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  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 $\Delta G=$ negative 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?

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  • $\begingroup$ 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. $\endgroup$ – orthocresol Sep 28 '15 at 5:06
  • $\begingroup$ Refer to this question for more info - chemistry.stackexchange.com/questions/28644/… $\endgroup$ – orthocresol Sep 28 '15 at 5:15
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  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.

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    $\begingroup$ 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 $\Delta G$. $\endgroup$ – orthocresol Sep 28 '15 at 5:07
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The sign of $\Delta G^0$ 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.

Chet

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