# Reaction quotient

In a chemical reaction, we can tell if the system has reached equilibrium by evaluating the equilibrium law expression and comparing it to $\displaystyle K$. If there is still too much of the reactants and not enough of the products, we say that the system must *shift right* (or the reaction must proceed to the right) to reach equilibrium. If it is the other way around, the system must *shift left*.

For example, a chemical reaction that begins with only reactants and no products must shift right because *some* products have to form, otherwise the chemical equation would be wrong.

When we evaluate the equilibrium law expression using the current concentrations (which may or may not be at equilibrium), we call the result the reaction quotient ($\displaystyle Q$). It is exactly the same as $\displaystyle K$ except that the system is not necessarily at equilibrium.

We compare $\displaystyle Q$ to $\displaystyle K$ to see where the system is relative to equilibrium:

- $\displaystyle Q = K \Rightarrow$ equilibrium
- $\displaystyle Q < K \Rightarrow$ left of equilibrium (must shift right to reach it)
- $\displaystyle Q > K \Rightarrow$ right of equilibrium (must shift left to reach it)

## Example

In the reaction N_{2(g)} + 3 H_{2(g)} ⇌ 2 NH_{3(g)}, the equilibrium constant is 0.40 at 500 ºC. The concentrations are currently 0.10 mol/L N_{2(g)}, 0.30 mol/L H_{2(g)}, and 0.20 mol/L NH_{3(g)}. Is this system at equilibrium?

Calculate the reaction quotient using the equilibrium law expression:

$\displaystyle Q = \frac{\left [ \text{NH}_{3} \right ]^{2}}{\left [ \text{N}_{2} \right ] \left [ \text{H}_{2} \right ]^{3}} = \frac{\left ( 0.20 \, \text{mol/L} \right )^{2}}{\left ( 0.10 \, \text{mol/L} \right ) \left ( 0.30 \, \text{mol/L} \right )^{3}} = 15$.

Since $\displaystyle Q > K$, the system is not at equilibrium. There is too much NH_{3(g)} and not enough N_{2(g)} and H_{2(g)}, so the system must shift left to reach equilibrium.