Given the rate equation... RATE = ([ Na 2 S 2 O 3 ]/time) = k [I - ] a [BrO 3 -

ID: 807795 • 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)

1. On doubling the cocentration of HCl the rate becomes 4 times faster

so the coeffecient is 2 (=c)

2.rate = conc of thiosulphate / time = K[I-] [BrO3-]2[HCl]2

0.01 / 50 X 75 = 2.66 X 10^-6 = K [0.1/50][0.4/50]2 [1/50]2 = 5.12 X 10^-11 X K

K = 0.519 X 10^5

3. total volume = 47.3

concentration of

0.0100 M Potassium iodide...9.7 mL = .01 X 9.7 / 47.3 =0.002

0.0010 M Sodium thiosulfate...6.0 mL = .001 X 6 / 47.3 = 1.26 X 10^-4

0.0400 M Potassium Bromate...10.7 mL = 0.04 X 10.7/ 47.3 = 0.009

0.100 M Hydrochloric acid...5.3 mL = 0.1 X 5.3 / 47.3 = 0.011

Rate = 1.26 X 10^-4 / time = K[I-] [BrO3-]2[HCl]2

1.26 X 10^-4 / time = 0.519 X 10^5 X 0.002 X 0.009X 0.009 X 0.011 X 0.001 = 0.101 X 10^-5

time = 1.26 / .0101 = 124.75sec

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