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

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

__Equilibrium Constant__**K**

_{P}**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**

**equation is derived by using**

__Van't Hoff__**.**

__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

**reaction equation.**__Van't Hoff__
Greater the value of

**Î”H**, the faster the equilibrium constant (^{0}**K**) changes with temperature (T)._{P}
Separating the variables and integrating,

∫ dlnK

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

^{0}is independent of temperature)
or, lnK

The integration constant can be evaluated and identified as Î”S_{P}= - (Î”H^{0}/R)(1/T) + C (integrating constant)^{0}/R using the relation Î”G

^{0}= Î”H

^{0}- TÎ”S

^{0}.

Thus the

**Equation is**__Van't hoff__**lnK**

_{P}= - (Î”H^{0}/R)(1/T) + Î”S^{0}/R
For a reaction 2A + B ⇆ 2C, Î”G

^{0}(500K) = 2 KJ mol^{-1}
Find the K

_{P}at 500K for the reaction 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 KJ mol

^{-1}= - 8.31 × 10^{-3}KJ mol^{-1}K^{-1}× 500K lnK_{P}
or, lnK

_{P}= 1/(8.31 × 0.5) = 0.2406**∴ K**

_{P}= 1.27####
__Plot of lnK___{P} vs 1/T :

__Plot of lnK__

_{P}vs 1/T :

__For exothermic reaction, Î”H__^{0}= (-) ve.Examples are the formation of ammonia from H

_{2 }and N2.

Plot of lnK_{P} vs 1/T |

__For endothermic reaction, Î”H__^{0}= (+) ve._{2 }and I2.

For the reaction, Î”H

^{0}= 0, lnK

_{P}is independent of T. Provided Î”S

^{0}does not change much with T.

Since for idel system, H is not a function of P and Î”H

^{0}= Î”H

and

**equation is dlnK**

__Van't Hoff___{P}/dT = Î”H/RT

^{2 }and the integrated equation is,

**lnK**

_{P}= (Î”H/R)(1/T) + Î”S/R^{0}and Î”G

^{0}.

If we consider the integrated

**at two temperature then, it becomes,**

__Van't Hoff equation__**ln(K**

_{P2}/K_{P1}) = (Î”H/R){(T_{2}-T_{1})/T_{1}T_{2})}Where K

_{P1}and K

_{P2}are the equilibrium constants of the reaction at two different temperature T

_{1}and T

_{2}respectively.

Thus determination of K

_{P1}and K

_{P2}at two temperature helps to calculate the value of Î”H of the reaction.

The above reaction is called

**Van't Hoff**reaction isobar since 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 = (Î”H^{0}- Î”Æ”RT)/RT^{2}
or, dlnK

Î”U_{C}/dT = Î”U^{0}/RT^{2}^{0}is standerd heat of reaction at constant volume. This is really the

**reaction isochore in differential form.**

__Van't Hoff__The integrated form of the equation at two temperature is,

**ln(K**

_{C2}/K_{C1}) = (Î”U^{0}/R){(T_{2}-T_{1})/T_{1}T_{2})}**For ideal system Î”U**

^{0}= Î”U.**.**

__Van't Hoff reaction isochore__

__Two important assumptions are:__(i) The reacting system is assumed to behave ideally and

(ii) 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 a and b also independent of pressure.

Hence Î”G

^{0}= - RT lnK_{P}
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}**Thus, Î”G = 0.**

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

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

**equations gives quantitative expression of**

__Van't Hoff__**. From the equation,**

__Le-Chatelier Principle__**lnK**

_{P}= -(Î”H/R)(1/T) + CIt is evident that for

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

__endothermic reaction__**(Î”Hã„‘ 0), with rise in temperature, a is decreased.**

__exothermic reaction__This change of

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

_{P}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 is always react 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.