Van't Hoff Equation

Van't Hoff Equation - Effect on temperature on Equilibrium Constant:

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 Gibbs - Helmholtz equation.
    The Gibbs - Helmholtz equation is,
ΔG0 = ΔH0 + T[d(ΔG0)/dT]P
Zero superscript is indicating the 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.
Separating the variables and integrating,
dlnKP = (ΔH0/R)(dT/T2)
Assuming that ΔH0 is independent of temperature.,br />or, lnKP = - (ΔH0/R)(1/T)+C(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/R)(1/T) + ΔS0/R
  • Problem 1:
    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.
  • Answer:
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:

  • 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 - Effect on temperature on Equilibrium Constant and Expression of Le -Chatelier Principle from Van't Hoff Equation
The plot of dlnKp vs T
    Since for ideal system, H is not a function of P and ΔH⁰ = ΔH and Van't Hoff equation is,
    dlnKp/dt = ΔH/RT²
    and the integrated equation is,
    lnKp = (ΔH/R)(1/T) + 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 temperature T₁ and T₂ respectively.
    Thus the determination of Kp₁ and Kp₂ at two temperature 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 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)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 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 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 3:
    Show that the equilibrium condition for any chemical reaction is given by ΔG = 0.
  • Answer:
Vant Hoff reaction isotherm is,
ΔG = - RT lnKa + RT lnQa
But when the reaction attains an equilibrium,
Qa = Ka
∴ ΔG = 0

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

Van't Hoff equations give a quantitative expression of 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 reaction.
    But for exothermic reaction (ΔHㄑ 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.

Van't Hoff Equation: Effect on temperature on Equilibrium Constant, Plot of lnKP vs 1/T and Expression of Le -Chatelier Principle from Van't Hoff Equation.

Inorganic Chemistry

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