Van't Hoff Equation

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 Van't Hoff equation is derived by using the Gibbs - Helmholtz equation.
    The Gibbs - Helmholtz equation is,
ΔG⁰ = ΔH⁰ + T[d(ΔG⁰)/dT]p
Zero superscripts are indicating stranded values

or,- (ΔH⁰/T² )= -(ΔG⁰/T² )+(1/T)[d(ΔG⁰)/dT]p
or, - (ΔH⁰/T² ) = [d/dT(ΔG⁰/T)]p

Again Van't Hoff isotherm is,

- RT lnKp = ΔG⁰

or, - R lnKp = ΔG⁰/T

Differentiating with respect to T at constant P,
- R [dlnKp/dT]p = [d/dT(ΔG⁰/T)]p

Comparing the above two equation we have,

dlnKp/dT = ΔH⁰/T²
    This is the differential form of Van't Hoff reaction equation.
    The greater the value of ΔH⁰, the faster the equilibrium constant (Kp) changes with temperature (T).
Separating the variables and integrating,
∫ dlnKp = (ΔH⁰/R)∫ (dT/T²)

Assuming that ΔH⁰ is independent of temperature.
or, lnKp = - (ΔH⁰/R)(1/T)+C
where C is integrating constant

The integration constant can be evaluated and identified as ΔS⁰/R using the relation,
ΔG⁰ = ΔH⁰ - TΔS⁰

Thus the Van't hoff Equation is,

lnKp = - (ΔH⁰/RT) + (ΔS⁰/R)

Problem
    For a reaction 2A + B ⇆ 2C, ΔG⁰(500 K) = 2 KJ mol⁻¹ find the Kp at 500 K for the reaction A + ½B ⇆ C.
Solution
ΔG⁰(500 K) for the reaction,
A + ½B ⇆ C; (2 KJ mol⁻¹)/2 = 1 KJ mol⁻¹.
The relation is ΔG⁰ = - 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, ΔH⁰ = (-) ve.
    Examples are the formation of ammonia from H₂ and N₂.
N₂ + H₂ ⇆ 2NH₃ ΔH⁰ = (-) ve
  • For endothermic reaction, ΔH⁰ = (+) ve.
    Examples are the dissociation of HI into H₂ and I₂.
HI ⇆ H₂ + I₂ ΔH⁰ = (+) ve
  • For the reaction, ΔH⁰ = 0.
    lnKp is independent of T. Provided ΔS⁰ does not change much with T.
Van't Hoff equation
Van't Hoff equation


    Since for ideal system, H is not a function of P and ΔH⁰ = ΔH and Van't Hoff equation is,
dlnKp/dt = ΔH/RT²

The integrated equation is,
lnKp = (ΔH/RT) + ΔS/R
    However, S of an ideal gas depends strongly on P, so ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS⁰ and ΔG⁰.
    If we consider the integrated Van't Hoff equation at two temperature then, it becomes,
ln(Kp₂/Kp₁) = (ΔH/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 ΔH of the reaction.
    The above relation is called Van't Hoff reaction isobar since P remains constant during the change of temperature.
Problem
    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?
Solution
Again Kp = Kc (RT)ΔƔ
or, lnKp = lnKc + ΔƔ lnR + ΔƔ lnT
Differentiating with respect to T,
dlnKp/dT = dlnKc/dT + ΔƔ/T
But dlnKp/dT = ΔH⁰/RT²
Hence, dlnKc/dT = ΔH⁰/RT² - ΔƔ/T
or, dlnKc/dT = (ΔH⁰ - ΔƔRT)/RT²

∴ dlnKc/dT = ΔU⁰/RT²
    ΔU⁰ 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(KC₂/KC₁) = (ΔU⁰/R){(T₂ - T₁)/T₁T₂)}
    For ideal system ΔU⁰ = ΔU. Since the reaction occurs in an equilibrium box at constant Volume this is called Van't Hoff reaction isochore.
  • Two important assumptions are
  1. The reacting system is assumed to behave ideally.
  2. ΔH is taken independent of temperature for a small range of temperature change.
    Due to the assumption involved ΔH and ΔU do not produce the precise value of these reactions. As ΔG is independent of pressure for ideal system ΔH and ΔU also independent of pressure.
Hence, ΔG⁰ = - RT lnKp
or, [dlnKp/dT]T = 0
Problem
    Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
Solution
Van't Hoff reaction isotherm is,
ΔG = - RT lnKa + RT lnQa
But when the reaction attains an equilibrium,
Qa = Ka

∴ ΔG = 0

Le -Chatelier principle from Van't Hoff equation

    Van't Hoff equations give a quantitative expression of the Le-Chatelier Principle.
lnKp = - (ΔH/R)(1/T) + C
    It is evident that for endothermic reaction (ΔH〉0), an increase of T increases the value of any of the reactions.
    But for exothermic reaction (ΔHㄑ 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.

Effect of temperature on equilibrium constant, Van't Hoff equation, expression of Le -Chatelier principle from Van't Hoff equation, problem solution

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