RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c 1A) You mix together in the p
ID: 879087 • Letter: R
Question
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)
3) You mix together in the proper manner the following VOLUMES:
0.0100 M Potassium iodide...14.7 mL
0.0010 M Sodium thiosulfate...5.6 mL
Distilled water...22.7 mL
0.0400 M Potassium Bromate...11.7 mL
0.100 M Hydrochloric acid...5.6 mL
Using the rate constant and rate equation from problem 2),
CALCULATE the time to turn blue...__________________ sec (c)
4) A plot of log Rate vs 1/T gives a best straight line which goes through....
1/T = 0.00300, log rate = -4.46
and 1/T = 0.00352, log rate = -6.00.
CALCULATE:
The Slope of the line..._____________ (d)
The Activation Energy (Ea) (Kj/mol)...________________(e)
Explanation / Answer
1. 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 = 3.78)
So the rate increases 3.78 times that of first experiment
Rate1 /Rate2 = {[HI]1 /[ HI2]}c = 1/3.78 = [0.1 X 10 / 0.1 X 20 ]c
3.78 = (2)c
log 3.78 = c log 2
0.577 = c X 0.3010
c = 1.91
The Experimental value of exponent c = 1.91
b. RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
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
................................................
c. RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
total volume =14.7 +5.6+22.7+11.7 + 5.6 = 60.3
i) 0.0100 M Potassium iodide...14.7 mL
so moles = molarity X volume = 0.01 X 14.7 /1000 =.147 / 1000
or concentration = moles / total volume = 0.000147X1000 / 60.3=0.0024 M
ii) 0.0010 M Sodium thiosulfate...5.6 mL
so moles = molarity X volume = 0.001 X 5.6 /1000 =
or concentration = moles / total volume = 0.0056 / 60.3 mL = 9.2 X 0^-5M
iii) Distilled water...22.7 mL
iv) 0.0400 M Potassium Bromate...11.7 mL
so moles = molarity X volume = 0.04 X 11.7 /1000 = 0.468 /1000
or concentration = moles / total volume = 0.468 / 60.3 mL =0.0077 M
v) 0.100 M Hydrochloric acid...5.6 mL
so moles = molarity X volume = 0.1 X 5.6 /1000 = 0.56 / 1000
or concentration = moles / total volume = 0.56 / 60.3 mL =0.0092 M
Rate constant = K = 47.1 ( from part b)
RATE = ([ Na2S2O3]/time) = k [I-] [BrO3-]0 [HCl]2 = 9.2 X 10-5/ time = 47.1 X 0.0024 X (0.0092)2
= 9.56 X 10^-6= 9.2 X 10-5/ time
Time = 9.6 seconds
.......................
d) For the given grap ( which is Arrehenius graph)
Slope =lnK2- lnK1 / 1/T2 - 1/T1 = (-6) - ( -4.46) / (0.00352 - 0.003) = -1.54 / 0.00052= -2961.53
Now , Slope = = -Ea / R
Ea = activation energy
Ea = 2961.53 X 8.314 =24622.16 J / mole = 24.622 KJ / mole
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