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 the temperature is changed, the value of equilibrium constant also changed.

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,
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,
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^{0}, the faster the equilibrium constant (K_{P}) changes with temperature (T).
Separating the variables and integrating,
∫ dlnK_{P} = (Î”H^{0}/R)∫ (dT/T^{2})
Assuming that Î”H^{0} is independent of temperature.,br />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,
∫ dlnK_{P} = (Î”H^{0}/R)∫ (dT/T^{2})
Assuming that Î”H^{0} is independent of temperature.,br />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 
 Problem 1:

For a reaction 2A + B ⇆ 2C, Î”G^{0}(500 K) = 2 KJ mol^{1}
Find the K_{P} at 500 K for the reaction A + ½B ⇆ C.
 Answer:
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 × 103 ) × (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 × 103 ) × (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_{2} and 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_{2} and I_{2}.
HI ⇆ H_{2} + I_{2}  Î”H^{0} = (+) ve 
 For the reaction, Î”H^{0} = 0.

lnK_{P} is independent of T. Provided Î”S^{0} does not change much with 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
∴ 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”U^{0} is 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”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 = 0
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:
Van't Hoff equations give a quantitative expression of LeChatelier Principle.
lnKp = (Î”H/R)(1/T) + C
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.