Is this reaction limited by temperature

C(s, graphite) +H2O (g) ---> CO (g) +H2 (G)
Enthalpy:110.53 kJ
Entropy:133.793 JK-1

Is it limited by temperature? (yes), why? And what is the temperature at which the reaction become product favored and is it product favored above or below this temperature?

We see that Gibb's Free Energy is at a certain temperature T (in Kelvins) is:
∆G = ∆H - T∆S
= (110.53 kJ) - T(0.133793 kJ/K)
= (110.53 - 0.133793T) kJ.

Then, we see that ∆G < 0, when the reaction is favorable, when:
110.53 - 0.133793T < 0 ==> T > 826 K.

So, yes, the reaction is limited by temperature. The reaction becomes product-favored when the temperature exceeds 826 K.

I hope this helps!

You can use the equation:
total entropy change = change in entropy of the system + change in entropy of the surroundings.

The change in entropy of the system is the entropy of the products minus the entropy of the reactants.
The change in entropy of the surroundings is negative enthalpy change divided by the temperature in Kelvin.

If the total change in entropy is positive, the reaction is feasible and the product will form.
The feasibility of the reaction depends on the temperature. For an exothermic reaction, if the temperature is too low, the value for the entropy change of the surroundings will be more negative than the positive value for the entropy change of the system. This makes the total entropy change negative, meaning the reaction is not feasible.

You can work out when the reaction is just about feasible by working out the temperature when you've set the total entropy change to zero.

This will be a lot clearer in a textbook or on wikipedia or some place.