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)

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

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.