IA - Activation Energy

Calculating activation energy of the bromidebromate reaction Raw data Temperatu Time Observati re (±1s) ons (±0.5°C) 25.0 920 Solution 35.0 370 went from 45.0 140 clear red 55.0 59 color to a 65.0 41 cloudy pink 75.0 20 Table 1: Temperature and time taken for reaction to go to completion

Analysis The experiment can be represented by the following reaction mechanism: 1. H2SO4 -> H+ + SO422. BrO3- + 5Br- + 6H+ -> 3Br2 + 3H2O 3. C6H5OH + Br2 -> C6H2Br3OH + HBr Sulphuric acid dissociates into H+ and SO42- in solution (reaction 1); when added to the bromide/bromate solution, the solution then dissociated into BrO 3- and Br-, which reacted to form bromine and water (reaction 2). Bromine usually acts as a strong oxidizing agent and reacts with methyl red, very quickly bleaching its color. However, in this experiment, the reaction was slowed by the presence of phenol, which reacts more readily with bromine to produce tribromophenol (reaction 3). Essentially, the phenol served to lengthen the time taken for the bromine to react with the indicator. Otherwise, the bromine-bromate reaction, which I am calculating the activation energy for, would complete too quickly to be measured. Bromine is in excess in this reaction, so when all the phenol has reacted, the remaining bromine reacts very quickly with methyl red; the solution turns colorless, indicating the endpoint of the reactions. Finding the rate constant Substance

Concentration (mol dm-3) 0.01 1

Volume (cm3 ± 1) 10 10

C6H5OH Bromide/bromate mixture H2SO4 0.5 5 Table 2: Concentrations and volumes of reagents

The activation energy of a reaction can be determined using the Arrhenius equation:

k¿Ae

−Ea RT

K is the rate constant, A is the pre-exponential factor (representing frequency of particle collisions), Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. A is almost completely independent of temperature, so it is assumed that A is the same for each reaction regardless of temperature. The rate constant can be derived from the rate expression, which in turn is derived from the chemical equation. The chemical equation and rate equation for the reaction between bromide and bromate are as follows: BrO3- + 5Br- + 6H+ -> 3Br2 + 3H2O The order of the bromide-bromate reaction are first order with respect to bromate, first order with respect to bromide, and second order with respect to H + (Capriola, 2015). Thus, the rate expression can be written: Rate of reaction = k [BrO3-] [Br-] [H+]2

−¿ −¿ + ¿¿ H ¿ ¿ Br ¿ ¿ Br O3¿ ¿ Rate of reaction=k ¿ −¿ −¿ +¿ H¿ ¿ ¿ Br ¿ ¿ Br O3¿ ¿ ¿ rate of reaction k= ¿ ¿

rate of reaction (1)(1) ( 0.500 )2

¿ rate of reaction÷ 0.25 ¿ rate of reaction× 4 mol

-3

9

dm s

-1

Rate = 1/time taken, so: Temperatu

Time

Rate of

re (s ± reaction (°C ± 1) 1) (s-1 ± ) 25 919 0.001088 35 369 0.002710 45 144 0.006944 55 59 0.016949 65 41 0.024390 75 20 0.050000 Table 3: Time and rate of reaction at each temperature The rate constant of a reaction is dependent on temperature, so k will be different at every temperature. k is calculated in Table 4 using the aforementioned rate expression. Temperatu re (°C ± 1) 25

Rate of reaction (s-1 ± ) 0.001088

k mol dm-9 s-1 0.001088

× 4=¿

35

0.002710

0.002710

45

0.006944

0.006944

× 4=¿ 0.01084 × 4=¿ 0.02778

55

0.016949

0.016949

65

0.024390

0.024390

× 4=¿ 0.06780 × 4=¿ 0.09756

75

0.050000

0.050000

× 4=¿ 0.20000

-3

0.00435

Table 4: Rate of reaction, temperature, and k Temperatu re (°C ± 1) 25

1/T (temperature) (1/K)

Rate constant (k) (mol-3 dm-9 s-1)

1 ÷ (25 + 273) = 0.00336

0.001088

35

1 ÷ (35 + 273) = 0.00325

0.002710

45

1 ÷ (45 + 273) = 0.00314

55

1 ÷ (55 + 273) = 0.00305

65

1 ÷ (65 + 273) = 0.00296

75

1 ÷ (75 + 273) = 0.00287

× 4=¿

0.00435

-2.69119

× 4=¿

0.09756 0.050000

-3.58344

× 4=¿

0.06780 0.024390

-4.52451

× 4=¿

0.02778 0.016949

-5.43758

× 4=¿

0.01084 0.006944

ln k (mol-3 dm-9 s1 )

-2.32729

× 4=¿

0.20000

-1.60944

Table 5: Calculating 1/T and ln k The Arrhenius equation can be re-arranged to give the following:

ln k =

−Ea +ln A R ×T

R is a constant, and A is assumed to be temperature independent. Thus, graphing ln k against 1/T should give a linear plot.

A graph of the natural log of the rate constant against temperature 0 -1 R² = 0.99

-2 ln k (mol-3 dm9 s-1) -3 -4 -5 -6

0

0

0

0

0

0

1/T (1/K)

The gradient of the line should be

−Ea R , and can be calculated using two

points on the graph, (x1, y1) and (x2, y2):

gradient=

y 2− y 1 x 2−x 1

−Ea y 2− y 1 = R x 2−x 1 ¿

−5.43759434−(−1.60437912 ) 0.003355703−0.002873563

¿−7950.419421744721 Ea =−7950.419421744721×−R ¿ 66067.98539469863 J mol-1

0

¿ 66.1 kJ mol

-1

Discussion The literature value for the activation energy of this reaction was found to be 52.4 kJ mol-1. Thus, the percentage error in the calculated value is:

¿ 52.4−66.1∨ ¿ =26.1 52.4 ¿ ¿ literature value−experimental value∨ =¿ literature value error =¿ An error of 26.1% is quite large, which means it is unlikely to be caused merely by random errors. It is more likely that systematic errors in the experimental methodology resulted in the large percentage error. This will be discussed in the evaluation section. In the analysis, a graph of ln k against 1/T had a linear line of best fit with a negative gradient. The correlation coefficient was calculated to be 0.9911, which is very close to 1, showing a very strong correlation. Mathematically, this trend shows that the rate constant increases with temperature. This is in accordance with kinetic molecular theory, which states that Kelvin temperature of a substance is proportional to the average kinetic energy of particles in the substance. If particles have greater average kinetic energy, a greater proportion of particles will have the energy necessary to initiate the reaction (overcome the activation energy barrier), so a greater proportion of particles will exceed the activation energy, increasing the rate of reaction. This is in line with the negative gradient of the graph of ln k against 1/T.

Evaluation Limitation

Effect

Improvement

Plot of ln k against 1/T was not completely linear due to random errors in recorded data. Assumed that rate of reaction of bromination of phenol is independent of temperature even though it is also part of the rate expression

The calculation of the gradient may not be accurate, leading to an inaccurate Ea value

Use technology to plot a line of best fit, and use its gradient to calculate activation energy.

Bromide/bromate rate of reaction values less accurate, calculation of the gradient and Ea value affected

Conduct experiments at same temperatures, but repeat for varying concentration of reagents and average the results to reduce the significance of any errors Run a trial ahead of time and allow sufficient time for it to react completely. Use color of solution as baseline to determine endpoint for

Difficult to determine when endpoint of reaction

Recorded time taken for each reaction to complete may be shorter or longer than it should have been, affecting the calculated rate of reaction, gradient,

Lids of water baths had to be kept open while monitoring the reaction in the flasks.

and thus Ea value The temperature of the water baths and thus the solutions may have been lower than recorded, making 1/T values bigger than expected. This would make the calculated Ea value bigger than it should have been

other reactions. Insulate the solution flasks to reduce heat loss.

Conclusion The aim of this experiment was to calculate a value for the activation energy of the bromide-bromate reaction. This was successfully done using empirical data to determine the time taken for the reaction to complete at different temperatures and then calculating rate of reaction to determine the rate constant, which was in turn used to calculate the activation energy. Using the Arrhenius equation, the activation energy was determined to be approximately 66.1 kJ mol-1, which was 26.1% away from the literature value of 52.4 kJ mol -1. http://adamcap.com/schoolwork/the-kinetics-of-the-bromate-bromide-reaction/