[Boxxxed]

Entropy increases with temperature so what I don't understand is S = q / t

Increasing the temperature will decrease the value of entropy according to the equation.

[rabolisk]

You're mixing two very different concepts here. S = q/t isn't even right. The proper form is dS = dq_rev/T. It has to do with change in entropy during a thermodynamic process.

[Boxxxed]

Is there a simple mathematical expression to show how entropy increases with increase in temperature?

[Compaq]

This famous equation shows the relationship pretty clearly:

dG = dH - TdS

where dG is the change in Gibb's Energy and dH is the enthalpy change.

For example, if dH>0 and dS<0, the rxn will be spontaneous at all temperatures. (the reaction is spontaneous if dG<0))

If both dH and dS <0, the rxn will be spontaneous at low temperatures.

The equation you wrote looked a lot like a (wrong) version of this when dG = 0 (chemical equilibrium), for example in phase changes.

Let's assume that water freezes at exactly 273K, then

H₂O(l) <---> H₂O(s), and dG=0  =>

dS = dH / T

[rabolisk]

This famous equation shows the relationship pretty clearly:

dG = dH - TdS

where dG is the change in Gibb's Energy and dH is the enthalpy change.

For example, if dH>0 and dS<0, the rxn will be spontaneous at all temperatures. (the reaction is spontaneous if dG<0))

If both dH and dS <0, the rxn will be spontaneous at low temperatures.

The equation you wrote looked a lot like a (wrong) version of this when dG = 0 (chemical equilibrium), for example in phase changes.

Let's assume that water freezes at exactly 273K, then

H₂O(l) <---> H₂O(s), and dG=0  =>

dS = dH / T

I don't think that answers his question at all...