How much Van’t Hoff equation – effect on temperature?

Van’t Hoff equationequilibrium of a chemical reaction is constant at a given temperature. At equilibrium solution, Kp values can be changed with the change of temperature for an endothermic and exothermic reaction.

van hoff equation equilibrium chemical solution

This will be evident from the study of Kp values at different temperatures of a chemical reaction.

N2 + O2 ⇆ 2NO

Temperature Kp × 104
2000° K 4.08
2200° K 11.00
2400° K 25.10
2600° K 50.30

The quantitative relation, known Van’t Hoff equation connecting equilibrium solution and temperature can be derived thermodynamically starting from Gibbs – Helmholtz equation.

The Gibbs – Helmholtz equation

    \[ \Delta G^{0}= \Delta H^{0}+ \left [ \frac{\partial \left (\Delta G^{0} \right )}{\partial T} \right ]_{P} \]

Zero superscripts are indicating stranded values.

    \[ or,-\frac{\Delta H^{0}}{T^{2}}= -\frac{\Delta G^{0}}{T^{2}}+ \left [ \frac{\partial \left (\Delta G^{0} \right )}{\partial T} \right ]_{P} \]

    \[ or,-\frac{\Delta H^{0}}{T^{2}}= \left [\frac{\partial }{\partial T}\left (\frac{\Delta G^{0}}{T} \right ) \right ]_{P} \]

Van’t Hoff isotherm
– RT lnKP = ΔG0
or, – R lnKP = ΔG0/T

Differentiating with respect to temperature at constant pressure

    \[ -R\left (\frac{\partial lnK_{P}}{\partial T} \right )_{P}= \left [ \frac{\partial }{\partial T} \left ( \frac{\Delta G^{0}}{T} \right )\right ]_{P} \]

Comparing the above two-equation

    \[ \frac{\partial lnK_{P}}{\partial T}= \frac{\Delta H^{0}}{T^{2}} \]

This is the differential form of Van’t Hoff equation.

Derive van’t Hoff general equation

The greater the value of standard enthalpy of a reaction, the faster the chemical reaction reaching an equilibrium point.

    \[ \frac{\partial lnK_{P}}{\partial T}= \frac{\Delta H^{0}}{T^{2}} \]

Separating the variables and integrating

    \[ \int dlnK_{P}= \frac{\Delta H^{0}}{R}\int \frac{dT}{T^{2}} \]

ΔH0 independent of temperature.

    \[ or, lnK_{P}= -\frac{\Delta H^{^{0}}}{R}\times \frac{1}{T}+ C \]

where C = integrating constant.

The integration constant can be evaluated and identified the value of ΔS0/R, using the relation

ΔG0 = ΔH0 – TΔS0

∴ Van’t hoff equation

    \[ lnK_{P}= -\frac{\Delta H^{0}}{RT}+\frac{\Delta S^{0}}{R} \]

Problem
The standard free energy of chemical reaction 2A + B ⇆ 2C at 500 K = 2 KJ mol-1. Calculate Kp at 500 K for the chemical reaction A + ½B ⇆ C.

Solution
Standard free energy at 500 K for the chemical reaction

A + ½B → C

= 2 kJ mol-1/2
= 1 kJ mol-1

ΔG0 = – RT lnKP

∴ 1 = – 8.31 × 10-3 × 500 × lnKP
or, lnKP = 1/(8.31 × 0.5)
= 0.2406

∴ KP = 1.27

Enthalpy of reaction endothermic or exothermic

Heat absorption and emission in a chemical reaction can be studied from the following analysis.

Reactants → Products

We may study the following possibilities

Endothermic reaction, ΔH0 = positive

Low enthalpy side → High enthalpy side

If a specific heat absorbed, the forward direction of the chemical reaction is the endothermic direction and the reverse one is the exothermic direction. The dissociation of hydrogen iodide into hydrogen and Iodine is an example of this type of chemical reaction.

HI ⇆ H2 + I2 ΔH0 = (+) ve

An exothermic reaction, ΔH0 = negative

High enthalpy side → Low enthalpy side

If heat emission, the forward direction of the chemical reaction is the exothermic direction and the reverse one is the endothermic direction.
The formation of ammonia from hydrogen and nitrogen is an example of this type of chemical reaction.

N2 + H2 ⇆ 2NH3 ΔH0 = (-) ve

The endothermic and exothermic reaction

Van’t Hoff equation drowns the relation between equilibrium constant and endothermic and exothermic reactions.

  1. For an endothermic reaction, ΔH0 〉 0 and the right-hand side of the equation positive. This leads to the fact that lnKp increases with increasing temperature.
  2. For an exothermic reaction, ΔH0 〈 0 and the right-hand side of the equation negative. This leads to the fact that lnKp decreases with increasing temperature.

Heat change of a chemical reaction in two temperature

For the ideal system, the heat change or thermodynamics enthalpy change is not a function of pressure.

∴ Standard enthalpy change = enthalpy change
or, ΔH0 = ΔH

    \[ \frac{\partial lnK_{P}}{\partial T}= \frac{\Delta H}{RT^{2}} \]

    \[ \therefore lnK_{P}=\frac{\Delta H}{RT}+\frac{\Delta S}{R} \]

However, the entropy of an ideal gas depends strongly on pressure and ΔS and ΔG per mole of reaction in the mixture differ quite substantially from ΔS0 and ΔG0.

The integrated form of the Van’t Hoff equation at two temperature

    \[ ln\left ( \frac{K_{P_{2}}}{K_{P_{_{1}}}} \right )= \frac{\Delta H}{R}\left ( \frac{T_{2}-T_{1}}{T_{1}T_{2}} \right ) \]

where KP1 and KP2 are the equilibrium constants of the reaction at two different temperatures T1 and T2 respectively.

Determination of KP1 and KP2 at two temperatures helps to calculate the value of change of enthalpy of the chemical reaction.

The above relation called Van’t Hoff reaction isobar since pressure remains constant during the change of temperature.

Problem
Show that the equilibrium solution for any chemical reaction given by ΔG = 0.

Solution
Van’t Hoff reaction isotherm

ΔG = – RT lnKa + RT lnQa.
When the solution attains the equilibrium,
Qa = Ka.

∴ ΔG = 0

Assumptions from Van’t Hoff equation

  1. The reacting system of the chemical reaction behaves ideally.
  2. ΔH has 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 independent of pressure for ideal system ΔH and ΔU also independent of pressure.

ΔG0 = – RT lnKP

    \[ or,\left (\frac{\partial lnK_{P}}{\partial T} \right )_{T}= 0 \]

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

KP = KC (RT)Δγ
lnKP = lnKC + ΔγlnR + ΔγlnT

Differentiating with respect to T,

    \[ \frac{\partial lnK_{P}}{\partial T}= \frac{\partial lnK_{C}}{\partial T}+\frac{\Delta \gamma }{T} \]

    \[ \frac{\partial lnK_{P}}{\partial T}= \frac{\Delta H^{0}}{RT^{2}} \]

    \[ or,\frac{\partial lnK_{C}}{\partial T}= \frac{\Delta H^{0}}{RT^{2}}-\frac{\Delta \gamma }{T} \]

    \[ \therefore \frac{\partial lnK_{C}}{\partial T}= \frac{\Delta U^{0}}{RT^{2}} \]

ΔU0 = standers heat of reaction at constant volume. This is the differential form of Van’t Hoff reaction isochore.

The integrated form of the equation at two temperature

    \[ ln\left ( \frac{K_{C_{2}}}{K_{C_{_{1}}}} \right )= \frac{\Delta U^{0}}{R}\left ( \frac{T_{2}-T_{1}}{T_{1}T_{2}} \right ) \]

For ideal system ΔU0 = ΔU. Since the reaction occurs at constant volume and the equation called Van’t Hoff reaction isochore.

The heat of a reaction and Le-Chatelier principal

Van’t Hoff equations give a quantitative expression of the Le-Chatelier principle.

    \[ lnK_{P} =-\frac{\Delta H}{R}\frac{1}{T}+ C \]

  1. Endothermic reaction, ΔH〉0, an increase in temperature increases the value of lnKp of the chemical reactions.
  2. An exothermic reaction, ΔHㄑ 0, with rising in temperature, lnKp decreased.

Change of Kp provides the calculation of the quantitative change of equilibrium point yield of products.

According to the Le – Chatelier Principle, whenever stress placed on any system in a state of the equilibrium point, the system always reacts in a direction to reduce the applied stress.

  1. The temperature increased in a chemical solution, the system at equilibrium point will try to move in a direction in which heat absorbed, which is endothermic reaction favors.
  2. When the temperature decreased in a chemical solution, the system at equilibrium point will try to move in a direction in which heat emitted, which is exothermic reaction favors.