IB Chemistry Higher Level Internal Assessment
Comparing the Activation Energy and Enthalpy of Neutralization to Find Total Bond Strengths of Product in Two Different Neutralization Reactions that Produce Sodium Chloride
Candidate Name: Sadha Satya Lotan Candidate Number: 002115 – 002115 – 0019 0019 School Name: Cita Hati Christian High School Session: May 2016
Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Background
Indonesia is an archipelago country surrounded by the seas and ocean causing a lot of the residents to exploit natural resources from the sea as their occupation to live. Other than fish, we can also exploit table salt from sea through evaporation of sea water. Table salt is substantial for our lives as it can preserve food while adding salty taste, resulting in a delicious food such as salted fish which is very common in my country, Indonesia. But now after I learnt Chemistry, I found out that table salt is primarily composed of NaCl (sodium chloride). And there are many chemical reactions capable of producing NaCl. My curiosity led me to deepen my understanding of the chemical reactions that produce NaCl, especially the minimum energy required for the reaction (activation energy) and the enthalpy change of the reaction. By finding out those two values, I can also find the total bond strength of products formed including NaCl. For every chemical reaction to occur, the molecules need to collide in the right orientation with enough energy to break the bonds between the reactants. The average minimum energy required is called activation energy. After bond breaking, the bonds are re-formed re-f ormed in different arrangements. The bond forming releases energy equivalent to the bond strengths of the formed products. The difference in the energy used for breaking the bonds and energy produced when bonds bonds are formed is called enthalpy of reaction. Fig 1. Enthalpy diagram of exothermic reaction
Being a salt, NaCl can be obtained through neutralization reaction between acid and base. Neutralization is an exothermic reaction that will also produce heat. The enthalpy of the reaction can be found out by measuring the temperature increase in the reacting solutions. According to the Arrhenius Equation 1, activation energy can be found by reacting them in different temperatures. The constant k stated below is the rate constant whose value can vary when there is a change in k is a rate constant that can vary if the experimental condition excluding the reactant concentration is changed, such as temperature.
−
[1]
Adding natural logarithm on [1]:
ln ln ln ln ln ln
[2]
Suppose the experiment was conducted in two different temperatures
[3]
[4]
and
:
1
Lower, S. (2009). Collision and Activation, the Arrhenius Law. Law . General Chemistry Virtual Textbook . Retrieved 31 January 2016, from http://www.chem1.com/acad/webte http://www.chem1.com/acad/webtext/dynamics/dynamics-3.html xt/dynamics/dynamics-3.html
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Subtracting [3] by [4]:
ln lnn ln ln × × ×−
[5]
Using the property of logarithm and factorization on [5]:
[6]
Applying some algebraic manipulation on [6]:
[7]
The equation to determine activation energy is shown in [7]. Based on [7], the value of needed. According to the Rate Law 2, equation to determine the reaction rate is: [8]
[][] … [][][][] ……
Suppose the experiment was conducted in temperatures Dividing [9] by [10]:
and
is
in constant concentration: [9] [10]
[11]
By using this method, the ratio of the rate of reaction needed to be identified where it would be done by using derivative on the curve of concentration of product against time. The curve was obtained by using data logger in the experiment. Although the reaction would be a neutralization reaction involving change in pH, it was not chosen because pH is a logarithmic measure of ion H + or OH - concentration where its value may drastically change when adding another reactant due to increase in volume, not because of the reaction. Another data logger considered to be suitable for this investigation was CO 2 (carbon dioxide) sensor to measure the increase in CO2 level as result of reaction. Therefore, the reactions for the investigation should be the ones that produce CO 2 while still producing NaCl, such as NaHCO 3 (sodium bicarbonate) or Na2CO3 (sodium carbonate) with HCl (hydrochloric acid) in the following chemical r eactions: NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l) Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l) The enthalpy of neutralization, activation energy and bond strength in the produced NaCl would be compared among the two reactions mentioned above. Research Question
In the reaction between NaHCO3 and HCl, and Na 2CO3 and HCl, how do the values of activation act ivation energy, enthalpy of neutralization and the total bond strength of products compare?
2
Chem.tamu.edu,. Rate Chem.tamu.edu,. Rate Laws. Laws. Retrieved 31 January 2016, from https://www.chem.tamu.edu/class/majors/tutorialnotefiles/ratecalcs.htm
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Hypothesis
The activation energy and enthalpy of neutralization in one reaction may differ from another, while the total bond strength in reaction using Na 2CO3 is higher than in NaHCO 3 because it has an extra molecule of NaCl as product when using Na 2CO3. Variables Independent Variable: o o
Activation energy: alkaline used, temperature of the reacting solution (15 oC, 20oC, 25oC) Enthalpy change: alkaline used
Dependent Variable: o o
Activation energy: initial rate of reaction – reaction – derivative derivative of the curve CO 2 level against time Enthalpy change: increase in temperature of the reacti ng solution Table 1. Controlled variables in this investigation
Controlled How? Variables Volume and Activation energy: concentration Used 50 ml of: NaHCO 3 0.2 M, of the reactants Na2CO3 0.1 M and HCl 0.2 M. Enthalpy of neutralization: Used 10.0 ml of NaHCO 3 1 M, Na2CO3 0.5 M and HCl 1 M. Environmental Activation energy and temperature enthalpy of neutralization: o 25 C, controlled by conducting the experiment in the laboratory with air conditioner set at 25 oC.
Initial temperature of solution (only for enthalpy of neutralization) Container of reacting mixture
Absence of catalyst and no shaking
Enthalpy of neutralization:
Controlled to be the same as environmental temperature, which is around 25-27 oC.
Why? Activation energy:
So that change in temperature would be the only factor to affect the initial rate of reaction. Enthalpy of neutralization: To allow fair increase in the solution temperature. Activation energy: Different environmental temperature may give different impacts in maintaining the temperature of reacting solution. Enthalpy of neutralization: To allow same amount of heat loss which might occur. Enthalpy of neutralization: To allow fair rate of heat transfer produced by reaction to the solution.
Activation energy:
Activation energy:
Used conical flask 250 ml for each trial. Enthalpy of neutralization: Used same size container made of Styrofoam.
To allow constant increase in the amount of CO2 produced because CO 2 sensor measures the CO2 concentration. Enthalpy of neutralization: To minimalize the heat loss and fair amount of heat loss absorbed by the container. Activation energy: So the rate of reaction will not be interfered. Enthalpy of neutralization: So that the rate of reaction will not be interfered causing the amount and rate of heat loss that may occur to be fair.
Activation energy and
enthalpy of neutralization: Unnecessary substance is not added and shaking is not done.
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Methodology Table 2. Apparatuses used in this investigation
No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.
Apparatuses Beaker 100 ml Conical flask 250 ml CO2 sensor sensor (ppm, (ppm, ±10 ) Digital balance (±0.01 g) Measuring cylinder 10.0 (±0.1) ml Measuring cylinder 100 (±0.5) ml Stirring rod Styrofoam cup Temperature sensor (oC, ±0.001) Thermometer Volumetric flask 250 ml
Quantity 34 30 1 1 3 3 6 16 1 2 6
Table 3. Materials used in this investigation
No. 1. 2. 3. 4. 5.
Materials Distilled water HCl 0.2 M HCl 1 M NaHCO3 Na2CO3
Quantity 2,000 ml 1,500 ml 250 ml 33.60 g 21.20 g
Procedures: 1. Preparing the required solution: 1.1. 250 ml of the alkaline solutions were prepared by dissolving the solid (for the mass, refer to Table 2) using 40-60 ml of distilled water in a 100 ml beaker, and then stirred with stirring rod and poured into a 250 ml volumetric flask. The beaker was filled with distilled water again and poured to the same volumetric flask until the 250 ml line is reached. Then the solution was shaken. Table 4. Mass of base solids
No 1. 2. 3. 4.
Alkaline NaHCO3 0.2 M NaHCO3 1 M Na2CO3 0.1 M Na2CO3 0.5 M
Mass (g) ±0.01 4.20 21.00 2.65 13.25
gg MM × LL×× 0.2 M × 1,250000 L × 84.007007 4.2003 00355 g ≈ 4.20 g
o
The value of the required mass noted in Table 4 was found through the following calculation:
o
For example, the mass required for NaHCO 3 of 0.2 M was: (Mr = 84.007)
o
The uncertainty in the solution concentration was:
o
The uncertainty of 0.1 M copper (II) sulfate solution was:
o
( )× (0.4.0210 0.25015) × 0.2 0.0.0006006
The concentration uncertainty of every solution was very small and therefore would be negligible.
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
2. Finding the activation energy: 2.1. 50 ml of NaHCO3 0.2 M was poured using measuring cylinder 100 ml into the conical flask 250 ml. 2.2. 50 ml of HCl 0.2 M was measured in another 100 ml measuring cylinder and poured into a beaker 100 ml. 2.3. The conical flask and beaker in previous steps were soaked in a basin containing cold water, and then the temperature was measured using thermometer. 2.4. Once the temperature reached 15oC, the conical flask and beaker were taken out. 2.5. The initial CO2 level in the flask was recorded by placing the CO 2 sensor on the opening of the flask and the data collection was started. 2.6. The CO2 sensor was removed, and then the prepared 50 ml of HCl 1 M was poured into the flask and the CO 2 sensor was immediately returned to the opening of the flask. 2.7. The data collection of the CO 2 level was run for 10 minutes. 2.8. Steps 2.1-2.7 were repeated until 5 trials, and then repeated at temperature 20 oC. It was then repeated again at temperature 25oC, but the conical flask and beaker did not need to be soaked in cold water wat er because the solution temperature was assumed to be the t he same s ame as room temperature. The whole process was then repeated again using Na 2CO3 0.1 M instead of NaHCO3 0.2 M. 3. Finding the enthalpy of neutralization: 3.1. 10.0 ml of NaHCO 3 1 M was measured using measuring cylinder 10.0 ml and poured into the Styrofoam. 3.2. 10.0 ml of HCl 1 M was prepared in another 10.0 ml measuring cylinder. 3.3. The initial solution temperature in the Styrofoam was recorded by placing the temperature sensor on the opening of the Styrofoam cover and the data collection was started. 3.4. The temperature sensor and the cover were removed, and the prepared 10.0 ml HCl 1 M was poured into the Styrofoam and the temperature sensor and cover were immediately returned to its original position. 3.5. The data collection of the change in temperature was run until the temperature reached the highest value and 30 seconds after that. 3.6. Steps 3.1-3.5 were repeated until 8 trials using NaHCO 3 1 M. The whole process was then repeated by using Na 2CO3 0.5 M instead of NaHCO 3 1 M.
Safety Precaution and Disposal: HCl is a strong acid which may irritate skin and eyes when contact occurs. Glove and safety googles are recommended to be used when handling HCl. After one trial ends, the conical flask containing the solution needed to be shaken to allow remaining reactants to react producing harmless NaCl. After that, the remaining solution in the conical flask was disposed to wash sink.
Data Collection Observation: When mixing the reactants, fizzy was observed. It indicated CO 2 gas was being produced. 5
Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Finding Activation Energy
Activation energy was obtained through calculation involving initial rate of reaction. Initial rate was obtained through derivative of graph CO 2 level against time. Below is example of the obtained result in graphical display using Data Logger.
Fig 2. Example of the obtained result to find activation energy in graphical graphical display – NaHCO NaHCO3 0.2 M at 15 oC trial 1
All of the data was copied and pasted into software Microsoft Excel 2010 (later will be shortened as Excel). The variable time was adjusted so that was when there had been the first increase in CO2 level. It was done by subtracting it with corresponding value. On the other hand, suppose the graph after time was adjusted had function . The shape of was similar to the shape of decreasing exponent function reflected to , however I could not get the function with equation I mentioned before by using Excel. Therefore, I manipulated the data by reflecting it to first which was done by multiplying with becoming . However, Excel does not provide exponential function where its y value is negative. Therefore, I manipulated the function again by translating it in y-axis direction as big as becoming . I only use data when adjusted time (26 data) to be plotted on the graph because I think 26 data are already enough to obtain the required function, so that initial rate of reaction can be determined while being fair to each of the trial because each trial may have different number of total data. The value of would be 13,000 for NaHCO 3 and 5,400 for Na 2CO3. They are different because it was observed that the CO 2 level produced after 10 minutes at the highest temperature tested (25oC) was also different where it was around 13,000 for NaHCO 3 and 5,400 for Na 2CO3.
0 1 0,4,8,…,100 o
Table 5. Example of adjusting the time – NaHCO NaHCO3 0.2 M at 15 C trial 1
Time (s) 0 4 8 … 64 68 72
CO2 level (ppm) 2.479553223 2.479553223 2.479553223 … 2.479553223 2.479553223 2.670288086
Adjusted time (s) 0
13,000 12997.32971 12 997.32971
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
76 80 … 168 172
148.2009888 316.2384033 … 3617.095947 3722.381592
4 8 … 96 100
12851.79901 12 851.79901 12683.7616 … 9382.904053 93 82.904053 9277.618408
72
72 1313,,000 13,000 ′
In Table 5, the first increase was detected at . Therefore, from time is adjusted by subtracting it by 72. The adjusted time and the function were plotted in graph on Excel. An exponential function equivalent to referred as can be determined using Excel.
o
Graph 1. Example of the plotted data – NaHCO NaHCO3 0.2 M at 15 C trial 1
14000
, or later
The first derivative of found through equation know that:
12000 0 0 0 , 3 1 + ) t ( f -
onwards, the
10000 8000
y = 13,025e -0.00346x R² = 0.99944
6000 4000 2000 0 0
50
could be . We
100
150
[12]
[13]
[14]
Adjusted time (s)
We had known that the function was a decreasing exponential function: [15]
×−×× − × − 0 ×
[16]
Through [14] we know that:
Since initial rate of reaction was to be found, therefore
[17]
:
For example, in NaHCO 3 0.2 M at 15 oC trial 1, initial rate was
[18] [19]
1313,,02525×× 0.00346 00346 45.1
ppm/s.
The complete data of initial rate of reaction will be shown in Appendix. The table below is showing only for NaHCO 3 0.2 M at 15 oC as the example of data.
13,13,012587−.−. 13,12,198294−.−. 12,864−.
0
o
Table 6. Example of raw data – initial initial rate of reaction in determining activation energy for NaHCO 3 0.2 M at 15 C
Sample NaHCO3 0.2 M
Temperature (oC) ±0.5
15
Trial 1 2 3 4 5
Curve Equation (
)
Rate at
(ppm/s)
45.1 53.8 52.9 64.7 48.1
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
15℃ 15273 15273K 288 K
The temperature unit will be converted into Kelvin by adding 273 on the value of temperature in Celcius, for example: The average of the initial rate of reaction r eaction will be calculated and shown in Table T able 7:
0
Table 7. Average initial rate in determining activation energy
Sample NaHCO3 0.2 M
Na2CO3 0.1 M
Temperature (K) ±0.5 288 293 298 288 293 298
Average Rate at (ppm/s) 52.9 73.4 133.0 15.8 22.9 28.5
∑ 1 ̅ .+.+.+.+. 52.9 ppm/s ln × × −×
The average initial rate is found through the formula: For example, in NaHCO 3 0.2 M at 288 K,
The activation energy is then found out using the formula
, where
is obtained using the three combinations possible from the three temperatures, 288 K & 293
K, 288 K & 298 K and 293 K & 298 K. The uncertainty in temperature will be negligible as the value is very small.
8888 ×29 × 2933 45,933 J/mol ≈ 45.9 kJ/mol ln 52.73.94 ×8.31× 2288293
Example of calculation (NaHCO3 0.2 M, 288 K & 293 K):
Table 8. Activation energy of the chemical reactions
Sample NaHCO3 0.2 M
Na2CO3 0.1 M
Combination of Temperatures (K) 288 & 293 288 & 298 293 & 298 288 & 293 288 & 298 293 & 298
Activation Energy (kJ/mol) 45.9 65.8 86.3 52.0 42.1 31.7
Average Activation Energy (kJ/mol)
There are only 3 data in activation energy, so the u ncertainty of its average: In NaHCO3 0.2 M, the uncertainty of average activation energy is
66.0 (±20.2)
42.0 (±10.2)
−
.−. 20.2 kJ/mol
Finding Enthalpy of Neutralization
The complete data of increase in temperature will be shown in Appendix. The table below is showing only for NaHCO NaHCO 3 1 M as the example of data.
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level Table 9. Example of raw data – initial initial and final temperature of solution to determine enthalpy of neutralization
Sample
NaHCO3 1 M
Trial 1 2 3 4 5 6 7 8
Initial Temperature (oC) ±0.001 25.680 25.751 25.651 26.083 25.857 25.850 25.852 25.774
Final Temperature (oC) ±0.001 26.131 26.200 26.119 26.339 26.225 26.206 26.234 26.108
The enthalpy of neutralization is found through the following formula:
× × × × ∆ ∆H × ×
Here, the mass used in the calculation will be the mass of water. Volume of solution is 20.0 (±0.2) ml, and by assuming the density of water is 1 g/ml, therefore the mass will be 20.0 (±0.2) g. The specific heat capacity used is also water’s specific heat capacity which is 4.18 J/goC. For example in NaHCO3 1 M at trial 1:
J/10.g℃×0× 26. 26.131 131 25.25.680℃ 80℃ 3,770.36 ≈ 3.77 kJ/mol ∆H 20.0 g×g × 4.1818 J/g℃ 1,000 L×1M
After finding the enthalpy of neutralization from each trial, the values will be averaged using similar formula as before. The results are shown in Table 10. Since 8 trials were conducted in finding enthalpy of neutralization; therefore its uncertainty will be determined using standard deviation because there is more data compared to in finding average activation energy. Table 10. Average enthalpy of neutralization and its standard deviation
Sample NaHCO3 1 M Na2CO3 0.5 M
Average Enthalpy of Neutralization (kJ/mol) 3.20 11.8
Standard Deviation (kJ/mol) 0.56 1.8
The calculation for the standard deviation is based on the following formula:
∑ = √
For example, the standard deviation of enthalpy of neutralization in NaHCO3 1 M is:
√ 3.203.77 3.203.75 3.203.91 3.202.14 8 3.203.08 3.202.98 3.203.19 3.202.79
0.5574594 kJ/mol ≈ 0.56 kJ/mol
Calculating Total Bond Strength of Product
Based on enthalpy diagram in Fig 1, the total bond strength in the formed products is equivalent to: . For example, in the reaction NaHCO 3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H 2O(l), total bond strength: . The uncertainty is obtained by adding the absolute uncertainty of both and : .
∆H 66.0 3.20 69.2 kJ/mol / m ol ∆H 20.2 0.56 20.76 kJ/mol ≈ 20.8 kJkJ/molmol
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Calculating Theoretical Value
The theoretical value of the total bond strength of product will be calculated to be compared with the experimental value. Table 11. Average bond and lattice enthalpy at 298 K 3
Bond NaCl O−H C O
Enthalpy (kJ/mol) 790 463 804
=
In the reaction NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l), total bond strength is of product is: kJ/mol
790 2 2 × 804 2 × 463 3,3,324 2 × 790 2 × 804 2 × 463 4,114
In the reaction Na 2CO3(aq) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l), total bond strength of product is: kJ/mol
The theoretical value is much greater than the experimental value. In using NaHCO 3, there is a difference:
,,−. 97.9%. In Na CO , the difference is: ,−. 98.7%. ×100% ,114114 ×100% 2
3
The theoretical value is not included in the chart for comparison due to the large differences. Chart 1. Comparison of activation energy, enthalpy of neutralization and bond strength in products in the reaction between alkaline and HCl 100 90 80
) l 70 o m / 60 J k ( 50 y g r 40 e n E 30
69.2 66
53.8
Enthalpy of Neutralization
42
20 10
Activation Energy
Bond Strength in Products
11.8 3.2
0 Sodium Bicarbonate
Sodium Carbonate
Discussion and Review
In finding the activation energy, a solution of concentration less than in finding enthalpy of neutralization was used to allow greater volume of solution, maintaining the temperature of the reacting solution because water is thermally stable due to the high specific heat capacity. Meanwhile, in finding enthalpy of neutralization, a more concentrated solution was used to allow equal amount of reaction (equal amount of heat released) happening despite smaller volume used, causing the temperature to rise greater to reduce error. A Styrofoam was used because it could act as insulator, minimalizing heat loss to the t he container or to the surrounding environment. In finding the activation energy, higher temperature caused the reaction rate to be greater. This is because temperature provides greater kinetic energy for the molecule to move and break the bonds, allowing more frequent successful collision compared to in lower temperature. As time 3
Aylward, G., & Findlay, T. (2008). SI Chemical Data (5th ed.). Queensland: John Wiley and Sons. Haynes, W. (2012). CRC handbook of chemistry and physics. physics . Boca Raton, Fla.: CRC.
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
passed, the rate in CO2 increase got reduced. This was because less reactant was present in the solution, lowering its concentration; less frequent collisions happened and rate of reaction decreased as the result. But even after ten minutes of mixing the reactants, the CO 2 level still increased. This could be caused by NaHCO 3 and Na2CO3 being weak alkaline that does not fully dissociate into OH- in water. After OH - ions reacted with HCl, the unreacted alkaline would dissociate to replace the lost OH- ions and would react with HCl again, so reaction time is longer. The error bar in Chart 1 was big for both enthalpy of neutralization and activation energy, indicating the results were imprecise. The experimental value was also inaccurate since it differed so much from the theoretical value, at about 98% difference. It was because the used theoretical value was measured when the reactants are in gaseous forms while the reactant in this experiment was aqueous. More of the errors will be discussed in Evaluation and Improvement . The mechanism of the reaction between Na 2CO3 and HCl is as following 4: Na2CO3(aq) + HCl (aq) NaCl(aq) + NaHCO3(aq) NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)
(1) (2)
It involved another reaction which is also being investigated. Putting that in mind, several phenomena could be explained. The rate of reaction when using NaHCO 3 was greater than when using Na2CO3. The rate of reaction was measured through the increase in CO 2 level. In the reaction using Na2CO3, it needed to undergo one extra step before producing CO 2 which did not happen in NaHCO 3, delaying CO 2 production. Furthermore, the concentration of Na 2CO3 was half of NaHCO3. This was done because the coefficient of HCl in reacting with Na 2CO3 is twice of when reacting with NaHCO 3, whereas the HCl concentration was tried to be kept constant, therefore the concentration of Na 2CO3 was half of NaHCO 3 to still allow the reaction completion. In the reaction using Na 2CO3, the enthalpy of neutralization was greater than using NaHCO3. This was because of the reaction mechanism which involves two steps, where both steps are neutralization which is exothermic and the second step is the same as the other investigated reaction. Based on Hess Law, the difference in their enthalpy of neutralization will correspond to the enthalpy of neutralization of first s tep in the reaction between Na 2CO3 and HCl. The activation energy in using NaHCO 3 was higher than in using Na 2CO3 which could be caused by the two steps mechanism of the reaction between Na 2CO3 with HCl. The first step which is exothermic will supply heat for the endothermic process or bonds breaking in the second step. Furthermore, the activation energy to initiate first step may also be not very high. These explained why Na 2CO3 has lower activation energy compared to NaHCO 3. The total bond strength in the reaction using NaHCO 3 was found to be higher than in Na2CO3 which was different from the hypothesis. The difference could be caused by random uncertainty since it still fell in the error bar. However, some other factors might cause this difference, such as the errors in finding activation energy or the reaction rate where human bias has huge impact in finding reaction rate. This can also be caused by experimental error in finding enthalpy of neutralization. There may be a lot of heat loss to the surrounding before the completion of reaction since the reaction proceeds ver y slowly because weak alkaline was used. 4
Clark, J. (2013). Acid-Base (2013). Acid-Base Indicators. Indicators. Chemguide.co.uk . Retrieved 24 January 2016, from http://www.chemguide.co.uk/physical/acidbaseeqia/indicators.html
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Conclusion
From the two investigated reactions that produce both NaCl and carbon dioxide, the research question, “In the reaction between Na 2CO3 and HCl, and NaHCO 3 and HCl, how do the values of activation energy, enthalpy of neutralization and the total bond strength of products compare?” can now be answered. answered. The activation energy when using NaHCO 3 is much higher than when using Na 2CO3. The enthalpy of neutralization when using NaHCO 3 is much lower than using Na 2CO3. The total bond strength of products formed using NaHCO 3 is higher than in using Na2CO3. These are possibly caused by the reaction mechanism where two steps were involved in Na 2CO3 which also produced NaHCO 3 in the first step as an intermediate product. Evaluation and Improvement Improvement
I was proud in conducting this investigation. I managed to apply my mathematical skills such as function transformation and derivative to help me in obtaining the required data. I could also make the required acid and alkaline solution on my own. I also managed to relate Arrhenius Equation with Rate Law. However, some experimental errors occurred and will be discussed. The temperature in finding activation energy could slightly change during the reaction. This may be caused by heat transfer from surrounding to the reacting solution or the exothermic reaction which releases heat and increased the solution temperature. This could be solved by conducting the experiment in an insulator where CO 2 sensor can fit in using it while preventing CO2 gas from escaping. Some CO2 gas may escape when adding HCl but before covering the flask with the sensor. A flask with longer neck or with two necks for CO 2 sensor and adding HCl could be used to reduce this error. Although Styrofoam as insulator was used in determining the enthalpy of neutralization, some heat loss to surrounding or absorbed by insulator could occur, causing the experimental value to be lower than the actual value. This could be solved by using a more concentrated solution and greater amount of solution. Furthermore, the reacting solution could be stirred to increase the rate of reaction so that heat loss can be reduced. Plot the graph of temperature against time and use extrapolation of graph to take ac count the heat loss. Inaccurate CO2 level because there is delay in detecting the increase of CO 2 level, which may cause the measured reaction rate to have different value from its actual value. The rate could be measured by measuring the volume of of CO 2 gas or the mass loss due to t o escaping CO2 gas. Determining bond strength of formed products in this case may not be accurate enough as not all of them are in gaseous forms. Using Born-Haber cycle to determine the lattice enthalpy based on the enthalpy of sublimation, atomization, electron affinity and ionization energy. The value of rate constant
was not found. Only the ratio
was used. Finding and
determining the order of the reaction is suggested to get the value of rate constant which will be used to determine the activation act ivation energy. This can be done by conducting reactions in different reactant concentration and compare the initial initia l rate of reaction. Furthermore, by plotting the graph of
ln
against , the activation energy can be found by finding its gradient. 12
Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
References
Aylward, G., & Findlay, T. (2008). SI Chemical Data (5th ed.). Queensland: John Wiley Wile y and Sons. Chem.tamu.edu,. Rate Chem.tamu.edu,. Rate Laws. Laws. Retrieved 31 January 2016, from https://www.chem.tamu.edu/class/majors/tutorialnotefiles/ratecalcs.htm Clark, J. (2013). Acid-Base (2013). Acid-Base Indicators. Indicators. Chemguide.co.uk . Retrieved 24 January 2016, from http://www.chemguide.co.uk/physical/acidbaseeqia/indicators.html Haynes, W. (2012). CRC handbook of chemistry and physics . Boca Raton, Fla.: CRC. James,. (2015). Enthalpy (2015). Enthalpy Diagram of Exothermic Exothermic Reaction. Reaction . Retrieved from http://sciencealevelhelp.co.uk/?p=775 Lower, S. (2009). Collision and Activation, the Arrhenius Law. Law . General Chemistry Virtual Textbook . Retrieved 31 January 2016, from http://www.chem1.com/acad/webtext/dynamics/dynamics-3.html
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Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
Appendix
Here are some documentations of the experiment.
Image 1. Cooling HCl and the alkaline
Image 2. Measuring CO2 level
Image 3. Measuring temperature increase
Below is the complete raw data of the initial rate of reaction to determine the activation energy. Table 12. Raw Data – Initial Initial rate of reaction to determine the activation energy
Sample
Temperature (oC) ±0.5
15
NaHCO3 0.2 M
20
25
15 Na2CO3 0.1 M 20
Trial 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3
−.
Curve Equation (
13,13,012587−. 13,12,198294−.−. 12,12,896456−.−. 13,13,015742−.−. 13,12,083472−.−. 13,13,146304−.−. 12,13,920685−.−. 13,5,320433−.−. 5,5,331823−.−. 5,5,330935−.−. 5,5,237713−.−. 5,300−.
)
Rate at
0
(ppm/s)
45.1 53.8 52.9 64.7 48.1 80.8 85.0 76.1 58.7 66.3 142.2 137.0 122.0 131.8 132.1 17.5 15.9 14.0 15.3 16.4 24.6 21.3 25.2 14
Sadha Satya Lotan/002115-0019 Lotan/002115-0019 Internal Assessment Assessment – – Chemistry Chemistry Higher Level
25
4 5 1 2 3 4 5
5,5,334744−.−. 5,5,342046−.−. 5,5,434985−.−. 5,320−.
20.3 23.0 31.5 26.8 26.2 30.0 28.2
Below is the raw data of initial and final temperature of solution to determine the enthalpy of neutralization. Table 8. Raw Data – Initial Initial and final temperature of solution to determine enthalpy of neutralization
Sample
NaHCO3 1 M
Na2CO3 0.5 M
Trial 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8
Initial Temperature (oC) ±0.001 25.680 25.751 25.651 26.083 25.857 25.850 25.852 25.774 26.271 25.760 25.719 25.679 25.899 25.855 25.590 25.779
Final Temperature (oC) ±0.001 26.131 26.200 26.119 26.339 26.225 26.206 26.234 26.108 26.783 26.393 26.481 26.426 26.542 26.592 26.497 26.503
15
