Question

Predict which of the organic compounds

(a) (CH3CH2)2O (g) or

(b) (CH3CH2)3O (g) or

(c) (CH3CH2)3O (l) has the greater standard molar entropy at 298˚K. Explain your reasoning.

Answer 1

Of the given organic compounds, (CH₃CH₂)₃O (g) has the greater standard molar entropy at 298^o K.

Reasoning:

Molar entropy is more for substances with more complex molecules. To some extent this is due to the mass since on the whole more complex molecules are heavier than simpler ones. The more atoms there are in a molecule, the more ways the molecule can change its shape by vibrating. In consequence there are more ways in which the energy can be distributed among the molecules. So therefore, comparatively (CH₃CH₂)₂O (g) has less molar entropy than (CH₃CH₂)₃O (g) and (CH₃CH₂)₃O (l).

The molar entropy generally increases as we move from solids to liquids to gases. In a solid, the molecules are only capable of restricted vibrations around a fixed point, but when a solid melts, the molecules, though still hampered by their mutual attraction, are much freer to move around. So, when a solid melts, the molar entropy of the substance increases. When a liquid vaporizes, the restrictions on the molecules’ ability to move around are relaxed almost completely and a further and larger increase in the entropy occurs. Therefore (CH₃CH₂)₃O (g) has the greater standard molar entropy.