contestada

Given the following values for the change in enthalpy (deltaH) and entropy (deltaS), which of the following processes can occur at 298 K without violating the second Law of Thermodynamics?
(a) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)
(b) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = -125j mol-2K-1)(-30 cal mol-1 K-1)
(c) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)
(d) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(-30 cal mol-1 K-1)

Respuesta :

Answer:

Option A and B

Explanation:

(a) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = Delta H – T * DS

Substituting the given values, we get –  

Delta G = -84 -298 *(125/1000) = -121.25  KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(b) DeltaH = -84 kj mol-2 (-20 kcal mol-1), DeltaS = -125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G =-84 -298 *(-125/1000) = -46.75 KJ/mol

Delta G is negative hence the process is spontaneous and will not violate the second law of thermodynamics

(c) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(+30 cal mol-1 K-1)

Delta G = 84 -298 *(125/1000) = +46.75 KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

(d) DeltaH = +84 kj mol-2 (+20 kcal mol-1), DeltaS = +125j mol-2K-1)(-30 cal mol-1 K-1)

Delta G = 84 -298 *(-125/1000) = + 121.25  KJ/mol

Delta G is positive hence the process is non-spontaneous and will violate the second law of thermodynamics

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