# In the following reaction, will the ratio between substances be 3:1?

In the reaction $$\ce{3 C2H2(g) <=> C6H6(g)}$$ will the ratio of $\ce{[C2H2(g)]}$ to $\ce{[C6H6(g)]}$ at equilibrium be 3:1?

How is the mole ratio related to the equilibrium (or lack thereof) of the reaction? How do I know if 3:1 is indeed the ratio? Thanks.

A chemical equilibrium appears to have a static composition, which is of course not true. The point in question is where the rates of the reaction equal in both directions. The equilibrium can be described by the equilibrium constant $K$, which is defined as the product of partial pressure $p_{\ce{B}}$, concentration $c_{\ce{B}}$, [...], chemical activity $a_{\ce{B}}$, in general $x_{\ce{B}}$ of the reactants and products, with respect to their stoichiometric number $\nu_{\ce{B}}$. $$K_x = \prod_{\ce{B}}(x_{\ce{B}})^{\nu_{\ce{B}}}$$ The standard equilibrium constant $K^\circ$ is given through the standard reaction Gibbs energy $\Delta_r G^\circ$ and the thermodynamic temperature $T$, $\mathcal{R}$ is the gas constant. $$K^\circ = \exp\left\{\frac{-\Delta_r G^\circ}{\mathcal{R}\cdot T}\right\}$$ Since all equilibrium constants are proportional to each other it follows for a constant temperature, $T = \mathrm{const.}$, that the location is dependent on the Gibbs energy: $$\prod_{\ce{B}}(x_{\ce{B}})^{\nu_{\ce{B}}} \propto \exp\left\{-\Delta_r G^\circ\right\}$$
In your case $$\ce{3 C2H2(g) <=> C6H6(g)}$$ you would probably choose a pressure based approach $$K_p = \frac{p(\ce{C6H6})}{p^3(\ce{C2H2})} \propto \exp\left\{-\Delta_r G^\circ\right\}$$