Entropy and the 2nd Law of Thermodynamics
Thermodynamic laws 1st Law – Energy is always conserved. It cannot be created or destroyed 2nd Law – The Entropy, (randomness or disorder), of the universe is always increasing. 3rd Law – States that the entropy of a pure crystal at 0 Kelvin is zero.
Entropy Defined as randomness or disorder. Units are J/K mole The more disorganization in the molecules or atoms means higher entropy. The entropy of a gas is much greater than a solid or liquid There is a natural tendency to increasing entropy.
Entropy So, any reaction or system in which disorder is increased will tend to be favored. Symbol for entropy is “S”, and we look at the change in entropy which is, ∆S. If ∆S is (+) then entropy is increasing – getting more random or disorder. If ∆S is (-) then entropy is decreasing – getting more ordered, or less random
Entropy The total entropy of the universe can be described as the sum or the entropies of a system and its surroundings. ∆Suniverse = ∆Ssystem + ∆Ssurroundings
∆Ssystem depends on positional entropy, for example, how many gas molecules form in a reaction. More gas molecules is more entropy. ∆Ssurroundings depends on heat. Is the process endothermic or exothermic. Exothermic process – heat is added to the surroundings so ∆Ssurroundings is (+) The magnitude of ∆S depends on the amount of heat that flows into the surroundings. Endothermic process – heat is take from the surroundings, so ∆Ssurroundings is (-) negative. Again, the magnitude of ∆S depends on the amount of heat that flows away from the surroundings.
∆Ssurroundings = -∆H T The negative sign in front of ∆H is included because ∆H is determined from the system’s point of view, whereas ∆S is determined from the surroundings point of view. Reactions with a ∆Suniverse that have a (+) value , or where entropy is increasing will be considered spontaneous. This means that will naturally proceed to make the products
∆Ssystem ∆Ssurroundings ∆Suniverse (+) Yes, always (-) No, reverse is Spontaneous ? (+) Yes, always (-) No, reverse is (endothermic) ? Yes, at high temps. (exothermic) Yes, at low temps.
#1 For CH4(g) + 2O2(g) CO2(g) + 2H2O(g) What will the ∆Suniv be for the above reaction and will it be spontaneous given: ∆Ssys = -38.5 J/K mole ∆H = -802.3 kJ/mole at 298K ∆Suniv = ∆Ssys + ∆Ssurr ∆Suniv = ∆Ssys + -(∆H) T = (-38.5 J/Kmol)+ -(-802,300 J/mol) 298K = (-38.5) + (2692) = + 2654 J/Kmol It is spontaneous because ∆Suniverse is increasing (+).
2) Is entropy increasing of decreasing in the following reactions? a. 2KClO3(s) 2KCl(s) + 3O2(g) Increasing – gas is formed and gases have higher entropy (more disorder) b. 2Ag(s) + Cl2(g) 2AgCl(s) Decreasing – product is all solid – and solids are more ordered and organized.
c. 2C8H18(l) + 25O2(g) 16CO2 (g) + 18H2O(g) Increasing – product side has more gas molecules (25 vs. 34 moles) **More complex molecules will have more entropy. Aqueous solutions with more ions also have more entropy
Calculating Entropy from a Reaction. We can calculate entropy like we calculate the enthalpy of a reaction, using the standard entropies from the table in the appendix ∆S = ∑ S products - ∑ S reactants