Problem 1. A student studied the clock reaction described in this experiment. Th

Question

Problem 1.  A student studied the clock reaction described in this experiment. They set up Reaction Mixture 4 by mixing 10. mL of 0.010 MNa2S2O3, 10. mL of 0.040 M KBrO3, and 20. mL of 0.10 M HCl using the procedure given. It took about 21 seconds for the color to turn blue.

Question 1. They found the concentration of each reactant in the reacting mixture by realizing that the number of moles of each reactant did not change when the reactant was mixed with the others, but that its concentration did. This, of course, follows the dilution equation.

The volume of the mixture was 50. mL (10. mL from water). What is the concentration of each reactant in the mixture?

Na2S2O3 M =

KBrO3 M =

HCl M =

Question 2. What is the relative rate of the reaction (1000/time)?

Question 3. What will be the form of the experimental rate law?

a. rate = k'[S2O32-]n[BrO3-]m[H+]p

b. time = k'[S2O32-]n[BrO3-]m[H+]p

c. rate = k'[S2O32-]n[BrO3-]n[H+]n

d. time = k'[S2O32-][BrO3-][H+]

e. rate = k'[S2O32-][BrO3-][H+]

Explanation / Answer

Q1)

Theory :

The redox reaction taking place:

BrO3- (aq) +6H+(aq) +6I- (aq)---->3I2 +Br-(aq) +3H2O(l)

2S2O32- (aq) +I2(aq) ------>S4O62-(aq) +2I-(aq)]*3

Net rxn: BrO3- (aq) +6H+(aq)+ 6S2O32- (aq) --->3S4O62-(aq)+Br-(aq) +3H2O(l)

I2 +starch --->blue -black color

So, mol BrO3-/mol S2O32-=1/6

mol BrO3-=M(BrO3-)*V(BrO3-)=0.040mol/L*0.010L=0.0004 mol    [ M=molarity,V=volume]

mol S2O32-=M(S2O32-)V(S2O32-)=0.010mol/L*0.010L=0.0001 mol

[BrO3-]=mol BrO3-/total volume=0.0004mol/0.050L=0.008M

[S2O32-]=mol S2O32-/total volume=0.0001mol/0.050L=0.002M

[HCl]=mol HCl/total volume=M(HCl)*V(HCl)/total volume=(0.020L*0.1mol/L)/0.050L=0.04M

Q2)relative rate=1000/time=1000/21 s=47.619 s^-1

Q3)

rate=k [S2O32-]^n*[BrO3- ]^m [HCl]^p

k=rate constant

n,m,p are the order of rxn with respect to BrO3- ,I- ,HCl

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