# Does this coupled reaction actually happen (decomposition of calcium carbonate)

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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• An other name for $\ce{CaO}$ than calcium oxide or quick lime is burnt lime; the process is known as calcination. Jan 2 at 21:34
• 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. Jan 18 at 12:54
• @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? Jan 20 at 0:45
• 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. Jan 20 at 17:57