What are they h-bond water acceptors in fentanyl, and is it souluable in water? Explain why or why not.
1 answer:
Here we have to get the atom, in the molecule fentanyl, which will behave as a hydrogen (H) bond acceptor from the donor water molecule.
The -O atom in the molecule will behave as the hydrogen bond acceptor from water.
The H-bond always forms between the proton (here water i.e. H₂O) and the proton acceptor atom in a molecule which is essentially is an electronegative element.
In the fentanyl molecule there are two nitrogen (N) atom and one oxygen (O) atom. We know the electronegativity of oxygen is more than the nitrogen. Thus it is expected that the O atom will be H-bond acceptor in the molecule.
The hydrogen bond with water molecule is shown in the figure.
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Answer: Option (a) is the correct answer.
Explanation:
The given data is as follows.
= 4.19
= 1.9
Heat of vaporization () at 1 atm and
is 2259 kJ/kg
= 0
Therefore, calculate the enthalpy of water vapor at 1 atm and as follows.
=
+
= 0 + 2259 kJ/kg
= 2259 kJ/kg
As the desired temperature is given and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and
is calculated as follows.
=
= 334.781 kJ/kg
Hence, enthalpy of water vapor at 10 bar and is calculated as follows.
=
= 2410.81 kJ/kg
Therefore, calculate the latent heat of vaporization at 10 bar and as follows.
=
= 2410.81 kJ/kg - 334.781 kJ/kg
= 2076.029 kJ/kg
or, = 2076 kJ/kg
Thus, we can conclude that at 10 bar and latent heat of vaporization is 2076 kJ/kg.