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Harlamova29_29 [7]
2 years ago
15

Consider the following molecules and the description of the bonding present in each: CH3CH2CH2CH3CH3CH2CH2CH3 (C−CC−C and C−HC−H

bonds only) CH3CH2CH2OH CH3CH2CH2OH (C−CC−C and C−HC−H bonds, and one C−OHC−OH bond) CH3CH2CHO CH3CH2CHO (C−CC−C and C−HC−H bonds, and one C=OC=O bond) Based on the general trends of intermolecular forces, rank these molecules in order of increasing boiling point.
Chemistry
1 answer:
denis23 [38]2 years ago
3 0

CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3

Explanation:

Boiling point trend of Butane, Propan-1-ol and Propanal.

Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.

In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.

Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.

In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol

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2.12×10²³ atoms.

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6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ
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Both reaction A and reaction B are non spontaneous.

Explanation:

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or, T> 350

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: \Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

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\Delta H-T\Delta S< 0

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or, T> -16

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T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
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