In: Chemistry
1. Indicate based on signs of enthalpy and entropy changes how temperature will affect spontaneity and/or calculate temperature at which spontaneity changes. (give example)
2. Explain how sign of ΔG, sign of E, and value of K indicate spontaneity from standard conditions.
Q1
In order to compare equilibirum vs. spontaneous/nonspontaneous reactions, we better use a criteria.
Recall that if dSuniverse > 0, this is spontaneous, if dSuniverse = 0, this is inequilbirium and if dSuniverse < 0 this is never possible.
Then, recall that
dSuniverse = dSsurroundings + dSsystem
dSsystem = Sproducts - Sreactants
dSsurroundings = Qsurroundings/T = -dHsystem/T
therefore
dSystem = -dHsystem/T + dSsystem
If we multiply by -T
dGrxn = dHrxn - T*dSrxn
Now, analysis of dG value... which is the "free energy" available for a process to follow
if dG <0 , this will be spontaneous
if dG = 0 , this is in equilibrium
if dG > 0, this will not be spontaneous
Now...
dG = dH - T*dS
Possible values are, dH = +/- and dS = +/-; T is always positive ( absolute value)
Analysis of cases:
Case 1.
if dH is positive (-) and dS is positive (+) --> this favours always a negative value of dG; spontaneous
Case 2.
if dH is positive (+) and dS is positive (-) --> this favours always a positive value of dG; not spontaneous
Case 3.
if dH is positive (+) and dS is positive (+) --> dG = dH - T*dS --> analysis must be done
if T is very low... then dH > T*dS; then this will be Positive value in dG; i.e. not spontaneous
if T is very high... then dH < T*dS; then this will be Negative value in dG; i.e. spontaneous
Case 4.
if dH is positive (-) and dS is positive (-) --> dG = dH - T*dS --> analysis must be done
if T is very low... then dH > T*dS; then this will be Negative value in dG; i.e. spontaneous
if T is very high... then dH < T*dS; then this will be Positive value in dG; i.e. non spontaneous
Q2
dG = -n*F*E°cell
dG = -RT*ln(K)
then, for spontaneous process:
dG < 0, K > 1, E°cell > 0
then, for non-spontaneous process:
dG > 0, K < 1, E°cell < 0