# How can a nonspontaneous reaction occur?

$$\frac{-\Delta G}{T}=\Delta S_{universe}$$

(This equation applies under isobaric and isothermal conditions.)

I understand that if $\Delta G$ is positive, the reaction is nonspontaneous, and adding that amount of energy to a system can drive that reaction to occur. However, if you’re increasing the Gibbs free energy of the system, then by that equation, the entropy of the universe would be decreasing, which would violate the Second Law of Thermodynamics. Can anyone explain this?

• Can you give an example of such are reaction? I think I might know where you confusion is but I'm not entirely sure.
– bon
Apr 19, 2016 at 13:12
• @bon How about an electrolysis reaction? I don’t know; I’m just thinking about nonspontaneous reactions in general... Apr 19, 2016 at 13:14
• In electrolysis you are putting in energy to drive the reaction forward. That energy has to come from somewhere and the process that creates in has a large, negative $\Delta G$.
– bon
Apr 19, 2016 at 13:17
• @bon I understand that, so that’s not really the focus of my question. I’m confused about the formula I posed above, really. My central confusion is that if you’re adding energy, then you’re increasing free energy, but you can’t decrease the universe’s entropy! Apr 19, 2016 at 13:23
• @lightweaver: The equation is true only when non-compression work is zero.
– user5764
Apr 19, 2016 at 14:48

When you add energy from somewhere, let's say a battery, you are creating a greater amount of entropy there (in the battery or whatever). That's all there is to it. The Second Law is not violated.

The electrolysis process is not non-spontaneous. Yes, it would not occur in the absence of the battery. But the fact that it does occur with the battery there must tell you that overall it is spontaneous. The fact that you are coupling a strongly spontaneous reaction (discharge of the batter) with a non-spontaneous reaction (the electrolysis) means that overall, the electrolytic process is spontaneous.

It's incorrect to treat the electrolysis as if it occurs without the battery.

• So you’re saying that when we’re adding energy to drive a nonspontaneous reaction, the equation for $\Delta G$ changes to incorporate the source of energy into that system? Apr 20, 2016 at 10:46
• You could look at it that way. To be more precise, the "system" that you are talking about should really include the battery as well. The equations follow. Apr 20, 2016 at 10:48
• So, just to verify your answer, in order for reactions that decrease entropy to occur, we need to supply them with energy. By doing that we increase entropy much more somewhere else and thus the overall entropy of the system is increased? Aug 16, 2016 at 15:49
• @ΚελεσίδηςΑντώνιος That depends on your definition of "system". The Second Law only applies to an isolated system. If you consider the reaction vessel to be the "system" (in this case a closed system, not an isolated system), and everything outside the reaction vessel to be the "surroundings", then the system and its surroundings make up an isolated system, and the overall entropy of the isolated system increases. However the entropy of the reaction vessel, a closed system, decreases. The entropy of the surroundings increases. Aug 16, 2016 at 17:46
• So the system, in which the entropy must and is increased, consists of the beaker with the reactants and the enviroment that supplies the necessary energy. I understand now. Thank you. Aug 16, 2016 at 18:40

Any reaction depends on two factors: enthalpy and entropy, and the summation of which is the change of the Gibbs energy by equation (for isobaric and isothermal process):

$$\Delta G = \Delta H - T \Delta S \qquad(1)$$ or, after manipulations we have

$$\frac{\Delta H-\Delta G}{T} = \Delta S \qquad(2)$$

The difference between equations (2) and one you've represented is the last one describes process of the hypothetical system (or real -> Universe) which always become to the initial state and there's no change of enthalpy: $$\Delta H = 0 \qquad(3)$$

If you will try to change Gibbs energy: $$\Delta G = G_2 - G_1 \qquad (4)$$ where$$G_i = H_i - TS_i \qquad (5)$$ or $$G_i = U_i + PV_i - TS_i\qquad (6)$$ you have to change internal energy or volume of the system which will effect on enthalpy and the change of it: $$\Delta H \neq 0 \qquad(7)$$ and you will get the equation (2)

Answering on your question: handle the change of enthalpy (7) of the reaction.

A non-spontaneous reaction does not proceed without the continuous supply of energy: A spontaneous reaction is one that, once started, continues without assistance. Under a particular set of conditions (e.g. room temperature), the reverse of a spontaneous reaction will always be non-spontaneous. For example, if A to B is spontaneous at 300 K, B to A is non-spontaneous at 300 K. However, a substantial increase in temperature (achieved by adding energy) often reverses spontaneity, so that (in our example) B to A is now spontaneous at (say) 1000 K and A to B non-spontaneous at 1000 K.

Electrolysis is an unusual tactic for forcing some non-spontaneous reactions to occur at room temperature. For example, the decomposition of water at room temperature is non-spontaneous, but a continuous supply of electricity causes continuous production of hydrogen and oxygen gases. Part of the electricity provides the energy to break the O-H bonds in water and part of the electricity is used to warm the water reservoir and the laboratory; the latter is essential for the entropy of the universe to increase without which the non-spontaneous reaction could not occur.

As one of the commenters near the top correctly said, the formula $$-\Delta G/T = \Delta S_{total}$$ only applies when non-compression work is zero. That is, the system is set up so that the only way it can exchange energy with the environment is to expand or contract against some ambient pressure, and to absorb or release heat. In that case, the enthalpy change of the system is equal to the heat absorbed from the environment: $$\Delta H = Q$$. This comes from the fact that the work done on the system is just $$W = -p\Delta V$$ at constant pressure. So $$\Delta H = \Delta(U+pV) = \Delta U + p\Delta V = Q + W + p\Delta V = Q -p\Delta V + p\Delta V = Q$$.

Then you can put that into the formula for total entropy:

$$\Delta S_{total} = \Delta S + \Delta S_{surroundings} = \Delta S - Q/T = \Delta S - \Delta H/T = -\Delta G/T$$

$$\Delta G$$ and $$\Delta S$$ are the Gibbs energy change and entropy change of the system only, while $$\Delta S_{total}$$ is the entropy change of "the universe", which figures into the most common way of stating the second law (i.e. total entropy never decreases).

In the electrolysis example, however, there is electrical work $$W_e$$ being done on the chemical system in addition to compression work. So total work is $$W = W_e - p\Delta V$$. When you put this into the formula for enthalpy change, $$\Delta H = Q + W + p\Delta V$$, the p-V terms cancel again and you're left with $$\Delta H = Q + W_e$$. Then substitute $$Q = \Delta H - W_e$$ in the formula for total entropy, and you get the equation that applies in this situation:

$$\frac{W_e- \Delta G}{T} = \Delta S_{total}$$

$$\frac{-\Delta G}{T}=\Delta S_\mathrm{universe}$$

(This equation applies under isobaric and isothermal conditions.)

Another requirement for this equation to hold is that the non-PV work is zero. When deriving it, you use the $$\Delta H = q$$, which is not the case when there is non-PV work.

I understand that if $$\Delta G$$ is positive, the reaction is nonspontaneous, and adding that amount of energy to a system can drive that reaction to occur. However, if you’re increasing the Gibbs free energy of the system, then by that equation, the entropy of the universe would be decreasing, which would violate the Second Law of Thermodynamics. Can anyone explain this?

The Gibbs energy is a state function, so no matter how you run a process, for the same starting and ending state you get the same change in Gibbs energy. "Adding energy" will not drive the reaction, but doing work on the system will. The battery does electrical work on the system if you place it outside of the system. In that case, you have an example of a process that goes forward even though $$\Delta G_\mathrm{sys}$$ is positive.