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

**Van’t Hoff equation** connecting 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 shifted in the direction that absorbs heat. Therefore, the extent of reaction in learning chemistry can be easily derived by Van’t Hoff relation. If the equilibrium shifted in the forward direction or the concentration of product molecule increases then more of the reactant molecule converted into product. Hence increases in the extent of reaction lead to increases in the equilibrium constant in chemical science which is related mathematically by Van’t Hoff equation.

### Derivation of Van’t Hoff Equation

At a given temperature, the equilibrium constant of the chemical reaction remains unaltered, but the value varies considerably when the temperature is changed. Therefore, the quantitative relation, known Van’t Hoff equation connecting equilibrium constant and temperature can be derived thermodynamically starting from Gibbs – Helmholtz free energy equation.

This is the differential form of Van’t Hoff equation. Hence from Van’t Hoff equation greater the value of ΔH^{0}, 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 shift of equilibrium constant with temperature. Integration of the above equation given, ln k_{p} = – (ΔH^{0}/RT) + c, where c = integration constant. The integration constant can be calculated from the thermodynamic entropy relation, ΔG^{0} = ΔH^{0} – TΔS^{0}. Therefore, the equation becomes, ln k_{p} = – (ΔH^{0}/RT) + (ΔS^{0}/R). The integrated form of the Van’t Hoff equation at two temperature

where K_{P1} and K_{P2} are the equilibrium constants of the reaction at two different temperatures T_{1} and T_{2} respectively. Determination of K_{P1} and K_{P2} 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: The standard free energy of chemical reaction 2A + B ⇆ 2C at 500 K = 2 k joule mol^{-1}. Calculate K_{p} at 500 K for the chemical reaction A + ½B ⇆ C.

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

### Van’t Hoff Reaction Isochore

The differential form of Van’t Hoff isochoric equation, dln k_{c}/dT = ΔU^{0}/RT^{2}, where ΔU^{0} = standard heat of reaction at constant volume. The integration from of this equation at two temperature limits

For the ideal system, ΔU^{0} = 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 Chemical Reaction

- If a specific heat absorbed or ΔH
^{0}=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. - If heat emission or ΔH
^{0}= negative, the forward direction of the chemical reaction is the exothermic direction and the reverse one is the endothermic direction. The formation of ammonia by bonding nitrogen and hydrogen atom is an example of an endothermic chemical reaction.

Endothermic and Exothermic Equilibrium

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

- For an endothermic reaction, ΔH
^{0}〉 0 and the right-hand side of the equation positive. This leads to the fact that lnk_{p}increases with increasing temperature. - For an exothermic reaction, ΔH
^{0}〈 0 and the right-hand side of the equation negative. This leads to the fact that lnk_{p}decreases with increasing temperature.

### Heat Change in the Chemical Reaction

For the ideal system, the heat change is not a function of pressure. Therefore, standard enthalpy change and enthalpy change are equal values for the ideal gases. However, 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 lnk_{a} + RT lnQ_{a}. When the chemical solution attains the equilibrium, Q_{a} = k_{a}. Therefore, ΔG = 0.

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

Van’t Hoff equation, ln k_{p} = -ΔH/RT + c give a quantitative expression of the Le-Chatelier principle with temperature and equilibrium constant.

- Endothermic reaction, ΔH > 0, an increase of temperature increases the value of k
_{p}of the reaction. - But for an exothermic reaction, ΔH < 0, with rising the temperature, k
_{p}decreased.

This change of k_{p} provides the calculation of the quantitative change of equilibrium yield of products. The above statement is in accordance with the Le – Chatelier Principle which states that whenever stress 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 equilibrium point will try to move in a direction in which heat absorbed by reacting or product element. Therefore the endothermic reaction favors.
- When the temperature decreased the system at equilibrium point will try to move in a direction in which heat emitted Therefore, the exothermic reaction favors.