Converting Kp & Kc — Convert between Kp and Kc - Chemistry 2

SECTION 1

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.

1864 • The Law of Mass Action

Cato Guldberg and Peter Waage proposed that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants, laying the groundwork for the equilibrium constant Kc.

1884 • Le Chatelier's Principle

Henri Le Chatelier described how equilibrium systems respond to changes in concentration, pressure, and temperature, deepening the understanding of dynamic equilibrium.

1901 • Kp for Gas-Phase Reactions

Chemists increasingly expressed gaseous equilibria using partial pressures, defining Kp as a convenient alternative to Kc for reactions involving gases.

1920s • Formal Kp–Kc Relationship

Using the ideal gas law (PV = nRT), scientists derived the precise mathematical relationship between Kp and Kc, enabling seamless conversion between the two constants.

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.

SECTION 2

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

Kc uses molar concentrations (mol/L) of gases and aqueous species in the equilibrium expression. Solids and pure liquids are excluded.

Kp — Pressure Constant

Kp uses partial pressures (typically in atm) of gaseous species. It applies only to reactions involving at least one gas-phase component.

Δn — Change in Moles of Gas

Δn equals the total moles of gaseous products minus the total moles of gaseous reactants. This value determines how Kp and Kc relate to each other.

R and T — Gas Law Constants

R is the ideal gas constant (0.0821 L·atm/mol·K) and T is the temperature in Kelvin. Both come directly from PV = nRT.

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KEY TAKEAWAY

Think of Kp and Kc like measuring the same room's size in square feet versus square meters. The room hasn't changed — you're just using different units. The conversion factor (RT)^Δn acts like the exchange rate between the two "currencies." When Δn = 0, both constants are equal because no net change in gas moles means pressure and concentration stay perfectly proportional.

SECTION 3

Visualizing the Kp–Kc Relationship

This diagram traces the logical path from the ideal gas law to the Kp–Kc conversion formula. Notice how the three scenarios at the bottom show how the sign of Δn determines whether Kp is greater than, equal to, or less than Kc.

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.

SECTION 4

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.

IDEAL GAS LAW REARRANGED

P = (n / V) × R × T = C × R × T

Here C = n/V is the molar concentration (mol/L), R = 0.0821 L·atm/(mol·K), and T is temperature in Kelvin.

MASTER CONVERSION FORMULA

Kp = Kc × (RT)^Δn

Δn = (sum of coefficients of gaseous products) − (sum of coefficients of gaseous reactants). Use R = 0.0821 L·atm/(mol·K) when pressures are in atm.

SOLVING FOR Kc

Kc = Kp / (RT)^Δn

Simply rearrange the master formula by dividing both sides by (RT)^Δn. This form is useful when you are given Kp and need Kc.

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Watch Your Units!

Always use the value of R that matches your pressure units. If Kp is expressed in atm, use R = 0.0821 L·atm/(mol·K). If Kp is in kPa, use R = 8.314 L·kPa/(mol·K). Mixing unit systems is one of the most common mistakes students make with this formula.

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.

SECTION 5

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.

Three reactions illustrating different Δn outcomes. Example A has Δn = −2 (fewer gas moles in products), Example B has Δn = 0 (balanced gas moles), and Example C shows how to handle solids — only CO₂(g) counts toward Δn.

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Common Pitfall

Students often forget to exclude solids and liquids when calculating Δn. Remember: if a species has (s) or (l) after its formula, it does not contribute to Δn. Only (g) species matter.

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

Additional examples for computing Δn

SECTION 6

Worked Example — Step by Step

Let's work through a complete conversion problem from start to finish.

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Problem Statement

For the reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), the value of Kc at 500 K is 0.040. Calculate Kp at 500 K. (Use R = 0.0821 L·atm/mol·K)

Converting Kc to Kp for the Haber Process

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Step 1 — Write the Balanced Equation and Identify Gas-Phase Species

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g). All three species are gases, so all three contribute to Δn.
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Step 2 — Calculate Δn

Δn = (moles of gaseous products) − (moles of gaseous reactants) = 2 − (1 + 3) = 2 − 4 = −2.

Δn = −2
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Step 3 — Calculate RT

RT = (0.0821 L·atm/mol·K) × (500 K) = 41.05 L·atm/mol.

RT = 41.05
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Step 4 — Calculate (RT)^Δn

(RT)^Δn = (41.05)⁻² = 1 / (41.05)² = 1 / 1685.1 = 5.934 × 10⁻⁴.

(RT)^Δn = 5.934 × 10⁻⁴
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Step 5 — Apply the Conversion Formula

Kp = Kc × (RT)^Δn = 0.040 × 5.934 × 10⁻⁴ = 2.4 × 10⁻⁵.

Kp = 2.4 × 10⁻⁵

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.

SECTION 7

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

Side-by-side comparison of Kc and Kp

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KEY TAKEAWAY

Kp and Kc are like Celsius and Fahrenheit — different scales describing the same temperature. You can always convert between them using the right formula. The crucial factor is Δn: when it's zero, the two scales happen to give the same number, just like 40° below zero is the same in both Celsius and Fahrenheit.

SECTION 8

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

How this lesson connects to more advanced topics

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.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

For the reaction H₂(g) + Cl₂(g) ⇌ 2 HCl(g), will Kp equal Kc, be greater than Kc, or be less than Kc? Explain your reasoning without doing any numerical calculation.

Show answer

Kp will equal Kc. The reaction has 2 moles of gaseous products (2 HCl) and 2 moles of gaseous reactants (1 H₂ + 1 Cl₂), so Δn = 2 − 2 = 0. When Δn = 0, (RT)⁰ = 1, and the conversion formula Kp = Kc × 1 simply gives Kp = Kc. Equal gas moles on both sides means the pressure and concentration expressions yield the same numerical value.

Question 2: BASIC CALCULATION

For the decomposition PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Kc = 0.0211 at 500 K. Convert this to Kp. (R = 0.0821 L·atm/mol·K)

Show answer

First, calculate Δn = (1 + 1) − 1 = +1. Then RT = 0.0821 × 500 = 41.05. Since Δn = 1, (RT)¹ = 41.05. Finally, Kp = Kc × (RT)^Δn = 0.0211 × 41.05 = 0.866. Notice Kp > Kc because Δn is positive, meaning the products have more gas moles.

Question 3: INTERMEDIATE

For the reaction 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), Kp = 3.0 × 10⁴ at 700 K. Find Kc. (R = 0.0821 L·atm/mol·K)

Show answer

Calculate Δn = 2 − (2 + 1) = −1. Rearrange the formula: Kc = Kp / (RT)^Δn. Compute RT = 0.0821 × 700 = 57.47. Then (RT)⁻¹ = 1/57.47 = 0.01740. So Kc = 3.0 × 10⁴ / 0.01740 = 3.0 × 10⁴ × 57.47 = 1.72 × 10⁶. Since Δn is negative, Kc is larger than Kp, which confirms our calculation.

Question 4: APPLIED

In an industrial setting, the water-gas shift reaction CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has Kp = 1.0 at 700 K. An engineer needs Kc to design a reactor using concentration data. What is Kc? What does this result tell you about this reaction's Δn?

Show answer

Calculate Δn = (1 + 1) − (1 + 1) = 0. When Δn = 0, (RT)⁰ = 1, so Kc = Kp / 1 = 1.0. This result tells the engineer that Kp and Kc are numerically identical for this reaction, meaning temperature does not create a discrepancy between the two constants. The equal number of gas moles on each side is the reason. In reactor design, this simplifies calculations because pressure and concentration data give the same equilibrium ratio.

Question 5: CRITICAL THINKING

Consider two reactions at the same temperature: Reaction 1 has Kc = 50 and Δn = +2, while Reaction 2 has Kc = 50 and Δn = −2. Without calculating exact values, explain how the Kp values of these two reactions compare to each other and to Kc. Then predict: if temperature increases, how does the conversion factor (RT)^Δn change for each reaction?

Show answer

For Reaction 1 (Δn = +2), Kp = Kc × (RT)², so Kp is much larger than 50 because (RT)² is a large number at typical temperatures. For Reaction 2 (Δn = −2), Kp = Kc × (RT)⁻² = Kc / (RT)², so Kp is much smaller than 50. The two Kp values are inverses of each other relative to Kc — one is amplified and the other is diminished by the same factor. If temperature increases, RT increases. For Reaction 1, (RT)² grows larger, pushing Kp even further above Kc. For Reaction 2, (RT)⁻² shrinks further, pushing Kp even further below Kc. This illustrates that the sign of Δn determines whether a temperature increase widens or narrows the Kp–Kc gap.

SUMMARY

Lesson Summary

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.