I was asked to find the hypothetical $\Delta H$ and $\Delta S$ of the decomposition of sodium carbonate and then asked to explain why it doesn't decompose.

I calculated the enthalpy and entropy change and from that I calculated the Gibbs free energy change. I was hoping to get a positive answer, but actually I got a negative answer, indicating that the process is actually spontaneous.

So is the reason that sodium carbonate doesn't decompose due to its kinetic properties, i.e. the reaction is too slow?

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    $\begingroup$ What is the decomposition pathway? Don't leave us in the dark! But in general if it's thermodynamically favoured but doesn't happen then it's a kinetic factor. $\endgroup$ Commented Feb 20, 2016 at 20:33
  • $\begingroup$ Your $\Delta G$ must be wrong. Sodium carbonate is thermodynamically stable. $\endgroup$ Commented Feb 20, 2016 at 20:36
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    $\begingroup$ I calculate a positive standard free energy change for the decomposition reaction. $\endgroup$ Commented Feb 20, 2016 at 21:31
  • $\begingroup$ Okay, thanks for all the help. I double checked my values I still get a negative value. The question probably gave the wrong values for the enthalpy and entropy change. $\endgroup$
    – Nanoputian
    Commented Feb 20, 2016 at 22:56
  • $\begingroup$ OK. How about telling us what you got for the free energies of formation of each of the products and the reactant? $\endgroup$ Commented Feb 21, 2016 at 0:13

1 Answer 1


From the values on the Wikipedia page for sodium carbonate, the Gibbs free energy of formation is negative and thus, formation is thermodynamically preferred over decomposition at $\pu{298K}$. Your given values seem to be wrong after all.

By the way, $\Delta H$ and $\Delta S$ should be the negative of those given on that page for your case, making $\Delta G$ a positive number.


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