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Kp relation

Relationship between Kp and Kc.

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K_{p} = K_{c} (R T)^{Δ n}

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Core idea

Overview

The Kp relation defines the mathematical link between equilibrium constants derived from partial pressures and those derived from molar concentrations. This relationship accounts for the work done by or on a gas system during a reaction where the total number of gaseous moles changes, assuming ideal gas behavior.

When to use: Apply this equation when converting between pressure-based (Kp) and concentration-based (Kc) constants for reactions involving gases. It requires the reaction temperature in Kelvin and the change in moles of gas, calculated as product gas moles minus reactant gas moles.

Why it matters: This allows scientists to predict equilibrium positions across different measurement units, which is essential for industrial processes like the Haber-Bosch ammonia synthesis. It clarifies why pressure changes only affect the equilibrium position when there is a net change in gaseous moles (dn ≠ 0).

Symbols

Variables

K_c = Equilibrium Kc, R = Gas Constant, T = Temperature, n = Delta n, K_p = Equilibrium Kp

K_{c}

Equilibrium Kc

V a r iab l e

R

Gas Constant

J / m o l K

T

Temperature

K

n

Delta n

V a r iab l e

K_{p}

Equilibrium Kp

V a r iab l e

Walkthrough

Derivation

Formula: Equilibrium Constant (Kp)

Equilibrium constant written using partial pressures for gaseous equilibria.

Gases behave ideally (A-Level approximation).
System is at dynamic equilibrium.
Partial pressures are expressed in consistent units.

Define Partial Pressure:

Partial pressure equals mole fraction times total pressure.

p_{A} = x_{A} P_{total}

State the Kp Expression:

Same structure as Kc, but uses partial pressures for gaseous species.

K_{p} = \frac{( p _{C} ) ^{c} ( p _{D} ) ^{d}}{( p _{A} ) ^{a} ( p _{B} ) ^{b}}

Result

K_{p} = \frac{( p _{C} ) ^{c} ( p _{D} ) ^{d}}{( p _{A} ) ^{a} ( p _{B} ) ^{b}}

Source: Edexcel A-Level Chemistry — Equilibria

Free formulas

Rearrangements

Solve for $K_{c}$

Make Kc the subject

K_{c} = K_{p} (R T)^{- n}

Exact symbolic rearrangement generated deterministically for Kc.

Difficulty: 2/5

Solve for $R$

Make R the subject

R = \frac{( \frac{K _{p}}{K _{c}} ) ^{\frac{1}{n}}}{T}

Exact symbolic rearrangement generated deterministically for R.

Difficulty: 3/5

Solve for $T$

Make T the subject

T = \frac{( \frac{K _{p}}{K _{c}} ) ^{\frac{1}{n}}}{R}

Exact symbolic rearrangement generated deterministically for T.

Difficulty: 3/5

Solve for $n$

Make dn the subject

n = \frac{\ln\left(\frac{K_p}{K_c} \right)}}{\ln\left(R T \right)}}

Exact symbolic rearrangement generated deterministically for dn.

Difficulty: 3/5

Solve for $K_{p}$

Make Kp the subject

K_{p} = K_{c} (R T)^{n}

Exact symbolic rearrangement generated deterministically for Kp.

Difficulty: 2/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph is a straight line passing through the origin with a slope equal to (RT) raised to the power of delta n. Because Kc is directly proportional to Kp, increasing Kc results in a constant, linear increase in Kp. For a chemistry student, this means that larger Kc values correspond to larger Kp values, indicating that the equilibrium position shifts toward products as the equilibrium constant increases. The most important feature is that the linear relationship means doubling Kc will always double Kp, provided

Graph type: linear

Why it behaves this way

Intuition

Visualize the relationship as converting between two different 'currencies' for measuring equilibrium: one based on the 'pressure' exerted by gas molecules and another based on their 'concentration'.

K_{p}

The equilibrium constant expressed in terms of partial pressures of gaseous reactants and products.

Indicates the extent to which a reaction proceeds towards products when pressures are used to quantify amounts; a larger Kp means more products at equilibrium.

K_{c}

The equilibrium constant expressed in terms of molar concentrations of reactants and products.

Indicates the extent to which a reaction proceeds towards products when concentrations are used to quantify amounts; a larger Kc means more products at equilibrium.

The ideal gas constant, a proportionality constant in the ideal gas law.

Converts between energy-related units and the product of pressure and volume, or the product of moles and temperature, for ideal gases.

Absolute temperature of the system in Kelvin.

A measure of the average kinetic energy of gas molecules, directly influencing their pressure and concentration at a given volume.

Δ n

The change in the total number of moles of gaseous species from reactants to products (moles of gaseous products - moles of gaseous reactants).

Represents the net change in the quantity of gas particles during the reaction, which dictates how the pressure-based and concentration-based equilibrium constants differ due to the ideal gas law.

Signs and relationships

(RT)^{Δ n}: The term (RT) arises from substituting partial pressures using the ideal gas law, $P_{i}$ = $C_{i}$ RT. Since Kp involves products of pressures and Kc involves products of concentrations, each $P_{i}$ term introduces an RT factor.

Free study cues

Insight

Canonical usage

This equation is used to convert between equilibrium constants expressed in terms of partial pressures (Kp) and molar concentrations (Kc) for gas-phase reactions.

Common confusion

Using the SI value of R (8.314) while expressing pressure in atmospheres or bars, which leads to a magnitude error of approximately 10^2 to 10^5.

Dimension note

While dn is a pure number, Kp and Kc are only truly dimensionless when activities or standard-state ratios are used; otherwise, they carry units of (pressure)^dn or (concentration)^dn.

Unit systems

$T$ K · Temperature must always be in Kelvin; Celsius will result in incorrect exponential scaling.

$d n$ dimensionless · Calculated as the sum of stoichiometric coefficients of gaseous products minus gaseous reactants.

$K c$ (mol dm^-3)^dn · In many contexts, Kc is treated as dimensionless by referencing to a standard state of 1 mol/dm^3.

$K p$ pressure^dn · Units depend on the pressure unit used (atm, bar, or Pa) raised to the power of dn.

One free problem

Practice Problem

For the synthesis of ammonia, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the value of Kc is 0.045 at 500 K. Using R = 0.0821 L·atm/(mol·K), calculate the value of Kp.

Equilibrium Kc0.045

Gas Constant0.0821 J/molK

Temperature500 K

Delta n-2

Solve for: $K p$

Hint: Calculate the change in moles (dn) by subtracting reactant gas moles (1+3) from product gas moles (2).

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Converting Kc to Kp for ammonia synthesis.

Study smarter

Tips

Always convert temperature to Kelvin by adding 273.15 to Celsius.
Ensure dn only counts coefficients of species in the gas phase.
Match the R constant to the pressure units, typically 0.0821 for atmospheres.
If dn is zero, Kp equals Kc because the (RT) term becomes 1.

Avoid these traps

Common Mistakes

Forgetting the sign of Δ n.
Using wrong R value.

Common questions

Frequently Asked Questions

Equilibrium constant written using partial pressures for gaseous equilibria.

Apply this equation when converting between pressure-based (Kp) and concentration-based (Kc) constants for reactions involving gases. It requires the reaction temperature in Kelvin and the change in moles of gas, calculated as product gas moles minus reactant gas moles.

This allows scientists to predict equilibrium positions across different measurement units, which is essential for industrial processes like the Haber-Bosch ammonia synthesis. It clarifies why pressure changes only affect the equilibrium position when there is a net change in gaseous moles (dn ≠ 0).

Forgetting the sign of Δ n. Using wrong R value.

Converting Kc to Kp for ammonia synthesis.

Always convert temperature to Kelvin by adding 273.15 to Celsius. Ensure dn only counts coefficients of species in the gas phase. Match the R constant to the pressure units, typically 0.0821 for atmospheres. If dn is zero, Kp equals Kc because the (RT) term becomes 1.

References

Sources

Atkins' Physical Chemistry
McQuarrie, Simon, Physical Chemistry: A Molecular Approach
Wikipedia: Equilibrium constant (specifically the section 'Relationship between Kp and Kc')
NIST CODATA
IUPAC Gold Book
McQuarrie & Simon, Physical Chemistry: A Molecular Approach
Brown, LeMay, Bursten, Chemistry: The Central Science
Edexcel A-Level Chemistry — Equilibria

Kp relation

Overview

Variables

Derivation

Define Partial Pressure:

State the Kp Expression:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Equilibrium constant

Frequently Asked Questions

Sources