E - The Reaction Quotient, Q (Lesson)

The Reaction Quotient, Q

In previous examples, ICE tables were used to calculate the concentration of all components of a chemical equilibrium when given the equilibrium constant K for the equilibrium process. In all cases, it was easy to determine the direction in which each reaction would shift to achieve equilibrium because the initial amount of at least one substance was zero. Therefore, in order to achieve equilibrium, the system would shift to that side. However, when confronted with a situation in which there are both reactants AND products present initially, how can the direction of shift be determined? Take for example the equilibrium shown below:

2 NO₂ (g) $LaTeX: \rightleftharpoons$ $\rightleftharpoons$ N₂O₄ (g)

What if, in this example, the initial concentration of both NO₂ (g) and N₂O₄ (g) was 0.5 M? How could the direction of shift be determined? To answer this question, a new parameter, referred to as the reaction quotient (Q), must be introduced. The reaction quotient is very similar to the equilibrium constant with the exception of one key concept: The reaction quotient is a ratio of all equilibrium components when the system is NOT at equilibrium.

When evaluating the figure below, we get a visual understanding of the difference between the reaction quotient, Q, and the equilibrium constant, K. Notice how K is used to evaluate the product/reactant ratio at equilibrium when concentrations are no longer changing whereas Q is used to evaluate the same ratio when the system has not yet reached equilibrium.

ReactionQGraph

With this in mind, the example mentioned above can be revisited. How can the direction of shift be determined? First, a Q expression must be written in the exact same fashion as an equilibrium expression is written:

$LaTeX: Q=\frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}$ $Q=\frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}$

Then, the reaction quotient can be calculated by substituting in the values of all substances involved. As mentioned above, the concentrations of both N₂O₄ and NO₂ were 0.5 M:

$LaTeX: Q=\frac{(0.5)}{(0.5)^{2}}=2$ $Q=\frac{(0.5)}{(0.5)^{2}}=2$

The value calculated in and of itself does not mean very much, but this value must be compared to the equilibrium constant, K, in order to determine the direction of shift. There are three possibilities:

Q < K and the reaction will proceed to the products. The equilibrium will shift to the right.
Q > K and the reaction will proceed to the reactants. The equilibrium will shift to the left.
Q = K and the reaction is already at equilibrium. There will be no shift in either direction.

At 418 K the equilibrium constant for the reaction example above is 0.32. Therefore, in this situation, Q > K, and the reaction will shift to the left. This means the concentration of NO₂ will increase in order to achieve equilibrium while the concentration of N₂O₄ will decrease.

You Try It!

In the following self-assessment activity, determine whether the mixture is in equilibrium; if it is not, determine the direction the reaction will shift. Click on the plus sign to check your answer!

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