# Why is the reaction quotient Q always equal to one at standard state?

I understand that at equilibrium, $$Q$$ is equal to $$K$$. This should not depend how I express concentrations (as amount of substance concentrations $$c$$ or as partial pressures $$p$$ or as molality $$b$$ or as mole fraction $$x$$) and which units I use (e.g. partial pressure in Pa or in atm) as long as I am consistent.

Various expressions rely on $$Q$$ being dimensionless (because we take its logarithm) and on being equal to one when all species are at standard state. For example, the equation for calculating the change in Gibbs free energy for a reaction should result in the standard Gibbs free energy for the standard state:

$$\Delta_r G = \Delta_r G^\circ + R T \ln Q = \Delta_r G^\circ \text{(for standard state)}$$

This requires that $$Q = 1$$ at standard state. The Nernst equation is another (related) example.

What I am wondering about is why $$Q$$ is always equal to one at standard state, no matter how I express concentrations when calculating reaction quotients.

For example, if I have a reaction like $$\ce{A(g) + B(g) -> C(g)}$$, doesn't the value of $$Q$$ change if I switch from concentration $$c$$ to partial pressure $$p$$, or use units of atm instead of Pa when measuring partial pressures? Come to think of it, doesn't the value of the equilibrium constant $$K$$ change as well when I change measures or units of concentration?

• This question is related, but not identical, to this question – Karsten Theis Apr 24 '19 at 10:58
• $Q$ is derived from activities and/or fugacities. You cannot express concentrations exactly as you want. You need to use the proper activity (concentration in molar) or fugacity (pressure in bar). Of course, these approximations are generally only valid at lower concentrations/pressures. – Zhe Apr 24 '19 at 12:58