Without doing a calculation, arrange the following group of molecules in order of decreasing standard molar entropy (S^o ): hex

Question

Without doing a calculation, arrange the following group of molecules in order of decreasing standard molar entropy (S^o ):

hexane (C6H14), benzene (C6H6), cyclohexane (C6H12)

a. C6H14 > C6H6 > C6H12
b. C6H6 > C6H12 > C6H14
c. C6H12 > C6H14 > C6H6
d. C6H14 > C6H12 > C6H6
e. C6H12 > C6H6 > C6H14
f. C6H6 > C6H14 > C6H12

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Cherry 3 years 2021-08-19T10:42:43+00:00 1 Answers 29 views 0

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  1. Bo
    0
    2021-08-19T10:44:01+00:00
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    Answer:

    d. C6H14 > C6H12 > C6H6

    Explanation:

    Standard molar entropy has to do with the number of atoms that are present in each of the species. The greater the number of atoms possessed by the species, the higher the value of the standard molar entropy due to a greater number of vibration modes.

    Hexane (C6H14) has the highest number of atoms followed by cyclohexane (C6H12) and lastly benzene (C6H6).

    Thus the order of decreasing molar entropy is;  C6H14 > C6H12 > C6H6.

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