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