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__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 temperature is changed, the value of equilibrium constant also changed. The quantitative relation, relation, Known as_{P}**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

_{P}= Î”G^{0}/T
Differentiating 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

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

**Î”H**is independent of temperature.^{0}
or, lnK

_{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 |

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**.
Î”G

^{0}(500K) for the reaction**A + ½B ⇆ C**is,
2 KJ mol-1/2 = (1 KJ mol-1)/2.

The relation is Î”G

^{0}= - RT lnK_{P}
or, 1 = - (8.31 × 10-3 ) × (500) lnK

_{P}
or, lnK

_{P}= 1/(8.31 × 0.5)
or, lnK

_{P}= 0.2406∴ K_{P} = 1.27 |

__Plot of lnK___{P}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
Since for

**ideal system**,**H**is not a function of P and**Î”H**=^{0}**Î”H**and Van't Hoff equation is,**dlnK**

_{P}/dT = Î”H/RT^{2}
and the integrated equation is,

**lnK**

_{P}= (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”S**and^{0}**Î”G**.^{0}
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

**K**and_{P}_{1}**K**are the equilibrium constants of the reaction at two different temperature_{P}_{2}**T**and_{1}**T**respectively._{2}
Thus determination of

**K**and_{P}_{1}**K**at two temperature helps to calculate the value of_{P}_{2}**Î”H**of the reaction.
The above relation is called

__since__**Van't Hoff reaction isobar****P**remains constant during the change of temperature.
Use

**Gibbs - Helmholtz equation**to derive the**Vant Hoff reaction isochore**. Under What condition do you expect a linear relationship between**logk and 1/T**?
Again

**K**_{P}= K_{C}(RT)Î”Æ”
or, lnK

_{P}= lnK_{C}+ Î”Æ” lnR + Î”Æ”lnT
Differentiating with respect to

**T**,
dlnK

_{P}/dT = dlnK_{C}/dT + Î”Æ”/T
But dlnK

_{P}/dT = Î”H^{0}/RT^{2}
Hence, dlnK

_{C}/dT = Î”H^{0}/RT^{2}- Î”Æ”/T
or, dlnK

_{C}/dT = (Î”H^{0}- Î”Æ”RT)/RT^{2}∴ dlnK_{C}/dT = Î”U^{0}/RT^{2} |

**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 small range of temperature change.

Due to assumption involved

**Î”H**and**Î”U**do not produce 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
Show that the equilibrium condition for any chemical reaction is given by

**Î”G = 0**.
Vant Hoff reaction isotherm is,

**Î”G = - RT lnK**

_{a}+ RT lnQ_{a}
But when the reaction attain equilibrium

**Q**=_{a}**K**_{a}∴ Î”G = 0 |

__Expression of Le -Chatelier Principle from Van't Hoff Equation:__**Van't Hoff**equations gives quantitative expression of Le-Chatelier Principle.

From the equation,

lnK_{P} = -(Î”H/R)(1/T) + C |

It is evident that for

**endothermic reaction**(**Î”H〉0**), increase of T increases the value of a of the reaction.
But for

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

**K**also provides the calculation of quantitative change of equilibrium yield of products._{P}
This above statement is in accordance with

**which states that whenever stress is placed on any system in a state of equilibrium, the system is always react in a direction to reduce the applied stress.**__Le - Chatelier Principle__
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.