### 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

Î”G⁰ = Î”H⁰ + T[d(Î”G⁰)/dT]p

Zero superscripts are indicating stranded values.

or,- (Î”H⁰/T² )= -(Î”G⁰/T² )+(1/T)[d(Î”G⁰)/dT]p

or, - (Î”H⁰/T² ) = [d/dT(Î”G⁰/T)]p.

Van't Hoff isotherm

- RT lnKp = Î”G⁰

or, - R lnKp = Î”G⁰/T.

Differentiating with respect to temperature at constant pressure

- R [dlnKp/dT]p = [d/dT(Î”G⁰/T)]p.

Comparing the above two-equation

dlnKp/dT = Î”H⁰/T²

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

Zero superscripts are indicating stranded values.

or,- (Î”H⁰/T² )= -(Î”G⁰/T² )+(1/T)[d(Î”G⁰)/dT]p

or, - (Î”H⁰/T² ) = [d/dT(Î”G⁰/T)]p.

Van't Hoff isotherm

- RT lnKp = Î”G⁰

or, - R lnKp = Î”G⁰/T.

Differentiating with respect to temperature at constant pressure

- R [dlnKp/dT]p = [d/dT(Î”G⁰/T)]p.

Comparing the above two-equation

dlnKp/dT = Î”H⁰/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 = Î”H⁰/T²

Separating the variables and integrating

∫ dlnKp = (Î”H⁰/R)∫ (dT/T²)

Î”H⁰ independent of temperature.

or, lnKp = - (Î”H⁰/R)(1/T) + C

where C = integrating constant.

Separating the variables and integrating

∫ dlnKp = (Î”H⁰/R)∫ (dT/T²)

Î”H⁰ independent of temperature.

or, lnKp = - (Î”H⁰/R)(1/T) + C

where C = integrating constant.

The integration constant can be evaluated and identified the value of Î”S⁰/R, using the relation

Î”G⁰ = Î”H⁰ - TÎ”S⁰.

∴ Van't hoff equation

lnKp = - (Î”H⁰/RT) + (Î”S⁰/R).

∴ Van't hoff equation

lnKp = - (Î”H⁰/RT) + (Î”S⁰/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.

Î”G⁰ = - RT lnKp.

∴ 1 = - 8.31 × 10⁻³ × 500 × lnKp

or, lnKp = 1/(8.31 × 0.5)

= 0.2406.

∴ Kp = 1.27.

= 1 kJ mol⁻¹.

Î”G⁰ = - 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, Î”H⁰ = 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 | Î”H⁰ = (+) ve |

- Heat emission, Î”H⁰ = 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 | Î”H⁰ = (-) ve |

- de-Broglie relation
- Emission hydrogen spectrum
- Rutherford model
- Bohrs model
- Crystalline solids
- Balancing chemical equation
- Heat capacity of gases

#### Endothermic and exothermic chemical reaction

Van't Hoff equation drowns the relation between equilibrium constant and endothermic and exothermic chemical reactions.- For an endothermic reaction, Î”H⁰ 〉 0 and the right-hand side of the equation positive. This leads to the fact that lnKp increases with increasing temperature.
- For an exothermic reaction, Î”H⁰ 〈 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, Î”H⁰ = Î”H

dlnKp/dt = Î”H/RT²

lnKp = (Î”H/RT) + Î”S/R

lnKp = (Î”H/RT) + Î”S/R

However, the entropy of an ideal gas depends strongly on pressure and Î”S and Î”G per mole of reaction in the mixture differ quite substantially from Î”S⁰ and Î”G⁰.

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

ln(Kp₂/Kp₁) = (Î”H/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 Î”G = 0.

Solution

Van't Hoff reaction isotherm

Î”G = - RT lnKa + RT lnQa.

When the reaction attains equilibrium point,

Qa = Ka.

∴ Î”G = 0

Î”G = - RT lnKa + RT lnQa.

When the reaction attains equilibrium point,

Qa = Ka.

∴ Î”G = 0

#### Assumptions from Van't Hoff equation

- The reacting system of the chemical reaction behaves ideally.
- Î”H has taken independent of temperature for a small range of temperature change.

Î”G⁰ = - RT lnKp

or, [dlnKp/dT]

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)Î”Æ”

or, lnKp = lnKc + Î”Æ” lnR + Î”Æ” lnT

Differentiating with respect to T,

dlnKp/dT = dlnKc/dT + Î”Æ”/T.

dlnKp/dT = Î”H⁰/RT²

or, dlnKc/dT = Î”H⁰/RT² - Î”Æ”/T

or, dlnKc/dT = (Î”H⁰ - Î”Æ”RT)/RT²

∴ dlnKc/dT = Î”U⁰/RT²

or, lnKp = lnKc + Î”Æ” lnR + Î”Æ” lnT

Differentiating with respect to T,

dlnKp/dT = dlnKc/dT + Î”Æ”/T.

dlnKp/dT = Î”H⁰/RT²

or, dlnKc/dT = Î”H⁰/RT² - Î”Æ”/T

or, dlnKc/dT = (Î”H⁰ - Î”Æ”RT)/RT²

∴ dlnKc/dT = Î”U⁰/RT²

Î”U⁰ = 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(K

_{c}₂/K_{c}₁) = (Î”U⁰/R){(T₂ - T₁)/T₁T₂)}.For ideal system Î”U⁰ = Î”U. 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 = - (Î”H/R)(1/T) + C.

- Endothermic reaction, Î”H〉0, an increase in temperature increases the value of lnKp of the chemical reactions.
- An exothermic reaction, Î”Hã„‘ 0, with rising in temperature, lnKp decreased.

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.

- 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.
- 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.