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

SECTION 1

Historical Context & Motivation

The concept of chemical equilibrium was not immediately obvious to early chemists, who often assumed reactions proceeded to completion. The realization that many reactions reach a dynamic balance — where forward and reverse rates equalize — emerged gradually throughout the nineteenth century. As chemists began to quantify equilibrium positions, they naturally confronted a practical question: should the equilibrium constant be expressed in terms of molar concentrations or partial pressures? This question is particularly acute for gas-phase reactions, where both representations are physically meaningful and experimentally accessible.

1864 • Law of Mass Action

Cato Guldberg and Peter Waage proposed the law of mass action, establishing that the rate of a reaction is proportional to the product of the active masses (concentrations) of the reactants, laying the foundation for equilibrium expressions.

1884 • Le Chatelier's Principle

Henri Louis Le Chatelier formalized his principle describing how systems at equilibrium respond to external perturbations, reinforcing the importance of quantifying equilibrium positions precisely.

1901 • van 't Hoff's Thermodynamic Treatment

Jacobus van 't Hoff received the first Nobel Prize in Chemistry in part for his work connecting thermodynamics to equilibrium, demonstrating how the ideal gas law bridges concentration and pressure descriptions of gas-phase equilibria.

1920s • Standardization of Kp and Kc Notation

As physical chemistry matured, the notation Kp (equilibrium constant in terms of partial pressures) and Kc (in terms of molar concentrations) became standard, along with the conversion formula relating them through the ideal gas constant R and temperature T.

The central question that this lesson addresses is both conceptual and practical: given that gas-phase equilibria can be described equally well in terms of concentrations or partial pressures, how do we convert between K_c and K_p? The answer lies in the ideal gas law, which provides a direct algebraic link between concentration (n/V) and pressure (P) for each gaseous species in the equilibrium mixture.

SECTION 2

Core Principles & Definitions

Before deriving the conversion formula, it is essential to establish clear definitions for the two forms of the equilibrium constant and the assumptions that underpin them. Both K_p and K_c describe the same equilibrium position for a gas-phase reaction, but they differ in the units used to express the activities of gaseous species. Understanding these foundational concepts ensures that the mathematical conversion is not merely a formula to memorize but a logical consequence of the ideal gas law.

Kc — Concentration-Based Constant

K_c is defined using equilibrium molar concentrations (mol/L) of gaseous species raised to their stoichiometric coefficients. It is the natural form when working with solution-phase analogies or when concentrations are measured directly.

Kp — Pressure-Based Constant

K_p is defined using equilibrium partial pressures (typically in atm or bar) of gaseous species raised to their stoichiometric coefficients. It is often more convenient experimentally because gas pressures are directly measurable with manometers.

Δn — Change in Moles of Gas

Δn is the difference between the total moles of gaseous products and the total moles of gaseous reactants (Δn = n_products − n_reactants). This value determines whether K_p and K_c are numerically equal or different.

Ideal Gas Assumption

The conversion K_p = K_c(RT)^Δn assumes ideal gas behavior for all gaseous species. At high pressures or low temperatures, fugacity corrections may be required.

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KEY TAKEAWAY

Think of K_p and K_c as two currencies describing the same economic value. Just as you need an exchange rate (say, dollars to euros) to convert between currency systems, you need the factor (RT)^Δn to convert between the concentration and pressure frameworks. When Δn = 0 (equal moles of gas on both sides), the exchange rate is exactly 1, meaning the two constants are numerically identical — analogous to a 1:1 currency peg.

SECTION 3

Visual Explanation

The following diagram illustrates the conceptual relationship between K_p and K_c. At the center sits the ideal gas law, which provides the mathematical bridge between concentration (n/V) and partial pressure (P) for each species. The conversion factor (RT)^Δn emerges naturally when the substitution P = (n/V)RT is applied to every gaseous species in the equilibrium expression.

The violet box represents K_c (concentration-based) and the cyan box represents K_p (pressure-based). The gold center box shows the ideal gas law bridge. The bottom row illustrates how the sign of Δn determines the relative magnitudes of the two constants.

As the diagram makes clear, the conversion between K_p and K_c is not arbitrary — it is a direct consequence of the ideal gas law. The three bottom boxes emphasize a critical insight: when Δn = 0, the conversion factor (RT)⁰ equals 1, so the two constants are numerically identical regardless of temperature. This occurs whenever the total moles of gaseous products equal the total moles of gaseous reactants.

SECTION 4

Mathematical Framework

We now derive the conversion formula rigorously. Consider a generic gas-phase equilibrium: aA(g) + bB(g) ⇌ cC(g) + dD(g). The equilibrium constant K_c is written in terms of molar concentrations, while K_p uses partial pressures. The ideal gas law for each species i states that P_i = [i]RT, where [i] is the molar concentration of species i. Substituting this into the K_p expression yields the conversion formula.

CONCENTRATION EQUILIBRIUM CONSTANT

Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

Square brackets denote molar concentrations (mol·L⁻¹) at equilibrium. The exponents a, b, c, d are stoichiometric coefficients.

PRESSURE EQUILIBRIUM CONSTANT

Kp = (Pc)ᶜ(Pd)ᵈ / ((Pa)ᵃ(Pb)ᵇ)

P_i denotes the partial pressure of species i at equilibrium, typically in atmospheres (atm) or bar.

Derivation via the Ideal Gas Law

From PV = nRT, we can write P_i = (n_i/V)RT = [i]RT. Substituting into K_p: K_p = ([C]RT)ᶜ([D]RT)ᵈ / (([A]RT)ᵃ([B]RT)ᵇ). Factoring out (RT) from each term, K_p = ([C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ) × (RT)⁽ᶜ⁺ᵈ⁻ᵃ⁻ᵇ⁾. Recognizing that the concentration quotient is K_c and defining Δn = (c + d) − (a + b), we arrive at the master conversion equation.

MASTER CONVERSION EQUATION

Kp = Kc × (RT)^Δn

R = 0.08206 L·atm·mol⁻¹·K⁻¹ (when pressures are in atm). T is the absolute temperature in Kelvin. Δn = (total moles of gaseous products) − (total moles of gaseous reactants). Only gaseous species are counted when computing Δn.

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Choice of R Matters

The value of R you use must be consistent with the pressure units in K_p. If K_p is expressed in atm, use R = 0.08206 L·atm·mol⁻¹·K⁻¹. If K_p is expressed in bar, use R = 0.08314 L·bar·mol⁻¹·K⁻¹. Never mix pressure units.

REARRANGED FOR Kc

Kc = Kp / (RT)^Δn = Kp × (RT)^(−Δn)

This rearrangement is useful when K_p is given and you need to find K_c.

SECTION 5

Detailed Breakdown by Δn

The value of Δn is the single parameter that determines how K_p and K_c relate to one another numerically. A careful tabulation of common reaction types alongside their Δn values illuminates this relationship and helps build intuition about when the two constants diverge significantly and when they coincide.

Reaction Example	Δn	Relationship
H₂(g) + I₂(g) ⇌ 2 HI(g)	2 − 2 = 0	K_p = K_c
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)	2 − 1 = +1	K_p = K_c × (RT)
N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)	2 − 4 = −2	K_p = K_c × (RT)⁻²
2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g)	2 − 3 = −1	K_p = K_c × (RT)⁻¹
2 NO₂(g) ⇌ N₂O₄(g)	1 − 2 = −1	K_p = K_c × (RT)⁻¹

Common reactions and their Δn values showing the Kp–Kc relationship

This graph plots log₁₀(K_p/K_c) versus Δn at T = 500 K. Each unit increase in Δn multiplies the ratio K_p/K_c by RT (≈ 41.03). The green point at Δn = 0 marks the identity K_p = K_c.

The graph above provides a powerful visual tool. Because log₁₀(K_p/K_c) = Δn × log₁₀(RT), the relationship is perfectly linear. At T = 500 K, RT ≈ 41.03 L·atm·mol⁻¹, so each unit increase in Δn shifts the logarithmic ratio by about 1.613. For a reaction with Δn = +2, K_p is roughly 1684 times larger than K_c. Conversely, for Δn = −2, K_p is about 1684 times smaller. This reinforces why getting the sign of Δn correct is absolutely essential.

SECTION 6

Worked Example

Consider the synthesis of ammonia via the Haber process at 500 K. Given that K_c = 0.500 for N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), find K_p at 500 K.

Converting Kc to Kp for the Haber Process

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Step 1 — Write the Balanced Equation and Identify Δn

The balanced equation is N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g). Count the total moles of gaseous products and reactants: products = 2 mol (NH₃), reactants = 1 + 3 = 4 mol (N₂ + H₂). Therefore, Δn = 2 − 4 = −2.

Δn = −2
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Step 2 — Identify Known Values

K_c = 0.500, T = 500 K, R = 0.08206 L·atm·mol⁻¹·K⁻¹ (since we want K_p in atm units).
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Step 3 — Calculate RT

RT = (0.08206 L·atm·mol⁻¹·K⁻¹)(500 K) = 41.03 L·atm·mol⁻¹.

RT = 41.03
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Step 4 — Calculate (RT)^Δn

(RT)^Δn = (41.03)⁻² = 1 / (41.03)² = 1 / 1683.5 = 5.940 × 10⁻⁴.

(RT)^Δn = 5.940 × 10⁻⁴
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Step 5 — Apply the Conversion Formula

K_p = K_c × (RT)^Δn = (0.500)(5.940 × 10⁻⁴) = 2.97 × 10⁻⁴.

Kp = 2.97 × 10⁻⁴
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Step 6 — Verify Reasonableness

Because Δn is negative (more moles of gas on the reactant side), we expect K_p < K_c. Indeed, 2.97 × 10⁻⁴ ≪ 0.500. The result is consistent with our earlier analysis: a negative Δn compresses the equilibrium constant when switching from concentration to pressure units.

SECTION 7

Common Pitfalls & Best Practices

The conversion between K_p and K_c is conceptually straightforward, but several common errors can undermine accurate calculations. The following table catalogues the most frequent mistakes and their remedies, so you can approach these problems with confidence.

Common Pitfall	Why It's Wrong	Correct Approach
Including solids or liquids in Δn	Only gaseous species contribute to partial pressure expressions; pure solids and liquids have activity = 1 and do not appear in Kp or Kc.	Count only species with (g) designation when computing Δn.
Using temperature in °C instead of K	The ideal gas law requires absolute temperature. Using Celsius gives wildly incorrect RT values.	Always convert to Kelvin: T(K) = T(°C) + 273.15.
Using the wrong value of R	R = 8.314 J·mol⁻¹·K⁻¹ gives RT in units incompatible with atm-based Kp.	Use R = 0.08206 L·atm·mol⁻¹·K⁻¹ for atm, or R = 0.08314 L·bar·mol⁻¹·K⁻¹ for bar.
Getting the sign of Δn wrong	Reversing (reactants − products instead of products − reactants) inverts the exponent, yielding the reciprocal of the correct answer.	Δn = Σ(moles gaseous products) − Σ(moles gaseous reactants). Always subtract reactants from products.
Forgetting to recalculate Δn when the equation is scaled	Multiplying an equation by a factor k changes Δn by the same factor, altering the conversion.	Recompute Δn each time the balanced equation is modified.

Summary of common errors in Kp–Kc conversions

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KEY TAKEAWAY

The K_p–K_c conversion is analogous to unit conversion in engineering: the physics does not change, but the numerical value must be adjusted by a dimensional factor. Just as an engineer converting a stress value from psi to pascals must use the correct conversion factor and verify the result is sensible (larger number in the smaller unit), you must use the correct R, the correct T (Kelvin!), and the correct Δn, then sanity-check the magnitude and direction of the change.

SECTION 8

Connection to the Thermodynamic Equilibrium Constant

In more advanced treatments, you will encounter the thermodynamic equilibrium constant K (sometimes denoted K° or K_thermo), which is defined in terms of activities rather than concentrations or pressures. Activities are dimensionless quantities that reduce to partial pressure divided by the standard pressure (P° = 1 bar) for ideal gases, or concentration divided by the standard concentration (c° = 1 mol/L) for ideal solutions. The thermodynamic K is therefore itself dimensionless, and it is this quantity that appears in the fundamental relation ΔG° = −RT ln K.

Feature	Kc and Kp	Thermodynamic K
Defined in terms of	Molar concentrations (Kc) or partial pressures (Kp)	Activities (dimensionless ratios to standard states)
Units	Generally have units (unless Δn = 0), though often treated as dimensionless by convention	Always dimensionless
Relation to ΔG°	Kp can be used in ΔG° = −RT ln Kp when pressures are referenced to 1 bar	Direct: ΔG° = −RT ln K
Ideal gas assumption	Required for the Kp = Kc(RT)^Δn conversion	Not required — fugacity coefficients can replace partial pressures
When they coincide	At low pressures and moderate temperatures (ideal gas regime)	K reduces to Kp (numerically) when P° = 1 bar and gases behave ideally

Comparison of Kc/Kp with the thermodynamic equilibrium constant

Understanding the distinction between K_p, K_c, and the thermodynamic K is crucial as you advance into physical chemistry and chemical thermodynamics. In most general chemistry courses, the distinction is minimal because problems typically assume ideal gas behavior, but the awareness of these deeper foundations will serve you well in kinetics, electrochemistry, and advanced equilibrium calculations involving non-ideal mixtures.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

For the reaction H₂(g) + I₂(g) ⇌ 2 HI(g), are K_p and K_c numerically equal? Explain why or why not.

Show answer

Yes, K_p and K_c are numerically equal for this reaction because Δn = 2 − (1 + 1) = 0. When Δn = 0, the conversion factor (RT)⁰ = 1, so K_p = K_c × 1 = K_c. Physically, this means the concentration-to-pressure conversion factors cancel perfectly between products and reactants because they have the same total number of moles of gas.

Question 2: BASIC CALCULATION

For PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), K_c = 0.042 at 250 °C. Calculate K_p.

Show answer

First, convert temperature to Kelvin: T = 250 + 273.15 = 523.15 K. Calculate Δn: products = 1 + 1 = 2, reactants = 1, so Δn = 2 − 1 = +1. Compute RT = (0.08206)(523.15) = 42.93 L·atm·mol⁻¹. Apply the formula: K_p = K_c × (RT)¹ = (0.042)(42.93) = 1.80. Since Δn > 0, we correctly find K_p > K_c.

Question 3: INTERMEDIATE

For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), K_p = 3.45 at 600 K. Find K_c.

Show answer

Determine Δn: products = 2, reactants = 2 + 1 = 3, so Δn = 2 − 3 = −1. Compute RT = (0.08206)(600) = 49.24. Since we need K_c, rearrange: K_c = K_p / (RT)^Δn = 3.45 / (49.24)⁻¹ = 3.45 × 49.24 = 169.9 ≈ 170. Because Δn is negative, K_c > K_p, which is confirmed by our result.

Question 4: APPLIED

In an industrial reactor, the decomposition of CaCO₃(s) ⇌ CaO(s) + CO₂(g) has K_p = 0.039 at 800 K. (a) What is K_c? (b) Why must you be careful about which species to include in Δn?

Show answer

(a) CaCO₃ and CaO are both solids, so they do not appear in the equilibrium expressions. The only gaseous species is CO₂, so Δn = 1 − 0 = +1. RT = (0.08206)(800) = 65.65. K_c = K_p / (RT)¹ = 0.039 / 65.65 = 5.94 × 10⁻⁴ mol/L. (b) This problem illustrates why you must only count gaseous species when computing Δn. If you mistakenly included CaCO₃(s) as a reactant, you would get Δn = 1 − 1 = 0 and incorrectly conclude K_c = K_p, which is wrong by a factor of 65.65.

Question 5: CRITICAL THINKING

Consider the equilibrium N₂O₄(g) ⇌ 2 NO₂(g). At 298 K, K_p = 0.113. (a) Find K_c. (b) If the equation is written as ½ N₂O₄(g) ⇌ NO₂(g), what is the new K_p' and the new K_c'? (c) Does the conversion formula still hold for the halved equation?

Show answer

(a) For the original equation, Δn = 2 − 1 = +1. RT = (0.08206)(298) = 24.45. K_c = K_p / (RT) = 0.113 / 24.45 = 4.62 × 10⁻³. (b) Halving the equation means K_p' = (K_p)^(1/2) =

(0.113)^{1/2}

= 0.336. The new Δn' = 1 − ½ = +½. K_c' = K_p' /

(RT)^{1/2}

= 0.336 /

(24.45)^{1/2}

= 0.336 / 4.945 = 0.0680. (c) Yes, the conversion formula K_p = K_c(RT)^Δn remains valid regardless of how the equation is scaled, provided Δn is recalculated for the scaled equation. This problem demonstrates that both K and Δn must be updated consistently when stoichiometric coefficients change.

SUMMARY

Lesson Summary

Summary

Gas-phase equilibria can be described using either Kc (molar concentrations) or Kp (partial pressures). The two are linked by the ideal gas law through the master equation Kp = Kc × (RT)^Δn, where Δn is the change in total moles of gas (products minus reactants), R = 0.08206 L·atm·mol⁻¹·K⁻¹ (for atm-based K_p), and T is in Kelvin.

When Δn = 0, K_p = K_c exactly. When Δn > 0, K_p > K_c; when Δn < 0, K_p < K_c. Common pitfalls include using Celsius instead of Kelvin, choosing the wrong R value, or counting solid or liquid species in Δn. Always verify that your result is reasonable: negative Δn should yield K_p < K_c, and positive Δn should yield K_p > K_c (at temperatures where RT > 1, which is essentially always in chemistry).