For this reaction, the standard state is a hypothetical one: All (gas) components are present as pure at standard pressure. Of course, you can create that state, but it would not react because each species is in its separate container.
You can still determine the standard Gibbs energy by extrapolation. The easiest way is to let a gas mixture come to equilibrium (so the Gibbs energy is zero), measure the concentrations, assume an ideal gas, and plug them into the equation below:
$$\Delta_\mathrm rG^⦵ = -RT\ln K \tag{1}$$
This may be derived from
$$\Delta_\mathrm rG = \Delta_\mathrm rG^⦵ + RT\ln Q \tag{2}$$
which is the "recipe" to calculate the Gibbs energy of reaction at any state from that of the standard state (assuming ideal gas or solution), combined with the fact that at equilibrium, $\Delta_\mathrm rG$ is equal to zero and $Q = K$.
[comment in the linked question] how can gases start out as pure component, they have to mix for the reaction to occur? Also, how can each of the products and reactants have 1 bar pressure, the pressure would depend on the no. Of moles of the different reactants and products(which are not the same for all reactions)?
The Gibbs energy of reaction is a state function. You can start with the standard state, mix the gases, let the reaction proceed for a while, and then separate the gases again. For each step, you have to measure the change in Gibbs energy, or calculate it. Then you add them all up to get the somewhat surprising net reaction of pure, separated substances reacting to make a bit more of the product, yielding pure, separated substances again. Curiously, you never actually have to do this experiment. The Gibbs energy of mixing is baked into the formulas above.