Equilibrium in Chemical Processes

Not every chemical reaction runs to completion. Place hydrogen and iodine vapour in a sealed flask, or seal colourless $\text{N}_2\text{O}_4$ gas in a bulb, and the reaction seems to stop partway — yet on the molecular scale nothing has stopped at all. Forward and reverse reactions are both still happening, at exactly matched rates, so the measurable amounts stop changing. This is chemical equilibrium: dynamic, not dead. This article extends the physical-equilibrium idea to reacting systems, states the law of mass action that Guldberg and Waage proposed, and shows how balancing forward and reverse rates hands us the constant ratio of concentrations that becomes the equilibrium constant, $K_c$ — the number the next article turns into real calculations.

The core insight

In a closed reversible reaction, chemical equilibrium is reached when forward and reverse rates become equal, and the law of mass action's rate expressions then combine into a fixed ratio of product to reactant concentrations, the equilibrium constant.

The core idea

A reversible reaction proceeds in both directions under the same conditions and is written with a double arrow:

\text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}

In a closed vessel, the forward reaction starts fast because reactant concentrations are highest. As C and D accumulate, the reverse reaction speeds up while the forward reaction slows as A and B are used up. Eventually the two rates coincide:

\text{Rate}_{\text{forward}} = \text{Rate}_{\text{backward}}

where: the forward rate depends on how much A and B remain, and the backward rate on how much C and D have formed. At this point chemical equilibrium is established — concentrations of A, B, C and D stop changing, even though molecules keep converting both ways. Equilibrium is dynamic, exactly as in physical equilibrium (liquid–vapour, solid–solution). The classic proof of this dynamism is the reaction between hydrogen and iodine vapour, studied by Max Bodenstein:

\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}

Whether you start with only $\text{H}_2$ and $\text{I}_2$ , only $\text{HI}$ , or a mixture of all three, the system settles at the same equilibrium composition at a given temperature. Reaching the identical state from opposite starting points is only possible if both reactions are genuinely running, in balance, throughout.

Figure

Two curves on a graph showing reactant concentration falling and product concentration rising then both flattening at the same time to mark equilibrium

Law of mass action. Proposed by Guldberg and Waage, this states that the rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to a power equal to its stoichiometric coefficient. For a general reaction

a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}

the forward rate is proportional to $[\text{A}]^a[\text{B}]^b$ and the backward rate is proportional to $[\text{C}]^c[\text{D}]^d$ . where: square brackets denote molar concentration, and $a, b, c, d$ are the balanced-equation coefficients, not arbitrary exponents. This law is the bridge between kinetics (rates) and the equilibrium constant developed below.

Characteristics of chemical equilibrium, paralleling physical equilibrium: it is attainable only in a closed system; it is dynamic, with both reactions continuing at equal, non-zero rates; measurable properties (concentration, colour, pressure) become constant; the same state is reached regardless of the direction approached from; and a catalyst speeds up how quickly equilibrium is reached but does not shift its final position.

Derivation of the equilibrium constant expression

Derivation AFrom equal rates to a constant ratio of concentrations

The goal is to show, using nothing beyond the law of mass action and basic algebra, that once forward and backward rates are equated at equilibrium, the concentration terms rearrange into a single fixed ratio — the equilibrium constant $K_c$ — for the general reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ , so that the definition used numerically in the next article is seen to fall directly out of matching rates rather than being an arbitrary rule.

Step 1. Write the forward and backward rate expressions from the law of mass action. Forward rate $= k_f[\text{A}]^a[\text{B}]^b$ , where $k_f$ is the forward rate constant. Backward rate $= k_b[\text{C}]^c[\text{D}]^d$ , where $k_b$ is the backward rate constant. Both $k_f$ and $k_b$ depend only on temperature, not on concentration.

Step 2. Apply the equilibrium condition. At equilibrium the forward and backward rates are equal, so $k_f[\text{A}]^a[\text{B}]^b = k_b[\text{C}]^c[\text{D}]^d$ , where all concentrations are now the fixed equilibrium values (often written $[\text{A}]_{eq}$ , and so on, though the subscript is usually dropped once the point is made).

Step 3. Isolate the concentration terms by dividing both sides by $k_b[\text{A}]^a[\text{B}]^b$ . This gives $\dfrac{k_f}{k_b} = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ . Because $k_f$ and $k_b$ are both constants at a fixed temperature, their ratio $k_f/k_b$ is itself a constant.

Step 4. Name the constant. Define $K_c = k_f/k_b$ , so

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

This is the equilibrium constant expression: a fixed ratio of product concentrations to reactant concentrations, each raised to its stoichiometric power, valid only at equilibrium and only at a given temperature. It is why the $\text{H}_2/\text{I}_2/\text{HI}$ system always settles at the same ratio $[\text{HI}]^2/([\text{H}_2][\text{I}_2])$ regardless of which side you start from — both starting points obey the same Step 2 equality.

Worked example

Problem: Colourless $\text{N}_2\text{O}_4$ gas is sealed in a glass bulb at fixed temperature for the reaction $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$ . The mixture develops a brown colour that deepens and then stops changing. Using the law of mass action, explain why the colour stabilises, and write the equilibrium constant expression for this reaction.

Step 1. Identify the two rate expressions. By the law of mass action, forward rate $= k_f[\text{N}_2\text{O}_4]$ (coefficient 1), and backward rate $= k_b[\text{NO}_2]^2$ (coefficient 2, so the concentration is squared).

Step 2. Track how the rates change with time. Initially only $\text{N}_2\text{O}_4$ is present, so only the forward reaction runs and the mixture starts turning brown as $\text{NO}_2$ forms. As $\text{NO}_2$ builds up, the backward rate ( $\propto[\text{NO}_2]^2$ ) rises while the forward rate ( $\propto[\text{N}_2\text{O}_4]$ ) falls as $\text{N}_2\text{O}_4$ is consumed.

Step 3. Apply the equilibrium condition. When $k_f[\text{N}_2\text{O}_4] = k_b[\text{NO}_2]^2$ , both concentrations stop changing and the colour intensity becomes constant — this is chemical equilibrium, sustained indefinitely in the sealed, constant-temperature bulb even though molecules keep interconverting.

Step 4. Write the equilibrium constant expression by dividing as in the derivation above:

K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}

The colour stabilises because forward and backward rates become equal, and the resulting fixed ratio is $K_c = [\text{NO}_2]^2/[\text{N}_2\text{O}_4]$ . (If the bulb were opened, escaping gas would break the closed-system condition and no steady colour could be sustained.)

Competitive stretch

For $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$ , Bodenstein found that starting from 1 mol $\text{H}_2$ + 1 mol $\text{I}_2$ in a 1 L flask at 717 K gives the same equilibrium ratio $[\text{HI}]^2/([\text{H}_2][\text{I}_2])$ as starting from 2 mol pure HI in an identical flask at 717 K. By Step 2 of the derivation, both flasks obey $k_f[\text{H}_2][\text{I}_2] = k_b[\text{HI}]^2$ at equilibrium, so both must give the same $K_c = k_f/k_b$ — confirming that $K_c$ at a fixed temperature is independent of starting composition, exactly as the algebra predicts.

Common traps

Believing a reaction "stops" at equilibrium — both reactions are still occurring, just at equal rates.
Assuming equilibrium means reactants and products are present in equal amounts — it only requires equal forward and backward rates; concentrations can differ greatly (as in the $\text{N}_2\text{O}_4$ / $\text{NO}_2$ case).
Confusing the law of mass action (a statement about rates) with the equilibrium constant expression (a statement about concentrations only) — the constant ratio emerges only after setting forward rate equal to reverse rate.
Writing exponents in $K_c$ from anything other than the balanced-equation coefficients — a wrongly balanced equation gives a wrong $K_c$ expression even if the concentrations are correct.

Ask the tutor: want to see the law of mass action turn into real numeric $K_c$ and $K_p$ calculations for the $\text{H}_2$ / $\text{I}_2$ / $\text{HI}$ system — ask the AI tutor to connect this article to "Equilibrium Constant."

If both reactions are still happening at equilibrium, why do concentrations look constant?

Because "constant" here means the net change is zero, not that nothing is happening. Every second, some A and B molecules are converting into C and D, and some C and D molecules are converting back into A and B — but because the forward rate and the backward rate have become exactly equal, these two flows cancel out. The concentrations you would measure in a lab (by colour, pressure or titration) hold perfectly steady, which is why it took careful experiments like Bodenstein's, approaching equilibrium from both directions, to reveal that the system was still dynamic underneath.

Why do we raise concentrations to the stoichiometric coefficient in the rate expression?

The law of mass action reflects how likely molecules are to collide and react. If a reaction needs two molecules of A to meet, doubling $[\text{A}]$ roughly quadruples the chance of a successful two-A collision, so the rate goes as $[\text{A}]^2$ , matching the coefficient 2 in the balanced equation. This is why the derivation writes forward rate as $k_f[\text{A}]^a[\text{B}]^b$ with the balanced-equation coefficients as exponents, and it is also why getting the balanced equation right comes before writing any equilibrium expression — the exponents are not optional or guessable.

Is K_c the same thing as the law of mass action?

No, though they are closely linked. The law of mass action is a statement about rates: forward rate depends on reactant concentrations, backward rate on product concentrations, each raised to a stoichiometric power. $K_c$ only appears afterward, when you set those two rates equal at equilibrium and divide — it is a ratio of concentrations (not rates) and equals $k_f/k_b$ . Students often skip the "setting rates equal" step and jump straight to writing products-over-reactants, which happens to give the right answer here but hides why the expression is valid only at equilibrium and only at a fixed temperature.