Question

solution               mL H2O2                    mL KI                 rate (mL O2/s)

1                          5.0                             10.0                          ?

2                          10.0                           10.0                          ?

3                           5.0                              20.0                        ?

Rate Law: since doubling either [H2O2] or [I-] doubles the reaction rate, what is the rate law?

Answer 1

The reaction is first order wrt H2O2 so doubling the [H2O2] will double the rate.

Rate = Constant x [H2O2]
Thus, Rate is directly proportional to the volume of H2O2.
Hence, doubling the volume of hard water would double the rate.

Detailed calculations.

3% H2O2 is about 0.9 Molar, & they called for 0.10Molar KI to be used
======================================...

Molarity of stock solution of appx H2O2 = 0.924 M
Molarity of stock solution of KI = 0.10 Molar

exp # ...mls H2O2......KI..... H2O
1.... ..... 5.0ml ... 10.0ml ... 15.0 ml
2.... ..... 10.0ml ...10.0ml ... 10.0 ml
3.... ..... 5.0ml ... 20.0ml ... 5.0 ml

Molarities of H2O2 :
exp # ......(original M) (dilution factor) = new Molarity
1.... ..... (0.924 M) (5.0 ml/30 ml) = 0.154 Molar
2.... ..... (0.924 M) (10.0 ml/30 ml) = 0.308 Molar
3.... ..... (0.924 M) (5.0 ml/30 ml) = 0.154 Molar

Molarities of KI:
exp # ......(original M) (dilution factor) = new Molarity
1.... ..... (0.10 M) (10.0 ml/30 ml) = 0.033 Molar
2.... ..... (0.10 M) (10.0 ml/30 ml) = 0.033 Molar
3.... ..... (0.10 M) (20.0 ml/30 ml) = 0.066 Molar

those should be your answers to your questions:
Initial molarity of H2O2 after dilution (M)
Initial Molarity of I- after dilution (M)
======================================...
Your data:

O2 mls... Exp1 ..... Exp 2 ....Exp 3

2.0 ....1.42 min ...1.22 ... 1.14
4.0 ... 1.08 ..... 0.46 ... 0.45
6.0 .....0.56 .... 0.16 .... 0.45
8.0 ... 1.10 .... 0.28 ..... 0.19
10.0 .. 1.14 ...... 0 .38 .... 0.40
14.0 .. 1.28 ..... 0.30 .... 0 .30

since no readings were taken in exp2 &3 after 14 ml's , that's where I'll stop
your question:
Initial rate of reaction (slope of tangent ) (MH2O2/s)

exp 1 total ml O2 / total seconds = 14.0 / 449 seconds = 0.0312 ml/sec
exp 2 total ml O2 / total seconds = 14.0 / 240 seconds = 0.0583 ml/sec
exp 2 total ml O2 / total seconds = 14.0 / 253 seconds = 0.0553 ml/sec

your answers are: the initial rates are:
exp 1 Rate = 0.0312 ml/sec
exp 2 Rate= 0.0583 ml/sec
exp 2 Rate = 0.0553 ml/sec
======================================...

rate data table:
exp # ... M H2O2...... M KI..... Rate
exp 1 .... 0.154 ... 0.33 .... 0.0312 ml/sec
exp 2 .... 0.308 ... 0.33 .... 0.0583 ml/sec
exp 2 . .. 0.154 .... 0.66.... 0.0553 ml/sec

your answes:
Order with respect of H2O2 (estimated @ "1", since doubling the Molarity of H2O2 doubled the rate between exp #1 & exp #2)
Order with respect to I- (estimated @ "1" , since doubling the Molarity of KI doubled the rate between exp #1 & exp #3)
======================================...

Order with respect to H2O2 (calculated)

rate exp #2 / rate exp #1 = (0.0583 ml/sec) / (0.0312 ml/sec)= 1.89
[H2O2] exp#2 / {H2O2 exp #1 = (0.308M] / (0.154) = 2

[H2O2]^? = log1.89 / log 2 = .276 / 0.301
your answer: 0.917 ie first order with respect to H2O2

--------------

Order with respect to KI (calculated)

rate exp #3 / rate exp #1 = (0.0553 ml/sec) / (0.0312 ml/sec)= 1.77
[KI] exp#3 / [KI] exp #1 = (0.66M] / (0.33) = 2

[KI]^? = log1.77 / log 2 = 0.248 / 0.301
your answer: 0.824 ie first order with respect to KI

======================================...

find Specific rate constants (specify the units):
exp 1
Rate = k [H2O2] [KI]
0.0312 = k [0.154] [0.33]
k= 0.614 (ml O2 / sec)(M)^-2

exp 2
Rate = k [H2O2] [KI]
0.0583 ml/sec = k [0.308] [0.33]
k= 0.574 (ml O2 / sec)(M)^-2


exp 3
Rate = k [H2O2] [KI]
0.0553 ml/sec = k [0.154] [0.66]
k= 0.544 (ml O2 / sec)(M)^-2
==============================

Average specific rate constant

(0.544 + 0.574 + 0.614) / 3 =

your answer:
0.577 (ml O2 / sec)(M)^-2