Explain why spontaneity depends on temperature?

1 Answer
Mar 19, 2016

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:

The following bonds were made:

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.