Explain why spontaneity depends on temperature?

Spontaneity is most easily defined by a thermodynamic equation you've probably seen before:

#\mathbf(DeltaG = DeltaH - TDeltaS)#

where:

• #DeltaG# is the Gibbs' free energy, which tells you when a reaction is spontaneous, non-spontaneous, or at equilibrium.
• #DeltaH# is the enthalpy, which is the heat flow #q# at a constant pressure.
• #DeltaS# is the entropy, which is directly proportional to the number of microstates that can pertain to a system, i.e. loosely speaking, a measure of "disorder".

Since #T# is a variable in the function for #DeltaG#, and since a reaction is spontaneous when #DeltaG < 0#, at equilibrium when #DeltaG = 0#, and nonspontaneous when #DeltaG > 0#, spontaneity depends on temperature.

EXAMPLE: ENTROPY

For example, let's consider this reaction:

#2"Al"(s) + 6"HCl"(aq) -> 2"AlCl"_3(aq) + 3"H"_2(g)#

You can see that there are more #"mol"#s of gas on the products side, so #color(green)(DeltaS > 0)#; the entropy of a gas is significantly higher than that of a solid or liquid.

So significant that in many General Chemistry classes, a general rule of thumb is used where the sign of the entropy can often be found from the #"mol"#s of gas on each side.

EXAMPLE: ENTHALPY

The following bonds were broken:

• #6xx("H"-"Cl")#, #DeltaH_"bond" ~~ "431 kJ/mol"#

The following bonds were made:

• #2xx("Al"-"Cl" xx 3)#, #DeltaH_"bond" ~~ "502 kJ/mol"#
• #3xx("H"-"H")# #DeltaH_"bond" ~~ "436 kJ/mol"#

Therefore, the total enthalpy of the reaction is approximately:

#color(green)(DeltaH_"rxn") = sum_i DeltaH_("broken",i) - sum_j DeltaH_("made",j)#

#= ("431 kJ/mol") - ("436 kJ/mol" + "502 kJ/mol")#

#= color(green)(-"507 kJ/mol")#

EXAMPLE: GIBBS' FREE ENERGY

Interestingly enough, since #DeltaH < 0# and #DeltaS > 0#, what we have is a situation where the reaction is spontaneous all the time.

Remember that #T > 0# if in #"K"#, and we know right now that #DeltaS > 0#. Therefore, we have:

#color(blue)(DeltaG = -|DeltaH| - |T|*|DeltaS| < 0)#

Therefore, this reaction is spontaneous for all temperatures.

However, had #DeltaH# been positive, then the spontaneity would depend on the temperature; if so, then:

• If #T# is high enough, the reaction is spontaneous because #|DeltaH| < |T|*|DeltaS|# and thus #DeltaG < 0#.
• If #T# is low enough, then the reaction is nonspontaneous because #|DeltaH| > |T|*|DeltaS|# and thus #DeltaG > 0#.
• If #DeltaH = TDeltaS#, the reaction is at equilibrium because #DeltaG = 0#.