The Equilibrium Constant
You leave a soda bottle tightly closed. Inside, CO₂ dissolves into the liquid and, at the same time, other CO₂ molecules escape into the air in the bottle. Even though everything looks still, it isn't. How can a system "not change" and still be full of activity? And in a chemical reaction, how can you tell whether it has already reached that point or still has a way to go? We'll find out using a single quantity: the equilibrium constant.
Dynamic Equilibrium
Many reactions are reversible: reactants form products, but the products can also turn back into reactants. They are written with a double arrow:
$$\text{N}_2\text{O}_4 \rightleftharpoons 2\ \text{NO}_2$$Imagine you start with only $\text{N}_2\text{O}_4$ (colorless). At first, the forward reaction is fast because there is a lot of reactant. As $\text{NO}_2$ (brown) appears, the reverse reaction begins to occur and speeds up. A moment comes when the two rates are equal.
Notice: at equilibrium the concentrations are constant, but not necessarily equal. They can end up as almost all reactants, almost all products, or a mixture.
The Expression for K_c
When you measure a system at equilibrium at a fixed temperature, a regularity appears: even if the initial concentrations change, a certain combination of the final concentrations always gives the same number. That combination is the equilibrium constant.
For a general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$:
Why are the coefficients exponents? Because if 3 molecules of $\text{H}_2$ are needed to react, there are many more possible combinations of collisions: the concentration of $\text{H}_2$ counts three times in the result.
Write the expression of the equilibrium constant $K_c$ for the synthesis of ammonia: $\text{N}2 + 3\ \text{H}2 \rightleftharpoons 2\ \text{NH}3$.
Write the expression of the equilibrium constant $K_c$ for the synthesis of ammonia: $\text{N}_2 + 3\ \text{H}_2 \rightleftharpoons 2\ \text{NH}_3$.
Calculating K_c from Data
If you are given the equilibrium concentrations, you just substitute and calculate.
How to interpret the value:
| Value of $K_c$ | What it indicates |
|---|---|
| Much greater than 1 | At equilibrium there are mostly products |
| Close to 1 | Comparable amounts of both |
| Much less than 1 | At equilibrium there are mostly reactants |
$K_c$ does not tell you how fast equilibrium is reached, only where it ends up.
At a certain temperature, the equilibrium $\text{PCl}5 \rightleftharpoons \text{PCl}3 + \text{Cl}2$ reaches these concentrations:…
At a certain temperature, the equilibrium $\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2$ reaches these concentrations: $[\text{PCl}_5] = 0.20\ \text{mol}/\text{L}$, $[\text{PCl}_3] = 0.10\ \text{mol}/\text{L}$ and $[\text{Cl}_2] = 0.10\ \text{mol}/\text{L}$. Calculate the value of $K_c$ at that temperature.
The Reaction Quotient Q
What happens if the system is not yet at equilibrium? You calculate the reaction quotient $Q$ with the same formula as $K_c$, but using the concentrations at that moment.
The logic is that of a scale: $Q$ measures how much product there is in proportion; if there is too much, the system sends it back; if there is too little, it makes more.
For the reaction $2\ \text{SO}2 + \text{O}2 \rightleftharpoons 2\ \text{SO}3$ at a certain temperature, $K_c = 4.0$. At a given moment the concentrations are…
For the reaction $2\ \text{SO}_2 + \text{O}_2 \rightleftharpoons 2\ \text{SO}_3$ at a certain temperature, $K_c = 4.0$. At a given moment the concentrations are $[\text{SO}_2] = 0.40\ \text{mol}/\text{L}$, $[\text{O}_2] = 0.20\ \text{mol}/\text{L}$ and $[\text{SO}_3] = 0.80\ \text{mol}/\text{L}$. a) Calculate the reaction quotient $Q$ with those values. b) Decide whether the system is at equilibrium and, if it is not, state which way it will shift (toward the products or toward the reactants) and why.
Back to the beginning: in the soda bottle, CO₂ passes into the liquid and out into the gas at the same rate; no change is noticeable, but the system is in dynamic equilibrium. In a reaction, $K_c$ tells you where that equilibrium ends up, and $Q$ tells you whether you have already arrived or which way it still has to go.