__Van't Hoff Equation - Effect on temperature on Equilibrium Constant:__

__Van't Hoff Equation - Effect on temperature on Equilibrium Constant:__

Equilibrium Constant

**K**of a reaction is constant at a given temperature but when the temperature is changed, the value of equilibrium constant also changed._{p}- The quantitative relation, relation, Known as Van't Hoff equation is derived by using Gibbs - Helmholtz equation.

- The Gibbs - Helmholtz equation is,

**Î”G**

^{0}= Î”H^{0}+ T[d(Î”G^{0})/dT]_{P}**Zero superscript**is indicating the stranded values.

or, - (Î”H

^{0}/T

^{2})= -(Î”G

^{0}/T

^{2})+(1/T)[d(Î”G

^{0})/dT]

_{P}

or, - (Î”H

^{0}/T

^{2}) = [d/dT(Î”G

^{0}/T)]

_{P}

Again Van't Hoff isotherm is,

- RT lnK_{P} = Î”G^{0} |

or, - R lnK

Differentiating with respect to T at constant P,

- R [dlnK

Comparing the above two equation we have,

_{P}= Î”G^{0}/TDifferentiating with respect to T at constant P,

- R [dlnK

_{P}/dT]_{P}= [d/dT(Î”G^{0}/T)]_{P}Comparing the above two equation we have,

dlnK_{P}/dT = Î”H^{0}/T^{2} |

- This is the differential form of

**Van't Hoff reaction equation**.

- Greater the value of

**Î”H**, the faster the

^{0}__equilibrium constant__(

**K**) changes with temperature (

_{P}**T**).

Separating the variables and integrating,

∫ dlnK

Assuming that

The integration constant can be evaluated and identified as Î”S

Î”G

Thus the Van't hoff Equation is,

∫ dlnK

_{P}= (Î”H^{0}/R)∫ (dT/T^{2})Assuming that

**Î”H**is independent of temperature.,br />or, lnK^{0}_{P}= - (Î”H^{0}/R)(1/T)+C(integrating constant)The integration constant can be evaluated and identified as Î”S

^{0}/R using the relation,Î”G

^{0}= Î”H^{0}- TÎ”S^{0}Thus the Van't hoff Equation is,

lnK_{P} = - (Î”H^{0}/R)(1/T) + Î”S^{0}/R |

- Problem 1:

- For a reaction

**2A + B ⇆ 2C,**

**Î”G**(500 K) =

^{0}**2 KJ mol**Find the

^{-1}**K**at 500 K for the reaction

_{P}**A + ½B ⇆ C**.

- Answer:

Î”G⁰(500 K) for the reaction,

The relation is Î”G⁰ = - RT ln

or, 1 = - (8.31 × 10-3 ) × (500) lnKP

or, lnKP = 1/(8.31 × 0.5)

or, lnKP = 0.2406

**A + ½B ⇆ C;**(2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.The relation is Î”G⁰ = - RT ln

**Kp**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:**

- For
**exothermic**reaction,**Î”H**=^{0}**(-) ve**.

- Examples are the formation of

**ammonia**from

**H**and

_{2}**N**.

_{2}N_{2} + H_{2} ⇆ 2NH_{3} | Î”H = ^{0}(-) ve |

- For
**endothermic**reaction,**Î”H**=^{0}**(+) ve**.

- Examples are the dissociation of

**HI**into

**H**and

_{2}**I**.

_{2}HI ⇆ H_{2} + I_{2} | Î”H = ^{0}(+) ve |

- For the reaction,
**Î”H**.^{0}= 0

**lnK**is independent of

_{P}**T**. Provided

**Î”S**does not change much with

^{0}**T**.

The plot of dlnKp vs T |

- 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 and the integrated equation is,

**lnKp = (Î”H/R)(1/T) + 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(K_{P}_{2}/K_{P}_{1}) = (Î”H/R){(T_{2} - T_{1})/T_{1}T_{2})} |

- Where

**Kp₁**and

**Kp₂**are the equilibrium constants of the reaction at two different temperature

**T₁**and

**T₂**respectively.

- Thus the determination of

**Kp₁**and

**Kp₂**at two temperature 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.

- Problem 2:

- Use

**Gibbs - Helmholtz equation**to derive the

**Vant Hoff reaction isochore**. What condition do you expect a linear relationship between

**logk and 1/T?**

- Answer:

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

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”U**is standers heat of reaction at constant volume. This is really the

^{0}**in differential form.**

__Van't Hoff reaction isochore__- The integrated form of the equation at two temperature is,

ln(K_{C}_{2}/K_{C}_{1}) = (Î”U^{0}/R){(T_{2} - T_{1})/T_{1}T_{2})} |

- For ideal system

**Î”U**=

^{0}**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 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”G

^{0}= - RT lnKp or, [dlnK

_{P}/dT]

_{T}= 0

- Problem 3:

- Show that the equilibrium condition for any chemical reaction is given by

**Î”G = 0**.

- Answer:

Vant Hoff reaction isotherm is,

Î”G = - RT lnKa + RT lnQa

But when the reaction attains an equilibrium,

Qa = Ka

Î”G = - RT lnKa + RT lnQa

But when the reaction attains an equilibrium,

Qa = Ka

**∴ Î”G = 0**__Expression of Le -Chatelier Principle from Van't Hoff Equation:__

__Expression of Le -Chatelier Principle from Van't Hoff Equation:__

Van't Hoff equations give a quantitative expression of Le-Chatelier Principle.

**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 reaction.

- But for exothermic reaction

**(Î”Hã„‘ 0)**, with rising in temperature, a is decreased. This change of

**Kp**also provides the calculation of quantitative change of equilibrium yield of products.

- This above statement is in accordance with 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.