Chemistry

Integrated van ’t Hoff Equation

Estimate how an equilibrium constant changes with temperature.

ln (\frac{K _{2}}{K _{1}}) = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

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Source & review

Difficulty

Intermediate

Read time

4 min

Prerequisites

Algebra, logarithms, and molar quantities

Review statusNeeds review

Source

OpenStax, Chemistry 2e, §16.4

Adapted from Teng-Jui Lin’s CHEM 145 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

FormuLab adapted notation and chemical symbols, added variable meanings, units, applicability conditions, and an original worked example.

Definition

Understanding the core concept

The integrated van ’t Hoff equation compares equilibrium constants at two absolute temperatures. Within the constant-enthalpy approximation, heating increases K for an endothermic reaction and decreases K for an exothermic reaction.

Variables & Units

Understanding each component

Symbol	Meaning	Units
$K_{1}$	Dimensionless equilibrium constant at T_1	$1$
$K_{2}$	Dimensionless equilibrium constant at T_2	$1$
$Δ H^{\circ}$	Standard reaction enthalpy change	$J mo l^{- 1}$
$R$	Molar gas constant	$J mo l^{- 1} K^{- 1}$
$T_{1}$	Initial absolute temperature	$K$
$T_{2}$	Final absolute temperature	$K$

Real-World Applications

Where this formula is used in practice

Temperature-dependent equilibrium

Estimate K at a nearby temperature.

Endothermic reactions

Predict the direction of the equilibrium-constant change on heating.

Worked Example

Step-by-step calculation with real numbers

Problem

An endothermic reaction has K_1 = 2.00 at 300 K and approximately constant standard enthalpy change +50.0 kJ/mol. Estimate K_2 at 310 K.

Given

K_{1} = 2.00

T_{1} = 300 K

T_{2} = 310 K

Δ H^{\circ} = 50000 J mo l^{- 1}

R = 8.314462618 J mo l^{- 1} K^{- 1}

Solution

ln (\frac{K _{2}}{K _{1}}) = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})

K_{2} = K_{1} exp [- \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]

K_{2} = 2.00 exp [- \frac{50000}{8.314462618} (\frac{1}{310} - \frac{1}{300})]

K_{2} \approx 3.818

Final Answer

K_{2} \approx 3.818

Formula Guide

Info

DifficultyIntermediate

Read Time4 min

PrerequisitesAlgebra, logarithms, and molar quantities

Tools

Source & review

StatusNeeds review

Last reviewedPending

Source

OpenStax, Chemistry 2e, §16.4

Adapted from Teng-Jui Lin’s CHEM 145 equation sheet. The linked textbook reference was used to check the equation and its physical context. Pending editorial review.

Adapted from Teng-Jui Lin, Equation Sheets, under CC BY 4.0.

FormuLab adapted notation and chemical symbols, added variable meanings, units, applicability conditions, and an original worked example.

Assumptions

• Assume standard reaction enthalpy is approximately constant over the temperature interval.
• Use the same reaction, phases, and standard-state definitions at both positive kelvin temperatures.

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