# Spontaneity of reversible reactions

1. If a reaction has positive $$\Delta G$$ then reverse reaction has negative $$\Delta G$$. How can both occur simultaneously, e.g., in a reversible reaction such as dimerisation of $$\ce{NO2}$$ to form $$\ce{N2O4}$$? Do forward and backward reactions occur spontaneously? How can both directions be spontaneous at the same time?

2. I read in some book that Haber's process to form ammonia is spontaneous at a temperature below $$170\ \mathrm{^\circ C}$$ (the author proved this using $$\Delta G < 0$$ for spontaneous process). This means that above this temperature $$\ce{N2}$$ and $$\ce{H2}$$ shouldn’t combine to form $$\ce{NH3}$$. But industrially Haber’s process is performed at temperature around $$400\ \mathrm{^\circ C}$$. Isn't this a contradiction?

• I just want to be really picky about your use of the symbol $\Delta G$. The symbol $\Delta G$ can be used at any stage of the reaction and in fact, gives no information about the final position of equilibrium. The correct symbol to use is $\Delta G^\circ$. (Actually it's not a circle, it should be the standard state symbol, but whatever.) The former varies as the reaction proceeds; the latter is a constant, defined to be equal to $\Delta G$ when $Q = 1$, that determines the position of equilibrium. Sep 28, 2015 at 5:06
• Refer to this question for more info - chemistry.stackexchange.com/questions/28644/… Sep 28, 2015 at 5:15
• If a reaction is not reversible, spontaneity depends on free energy being positive or negative. If a reaction is reversible, it can be spontaneous in both direction until equilibrium is reached. Mar 8, 2022 at 16:08

1. You seem to be mixing up thermodynamic and kinetic considerations. All reactions occur spontaneously all the time, and their backward reactions too. If left alone for long enough, a system will eventually reach the point where the forward and backward reactions go at the same rate (and it's not like they both stop altogether!) - that's an equilibrium. Where that equilibrium will be is judged by $\Delta G$; if it is negative, then the equilibrium will be closer to the products.

2. As for ammonia, that's where the kinetics kicks in. At room temperature $\Delta G$ is in our favor, but the equilibrium would take forever to reach. We raise the temperature so as to make the reaction faster, but that also shifts the equilibrium in the wrong direction. So we raise the pressure, too. This does the trick: now the reaction goes the way we need, and at reasonable rate.

• Great explanation but I just want to be picky (same thing with the question text) - the correct symbol in this context is $\Delta G^\circ$ and not $\Delta G$. Sep 28, 2015 at 5:07

The sign of $$\Delta G^\circ$$ is not an indication of whether a reaction will occur spontaneously. For a reversible reaction, if you start with a mixture of only reactants, the reaction will proceed to produce products until the system reaches equilibrium. If you start with a mixture of only the products (i.e., the products of the forward reaction), the reverse reaction will proceed to produce the reactants (of the forward reaction) until the system reaches equilibrium.

$$\Delta G^⦵$$ is the free energy change, for the reaction as written, of the reaction of the reactants and products in their standard states to their concentrations at equilibrium. The reaction written one way will be the negative of the reverse.

$$\Delta G$$ is the free energy change for the actual concentrations involved. At equilibrium $$\Delta G = 0,$$ the activities are the equilibrium activities, and $$\Delta G^⦵ = -RT\ln K_\mathrm{eq}$$.

As Chet Miller states, if there are only reactants, regardless of the reaction direction $$\Delta G$$ will be negative and the reaction will proceed to equilibrium. The extent of the reaction is determined by the equilibrium constant. Of course, the reactants must be properly mixed, have sufficient activation energy or catalysis, and there must be a possible mechanism.