Consider the data below representing a run for the Clock Reaction you\'ll be per
ID: 475999 • Letter: C
Question
Consider the data below representing a run for the Clock Reaction you'll be performing in lab. Calculate the rate of reaction in M/s if the volumes of reagents indicated are combined and 219 seconds elapse before the color appears. 0.50 mL 0.00050 M Na2S2O3 0.50 mL 0.010 M KI 0.50 mL 0.040 M KBrO3 0.25 mL 0.100 M HCl 0.75 mL DI H2O 1-2 drops 1% starch indicator (negligible volume) A) 2.3 x 10-6 M/s B) 9.1 x 10-7 M/s C) 4.6 x 10-7 M/s D) 3.8 x 10-7 M/s E) 7.6 x 10-8 M/s. I know the answer is not C.
The answer I was given (which has the correct answer): The reaction that occurs between the iodine ion and bromate ion will be studied, which is as follows: 6I- (aq) + BrO3 - (aq) + 6H+ (aq) > 3I2(aq) + Br- (aq) + 3H2O(1) E) 7.6 x 10^-8 M/s concentration of [Na2S2O3] in the final solution = 0.0005 M x 0.5 ml/2.5 ml = 1 x 10^-4 M total volume = 2.5 ml now calculate the concentration of I2: So, [I2] = 1 x 10^-4/2 = 5 x 10^-5 M Thus, the rate of reaction = d[I2]/3dt = 5 x 10^-5/3 x 219 = 7.6 x 10^-8 M/s
Can someone better explain the steps of this problem? How is [I2] calculated from the answer of 9Na2s203?
Explanation / Answer
For the given reaction, rate of reaction would be,
E) 7.6 x 10^-8 M/s
Details:
concentration of [Na2S2O3] in the final solution = 0.0005 M x 0.5 ml/2.5 ml = 1 x 10^-4 M
So, [I2] = 1 x 10^-4/2 = 5 x 10^-5 M
Thus, the rate of reaction = d[I2]/3dt = 5 x 10^-5/3 x 219 = 7.6 x 10^-8 M/s