A Comparison Between The Enthalpies Calculated Using Conventional Methods And Computational Methods

Indian Institute of Science Education & Research, Kolkata

Semester IV – April 2009

Chemistry Project Report Title: A comparison between the enthalpies calculated using conventional methods and computational methods

Indian Institute of Science Education & Research, Kolkata Semester IV

April 2009

By: Harsh Purwar (07MS-76) Satyam Singhal (07MS-66) Jyothi V. Nair (07MS-72) 1

Indian Institute of Science Education & Research, Kolkata

Semester IV – April 2009

Introduction: Experiments have always formed the basis of human quest in science. What keeps on changing is the sophistication of the experimental techniques for a still greater precision. One way of carrying out the experiment is the use of conventional methods which involve manual experimentation have limited accuracy since the systematic and random errors can never be fully removed. This is the age of computers wherein much advanced and complex experiments are carried out using the computer programmes and soft-wares. These techniques of experimentation are practically foolproof with negligible error involved. Here in this experiment we wish to establish this fact convincingly by comparing the efficiencies of both the methods for a particular experiment.

Aim of the Experiment: To compare the enthalpy of hydration of an endothermic and an exothermic reaction using the conventional experimental techniques (involving titrations) and computational methods (using Gaussian).

Chemical Reactions: • Exothermic Reaction: ‫ܽܥ‬ሺܱ‫ܪ‬ሻଶ ሺ‫ݏ‬ሻ ⟶ ‫ܽܥ‬ଶା ሺܽ‫ݍ‬ሻ + 2ܱ‫ ିܪ‬ሺܽ‫ݍ‬ሻ • Endothermic Reaction: ܰ‫ܪ‬ସ ܱܰଷ ሺ‫ݏ‬ሻ + ‫ܪ‬ଶ ܱሺܽ‫ݍ‬ሻ ⟶ ܰ‫ܪ‬ସ ܱ‫ܪ‬ሺܽ‫ݍ‬ሻ + ‫ܪ‬ା ሺܽ‫ݍ‬ሻ + ܱܰଷି ሺܽ‫ݍ‬ሻ

Theory: The experiment involves determining the equilibrium concentration of OH- ions in a solution containing excess of Ca(OH)2 by titrating it with about 0.01 M HCl. ‫ܽܥ‬ሺܱ‫ܪ‬ሻଶ ሺ‫ݏ‬ሻ ⟶ ‫ܽܥ‬ଶା ሺܽ‫ݍ‬ሻ + 2ܱ‫ ିܪ‬ሺܽ‫ݍ‬ሻ Moles of hydroxide = moles of HCl = Volume of HCl in litres X [HCl] Moles of Ca(OH)2 = 1/2 moles of hydroxide Molar solubility of Ca(OH)2 = moles of Ca(OH)2 / Volume in litres The solubility product, Ksp is then calculated as 2

Indian Institute of Science Education & Research, Kolkata

Semester IV – April 2009

ଶା

ି ଶ

‫ܭ‬௦௣ = [‫ = ] ܪܱ[] ܽܥ‬4‫ݏ‬

ଷ

∆‫்ܩ‬଴ = −ܴ݈ܶ݊‫ܭ‬௦௣ ∆‫்ܩ‬଴ = ∆‫ܪ‬଴ − ܶ∆ܵ଴ Similar theory is applicable to the other reaction as well.

Experiment: Part I 1. Standard oxalic acid solution was prepared by weighing. 2. Sodium hydroxide (NaOH) solution was prepared and was standardized using oxalic acid solution. 3. HCl solution was prepared and was standardized with standardized NaOH solution. 4. Calcium hydroxide solution was then drawn off from a beaker containing excess of solid Ca(OH)2 that was being stirred for 24 hours. 5. Temperature of the sample was recorded using a thermometer. 6. Sample was then filtered and diluted accordingly and was titrated with standardized HCl using appropriate indicator. 7. Another Ca(OH)2 solution was prepared at a higher temperature (100° C) and steps 6 & 7 were repeated.

Part II Similar procedure was followed for ammonium nitrate involving titration with HCl. The only difference was that the second sample was made at a lower temperature (0° C instead of 100° C) because this reaction being an endothermic reaction. That is the solubility of NH4NO3 increases with the rise in temperature, and so if the sample would have been prepared at a higher temperature then during the reaction due to cooling of the sample NH4NO3 would have been precipitated.

Part III Next the software Guassian is used to calculate the ∆‫ܪ‬ଶଽ଼ values for both the above reactions using the standard protocol involving: • ‫ܧ‬௘଴ is calculated by single point energy calculation using LSDA / B3LYP method and 6311+g(2df,2p) basis set for various entities in the reactions. • Geometry optimization + frequency calculations were done using LSDA / B3LYP method and 6-31+g(d) basis set.

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Indian Institute of Science Education & Research, Kolkata

Semester IV – April 2009

Results: Part I The temperatures of the two samples of Ca(OH)2 are 33° C (R.T.) and 100° C. Table # 1: Preparation of Oxalic acid solution. Amount of Oxalic acid taken (g) 0.4027

Volume of the solution (mL) 100

Strength of the solution (M) 0.045

Table # 2: Standardization of NaOH solution. S. No.

1 2

Volume of Oxalic acid solution (mL) 10 10

Burette Reading Initial Final 0 0

16.8 16.7

Volume of NaOH solution (mL) 16.75

Strength of NaOH solution (M) 0.054

Table # 3: Standardization of HCl solution. S. No.

1 2

Volume of HCl solution (mL) 10 10

Burette Reading Initial Final 0 7.0

6.8 14.0

Volume of NaOH solution (mL) 6.9

Strength of HCl solution (M) 0.037

Table # 4: Titration of Ca(OH)2 solution prepared at room temperature. S. No.

1 2

Volume of Ca(OH)2 solution (mL) 10 10

Burette Reading Initial Final 0 11

10.8 21.9

Volume of HCl solution (mL) 10.85

Molar solubility of Ca(OH)2 (s) in the solution prepared at 33° C is given by, ‫=ݏ‬

[‫݈݋ܸ × ]ܮܥܪ‬. ‫ ݈ܥܪ ݂݋‬0.037 × 10.85 = = 0.020 ‫ܯ‬ 2 × ܸ‫݈݋‬. ‫݊݋݅ݐݑ݈݋ݏ ݂݋‬ 2 × 10 4

Indian Institute of Science Education & Research, Kolkata

Semester IV – April 2009

Table # 5: Titration of Ca(OH)2 solution prepared at 100° C. S. No.

Volume of Ca(OH)2 solution (mL) 10 10

1 2

Burette Reading Initial Final 0 6.0

Volume of HCl solution (mL)

6.0 12.1

6.05

Solubility of Ca(OH)2 in solution (M) 0.011

Molar solubility of Ca(OH)2 (s) in the solution prepared at 100° C is given by, ‫=ݏ‬

[‫݈݋ܸ × ]ܮܥܪ‬. ‫ ݈ܥܪ ݂݋‬0.037 × 6.05 = = 0.011 ‫ܯ‬ 2 × ܸ‫݈݋‬. ‫݊݋݅ݐݑ݈݋ݏ ݂݋‬ 2 × 10

Part II The temperatures of the two samples of NH4NO3 are 33° C (R.T.) and 4° C. Table # 6: Titration of NH4NO3 solution prepared at room temperature. S. No.

1 2

Volume of NH4NO3 solution (mL) 10 10

Burette Reading Initial Final 0 2.5

Volume of HCl solution (mL)

1.9 4.4

1.9

Strength of NH4NO3 solution (M) 0.00703

Table # 7: Titration of NH4NO3 solution prepared at 4° C. S. No.

1 2

Volume of NH4NO3 solution (mL) 10 10

Burette Reading Initial Final 0 0

1.8 1.8

Volume of HCl solution (mL) 1.8

Strength of NH4NO3 solution (M) 0.00666

Part III Table # 8: Computational Data. S. No. 1

Entity

Ca(OH)2

Electronic energies at 0° K ሺ‫ܧ‬௘଴ ሻ (Kcal/mole) -829.3791467 5

Thermal energy corrections (Kcal/mole) 0.031191

Energies including the corrections -829.3479557

Indian Institute of Science Education & Research, Kolkata

2 3 4 5 6

2+

Ca OHNH4NO3 NH4+ NO3-

Semester IV – April 2009

-676.9057451 -75.82967035 *** -56.92138592 -280.4598192

0.002360 0.011636 *** 0.053741 0.018000

-676.9033851 -75.81803435 *** -25.57884592 -271.7486892

Calculations: Part I Molar solubility for Ca(OH)2 at 306° K is: sଷ଴଺ = 0.020 M So, we have ଷ଴଺ ‫ܭ‬௦௣ = 4‫ ݏ‬ଷ = 3.2 × 10ିହ ଷ଻ଷ ‫ܭ‬௦௣ = 5.324 × 10ି଺

Since, we know that ∆‫ ܩ‬଴ = −ܴ݈ܶ݊‫ܭ‬௦௣ ଴ ∆‫ܩ‬ଷ଴଺ = −8.314 × 306.15 × ݈݊ሺ3.2 × 10ିହ ሻ

= −2545.3311 × ሺ−10.35ሻ = 2.634 × 10ସ ‫ܬ‬/݉‫݈݋‬ Similarly, ଴ ∆‫ܩ‬ଷ଻ଷ = 3.767 × 10ସ ‫ܬ‬/݉‫݈݋‬

So now we have two simultaneous equations in ∆‫ܪ‬଴ and ∆ܵ ଴ as, ଴ ଴ ଴ ∆‫ܩ‬ଷ଴଺ = ∆‫ܪ‬ଷ଴଺ − ܶ∆ܵଷ଴଺ ⋯ ⋯ ⋯ ሺ1ሻ ଴ ଴ ଴ ∆‫ܩ‬ଷ଻ଷ = ∆‫ܪ‬ଷ଻ଷ − ܶ∆ܵଷ଻ଷ ⋯ ⋯ ⋯ ሺ2ሻ

Here we neglect the difference between the enthalpy at two different temperatures i.e. ଴ ଴ ∆‫ܪ‬ଷ଻ଷ and ∆‫ܪ‬ଷ଴଺ and also between the entropies at the two temperatures as the differences are small and insignificant. Solving the two equations, 2.634 × 10ସ = ∆‫ܪ‬଴ − 306.15 × ∆ܵ ଴ 6

Indian Institute of Science Education & Research, Kolkata ସ

Semester IV – April 2009 ଴

3.767 × 10 = ∆‫ ܪ‬− 373.15 × ∆ܵ ଴ we get, standard change in enthalpy for the hydration of Ca(OH)2, ∆‫ܪ‬଴ = −14.272 ‫ܬܭ‬/݉‫݈݋‬

Part II Molar solubility of NH4NO3 at 306° K, ‫ = ݏ‬0.00703 ‫ܯ‬ So, we have ଷ଴଺ ‫ܭ‬௦௣ = 4‫ ݏ‬ଷ = 1.389 × 10ି଺ ଶ଻ଷ ‫ܭ‬௦௣ = 1.182 × 10ି଺

Since, we know that ∆‫ ܩ‬଴ = −ܴ݈ܶ݊‫ܭ‬௦௣ ଴ = −8.314 × 306.15 × ݈݊ሺ1.389 × 10ି଺ ሻ ∆‫ܩ‬ଷ଴଺

= −2545.3311 × ሺ−13.487ሻ = 3.433 × 10ସ ‫ܬ‬/݉‫݈݋‬ Similarly, ଴ ∆‫ܩ‬ଶ଻ଷ = 3.099 × 10ସ ‫ܬ‬/݉‫݈݋‬

The two equations involving enthalpy and entropy, 3.433 × 10ସ = ∆‫ܪ‬଴ − 306.15 × ∆ܵ ଴ 3.099 × 10ସ = ∆‫ܪ‬଴ − 273.15 × ∆ܵ ଴ we get, standard change in enthalpy for the hydration of NH4NO3, ∆‫ܪ‬଴ = 65.3 ‫ܬܭ‬/݉‫݈݋‬

Part III Standard change in enthalpy for the hydration of Ca(OH)2 can be calculated using the data obtained from Gaussian by frequency and single point energy calculations listed in table # 8 as, ଴ ∆‫ܪ‬ଶଽ଼ = ∆‫ܧ‬௘଴ + ∆ሺcorrection termsሻ

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Indian Institute of Science Education & Research, Kolkata

Semester IV – April 2009

Here ∆ on the right side signifies difference between the products and reactants. The governing chemical equation here written as, ‫ܽܥ‬ሺܱ‫ܪ‬ሻଶ ሺ‫ݏ‬ሻ ⟶ ‫ܽܥ‬ଶା ሺܽ‫ݍ‬ሻ + 2ܱ‫ ିܪ‬ሺܽ‫ݍ‬ሻ So the enthalpy can be written as, ଴ ∆‫ܪ‬ଶଽ଼ = 0.808501888 ‫݁݁ݎݐݎܽܪ‬/ܲܽ‫݈݁ܿ݅ݐݎ‬

= 2.123 × 10ଷ ‫ܬܭ‬/݉‫݈݋‬

Conclusion: The enthalpy of hydration of Ca(OH)2 calculated via experimentation is found to be -14.272KJ/mol. The enthalpy of hydration of NH4NO3 calculated via experimentation is found to be 65.3 KJ/mol. The enthalpy of hydration of Ca(OH)2 calculated via computational method is found to be 2.123 × 10ଷ KJ/mol.

Discussion: Seeing the results obtained we can firmly conclude that computational methods are much superior compared to conventional methods when it comes to experimentation. But still the conventional methods continue to hold their importance because they give better visualization and establish theoretical principles more convincingly. Moreover they may not be accurate in absolute terms but on a relative scale they are still pretty helpful and of great use. So the need of the hour is to take the middle path and have a balanced approach.

References: Our very own acquired knowledge along with the big creativity bank we have with us. • World Wide Web (Google search) • Wikipedia: http://en.wikipedia.org/wiki •

Acknowledgement: 8

Indian Institute of Science Education & Research, Kolkata

Semester IV – April 2009

• Prof. Sanjib Bagchi, HOD Chemistry • Dr. Srikanth • Mr. Saroj Naik, Lab Assistant

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