Question

Explain why spontaneity depends on temperature?

Answer 1

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`.