Historical Context & Motivation
When chemists first began studying reversible reactions in the 1800s, they needed a way to describe how far a reaction proceeds before reaching a state of balance. Early pioneers realized that the concentrations of reactants and products at equilibrium follow predictable mathematical patterns. However, a problem arose when dealing with gaseous reactions: should chemists express equilibrium in terms of molar concentrations or partial pressures? Both approaches are valid, but each gives a different numerical value for the same reaction. This created a need for a reliable formula to convert between the two expressions.
The central question this lesson addresses is straightforward: if you know the equilibrium constant in one form, how do you find it in the other? The answer hinges on a surprisingly simple connection between concentration and pressure provided by the ideal gas law.
Core Principles & Definitions
Before converting between Kp and Kc, you need to understand what each constant represents and why they can differ numerically. Both constants describe the same equilibrium state, but they use different units to measure the amounts of reactants and products.
Kc — Concentration Constant
Kp — Pressure Constant
Δn — Change in Moles of Gas
R and T — Gas Law Constants
Visualizing the Kp–Kc Relationship
The diagram above reveals the core logic. Since the ideal gas law tells us that partial pressure equals concentration multiplied by RT, we can substitute this relationship into any equilibrium expression. Every factor of concentration in Kc picks up one factor of RT when converted to a pressure-based expression. The net number of such factors is exactly Δn, which is why the exponent on (RT) is Δn. If there are equal moles of gas on both sides of the equation, all the RT factors cancel out, and Kp equals Kc exactly.
Mathematical Framework
The derivation begins with the ideal gas law and ends with a clean conversion equation. Understanding each step will help you remember the formula rather than just memorizing it.
Notice that when Δn = 0, the term (RT)0 equals 1, so Kp = Kc. This makes intuitive sense: if the number of gas moles doesn't change during the reaction, there's no "extra" pressure effect to account for.
Calculating Δn — The Critical Step
The most important step in any Kp–Kc conversion is correctly calculating Δn. A mistake here will throw off the entire calculation. Only count species that are in the gas phase; solids (s) and liquids (l) do not appear in either Kp or Kc expressions and must be ignored when computing Δn.
| Reaction | Gaseous Products (moles) | Gaseous Reactants (moles) | Δn |
|---|---|---|---|
| 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) | 2 | 3 | −1 |
| PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) | 2 | 1 | +1 |
| CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) | 2 | 2 | 0 |
Worked Example — Step by Step
Let's work through a complete conversion problem from start to finish.
Notice that Kp is much smaller than Kc here. This makes sense because Δn is negative: the reaction produces fewer moles of gas than it consumes, so expressing the equilibrium in terms of pressure "shrinks" the constant's value.
Comparing Kp and Kc — When to Use Which
Both Kp and Kc describe the same equilibrium, so choosing between them depends on the information you have and what's most convenient for the problem at hand.
| Feature | Kc | Kp |
|---|---|---|
| Quantity used | Molar concentration (mol/L) | Partial pressure (atm, kPa, etc.) |
| Applies to | Any phase (g, aq) | Gas-phase reactions only |
| Preferred when | Concentrations are given or needed | Pressures are given or gas behavior is the focus |
| Temperature dependence | Changes with temperature | Changes with temperature |
| Equal when | Δn = 0 | Δn = 0 |
Connecting to Thermodynamics & Beyond
The Kp–Kc relationship is your first glimpse into deeper thermodynamic ideas. In more advanced chemistry courses, you'll encounter the connection between equilibrium constants and the Gibbs free energy (ΔG°). The equation ΔG° = −RT ln K directly ties the spontaneity of a reaction to its equilibrium constant. The version of K used in this equation is actually a dimensionless "thermodynamic equilibrium constant" related to activities, but for ideal gases, it is numerically equal to Kp when pressures are measured relative to 1 atm.
| Concept | What You Know Now | What's Coming Next |
|---|---|---|
| Equilibrium constant | Kp and Kc describe product-to-reactant ratios at equilibrium | K is connected to ΔG° via ΔG° = −RT ln K |
| Ideal gas assumption | Kp = Kc(RT)^Δn assumes ideal gas behavior | Real gases use fugacity instead of pressure for more accurate results |
| Temperature effects | T appears in the conversion formula and equilibrium shifts | The van 't Hoff equation quantifies how K changes with temperature |
For now, the key point is that mastering the Kp–Kc conversion gives you a solid foundation. The skills you practice here — identifying gas-phase species, computing Δn, and handling exponents — will directly carry over when you study reaction energetics and advanced equilibrium theory.
Practice Problems
Lesson Summary
The equilibrium constant Kc is expressed in molar concentrations, while Kp uses partial pressures. They are connected by the formula Kp = Kc × (RT)^Δn, which is derived directly from the ideal gas law. The key variable is Δn, the difference between moles of gaseous products and moles of gaseous reactants. Only gas-phase species count toward Δn — solids and liquids are excluded.
When Δn = 0, Kp equals Kc. When Δn > 0, Kp is larger than Kc. When Δn < 0, Kp is smaller than Kc. Always use R = 0.0821 L·atm/(mol·K) when pressures are in atmospheres, and remember that T must be in Kelvin. Mastering this conversion prepares you for advanced topics like Gibbs free energy and thermodynamic equilibrium.