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I found the following example of a coupled reaction to drive the decomposition of calcium carbonate. I get the calculation part of it, that the changes in Gibbs energy sum to a negative amount. But does this reaction actually happen when you place the reactants together? Would it happen with nothing more than, say, a brief flame to ignite the coal? Or does it require more complicated steps to make it happen?

From Chemistry LibreTexts 19.8 Coupled Reactions

Many chemicals' reactions are endergonic (i.e., not spontaneous ($\Delta{}G > 0$)) and require energy to be externally applied to occur. However, these reaction can be coupled to a separate, exergonic (thermodynamically favorable $\Delta{}G < 0$) reactions that 'drive' the thermodynamically unfavorable one by coupling or 'mechanistically joining' the two reactions often via a share intermediate. Since Gibbs Energy is a state function, the $\Delta{}G$ values for each half-reaction may be summed, to yield the combined $\Delta{}G$ of the coupled reaction.

One simple example of the coupling of reaction is the decomposition of calcium carbonate:

$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)} \;\;\;\;\;\;\; \Delta G^o = \pu{130.40 kJ/mol}$$

The strongly positive $\Delta{}G$ for this reaction is reactant-favored. If the temperature is raised above $\pu{837 ºC}$, this reaction becomes spontaneous and favors the products. Now, let's consider a second and completely different reaction that can be coupled ot this reaction. The combustion of coal released by burning the coal $\Delta{}G^\circ = \pu{−394.36kJ/mol}$ is greater than the energy required to decompose calcium carbonate ($\Delta{}G^\circ{} = \pu{130.40kJ/mol}$).

$$ \ce{C(s) + O2 <=> CO2(g)} \;\;\;\;\;\;\; \Delta{}G^\circ = \pu{-394.36 kJ/mol}$$

If reactions 19.8.1 and 19.8.2 were added

$$\ce{CaCO3(s) + C(s) + O2 <=> CaO(s) + 2CO2 (g)} \;\;\;\; \Delta{}G^\circ{} = \pu{-263.96 kJ/mol}$$

and then Hess's Law were applied, the combined reaction (Equation 19.8.3) is product-favored with $\Delta{}G^\circ = \pu{−263.96 kJ/mol}$. This is because the reactant-favored reaction (Equation 19.8.2) is linked to a strong spontaneous reaction so that both reactions yield products. Notice that the $\Delta{}G$ for the coupled reaction is the sum of the constituent reactions; this is a consequence of Gibbs energy being a state function:

$$ \Delta{}G^\circ = (\pu{130.40 kJ/mol}) + (\pu{-394.36 kJ/mol}) = \pu{-263.96 kJ/mol}$$

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    $\begingroup$ Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. $\endgroup$
    – Buttonwood
    Jan 2 at 21:29
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    $\begingroup$ An other name for $\ce{CaO}$ than calcium oxide or quick lime is burnt lime; the process is known as calcination. $\endgroup$
    – Buttonwood
    Jan 2 at 21:34
  • $\begingroup$ The question here is not entirely clear. Do you understand the concept of free energy change during a reaction? When it is negative, the reaction proceeds spontaneously to products. The details of the reaction are somewhat secondary, but you might speed it up (if just initially) by igniting the coal. The assumption is constant T and p, and no exchange of matter with the surroundings. $\endgroup$
    – Buck Thorn
    Jan 18 at 12:54
  • $\begingroup$ @BuckThorn Yes that's exactly the point of the question--can you assume that a negative change in G means the reaction will actually take place. And if not, then what makes coupled reactions coupled? $\endgroup$
    – user796099
    Jan 20 at 0:45
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    $\begingroup$ This might be helpful: iubmb.onlinelibrary.wiley.com/doi/full/10.1002/bmb.5 I think the answer is no, they are not coupled unless you have a way of channeling the free energy from one reaction to drive the second. $\endgroup$
    – Buck Thorn
    Jan 20 at 17:57

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Those are not Coupled reactions; the burning of carbon simply supplies the energy to decompose the carbonate. Another way of looking at it would be if you were the fireman on an old steam locomotive would you prefer to shovel pure coal or a mix of coal and limestone into the firebox for several hours trying to keep the train on time. The reactions are independent except for the increased CO2 inhibiting carbonate decomposition.

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  • $\begingroup$ Okay, thanks. So what makes reactions coupled? Also, I assume I would prefer pure coal, but I don't really see what you are getting at with that example. Can you spell it out a little? $\endgroup$
    – user796099
    Jan 20 at 0:52
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    $\begingroup$ My idea of coupled reactions is when the overall reaction is broken into lower energy steps and the products of one step are the reactants of a subsequent step. The best examples are cell respiration [Krebs cycle] and photosynthesis. Many catalytic processes fit this description but are hard to analyse because of increased reaction rates and that the consecutive reactions are happening more or less together. One reaction that has always bothered me is the corrosion of Fe to rust, Fe2O3. This is usually written as a composite equation combining water and O2. It probably isn't. [out of space] $\endgroup$
    – jimchmst
    Jan 23 at 1:46

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