Question

What is the relation b/w Spontaneity and entropy?

Answer 1

∆H= enthalpy, heat, total kinetic energy (1st law of thermodynamics)

 

Negative: exothermic

Positive: endothermic

 

Equation(s):

 

∆H= H (products) – H (reactants)

 

Units= kJ

 

∆S= entropy, disorder (2nd law of thermodynamics)

 

Negative: disorder decreases

Positive: disorder increases

 

Equation(s): ∆S= S (products) – S (reactants)∆S= ∆H-∆G/T

 

Units=J/K

 

∆ G= free energy, spontaneity

 

Negative: spontaneous

Positive: non-spontaneousEqual to Zero: at equilibrium/ becomes K

 

Equation(s):

 ∆ G= ∆ H- T ∆ S, T being temp (Kelvin)(relations of temperature: decrease in temp. = ∆ G is positive; increase in temp. = ∆ G is negative)∆G= -RT ln KR being gas constant= 8.3145 J/K*mol and K being equilibrium constant equation (equation only for concentration at equilibrium)

∆ G= G (products) – G (reactants)

 

Units: kJ

 

 Relationship between ∆ H and ∆ S

                              CASE                             RESULT∆ S positive, ∆ H negativeSpontaneous at all temperatures∆ S positive, ∆H positiveSpontaneous at high temperatures (importance of exothermic process)∆ S negative, ∆ H negativeSpontaneous at low temperatures (dominance of exothermic process)∆ S negative, ∆ H positiveProcess not spontaneous at any temperature (reverse process is spontaneous at all temperatures)

 

 

 

          Negative and Positive Relationships between ∆H, ∆S and ∆G

            Negative (-)             Positive (+)∆ H (enthalpy)ExothermicEndothermic∆ S (entropy)Disorder increasesDisorder decreases∆ G (free energy)Non-spontaneousSpontaneous

 

Answer 2

" If heat flows into the surrounding the random motions of the moleculesinthe surrounding increase. thus the entropy of surrounding increases.
The second law of thermodynamics states that the total entropy of the universe always increase s for a spontaneous process "