# Van't Hoff equation-chemical equilibrium

### How much Van't Hoff equation - effect on temperature?

Van't Hoff equation proposed equilibrium of a chemical reaction is constant at a given temperature. The equilibrium constant, Kp values can be changed with the change of temperature.

This will be evident from the study of Kp values at different temperatures of a chemical reaction.
N₂ + O₂ ⇆ 2NO

 Temperature Kp × 10⁴ 2000° K 4.08 2200° K 11.00 2400° K 25.10 2600° K 50.30

The quantitative relation, known Van't Hoff equation connecting chemical equilibrium and temperature can be derived thermodynamically starting from Gibbs - Helmholtz equation.

The Gibbs - Helmholtz equation
ΔG⁰ = ΔH⁰ + T[d(ΔG⁰)/dT]p
Zero superscripts are indicating stranded values.

or,- (ΔH⁰/T² )= -(ΔG⁰/T² )+(1/T)[d(ΔG⁰)/dT]p
or, - (ΔH⁰/T² ) = [d/dT(ΔG⁰/T)]p.

Van't Hoff isotherm

- RT lnKp = ΔG⁰

or, - R lnKp = ΔG⁰/T.

Differentiating with respect to temperature at constant pressure
- R [dlnKp/dT]p = [d/dT(ΔG⁰/T)]p.

Comparing the above two-equation

dlnKp/dT = ΔH⁰/T²

This is the differential form of Van't Hoff equation.

#### The integrated form of van't Hoff equation

The greater the value of standard enthalpy of a reaction, the faster the chemical reaction reaching an equilibrium point.

dlnKp/dT = ΔH⁰/T²

Separating the variables and integrating
∫ dlnKp = (ΔH⁰/R)∫ (dT/T²)

ΔH⁰ independent of temperature.
or, lnKp = - (ΔH⁰/R)(1/T) + C
where C = integrating constant.

The integration constant can be evaluated and identified the value of ΔS⁰/R, using the relation
ΔG⁰ = ΔH⁰ - TΔS⁰.

∴ Van't hoff equation

lnKp = - (ΔH⁰/RT) + (ΔS⁰/R).

Problem
The standard free energy of chemical reaction 2A + B ⇆ 2C at 500 K = 2 KJ mol⁻¹. Calculate Kp at 500 K for the chemical reaction A + ½B ⇆ C.

Solution
Standard free energy at 500 K for the chemical reaction A + ½B ⇆ C
= (2 kJ mol⁻¹)/2
= 1 kJ mol⁻¹.

ΔG⁰ = - RT lnKp.
∴ 1 = - 8.31 × 10⁻³ × 500 × lnKp
or, lnKp = 1/(8.31 × 0.5)
= 0.2406.

∴ Kp = 1.27.

#### Heat absorption and emission in a chemical reaction

Heat absorption and emission in a chemical reaction can be studied from the following analysis.

Reactants → Products

We may study the following possibilities
• Heat absorption, ΔH⁰ = positive.
Low enthalpy side → High enthalpy side.

If heat absorbed, the forward direction of the chemical reaction is the endothermic direction and the reverse one is the exothermic direction. The dissociation of hydrogen iodide into hydrogen and Iodine is an example of this type of chemical reaction.

 HI ⇆ H₂ + I₂ ΔH⁰ = (+) ve
• Heat emission, ΔH⁰ = negative.
High enthalpy side → Low enthalpy side.

If heat emission, the forward direction of the chemical reaction is the exothermic direction and the reverse one is the endothermic direction.
The formation of ammonia from hydrogen and nitrogen is an example of this type of chemical reaction.

 N₂ + H₂ ⇆ 2NH₃ ΔH⁰ = (-) ve
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#### Endothermic and exothermic chemical reaction

Van't Hoff equation drowns the relation between equilibrium constant and endothermic and exothermic chemical reactions.
1. For an endothermic reaction, ΔH⁰ 〉 0 and the right-hand side of the equation positive. This leads to the fact that lnKp increases with increasing temperature.
2. For an exothermic reaction, ΔH⁰ 〈 0 and the right-hand side of the equation negative. This leads to the fact that lnKp decreases with increasing temperature. Van't Hoff equation

#### Heat change of a chemical reaction in two temperature

For the ideal system, the heat change or enthalpy change is not a function of pressure.

∴ Standard enthalpy change = enthalpy change
or, ΔH⁰ = ΔH

dlnKp/dt = ΔH/RT²

lnKp = (ΔH/RT) + ΔS/R

However, the entropy of an ideal gas depends strongly on pressure and ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS⁰ and ΔG⁰.

The integrated form of the Van't Hoff equation at two temperature

ln(Kp₂/Kp₁) = (ΔH/R){(T₂ - T₁)/T₁T₂)}

where Kp₁ and Kp₂ are the equilibrium constants of the reaction at two different temperatures T₁ and T₂ respectively.

Determination of Kp₁ and Kp₂ at two temperatures helps to calculate the value of change of enthalpy of the chemical reaction.

The above relation called Van't Hoff reaction isobar since pressure remains constant during the change of temperature.

Problem
Show that the equilibrium point for any chemical reaction given by ΔG = 0.

Solution
Van't Hoff reaction isotherm
ΔG = - RT lnKa + RT lnQa.
When the reaction attains equilibrium point,
Qa = Ka.

∴ ΔG = 0

#### Assumptions from Van't Hoff equation

1. The reacting system of the chemical reaction behaves ideally.
2. ΔH has taken independent of temperature for a small range of temperature change.
Due to the assumption involved ΔH and ΔU do not produce the precise value of these reactions. As ΔG independent of pressure for ideal system ΔH and ΔU also independent of pressure.

ΔG⁰ = - RT lnKp
or, [dlnKp/dT]T = 0.

Problem
Use Gibbs - Helmholtz equation to derive the Van't Hoff reaction isochore. What condition do you expect a linear relationship between log k and 1/T?

Solution
Again Kp = Kc (RT)ΔƔ
or, lnKp = lnKc + ΔƔ lnR + ΔƔ lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + ΔƔ/T.

dlnKp/dT = ΔH⁰/RT²
or, dlnKc/dT = ΔH⁰/RT² - ΔƔ/T
or, dlnKc/dT = (ΔH⁰ - ΔƔRT)/RT²

∴ dlnKc/dT = ΔU⁰/RT²

ΔU⁰ = standers heat of reaction at constant volume. This is the differential form of Van't Hoff reaction isochore.

The integrated form of the equation at two temperature
ln(Kc₂/Kc₁) = (ΔU⁰/R){(T₂ - T₁)/T₁T₂)}.

For ideal system ΔU⁰ = ΔU. Since the reaction occurs at constant volume and the equation called Van't Hoff reaction isochore.

### The heat of a reaction and Le-Chatelier principal

Van't Hoff equations give a quantitative expression of the Le-Chatelier principle.
lnKp = - (ΔH/R)(1/T) + C.
1. Endothermic reaction, ΔH〉0, an increase in temperature increases the value of lnKp of the chemical reactions.
2. An exothermic reaction, ΔHㄑ 0, with rising in temperature, lnKp decreased.
Change of Kp provides the calculation of the quantitative change of equilibrium point yield of products.

According to the Le - Chatelier Principle, whenever stress placed on any system in a state of the equilibrium point, the system always reacts in a direction to reduce the applied stress.
1. The temperature increased in a chemical reaction, the system at equilibrium point will try to move in a direction in which heat absorbed, which is endothermic reaction favors.
2. When the temperature decreased in a chemical reaction, the system at equilibrium point will try to move in a direction in which heat emitted, which is exothermic reaction favors.

Study online Van't Hoff equation-chemical equilibrium, heat absorption, and emission, endothermic and exothermic chemical reaction, heat change