Question

True or False or short answer. You must provide reasoning.

a. If DSsys < 0 for a spontaneous process, then a smaller positive change in Ssurr must occur.

b. The combustion of ethane is spontaneous, but DS° is -620.21 J/K. Explain why this does not violate the second law of thermodynamics.

Answer 1

According to the second law of thermodynamics, for a spontaneous process

∆Stotal = ∆Ssys + ∆Ssurr > 0

a) It is given that ∆S sys < 0, then ∆Ssurr must have a larger positive magnitude in order for ∆Stotal to be positive. This means |∆Ssurr| > |∆Ssys|.

Therefore, statement (a) is FALSE.

b) For combustion of ethane, ∆H° < 0 (combustion involves release of large amounts of heat).

Also ∆Hsurr = - ∆Hsys

So, ∆Ssurr = ∆Hsurr/T > 0

In conclusion, if ∆Ssys is neagtive for combustion of ethane, then ∆Ssurr must have magnitude large positive enough such that ∆Stotal is positive. So, this does not violate the second law of thermodynamics.