Harcourt-Essen Reaction

P6 – Harcourt EssenReaction
Rafia Aslam
1327166
06/03/2014

Abstract
The aim of the investigation was to determine the effect of changes in concentration and
temperature on the rate of this particular Harcourt-Essen reaction. By varying the concentration of
one reactant fortwo outof fourruns, the individual andoverall rate orders were determined. Graph
analysis was conducted using data obtained from the completion of one run, to calculate the
changes in [H2O2] and [I2] to determine the rate coefficient and its dependence on temperature.
Introduction
The rate of a reaction is defined by the change in the concentration of reactants or products, per
unit time. It is the gradient of a concentration-time graph.
Experimental data leads to the derivation of the rate law and reaction orders.
The species in the rate law are present in the rate determining step, as
the rate of the reaction is dependent upon the changes in the concentrations
of these species. It does not include intermediates. The following graphs
show the relationship between the concentration of a particular species upon
the reaction rate: Figure 1; Conc v. Time graph1
Figure 2; (a) Rate = k, 0 order, (b) Rate = k[R], 1st order (c) Rate = k[R]2/more, +2nd order2
The mechanism of the Harcourt Essen reaction consists of a series of elementary steps.
Consequently, the overall reaction equation is the sum of these elementary steps.
2H3O+
+ 2I-
+ H2O2  4H2O + I2
________________________
H2O2 + H3O+
 H3O2
+
+ H2O fast
H3O2
+
+ I-
 H2O + HOI slow
HOI + I-
 HO-
+ I2 fast
HO-
+ H3O  2H2O fast
All three reactants are colourless. As H2O2 oxidises I-
to I2, the presence of starch allows for the
formationof a deep blue iodine-starch complex resulting from the addition of an iodine molecule
into the amylose coil of starch.3
Na2S2O3 is then used to reduce I2 back to I-
ions and the solution
regains its initial colourless appearance.
I2 + 2S2O3
2-
 2I-
+ S4O6
2-
fast
Once all S2O3
2-
added has reacted, the presence of excess I2 causes the deep blue colour to return.
Another addition of S2O3
2-
is immediately added and the colour disappears again. Using the total
volume of thiosulphate added, [H2O2]0, [I2]t and subsequently [H2O2]∞ are determined.

As the Harcourt-Essenreactioncanvary interms of its reactants, for instance, by using KIO3 instead
of KI, or Na2S2O8 as the oxidising agent, or even changing the acid from H2SO4 to CH2(COOH)2, the
steady-state and pre-equilibrium approximations work differently and give different rate laws.4
In
this case, the latter is used as the rate determining step is not the first, but the second step of the
mechanism. It is assumed that all the fast reactions are at equilibrium and hence the rate of the
overall reaction is equal to the rate of this step.
H2O2 + H3O+
⇌ H3O2
+
+ H2O fast
H3O2
+
+ I-
 H2O + HOI slow
Rate = -d[H2O2]/dt = k[H3O2
+
][I-
]  H3O2
+
= intermediate
Keq = [H3O2
+
][H2O]/[H2O2][H3O+
]
[H2O] are [H3O+
] ≈ large and constant:
Keq = [H3O2
+
]/[H2O2]  [H3O2
+
] = Keq[H2O2]
-d[H2O2]/dt = k[I-
]. Keq[H2O2] hence
-d[H2O2]/dt = k’[I-
][H2O2] (where k’ = Keq x k = rate coefficient)
The rate law suggests that the reaction is second order overall and first order with respect to each
reactant. This is later investigated.
As step 3 of the mechanism is presumed to be at equilibrium, the iodine formed immediately
convertsback intoiodide ions,thus[I-
] ≈constant. Thissuggeststhat at the end of the reaction, [I-
]∞
> [H2O2]∞.The “Isolation Method” is an accurate way to determine the orders with respect to each
reactant where the concentrations of all reactants need to be measured simultaneously.5
Thus:
Rate = -d[H2O2]/dt = k”[h2o2] (where k” = k’[I-
] = pseudo 1st
order rate coefficient)
Throughintegration, ln[H2O2]t =-k”t+ ln[H2O2]0 isplottedtodetermine k”,andsubsequentlyk’. (See
Appendix C-1)
Experimental Method
Chemicals/Apparatus Error
1.0164M Na2S2O3 ± 2%
8g/l KI ± 10%
4 vol (nominal) H2O2 ± 10%
1:5 acid:water H2SO4 ± 10%
1% starch solution ± 10%
25 ml burette ± 0.06 ml
10 ml burette ± 0.04 ml
Pasteur pipette ± 10%
200-1000 μl Micropipette ± 100 μl
Run 1 Run 2 Run 3 Run 4
KI (ml) 120 180 120 120
H2O (ml) 120 60 120 120
H2SO4 (ml) 10 10 10 10
Starch (drops) 5 5 5 5

Temperature (K) 303 303 308 313
4 solutions were made according to the table above. The temperature was controlled via the hot
plate andstill headanda magneticstirrerwasaddedto each flask for gentle stirring. 25 cm3
of H2O2
was boiled and 1.5 cm3
of Na2S2O3 was added via a micropipette to conical flask 1, after 10-15
minutes of climatisation. Boiling H2O2 was added and the stirring speed was increased. The time
taken for a dark blue-brown colour to appear was recorded along with temperature. Immediately
another 1.5 cm3
of Na2S2O3 was added and the process until 9 cm3
of Na2S2O3 had been added. The
procedure wascarriedout forruns 1-3 whilstrun4 was run tocompletionuntil nomore blue-brown
colour appeared.
Results and Discussion
Part 1
Table 1; Data values for [I-], k” and k’ for all runs (See Appendices T3-6, G1-4 and C5 for full results and calculations)
The volume of KIwas increased by a factor of 1.5 from 120 cm3
in Run 1 to 180 cm3
inRun 2. The rate
coefficient, k’, should be constant for both reactions yet mine varied by 0.013802 mol-1
dm3
s-1
. This
was due to the fact that there were errors associated with the apparatus and chemical solutions
themselves, as well as errors stemming from the volumetric make-up of the solutions.
Rate = k’[H2O2][I-
] where [H2O2]0 = 0.0320 M(See Appendix C2)
 Initial Rate1 = k’[H2O2][I-
] = 0.08044 x 0.0320 x 0.02101 = (5.4 ± 0.8) x 10-5
moldm-3
s-1
 Initial Rate2 = k’[H2O2][I-
] = 0.06664 x 0.0320 x 0.03152 = (6.7 ± 1.0) x 10-5
moldm-3
s-1
The rate of Run 2 is faster than Run 1 by a factor of 1.2. This deviation from the expected value of
1.5 is the result of experimental and technique errors. However it does confirm that the rate of
reaction is directly proportional, and hence, 1st
order with respect to [I-
]. Increasing the
concentrationof a speciesincreasesthe numberof particles sothe likelihoodof successful collisions
is greater than at a reduced concentration.
The plot of ln{[H2O2]0/[H2O2]t} againsttimeislinearandthusshows that the reaction is also 1st
order
with respect to [H2O2]. (See Appendices G1-4)
Hence the overall order of the reaction is 2nd
order.
Part 2
Table 2; Data values for average weighted temperature, time inverse and logarithmic k’ for runs 1, 3 and 4 (See
Appendix G5 for graph)
Run Experimental
Temperature (K)
k” (s-1
) [I-] (moldm-3
) k’ (mol-1
dm3
s-1
)
1 303 0.00169 0.02101 0.08044
2 303 0.00210 0.03152 0.06664
3 308 0.00227 0.02101 0.10804
4 313 0.00277 0.02101 0.13184
Run Weighted
Temperature
1/T
(K-1
)
ln k’

The difference between runs 1, 3 and 4 was the temperature change. (See Appendix C6 for error
calculation on rate)
Rate1 = (5.4 ± 0.8) x 10-5
moldm-3
s-1
Rate3 = (7.3 ± 1.0) x 10-5
moldm-3
s-1
Rate4 = (8.9 ± 1.0) x 10-5
moldm-3
s-1
The resultsshowthat the k’ increases withtemperature,resultinginanincreasedrate of reaction. At
higher temperatures, there are more particles with a kinetic energy greater than the activation
energy, Ea, which in turn increases the likelihood of successful collisions between the species
involved.
The Arrhenius equation demonstrates the link between Ea, temperature and k’:
k’ = Ae-Ea/RT
 ln k’ = (-Ea/R)(1/T) + lnA
(Ea = minimumenergy needed for reaction to proceed;A=pre-exponential factor- number of collisions occurring)
Gradient = -Ea/R = -5017.26923
-Ea = -5017.26923 x 8.3145 = -41716 kJ/mol
y-intercept on graph = ln A = 13.98143
A = e13.98143
= 1.18 x 106
dm3
mol-1
s-1
Conclusion
The Harcourt-Essen reaction is 2nd
order overall and 1st
order with respect to peroxide ions and
iodide ions,whererate = k’[H2O2][I-
].Thereisapositive correlationbetweentemperature andrate of
reaction. Asprecisionwasvital forthisinvestigation,itcanbe saidthat volumetrictechnique needed
to be controlled in order to reduce the effect of random errors, as that had an effect on the
corresponding values obtained for the rate coefficient.
Evaluation
The experimentcouldhave been further improved if more runs had been conducted, in particular,
where the initial concentrationof iodide ionswasvariedtoagreaterextent.Thiswouldhave ledtoa
more reliable calculation of the orders with respect to each reactant. As only the integration and
isolation methods had been used, a wider range of runs would have enabled the utilisation of
differential and half-life methods to determine a more accurate value for k’ and individual orders.
(K)
1 304.40343 0.00329 -2.52028
3 308.86535 0.00324 -2.22522
4 313.54805 0.00319 -2.02615

References
1. J. Clark, Finding Orders of Reaction Experimentally, 2011, accessed on 28th March 2014,
< http://www.chemguide.co.uk/physical/basicrates/experimental.html#top>
2. T. Lister, J. Renshaw, Understanding Chemistry for Advanced Level, Stanley Thornes Ltd, UK, 3rd edn, 2000
3. C. Ophardt, Starch and Iodine, accessed on 30th March 2014,
<http://chemwiki.ucdavis.edu/Biological_Chemistry/Carbohydrates/Case_Studies/Starch_and_Iodine>
4. M. A. Atuori, A. G. Brolo, A. L. M. L. Mateus, J. Chem. Educ., 1989, 66, 852
5. D. N. Blauche, Isolation Method, 2000-2014, accessed on 2nd April 2014,
< http://www.chm.davidson.edu/vce/kinetics/IsolationMethod.html>
6. Gama Group a.s., Pasteur Pipettes, 2007-2014, accessed on 5th April 2014,
<http://www.gama.cz/en/pasteur-pipettes>
7. Chemspider, Potassium Iodide, Royal Society of Chemistry, UK, 2014, accessed on 20th March 2014,
< http://www.chemspider.com/Chemical-Structure.4709.html>
Appendices
T3; Results for Run 1
Time (s) Total vol. of
S2O3
2-
added
(cm3
)
Temperature
of reaction
mixture (K)
[I2](t)
(moldm-3
)
[C]t
(moldm-3
)
ln {[C]0/[C]t}
56 1.5 304.6 0.00275 0.02925 0.08999
75 2 304.8 0.00367 0.02834 0.12165
95 2.5 304.7 0.00457 0.02743 0.15422
113 3 304.7 0.00548 0.02653 0.18777
131 3.5 304.4 0.00638 0.02562 0.22236
152 4 304.6 0.00728 0.02473 0.25805
173 4.5 304.6 0.00817 0.02383 0.29493
195 5 304.6 0.00907 0.02294 0.33308
219 5.5 304.6 0.00996 0.02205 0.37261
244 6 304.6 0.01084 0.02116 0.41361
272 6.5 304.6 0.01172 0.02028 0.4562
297 7 304.4 0.0126 0.0194 0.50054
325 7.5 304.3 0.01348 0.01853 0.54676
355 8 304.2 0.01435 0.01765 0.59505
385 8.5 304.1 0.01522 0.01678 0.6456
412 9 304.1 0.01609 0.01592 0.69865

G1; Plot of ln ([C]0/[C]t) vs. Time for Run 1
50 100 150 200 250 300 350 400 450
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
ln [C(0)/C(t)]
Linear Fit of ln [C(0)/C(t)]ln[C(0)/C(t)]
Time (s)
Equation y = a + b*x
Weight No Weighting
Residual Sum of
Squares
1.70224E-4
Pearson's r 0.99985
Adj. R-Square 0.99967
Value Standard Error
ln [C(0)/C(t)] Intercept -0.00201 0.00194
ln [C(0)/C(t)] Slope 0.00169 7.93935E-6
Run 1
S2O3
2-
added
(cm3
)
Temperature
of reaction
mixture (K)
[I2](t)
(moldm-3
)
[C]t
(moldm-3
)
ln {[C]0/[C]t}
52 1.5 303.3 0.00275 0.02925 0.08999
68 2 303.3 0.00367 0.02834 0.12165
82 2.5 303.4 0.00457 0.02743 0.15422
97 3 303.3 0.00548 0.02653 0.18777
113 3.5 303.3 0.00638 0.02562 0.22236
130 4 303.4 0.00728 0.02473 0.25805
146 4.5 303.6 0.00817 0.02383 0.29493
170 5 303.6 0.00907 0.02294 0.33308
182 5.5 303.5 0.00996 0.02205 0.37261
201 6 302.8 0.01084 0.02116 0.41361
222 6.5 302.8 0.01172 0.02028 0.4562
243 7 302.8 0.0126 0.0194 0.50054
268 7.5 302.8 0.01348 0.01853 0.54676
290 8 302.8 0.01435 0.01765 0.59505
317 8.5 302.8 0.01522 0.01678 0.6456
343 9 302.8 0.01609 0.01592 0.69865

50 100 150 200 250 300 350
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
ln [C(0)/C(t)]
Linear Fit of ln [C(0)/C(t)]
ln[C(0)/C(t)]
Time (s)
Weight No Weighting
Residual Sum of
Squares
3.46888E-4
Pearson's r 0.99968
ln [C(0)/C(t)] Intercept -0.01617 0.00286
ln [C(0)/C(t)] Slope 0.0021 1.40845E-5
Run 2
Time (s)
Total vol. of
S2O3
2-
added
(cm3
)
Temperature
of reaction
mixture (K)
[I2](t)
(moldm-3
)
[C]t
(moldm-3
)
ln {[C]0/[C]t}
40 1.5 308.8 0.00275 0.02925 0.08999
50 2 308.8 0.00367 0.02834 0.12165
65 2.5 308.8 0.00457 0.02743 0.15422
77 3 308.8 0.00548 0.02653 0.18777
94 3.5 308.8 0.00638 0.02562 0.22236
110 4 308.6 0.00728 0.02473 0.25805
124 4.5 308.8 0.00817 0.02383 0.29493
141 5 308.8 0.00907 0.02294 0.33308
160 5.5 308.8 0.00996 0.02205 0.37261
178 6 308.8 0.01084 0.02116 0.41361
196 6.5 308.8 0.01172 0.02028 0.4562
217 7 308.8 0.0126 0.0194 0.50054
238 7.5 308.8 0.01348 0.01853 0.54676
257 8 308.8 0.01435 0.01765 0.59505
282 8.5 308.6 0.01522 0.01678 0.6456
305 9 309.6 0.01609 0.01592 0.69865

0 50 100 150 200 250 300 350
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
ln[C(0)/C(t)]
Linear Fit of ln[C(0)/C(t)]ln[C(0)/C(t)]
Time (s)
Weight No Weighting
Residual Sum of
Squares
1.92321E-4
Pearson's r 0.99983
ln[C(0)/C(t)] Intercept 0.00835 0.00202
ln[C(0)/C(t)] Slope 0.00227 1.13225E-5
Run 3
S2O3
2-
added
(cm3
)
Temperature
of reaction
mixture (K)
[I2](t)
(moldm-3
)
[C]t
(moldm-3
)
ln {[C]0/[C]t}
37 1.5 312.9 0.00275 0.02925 0.08999
49 2 313.2 0.00367 0.02834 0.12165
62 2.5 313.2 0.00457 0.02743 0.15422
77 3 313.3 0.00548 0.02653 0.18777
95 3.5 313.3 0.00638 0.02562 0.22236
107 4 313.4 0.00728 0.02473 0.25805
122 4.5 313.4 0.00817 0.02383 0.29493
137 5 313.5 0.00907 0.02294 0.33308
153 5.5 313.5 0.00996 0.02205 0.37261
171 6 313.5 0.01084 0.02116 0.41361
187 6.5 313.5 0.01172 0.02028 0.4562
205 7 313.6 0.0126 0.0194 0.50054
221 7.5 313.6 0.01348 0.01853 0.54676
241 8 313.7 0.01435 0.01765 0.59505
265 8.5 313.7 0.01522 0.01678 0.6456
283 9 313.8 0.01609 0.01592 0.69865
306 9.5 313.8 0.01695 0.01505 0.75448
332 10 313.7 0.01782 0.01419 0.81339
356 10.5 313.9 0.01867 0.01333 0.87577
385 11 313.9 0.01953 0.01248 0.94205
410 11.5 313.8 0.02038 0.01162 1.01279
443 12 313.9 0.02123 0.01078 1.08864
476 12.5 314 0.02208 0.00993 1.17042

512 13 313.9 0.02292 0.00909 1.25917
548 13.5 313.9 0.02376 0.00825 1.35621
587 14 313.8 0.0246 0.00741 1.4633
632 14.5 313.8 0.02543 0.00657 1.58282
685 15 313.8 0.02626 0.00574 1.71807
742 15.5 313.9 0.02709 0.00491 1.87394
806 16 313.8 0.02792 0.00409 2.05798
883 16.5 313.9 0.02874 0.00326 2.28284
971 17 313.9 0.02956 0.00244 2.57224
1087 17.5 313.8 0.03038 0.00163 2.97941
1212 18 313.8 0.03119 8.12E-4 3.67426
1456 18.5 313.7 0.03201 0 --
0 200 400 600 800 1000 1200 1400
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
ln[C(0)/C(t)]
Linear Fit of ln[C(0)/C(t)]
ln[C(0)/C(t)]
Time (s)
Weight No Weighting
Residual Sum of
Squares
0.2618
Pearson's r 0.99483
ln[C(0)/C(t)] Intercept -0.08079 0.0255
ln[C(0)/C(t)] Slope 0.00277 4.99113E-5
Run 4

G5; Plot of ln k’ vs. Time showing temperature dependence of rate for data collected from Runs 1,3 and 4
0.00318 0.00320 0.00322 0.00324 0.00326 0.00328 0.00330
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
ln(k')
Linear Fit of Sheet1 ln(k')ln(k')
1/T (K
-1
)
Weight No Weighting
Residual Sum of
Squares
0.00212
Pearson's r -0.99141
ln(k') Intercept 13.98143 2.14284
ln(k') Slope -5017.26923 662.02523
C1; Calculation of ln{[C]0/[C]t}
-d[H2O2]/dt = k”[h2o2]
∫ d[H2O2]/[H2O2] = ∫ k”t dt  Conc limits = [C]0 and [C]t (where [C]0 > [C]t) & Time limits = 0 and t
ln[H2O2]0 - ln[H2O2]t = 0-(-k”t) = k”t
ln{[C]0/[C]t} = k”t
C2; Calculation of initial concentration of H2O2, [C]0
Through conservation of mass, the number of moles of product formed = number of moles of
reactant lost:
[H2O2]0 = [H2O2]t + [I2]t hence
[H2O2]t = [H2O2]0 – [I2]t
When t = ∞, all of the iodine has been reduced to iodide ions via thiosulphate ions, and no more
blue-browncolourappears.Atthispoint,[I2] =[H2O2]0 andthe reaction has been run to completion.
2H3O+
+ 2I-
+ H2O2  4H2O + I2
I2 + 2S2O3
2-
 2I-
+ S4O6
2-

[I2]t = ½[S2O3]2-
Run 4 was run to completion and 18.5 cm3
of 1.0164 M S2O3
2-
was added.
C = N/V
(18.5/1000) x 1.0164 = 0.0188034 moles of S2O3
2-
0.0188034/2 = 9.4017x10-3
(2:1 ratio of S2O3
2-
: H2O2)
Total volume of reaction mixture = VKI + Vacid + Vwater + Vstarch + Vperoxide + Vthiosulphate
120 + 120 + 10 + 5(0.05) + 25 + 18.5 = 293.75 cm3
(NB:Each dropof starch is roughly 0.05 ml.6
)
(9.4017x10-3
x 1000) / 293.75 = 0.0320 M = [H2O2]0
C3; Calculation of [I2]t
Example: Run 1, [I2] when t = 152 s
Volume of thiosulphate added = 4 cm3
(4/1000) x 1.0164 x 0.5 = 2.0328 x 10-3
moles
Total volume of reaction mixture at this point = 279.25 cm3
(2.0328 x 10-3
) / (279.25/1000) = 7.28 x 10-3
M
C4; Calculation of [C]t
[C]t = [C]0 – [I2]t
For run 1 at 152 s:
0.0320 – (7.279 x 10-3
) = 0.0247 M
C5; Calculation of rate coefficient k’
Example: Run 1
Mass = Density x Volume
8 x 120/1000 = 0.96g
Mr of KI = 166.002808.7
Mass/Mr = 0.96/166.002808 = 5.78303x10-3
moles
Ans x 1000 / 275.25 = 0.02101 M
k” = k’ [I-
]
k” = 0.00169 s-1

k’ = 0.00169/0.02101 = 0.08044 mol-1
dm3
s-1
Error Propagation
Total volume = 120 cm3
hence δV  √{(4 x δ 25 ml burette)2
+ (2 x δ 10 ml burette)2
} =
√{(4x0.06)2
+ (2x0.04)2
} ≈ ± 0.253 cm3
δm/0.96 g = √{(0.8/8)2
+ (0.253/120)2
} ≈ 0.100
Hence δm = 0.96 x 0.100 ≈ ± 0.0960 g
0.0960/166.002808 = ± 5.78303x10-4
(error on no. Moles)
δVtotal = √(2x0.242
+ 2x0.082
+ 0.042
+ 0.062
+ 0.0252
) ≈ 0.366
δ[I-
]/0.02101 M= √{(5.783e-4/5.783e-3)2
+ (0.366/275.25)2
} ≈ 0.100
Hence δ[I-
] = 0.100x 0.02101 ≈ 2.101x10-3
≈ ± 2x10-3
M
δk” = ± 7.94x10-6
s-1
δk’ = √{(7.94e-6/0.00169)2
+ (2.1e-3/0.02101)2
} ≈ 8.049x10-3
≈ 8.05x10-3
mol-1
dm3
s-1
Thus k’ = 0.080 ± 0.008 mol-1
dm3
s-1
C6; Calculation of error on Rate of Reaction
Example Run 1
δR/(5.41x10-5
) = √{(δ[H2O2]0/[H2O2])2
+ (δ[I-
]/[I-
])2
+ (δk’/k’)}
δR/(5.41x10-5
) = √{(6.4x10-4
/0.0320)2
+ (2.1x10-3
/0.02101)2
+ (8.05x10-3
/0.08044)}
= 0.1428....
δR = Ans x 5.41x10-5
≈ 7.728...x10-6
≈ 0.77x10-5
moldm-3
s-1
≈ 0.8x10-5
moldm-3
s-1

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