### Van't Hoff equation - effect on temperature

Equilibrium constant Kp of a reaction is constant at a given temperature but when the temperature is changed, the value of equilibrium constant also changed.

- The quantitative relation, relation, known as

**is derived by using the Gibbs - Helmholtz equation.**

*Van't Hoff equation*- The Gibbs - Helmholtz equation is,

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

Again Van't Hoff isotherm is,

- RT lnKp = Î”G⁰

or, - R lnKp = Î”G⁰/T

Differentiating with respect to T at constant P,

- R [dlnKp/dT]p = [d/dT(Î”G⁰/T)]p

Comparing the above two equation we have,

dlnKp/dT = Î”H⁰/T²

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

Again Van't Hoff isotherm is,

- RT lnKp = Î”G⁰

or, - R lnKp = Î”G⁰/T

Differentiating with respect to T at constant P,

- R [dlnKp/dT]p = [d/dT(Î”G⁰/T)]p

Comparing the above two equation we have,

dlnKp/dT = Î”H⁰/T²

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

- The greater the value of Î”H⁰, the faster the equilibrium constant (Kp) changes with temperature (T).

Separating the variables and integrating,

∫ dlnKp = (Î”H⁰/R)∫ (dT/T²)

Assuming that Î”H⁰ is independent of temperature.

or, lnKp = - (Î”H⁰/R)(1/T)+C

where C is integrating constant

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

Î”G⁰ = Î”H⁰ - TÎ”S⁰

Thus the Van't hoff Equation is,

lnKp = - (Î”H⁰/RT) + (Î”S⁰/R)

∫ dlnKp = (Î”H⁰/R)∫ (dT/T²)

Assuming that Î”H⁰ is independent of temperature.

or, lnKp = - (Î”H⁰/R)(1/T)+C

where C is integrating constant

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

Î”G⁰ = Î”H⁰ - TÎ”S⁰

Thus the Van't hoff Equation is,

lnKp = - (Î”H⁰/RT) + (Î”S⁰/R)

Problem

- For a reaction 2A + B ⇆ 2C, Î”G⁰(500 K) = 2 KJ mol⁻¹ find the Kp at 500 K for the reaction A + ½B ⇆ C.

Î”G⁰(500 K) for the reaction,

A + ½B ⇆ C; (2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.

The relation is Î”G⁰ = - RT lnKp

or, 1 = - (8.31 × 10-3 ) × (500) lnKP

or, lnKP = 1/(8.31 × 0.5)

or, lnKP = 0.2406

∴ KP = 1.27

A + ½B ⇆ C; (2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.

The relation is Î”G⁰ = - RT lnKp

or, 1 = - (8.31 × 10-3 ) × (500) lnKP

or, lnKP = 1/(8.31 × 0.5)

or, lnKP = 0.2406

∴ KP = 1.27

### Plot of lnKP vs 1/T from Van't Hoff equation

- For exothermic reaction, Î”H⁰ = (-) ve.

- Examples are the formation of ammonia from H₂ and N₂.

N₂ + H₂ ⇆ 2NH₃ | Î”H⁰ = (-) ve |

- For endothermic reaction, Î”H⁰ = (+) ve.

- Examples are the dissociation of HI into H₂ and I₂.

HI ⇆ H₂ + I₂ | Î”H⁰ = (+) ve |

- For the reaction, Î”H⁰ = 0.

- lnKp is independent of T. Provided Î”S⁰ does not change much with T.

Van't Hoff equation |

- Since for ideal system, H is not a function of P and Î”H⁰ = Î”H and Van't Hoff equation is,

dlnKp/dt = Î”H/RT²

The integrated equation is,

lnKp = (Î”H/RT) + Î”S/R

The integrated equation is,

lnKp = (Î”H/RT) + Î”S/R

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

- If we consider the integrated Van't Hoff equation at two temperature then, it becomes,

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.

- Thus the determination of Kp₁ and Kp₂ at two temperatures helps to calculate the value of Î”H of the reaction.

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

- 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?

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

But dlnKp/dT = Î”H⁰/RT²

Hence, 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

But dlnKp/dT = Î”H⁰/RT²

Hence, dlnKc/dT = Î”H⁰/RT² - Î”Æ”/T

or, dlnKc/dT = (Î”H⁰ - Î”Æ”RT)/RT²

∴ dlnKc/dT = Î”U⁰/RT²

- Î”U⁰ is the standers heat of reaction at constant volume. This is really the Van't Hoff reaction isochore in differential form.

- The integrated form of the equation at two temperature is,

ln(K

_{C}₂/K_{C}₁) = (Î”U⁰/R){(T₂ - T₁)/T₁T₂)}- For ideal system Î”U⁰ = Î”U. Since the reaction occurs in an equilibrium box at constant Volume this is called Van't Hoff reaction isochore.

- Two important assumptions are

- The reacting system is assumed to behave ideally.
- Î”H is taken independent of temperature for a small range of temperature change.

- Due to the assumption involved Î”H and Î”U do not produce the precise value of these reactions. As Î”G is independent of pressure for ideal system Î”H and Î”U also independent of pressure.

Hence, Î”G⁰ = - RT lnKp

or, [dlnKp/dT]

Problemor, [dlnKp/dT]

_{T}= 0- Show that the equilibrium condition for any chemical reaction is given by Î”G = 0.

Van't Hoff reaction isotherm is,

Î”G = - RT lnKa + RT lnQa

But when the reaction attains an equilibrium,

Qa = Ka

∴ Î”G = 0

Î”G = - RT lnKa + RT lnQa

But when the reaction attains an equilibrium,

Qa = Ka

∴ Î”G = 0

### Le -Chatelier principle from Van't Hoff equation

**give a quantitative expression of the Le-Chatelier Principle.**

*Van't Hoff equations*lnKp = - (Î”H/R)(1/T) + C

- It is evident that for endothermic reaction (Î”H〉0), an increase of T increases the value of any of the reactions.

- But for exothermic reaction (Î”Hã„‘ 0), with rising in temperature, a is decreased. This change of Kp also provides the calculation of the quantitative change of equilibrium yield of products.

- This above statement is in accordance with the Le - Chatelier Principle which states that whenever stress is placed on any system in a state of equilibrium, the system always reacts in a direction to reduce the applied stress.

- Thus when T is increased, the system in equilibrium will try to move in a direction in which heat is absorbed that is endothermic reaction is favors.