= Solution
Introduce an <extent of reaction> $\xi$ increasing in the dissociation direction. The allowed particle-number changes conserve both carbon and <oxygen> atoms:
$$
dN_c=-2\,d\xi,\qquad dN_{\rm CO}=2\,d\xi,\qquad dN_{\rm O_2}=d\xi.
$$
At fixed <temperature> and <volume>, the resulting <Helmholtz free energy> change is
$$
dF=\left(-2\mu_c+2\mu_{\rm CO}+\mu_{\rm O_2}\right)d\xi.
$$
At an interior <chemical equilibrium>, both forward and backward changes are allowed, so the coefficient must vanish:
$$
\boxed{2\mu_{\rm CO_2}=2\mu_{\rm CO}+\mu_{\rm O_2}.}
$$
Equality of the three individual <chemical potentials> would be wrong: a reaction exchanges species in its stoichiometric proportions rather than <molecule> for <molecule>. Stability is also explicit here, since
$$
\frac{d^2F}{d\xi^2}=k_BT\left(\frac4{N_c}+\frac4{N_{\rm CO}}+\frac1{N_{\rm O_2}}\right)>0
$$
when all three species are present. Thus the equilibrium composition minimizes the free <energy> along the conserved-composition constraint.
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