Equilibrium reaction where the number of moles of reactants and products are the same
A simple example:
If $T &gt; \pu{100 °C},$ and $p \leq \pu{1 atm},$ then water will be in the gas phase and:
$$\ce{CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)}$$
Note, however, that while this reaction's equilibrium will be highly insensitive to pressure changes (within the regime that all molecules remain in the gas phase), it will not be entirely insensitive to pressure.
That's because, even though there are the same number of moles of gas on the LHS and RHS, these are real gases, and thus their molar volumes will differ slightly. This is a small effect compared to the effect of actually having different numbers of moles of gases on the LHS and RHS.