ( 53/14).
So the rate increaes four times that of first experiment
This indicates that rate of reaction = constant X [HCl]2
The Experimental value of exponent c = 2
b. RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
Note:
1. 10.0 mL of .0100 M Potassium Iodide, so moles = molarity X volume = 0.01 X 10 /1000 =
or concenration = moles / total volume
= 0.0001 X1000 / 50 mL
=0.002 M
2. 10.0 mL of .00100 M Sodium thiosulfate; so moles = molarity X
volume = 0.01 X 10 /1000 =
or concenration = moles / total volume
= 0.00001 X1000 / 50 mL =0.0002M
3.10.0 mL of water
4. 10.0 mL of .0400 M Potassium Bromate ; so moles = molarity X
volume = 0.04 X 10 /1000 = 0.0004
or concenration = moles / total volume
= 0.0004 X1000 / 50 mL
=0.008
5. 10.0 mL of .100 M HCl ; so moles = molarity X volume = 0.1 X 10
/1000 = 0.001
or concenration = moles / total volume = 0.001 X1000 / 50 mL =0.02M
RATE = ([ Na2S2O3]/time) = k [I-] [BrO3-]0 [HCl]2 = 0.0002 / 53 = K X 0.0002 X (0.02)2
3.77 X 10^-6 = K X8X10^-8
K = 47.1 moles ^-2 X L-1 X second^-1
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