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# Van’t Hoff Equation

## Van’t Hoff Equation-Temperature Effect on Equilibrium

Van’t Hoff equation connecting chemical equilibrium constant and temperature by thermodynamics relation of Gibbs-Helmholtz free energy equation. The standard chemical equilibrium constant of the endothermic and exothermic reaction depends on the temperature. When the temperature of the system in equilibrium increases, the equilibrium of the chemical reaction shifted in the direction that absorbs heat. The extent of reaction in learning chemistry can be easily derived by Van’t Hoff equation relation.

If the equilibrium shifted in the forward direction or the concentration of the product molecule increases then more of the reactant molecule is converted into a product. Hence the increases in the extent of reaction lead to increases in the equilibrium constant in physical chemistry which is related mathematically by Van’t Hoff equation. ## Derivation of Van’t Hoff Equation

At a given temperature according to Vant Hoff, the equilibrium constant of the chemical reaction remains unaltered, but the value varies considerably when the temperature is changed. The quantitative relation, known Van’t Hoff equation connecting equilibrium constant and temperature can be derived thermodynamically starting from Gibbs – Helmholtz free energy equation. The above relation is the differential form of the Van’t Hoff equation but the greater the value of ΔH0, the faster the equilibrium constant changes with temperature.

### Integration form of van’t Hoff Equation

The integration of Van’t Hoff isotherm enables us to calculate numerically the shift of equilibrium constant with temperature. Integration of the above equation given, ln kp = – (ΔH0/RT) + c, where c = integration constant. The integration constant can be calculated from the thermodynamic entropy relation, ΔG0 = ΔH0 – TΔS0. The equation becomes, ln kp = – (ΔH0/RT) + (ΔS0/R). Hence the integrated form of the Van’t Hoff equation at two temperatures where KP1 and KP2 are the equilibrium constants of the Van’t Hoff equation at two different temperatures T1 and T2 respectively. The determination of KP1 and KP2 at two temperatures helps to calculate the value of change of enthalpy of the chemical reaction. The above relation is called Van’t Hoff reaction isobar since pressure remains constant during the change of temperature.

Problem: The standard free energy of chemical reaction 2A + B ⇆ 2C at 500 K = 2 k joule mol-1. Calculate Kp at 500 K for the chemical reaction A + ½B ⇆ C by Van’t Hoff equation.

Solution: Standard free energy at 500 K for A + ½B → C = 1 kJ mol-1. Therefore, the calculated kp from the Van’t Hoff relation = 1.27.

### Van’t Hoff Reaction Isochore

The differential form of Van’t Hoff isochoric equation, dln kc/dT = ΔU0/RT2, where ΔU0 = standard heat of reaction at constant volume. The integration form of Van’t Hoff equation at two temperature limits For the ideal system, ΔU0 = ΔU, and two important assumptions are given from Van’t Hoff isochoric equation. The reacting system of the chemical reaction behaves ideally. ΔH has taken independent of temperature for a small range of temperature change.

## Endothermic and exothermic reactions

If a specific heat is absorbed or ΔH0 =positive, 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.

When heat emission or ΔH0 = negative. The forward direction of the chemical reaction is the exothermic direction and the reverse one is the endothermic direction. When the nitrogen atom chemical bonding with the hydrogen atom to form an ammonia molecule, it is an endothermic reaction.

### Endothermic and Exothermic Equilibrium

Vant Hoff drowns the relation between equilibrium constant with temperature in endothermic and exothermic reactions. 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. 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 in Chemical Reaction

For the ideal system, the heat change is not a function of pressure. But the standard enthalpy change and normal enthalpy change are equal values for the ideal gases. The entropy of an ideal gas depends strongly on pressure, entropy and free energy per mole of reaction in the mixture differ quite substantially from standard entropy and free energy.

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

Solution: From the Van’t Hoff reaction isotherm, ΔG = – RT lnka + RT lnQa. When the chemical solution attains the equilibrium, Qa = ka. Therefore, ΔG = 0.

### Le-Chatelier Principle and Van’t Hoff Equation

Van’t Hoff equation, ln kp = -ΔH/RT + c give a quantitative expression of the Le-Chatelier principle with temperature and equilibrium constant. Endothermic reaction, ΔH > 0, an increase in temperature increases the value of kp of the reaction. But for an exothermic reaction, ΔH < 0, with rising the temperature, kp decreased.

This change of kp provides the calculation of the quantitative change of equilibrium yield of products. Hence the above statement is in accordance with Le-Chatelier Principle. According to Le Chatelier whenever stress is placed on any system in a state of the equilibrium point, the system always reacts in a direction to balancing the applied stress.

The temperature increased the system at the equilibrium point will try to move in a direction in which heat is absorbed by reacting or product chemical element. Therefore the endothermic reaction favors. When the temperature decreased the system at the equilibrium point will try to move in a direction in which heat is emitted, the exothermic reaction favors according to Van’t Hoff equation and Le Chatelier principle.