B. To measure the strength of hydrogen bond formed between ethanol molecules
The same steps in part A were repeated using same volume of ethanol but 20cm3 and 30cm3 of cyclohexane respectively.
C. To investigate the formation of hydrogen bonds between molecules of ethyl ethanoate and trichloromethane
1. 10cm3 of ethyl ethanoate was measured using a measuring cylinder and poured into an insulated 50cm3 beaker
2. The temperature of ethyl ethanoate was measured using a thermometer.
3. 10cm3 of trichloromethane was added to the ethanol in the beaker and mixed well.
4. The highest temperature of the mixture was measured using a thermometer.
D. To measure the strength of hydrogen bonds formed between molecules of ethyl ethanoate and trichloromethane
1. The same steps in part C were repeated using same volume of ethyl ethanoate but 20cm3 and 30cm3 of trichloromethane respectively.
2. The same steps in part C were repeated using same volume of trichloromethane but 20cm3 and 30cm3 of ethyl ethanoate respectively.
Relevant Physical Data
Specific heat capacity of glass=0.78 KJ/kg/K
Results
A. Table 1 shows the temperature change after mixing 10cm3 ethanol with 10 cm3 cyclohexane:
B. Table 2 shows the temperature change after mixing 10 cm3 ethanol with 20 cm3 and 30 cm3 cyclohexane respectively:
C. Table 3 shows the temperature change after mixing 10cm3 ethyl ethanoate with 10 cm3 trichloromethane:
D. Table 3 shows the temperature change after mixing ethyl ethanoate with trichloromethane in excess and trichloromethane with ethyl ethanoate in excess respectively:
*Data obtained from student Lam Kung Hung
Calculation
Part A and B:
A. Energy absorbed to break hydrogen bond
= energy absorbed from ethanol (E1) + energy absorbed from cyclohexane (E2)
Energy = mc△T
E1 = [10 x0.81÷1000]．[2.44]．[3.5] kJ
= 0.069174 kJ
E2 = [10x0.78÷1000]．[1.83]．[3.5] kJ
= 0.049959 kJ
E = E1 + E2
= 0.119133 kJ
Bond enthalpy/Strength of hydrogen bond of ethanol
= E ÷ no. of moles of ethanol
= E x 46/(10x0.81) =0.677 kJmol-1
B(20cm3): E1 = [10 x0.81÷1000]．[2.44]．[3.5] kJ
= 0.069174 kJ
E2 = [20x0.78÷1000]．[1.83]．[3.5] kJ
= 0.099918 kJ
E = E1 + E2
= 0.169092 kJ
Bond enthalpy/Strength of hydrogen bond of ethanol
= E ÷ no. of moles of ethanol
= E x 46/(10x0.81) =0.960 kJmol-1
B(30cm3):
E1 = [10 x0.81÷1000]．[2.44]．[3.5] kJ
= 0.069174 kJ
E2 = [30x0.78÷1000]．[1.83]．[3.5] kJ
= 0.149877 kJ
E = E1 + E2
= 0.219051 kJ
Bond enthalpy/Strength of hydrogen bond of ethanol
= E ÷ no. of moles of ethanol
= E x 46/(10x0.81) =1.24 kJmol-1
Part C and D:
C. Energy released to form hydrogen bond
= energy released from ethyl ethanoate (E1) + energy released from trichloromethane (E2)
Energy = mc△T
E1 = [10 x1.48÷1000]．[0.98]．[8] kJ
= 0.116032 kJ
E2 = [10x0.90÷1000]．[1.92]．[8] kJ
= 0.13824 kJ
E = E1 + E2
= 0.254272 kJ
Bond enthalpy/Strength of hydrogen bond of product
= E ÷ no. of moles of product
= E x [(119.5+88)/ (10x1.48+10x0.9)]=2.22 kJmol-1
D.(20 cm3 trichloromethane):
E1 = [10 x1.48÷1000]．[0.98]．[9] kJ
= 0.130536 kJ
E2 = [20x0.90÷1000]．[1.92]．[9] kJ
= 0.31104 kJ
E = E1 + E2
= 0.441576 kJ
Bond enthalpy/Strength of hydrogen bond of product
= E ÷ no. of moles of product
= E x [(119.5+88)/ (10x1.48+10x0.9)]=3.85 kJmol-1
D.(30 cm3 trichloromethane):
E1 = [10 x1.48÷1000]．[0.98]．[8] kJ
= 0.116032 kJ
E2 = [30x0.90÷1000]．[1.92]．[8] kJ
= 0.41472 kJ
E = E1 + E2
= 0.530752 kJ
Bond enthalpy/Strength of hydrogen bond of product
= E ÷ no. of moles of product
= E x [(119.5+88)/ (10x1.48+10x0.9)]=4.63 kJmol-1
D.( 20 cm3 ethyl ethanoate):
E1 = [20 x1.48÷1000]．[0.98]．[7] kJ
= 0.203056 kJ
E2 = [10x0.90÷1000]．[1.92]．[7] kJ
= 0.12096 kJ
E = E1 + E2
= 0.324016 kJ
Bond enthalpy/Strength of hydrogen bond of product
= E ÷ no. of moles of product
= E x [(119.5+88)/ (10x1.48+10x0.9)]=2.82 kJmol-1
D.(30 cm3 ethyl ethanoate):
E1 = [30 x1.48÷1000]．[0.98]．[6.5] kJ
= 0.282828 kJ
E2 = [10x0.90÷1000]．[1.92]．[6.5] kJ
= 0.11232 kJ
E = E1 + E2
= 0.486336 kJ
Bond enthalpy/Strength of hydrogen bond of product
= E ÷ no. of moles of product
= E x [(119.5+88)/ (10x1.48+10x0.9)]=3.45 kJmol-1
∴Hydrogen bond between ethanol molecules is 1.24 kJ mol-1.
Hydrogen bond between ethyl ethanoate and trichloromethane is 4.24 kJ mol-1.
Precaution
1.> Since both ethanol and cyclohexane are easily vaporize, the volume of substances added may be affected, so the transfer process and the experiment should be done as fast as possible.
2.> The ethanol - cyclohexane mixture should be stirred gently & carefully with the thermometer, to ensure heat is evenly distributed.
Answer to Study Questions
What (theoretical) assumption(s) is/are taken in estimating the strength of hydrogen bond in the two cases?
Hydrogen bond is defined as the intermolecular force formed between H atom attached to F, O, or N and the lone pair electrons on F, O or N. The more electropositive (δ+) on H, the stronger is the hydrogen bond.
In this experiment, there are 2 assumptions made - it is assumed that no other interactions other than hydrogen bonds and hydrogen bonds between ethanol molecules are broken.
Other than hydrogen bonds, there are still relatively weaker van der Waals’ forces such as dipole-dipole interactions between polar ethanol molecules that may be broken in the practical. Meanwhile, there may be dipole-induced dipole interactions and instantaneous dipole-induced dipole interactions formed when polar ethanol and non-polar cyclohexane are mixed. Energy is then released and this compensate a bit of that absorbed for bond-breaking. In addition, there may be dipole-dipole interactions formed when polar ethyl ethanoate and polar trichloromethane are mixed. The bond enthalpy calculated accounts for the formation of these interactions also other than that of hydrogen bonds.
It should be, moreover, noticed that among ethanol molecules, hydrogen bonds can be present due to the presence of lone pair electrons on O attached to H:
However, among cyclohexane molecules, there is only relatively much weaker van der Waal's force but no H-bond:
After mixing these two substances, hydrogen bonds formed between ethanol molecules are broken because cyclohexane molecules separate ethanol molecules and hence interrupt its hydrogen bonds from forming:
To ensure that all the hydrogen bonds are broken, excess cyclohexane is added.
However, in this experiment, hydrogen bond strength obtained is not reliable because we can't ensure that all hydrogen bonds are broken. As a consequence, the bond enthalpy of hydrogen bonds in ethanol calculated as well as the bond strength is underestimated.
Discussion
1.Why does it not matter which liquid (ethyl ethanoate or chloroform) is used in excess?
In part D, it does not matter which liquid (ethyl ethanoate or chloroform) is used in excess and the results are similar when either of the liquids is in excess. This is because the number of hydrogen bonds formed is the same no matter which liquid is in excess. The ethyl ethanoate to trichloromethane ratio required is 1:1 in either situations and there is 1 H-bond per molecule formed. Adding excess is just to make sure there is no remaining molecules that have not form any H-bond and the bond enthalpy of formation can be correctly determined.
2.Why/How hydrogen bonds are formed in the mixture of chloroform and ethyl ethanoate?
Consider a chloroform molecule (HCCl3). The 3 chlorine atoms are highly electronegative, which withdraw electrons and thus build up a strong partial positive charge on H atom (-Hδ+).
Consider an ethyl ethanoate molecule (CH3COOCH2CH3). The O atom in C=O group can withdraw the σ electrons by inductive electrons and withdrawing the π electrons by the resonance effect. Therefore, a strong partial negative charge on the O atom is built up on the O atom of the C=O group (C=Oδ-) .
Similar, hydrogen bond is formed between the H atom of chloroform and the O atom of the C=O group of ethyl ethanoate molecule.
3.Practical assumptions and errors:
A. In the calculation, we assume that the heat capacity of the beaker is negligible. However, actually the heat capacity of the beaker is not equal to zero and it cannot ensure no heat was loss to surrounding. Therefore, the result is much smaller from the real one.
B. The beaker used in the experiments must be kept dry. Otherwise, there will be some reaction between the organic compounds with water (such as hydrogen bond).This overestimates the bond enthalpy.
C. When we use the thermometer as the stirrer, we should use it very carefully to prevent any liquid spit out to the outside of the beaker before well mixing. Otherwise, the mass of liquids will decrease and the result will be smaller than the real one.
D. In part D, the theoretical results when either of the two liquids is in excess should be almost the same but the experimental results are not. This is probably due to the fact that the results are obtained from different people and there are human errors. To improve, the data should be obtained from the same person.
4.Improvements:
If we want to have a more reliable result, we can improve the experiment by several methods:
< 1.> Repeat the experiment by adding more cyclohexane until the temperature becomes constant which indicates hydrogen bonds are no longer
present.
< 2.> Since ethanol is a polar molecule whereas cyclohexane is non-polar, they are immiscible. To ensure all hydrogen bonds are broken, it's better to use 2 substances which are miscible, with only ethanol can form hydrogen bond.
< 3.> Ethanol used should be as pure as possible, say, 99% to eliminate the effect of hydrogen bonds formed between water molecules.
When hydrogen bonds are broken, energy is absorbed, temperature drops. Since hydrogen bond is much stronger than van der Waal's force, the slightly increase in temperature caused by formation of van der Waal's force between ethanol & cyclohexane molecules can be compensated by the highly decrease in temperature caused by breaking of hydrogen bond among ethanol molecules. Hence, the approximate value of hydrogen bond strength of ethanol can be obtained by this calorimetric method.
Conclusion:
From this experiment, we can conclude that hydrogen bond is present between ethanol molecules due to the presence of hydroxyl group, ──OH. Hydrogen atom and very electronegative atom (such as oxygen) are necessary for the formation of hydrogen bonds.
The approximate strength of hydrogen bond of ethanol can be measured by using simple calorimetric method but improvements can be made to obtain more reliable results..
Hydrogen bond between ethanol molecules is 1.24 kJ mol-1.
Hydrogen bond between ethyl ethanoate and trichloromethane is 4.24 kJ mol-1.
Reference:
Hands-on Science (H-Sci) Project: Chemical Safety Database of The Physical and Theoretical Chemistry Laboratory of Oxford University:
Physical Chemistry I by Leung and Lee- Section 17.2
~End of Lab Report 8~
