In which direction must the reaction proceed to reach equilibrium?

Question

K= $1.2*10^3$ at $400^@C$ for the reaction to produce the highly toxic nerve gas phosgene:

CO(g) + $Cl_2$ (g) $->$ $COCl_2$ (g)

If the concentration of all 3 species is each 0.050 M, in which direction must the reaction proceed to reach equilibrium

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Answer 1 · 2018-05-07T12:19:03

The reaction will shift to the right in order to achieve equilibrium.

This is our equilibrium:

$CO(g) + Cl_2(g) rightleftharpoons COCl_2(g)$

From this, we can determine that the expression for $K_c$ is this:

$K_c = ([COCl_2])/([CO] xx [Cl_2]) = 1.2 xx 10^3$

Right now, all of our concentrations have a concentration of $"0.050 M"$ . So, the value of our $Q_c$ expression would be:

$([COCl_2])/([CO] xx [Cl_2]) = ("0.050 M")/("0.050 M" xx "0.050 M") = 2.0 xx 10$

The value of $Q_c$ is much less than $K_c$ , which means that the concentration of products will be much higher than reactants at equilibrium.

So, to produce more products and reduce the concentration of reactants, the forward/right reaction would be favoured.