The Van’t Hoff equation concerns how the equilibrium constant (K_{eq}) changes with the change in temperature. This equation was proposed by the Dutch chemist Jacobus Henricus van’t Hoff in 1884.

In thermodynamics, the Van’t Hoff equation is used to explore the changes in the state function.

## Derivation of Van’t Hoff equation

The derivation of this equation starts from the following equations:

ΔG^{º} = – RT lnK → (i)

lnK = – ΔG^{º}/RT → (ii)

**Step 1: Differentiate the equation (ii) for lnK **

In the first step, we differentiate the lnK with temperature. Hence, it will give,

dlnK/dT = – 1/R d((ΔG^{º}/T)/dT

It is rearranged into the following:

d((ΔG^{º}/T)/dT = -R dlnK/dT

**Step 2: Use the Gibbs-Helmholtz equation (i) and rearrange**

In this step, we use the Gibbs-Helmholtz equation in the following form:

d((ΔG^{º}/T)/dT = – ΔH_{r}^{º}/T^{2}

Where,

- ΔG
^{º }is the standard Gibbs free energy - T is the absolute temperature
- ΔH
_{r}^{º}= Standard reaction enthalpy

Now combine this equation with the expression we get in step 1,

R dlnK/dT = ΔH_{r}^{º}/T^{2}

It rearranged into the following equation,

d(lnK)/dT = ΔH_{r}^{º}/RT^{2}

Hence, this equation is known as Van’t Hoff equation:

**Alternating form of Van’t Hoff equation**

The Van’t Hoff equation can also be written as:

dlnK/d(1/T) = – ΔH_{r}^{º}/R

**The integrated form of the Van’t Hoff equation **

Van’t Hoff equations can be written in the following form:

dlnK= ΔH_{r}^{º}/RT^{2} × dT

By integrating the above equation we get:

lnK = – ΔH_{r}^{º}/RT+ I

Where,

- I = constant of integration

According to the above equation, lnK varies with temperature linearly. However, the slope is equal to – ΔH_{r}^{º}/R.

- The plot of lnK versus 1/T when ΔH
_{r}^{º }is negative

- The plot of lnK versus 1/T when ΔH
_{r}^{º }is positive

## Application of the Van’t Hoff equation

There are the following conclusions drawn from Van’t Hoff equations:

### For an endothermic reaction

For the endothermic reaction ΔH_{r}^{º}> 0. Therefore, the right-hand side of the equation is positive. Hence, ‘lnK’ increases with increasing temperature.

### For exothermic reactions

For the exothermic reaction ΔH_{r}^{º}< 0. Therefore, the right-hand side of the equation is negative. So, lnK decreases with increasing temperature.

**Related Resources**

- Internal Energy vs. Enthalpy
- Endothermic vs. Exothermic reactions
- Enthalpy: Methods to calculate the enthalpy

## Concepts Berg

**Is Van’t Hoff factor always 1?**

The Van’t Hoff factor for the non-ionizing molecule is one. However, for the ionic compound, the value of Van’t Hoff factor equals the number of ions that dissociate.

**How do you determine Van’t Hoff?**

The Van’T Hoff factor can be determined by the values of calculated molar mass and experimentally observed molar mass,

**What is Vant Hoff’s rule of N?**

The N is the number of ions produced by the complete dissociation of one mole of the substance.

**What is meant by Van’t Hoff’s rule?**

Van’t Hoff’s rule states that any change in temperature shifts the equilibrium in that direction which nullifies its effect.

**Which has the highest Van t Hoff factor?**

The aluminum sulfate Al_{2}(SO_{4})_{3 }has the highest value of the Vant’t Hoff factor.

**What is the van ‘t Hoff factor of H _{2}O?**

Although water undergoes self-ionization, but is dissociation is very weak. Therefore, it has a value of Van’t Hoff factor one.

**What is Van’t Hoff isotherm?**

According to Van’t Hoff isotherm, at the constant temperature the osmotic pressure of the liquid is inversely proportional to the volume.

**References**

- Physical Chemistry by Peter f. Atkins
- A textbook of physical chemistry Thermodynamic and Chemical equilibrium by K.L Kapoor
- Chap 9 (Binghamanton.edu)