Question #0ef97

Jan 26, 2015

Your reaction is at equilibrium.

In order to determine whether or not a reaction is at equilibrium, you must calculate the reaction quotient, or ${\text{Q}}_{c}$. The value of ${\text{Q}}_{c}$ will tell you in which direction the reaction will progress if equilibrium has not yet been reached.

${Q}_{c}$ expressed the ratio of products to reactants at a given instant. If the value you obtain for ${Q}_{c}$ is smaller than ${K}_{e q}$, the equilibrium constant, there are more reactants than products, which will cause the equilibrium to shift to the right, favoring the formation of more products.

If ${Q}_{c}$ is bigger than ${K}_{e q}$, there are more products than reactants, which will cause the equilibrium to shift ot the left, favoring the formation of more reactants.

Finally, if ${Q}_{c}$ is equal to ${K}_{e q}$, the reaction is at equilibrium and no shift will take place.

So, let's calculate ${Q}_{c}$ and compare it with ${K}_{e q}$. Remember that you must express the concentrations of the species in $\text{mol/L}$, not $\text{mmol/L}$

${Q}_{c} = \frac{\left[K\right] \cdot \left[L\right]}{\left[J\right]} = \frac{1.10 \cdot {10}^{- 3} M \cdot 125 \cdot {10}^{- 3} M}{11.0 \cdot {10}^{- 3} M}$

${Q}_{c} = \frac{0.125 \cdot 0.0011}{0.011} = 0.0125$

As you can see, ${Q}_{c}$ is equal to ${K}_{e q}$, which means the reaction is at equilibrium and no shift will take place.