Sample Rate Law and Activation Energy (Ea)
Calculations
 2013, Sharmaine S. Cady
East Stroudsburg University
The following set of rate data was collected for the reaction between I- and H2O2.
Table 1. Reagent Volumes for Kinetic Trials at 21.0 °C
Solution A
Kinetic
Trial
Deionized
water, mL
1
2
3
4
5
6
7
8
4.0000
3.0000
2.0000
1.0000
2.0000
0.0000
5.0000
2.0000
Buffer,
0.500 M
HAc and
0.4998 M
NaAc, mL
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
Solution B
0.3001 M
KI, mL
0.0201 M
Na2S2O3,
mL
Starch
0.0998 M
H2O2, mL
1.0000
2.0000
3.0000
4.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
5 drops
5 drops
5 drops
5 drops
5 drops
5 drops
5 drops
5 drops
3.0000
3.0000
3.0000
3.0000
5.0000
7.0000
2.0000
4.0000
Time, s
65.00
31.00
21.00
15.00
37.00
27.00
96.00
49.00
The rate of the reaction for each trial is calculated by taking the [I2] for each trial and dividing by the
time in seconds.
1
How to find [I2]
The number of moles of I2 present at the start of the reaction is 0. As the reaction proceeds, the redox
reaction produces I2. The number of moles I2 present at the time the blue-black color appears can be
determined from the stoichiometry between I2 and the amount of S2O32- at the start of the reaction.
The equation for the number moles of I2 present when the blue black color appears is given on the
following page.
stoichiometric ratio from
reaction equation
calculated Na2S2O3 M
volume Na2S2O3
micropipetted
Rate laws use concentrations in units of molarity (M). To calculate the [I2] in units of molarity, the
number of moles of I2 formed must be divided by the total volume of the reaction mixture in units of
liters (L). Summing the volumes of all reagents gives 10.0000 mL as the total reaction volume:
The two equations can be combined to determine the [I2]:
Since the volume of NaS2O3 micropipetted is the same for all trials, [I2] is the same for all trials. The
molarity of the thiosulfate solution has 3 significant digits, the volume pipetted has 5, and the total
reaction volume has 6. Therefore, the value of [I2] is rounded off to the lower number, which is 3
significant digits.
2
How to find rate (M/s)
[I2] is divided by the time for each trial. The data from Trial 1 is used in the following calculation.
Since all the times have 4 significant digits and [I2] has 3, the rates are rounded off to the lower
number, which is 3 significant digits. Note the units on the rate.
This calculation is repeated for the next seven trials and their measured time values.
How to find the reaction order with respect to [I-]
To determine how the rate depends on [I-], select the trials where the volume of H2O2 is held constant.
This gives a constant [H2O2], which means the rate only depends on [I-]. For this experiment, trials 1-4
are used because the volume of H2O2 remains constant at 3.0000 mL. Determine the initial
concentration of [I-]0 from the volume of KI micropipetted, reagent concentration, and the total volume
of the reaction mixture. This is a dilution calculation and the following equation may be used:
For this experiment, Mc = original concentration KI (0.3001 M), Vc= volume micropipetted, MD = [KI]0,
and VD = reaction solution volume (10.0000 mL). Substituting the values for trial 1 into the above
equation and rearranging to solve for MD ([KI]0) yields:
(
)(
)
(
)
(
)(
)
(
)
This is also [I-]0 since, given the formula stoichiometry, there is 1 mole of I- for every mole of KI. This
calculation is repeated for trials 2-4 where the Vc volume is the only variable that changes.
The log values of the rate and [I-]0 for each trial are computed. To express the log values with the
correct number of significant digits, count the number of significant digits in the [I-]0, and express the log
value with the same number of decimal places. The log value for the rate is expressed in a similar
manner. Log values have no units.
3
Table 2. Log Values for [I-]0 and Rate
[I-]0, M
0.03001
0.06002
0.09003
0.1200
Trial
1
2
3
4
Log [I-]0
-1.5227
-1.2217
-1.0456
-0.9208
Rate, M/s
1.55 x 10-5
3.26 x 10-5
4.81 x 10-5
6.73 x 10-5
Log Rate
-4.810
-4.487
-4.318
-4.172
Note the [I-]0 values have 4 significant digits and the log values have 4 decimal places. The rate has 3
significant digits and the log values have 3 decimal places.
The log rate vs. log [I-]0 is plotted to obtain a linear plot with given slope. The slope rounded off to the
nearest whole integer is the reaction order with respect to [I-].
Log Rate vs. Log [I-]0
-1.6000
-1.4000
-1.2000
-1.0000
-0.8000
-0.6000
-0.4000
-0.2000
-4.100
0.0000
-4.200
y = 1.048x - 3.2142
R² = 0.9993
-4.300
log rate
-4.400
-4.500
-4.600
-4.700
-4.800
log [I-]0
4
-4.900
How to find the reaction order with respect to [H2O2]
To determine how the rate depends on [H2O2], select the trials where the volume of I- is held constant.
This gives a constant [I-], which means the rate only depends on [H2O2]. For this experiment, trials 1 and
5-7 are used because the volume of KI remains constant at 1.0000 mL. Determine the initial
concentration of [H2O2]0 from the volume of H2O2 micropipetted, reagent concentration, and the total
volume of the reaction mixture. This is a dilution calculation and the following equation may be used:
For this experiment, Mc = original concentration KI (0.0998 M), Vc= volume micropipetted, MD = [H2O2]0,
and VD = reaction solution volume (10.0000 mL). Substituting the values for trial 1 into the above
equation and rearranging to solve for MD ([KI]0) yields:
(
)(
)
(
)
(
)(
)
(
)
This calculation is repeated for trials 5-7 where the Vc volume is the only variable that changes.
The log values of the rate and [H2O2]0 for each trial are computed.
Table 3. Log Values for [H2O2]0 and Rate
Trial
1
2
3
4
[H2O2]0, M
0.0299
0.0499
0.0699
0.0200
Rate, M/s
1.55 x 10-5
2.73 x 10-5
3.74 x 10-5
1.05 x 10-5
Log [H2O2]0
-1.524
-1.302
-1.156
-1.699
Log Rate
-4.810
-4.564
-4.427
-4.979
Note the [H2O2]0 values have 3 significant digits and the log values have 3 decimal places. The rate has 3
significant digits and the log values have 3 decimal places.
5
The log rate vs. log [H2O2]0 is plotted to obtain a linear plot with given slope. The slope rounded off to
the nearest whole integer is the reaction order with respect to [H2O2].
Log Rate vs. Log [H2O2]0
-1.800
-1.600
-1.400
-1.200
-1.000
-0.800
-0.600
-0.400
-4.300
-0.200
0.000
-4.400
y = 1.0255x - 3.2399
R² = 0.999
-4.500
log rate
-4.600
-4.700
-4.800
-4.900
-5.000
log [H2O2]0
-5.100
How to find k’
Since the slopes rounded off to the nearest whole integer are both 1 for the two reactants, the rate law
becomes
The reaction is first order with respect to [I-] and [H2O2] and second order overall. The computed rates
for trials 1-8 and the computed initial concentrations of I- and H2O2 are substituted into the above
equation. These are the same values used to obtain the log values for the plots. The equation is then
solved for k’.
6
Table 4. Calculated k’ Values
Trial
1
2
3
4
5
6
7
8
[I-]0 , M
0.03001
0.06002
0.09003
0.1200
0.03001
0.03001
0.03001
0.03001
Rate, M/s
1.55E-05
3.24E-05
4.79E-05
6.70E-05
2.72E-05
3.72E-05
1.05E-05
2.05E-05
[H2O2]0, M
0.0299
0.0299
0.0299
0.0299
0.0499
0.0699
0.0200
0.0399
AVERAGE k’
STD DEV
k’, M-1s-1
0.0172
0.0180
0.0178
0.0186
0.0181
0.0178
0.0175
0.0171
0.0178
0.0005
Substitution of the values from Trial 1 yields the following for k’:
⁄
(
(
)(
)(
)
)
The rate and [H2O2]0 both have 3 significant figures and [I-]0 has 4. The answer is rounded off to the
lower number of significant digits, which is 3.
This calculation is repeated for the next seven trials and their measured rates and concentrations.
Using Excel to compute the average and standard deviation
Enter the eight values for k’ into the cells for the first column. Click in the cell below the last entry.
Click on Formulas on the menu ribbon and then Insert Function.
Select AVERAGE in the function box. Then OK.
Make sure the dialog box shows the correct range for the four entered values. Click OK.
Click the cell below the average value.
Click on Insert Function. Select STDEV in the function box. Make sure the dialog box shows the correct
range of entered values. Click OK.
The standard deviation is rounded off to 1 significant figure.
7
How to find Ea
The rate and k’ are calculated for the two kinetic runs performed at 0 °C and 30 °C. The calculations are
the same as those used for Trial 4 at room temperature. The reciprocal of the temperatures in Kelvin
units are calculated. The A graph of ln k’ vs. 1/T is plotted. The average k’ from the room temperature
trials is used for plotting purposes.
Table 5. Rate and Temperature Data
T, K (°C + 273.15)
273.2
294.2
309.2
time, s
75.00
15.00
4.00
k’, M-1s-1
3.73 x 10-3
1.78 x 10-2
6.99 x 10-2
Rate, M/s
1.34 x 10-5
-2.51 x 10
1/T
3.661 x 10-3
3.399 x 10-3
3.245 x 10-3
ln k’
-5.592
-4.029
-2.661
Ea is calculated by sustituting the slope into the following equation:
(
)(
)
The k’ has 3 significant digits and T has 4; therefore the slope has 3 significant digits. The red numbers
are not significant but are retained until the desired answer is obtained. Ea is rounded off to 3 significant
digits and expressed in units of kJ/mol. Note that the slope is negative and yields a positive value for Ea.
8
ln k' vs. 1/T
0.000
3.200E-03 3.250E-03 3.300E-03 3.350E-03 3.400E-03 3.450E-03 3.500E-03 3.550E-03 3.600E-03 3.650E-03 3.700E-03
-1.000
y = -6926.6x + 19.699
R² = 0.988
ln k'
-2.000
-3.000
-4.000
-5.000
-6.000
1/T, K-1
9