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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.

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No. The location of the equilibrium is dependent on many factors, most prominently temperature and pressure.

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 $$K_p = \frac{p(\ce{C6H6})}{p^3(\ce{C2H2})} \propto \exp\left\{-\Delta_r G^\circ\right\}$$

And you can see the temperature and pressure dependency in the original definition: \begin{align} G(p,T) &= H - TS\\ H(p,T) &= U + pV\\ \end{align}

So you will only know the product ratio by very accurate calculations or empirical measurement.

For the system in question, I believe it has been done and the results may be found in W. J. Taylor, et. al., J. Res. Natl. Bur. Stand., Vol. 37, No. 2, p. 95.

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