# How can you determine if a reaction is non-spontaneous?

Apr 11, 2017

By looking at the Gibbs-free energy

#### Explanation:

A reaction's spontaneity (i.e. if it occurs) can be deduced from the Gibbs-free energy values. I assume that your level of chemistry is already good enough for you not needing a definition for Gibbs free energy, hence I will go immediately into the calculations:

∆G=∆H-T∆S <-- This is the most common way to approach this

∆G = Gibbs free energy
∆H = Enthalpy change
T = Temperature in kelvin
∆S = Entropy change

Every reaction has a certain ∆H and ∆S value, however, the Temperature can be variable. Take the following reaction as an example:

$C a C {O}_{3} \to C a O + C {O}_{2}$ <- Where ∆H = 177 kJ and a ∆S = 0,161kJ

NOTE: The ∆H and ∆S value can be found in digital textbooks, data-sites (like webbook.nist.gov) or they are given in the question. (If you need further clarification, then just comment below)

These values can now be plugged into the Gibbs-free energy equation:

∆G=177-T*0,161

Now we have to find the temperature for which this reaction is spontaneous/non-spontaneous/equilibrium. For this you have to know that:

When...
...∆G is positive, the reaction is non-spontaneous
...∆G is negative, the reaction is spontaneous
...∆G is equal to 0, the reaction is at an equilibrium

Hence, now you can solve for whatever you want. E.g. non-spontaneous (as was your question):

$0 < 177 - T \cdot 0 , 161$
$T < 1099 , 38 K$ <-- anything below that temperature will not permit the reaction to occur

On the other hand, if you are looking for the ∆G at a specific temperature, simply plug in your temperature in Kelvin and see what your answer for ∆G is.

Hope it helped :)