In liquid water, water molecules are continually forming and breaking hydrogen b
ID: 479374 • Letter: I
Question
In liquid water, water molecules are continually forming and breaking hydrogen bonds. In water, a hydrogen bond has a bond energy of about 0.2 eV. Two water molecules are shown forming a hydrogen bond (dashed line) in the figure at the right. Is the formation of the hydrogen bond exothermic (releases energy, reducing the internal energy of the molecules) or endothermic (absorbs energy, increasing the internal energy of the molecule)? Let's see whether hydrogen bonds can account for the energy needed to boil water. Assume the model that when water boils, the energy that has to be put in is the energy needed to break the hydrogen bonds. It takes about 2.3 times 10^6 J to boil 1 kg of water. First, figure out how many water molecules there are in 1 kg of water. Do you expect that there would be more hydrogen bonds than water molecules or fewer? Explain your reasoning. Next, figure out in our model of boiling, that is, if boiling is breaking hydrogen bonds, how many hydrogen bonds are broken in order to boil 1 kg of water? Number of bonds Explain your reasoning. Does your calculation support our simple model or not? Why?Explanation / Answer
Ans. A. Formation of hydrogen- bond is exothermic.
The weak attractive interaction between a highly electronegative atom (F, O, N) and a H-atom covalently linked to any such highly electronegative atoms is called hydrogen bond. Formation of a chemical bond is exothermic, so is the formation of H-bonds.
Ans. C. 1 kg = 1000.0 g
Number of moles of water in 1000.0 g sample = Mass / molar mass
= 1000.00 g/ (18.0 g mol-1) ; [1 dalton = 1 g mol-1]
= 55.56 mol
Number of water molecules in 1000.0 g sample = number of moles x Avogadro number
= 55.56 mol x (6 x 1023 molecules/ mol)
= 3.33 x 1025 molecules
Ans. D. All the three atoms (2 H-atoms, 1 O-atom) in a water molecule forms hydrogen bond with neighboring molecules.
That is, one water molecule forms 3 H-bonds. So, number of H-bonds in greater than that of number of water molecules.
Ans. E. Number of H-bonds = 3 x number of water molecules
= 3 x 3.33 x 1025
= 9.99 x 1025
Ans. F.
Total bond enthalpies of all H-bonds in 1 kg water sample =
(Total number of H-bonds x bond enthalpy of 1 H-bond)
= 9.99 x 1025 x 0.2 eV ; [1 eV = 1.60 x 10-19 J]
= 9.99 x 1025 x 0.2 x (1.60 x 10-19) J
= 3.1968 x 106 J
Yes, the calculation supports our simple calculation because it gives a value bear to the standard value. Inclusion of other factor like temperature, exact values upto three decimal points, etc. shall produce a value much closer to the standard.
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