Given the rate equation... RATE = ([ Na 2 S 2 O 3 ]/time) = k [I - ] a [BrO 3 -
ID: 807755 • Letter: G
Question
Given the rate equation...
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... 75 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... 20 sec
Calculate:
The Experimental value of exponent c..._______________ (a)
2) If the exponents a, b, and c have the values:
a = 1; b = 2; c = 2
CALCULATE:
Rate constant k for data in 1B...______________ (b)
3) You mix together in the proper manner the following VOLUMES:
0.0100 M Potassium iodide...9.7 mL
0.0010 M Sodium thiosulfate...6.0 mL
Distilled water...15.6 mL
0.0400 M Potassium Bromate...10.7 mL
0.100 M Hydrochloric acid...5.3 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 = -5.00
and 1/T = 0.00365, log rate = -6.00.
CALCULATE:
The Slope of the line..._____________ (d)
The Activation Energy (Ea) (Kj/mol)...________________(e)
Explanation / Answer
Concentration of each species in the mixture A is calculated to be: (total volume is 50mL)
[Na2S2O3] = 2x10-4 M, [I-] = 2x10-3 M, [BrO3-] = 8x10-3 M, [HCl] = 2x10-2 M
Concentration of each species in the mixture B is calculated to be: (total volume is 50mL)
[Na2S2O3] = 2x10-4 M, [I-] = 2x10-3 M, [BrO3-] = 8x10-3 M, [HCl] = 4x10-2 M
RATE = ([ Na2S2O3]/time) = k [I-]a [BrO3-]b [HCl]c
Therefore, substituting the appropriate values of mixture A and B and then dividing rate of mixture A by mixture B, and simplifying it yields,
20/75 = (1/2)c => c = 1.91
2) Rate constant k for mixture B:
(2x10-4 /20) = k(2x10-3 * 64x10-6 * 16x10-4 ) => k = 48,828.125 M-4s-1
3) time calculation:
Concentration of each species in the mixture A is calculated to be: (total volume is 47.3mL)
[Na2S2O3] = 1.27x10-4 M, [I-] = 2.05x10-3 M, [BrO3-] = 2.26x10-3 M, [HCl] = 1.12x10-2 M
Substituting the values and calculating for t gives, time = 1980.3 s
4) Slope of the line = (-5 - (-6))/(0.003 - 0.00365) = -1538.5
Activation Energy = slope*2.303*R = -1538.5*2.303*8.314 = 29.46 kJ/mol
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