Van’t Hoff’s Equation- Equation, Derivation, Graphical representation, Practice Problems and FAQs

Do you know Van’t Hoff? Jacobus Henricus van 't Hoff was a Dutch chemist to claim the first Nobel Laureate in Chemistry and a very prominent theoretical chemist of his time. His groundbreaking work paved the way for modern chemical affinity, chemical equilibrium, chemical kinetics, and chemical thermodynamics theories.

He derived a relation connecting the change in the equilibrium constant with temperature from Gibbs- Helmholtz thermodynamic relations, which became known as the van’t Hoff equation. The equation is the change in the equilibrium constant, K_{eq}, of a chemical reaction to the change in temperature, T, assuming a constant enthalpy change over the temperature region. The Van 't Hoff equation has been widely used to investigate changes in thermodynamic state functions. The Van 't Hoff plot, which is derived from this equation, is particularly useful for assessing a chemical reaction's change in enthalpy and entropy.

Table of Contents

• Van’t Hoff’s Equation
• Derivation of Van’t Hoff’s equation
• Graphical Representation
• Practice Problems
• Frequently Asked Questions

Van’t Hoff’s Equation

Van’t Hoff equation is a relationship between the change in the equilibrium constant of a reversible reaction with the change in the temperature at constant enthalpy of the reactions. \frac{dlnK_}{dT}=\frac{\mathrm{\Delta}H^o}{\ RT^2} Applying between two temperatures,T1 and T2 with equilibrium constants K1 and K2 respectively Van't Hoff equation becomes on integration as, log\ \frac{K_1}{K_2}=\frac{\mathrm{\Delta}H^o}{2.303\ R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) This equation is referred to as the Van’t Hoff equation. It is used to quantify the effect of temperature on the value of the equilibrium constant.

Derivation of Van’t Hoff’s equation

A reversible reaction will proceed spontaneously in the forward direction only if it leads to a lowering of Gibbs free energy i.e., \mathrm{\Delta G} < 0. Consider a reversible reaction as given below. aA(aq)+bB(aq)\rlhar cC(aq)+dD(aq) The Gibbs free energy change for a reaction is given by \mathrm{\Delta G}=\mathrm{\Delta}G^o+RTln\ Q. Here, \mathrm{\Delta G} is the Gibbs free energy change of the reaction, \mathrm{\Delta}G^o is the standard Gibbs free energy change of the reaction, and Q is the reaction quotient. \mathrm{\Delta G}=\mathrm{\Delta}G^o+2.303\ RTlog_{10}\ Q Since at equilibrium, the reaction does not shift in either direction (forward or backward). Thus,\mathrm{\Delta G}=0 and Q=K_{eq} where K_{eq}=[C]c[D]d[A]a[B]b Here, K_{eq} refers to K_P if gas is there and K_C if an aqueous solution is given

On substituting this in the above equation, we get the following, \mathrm{\Delta}G^o=-2.303\ RTlog_{10}\ K_{eq} or \mathrm{\Delta}G^o=-RTln\ K_{eq}…………..(1) Now, we know that standard Gibbs free energy change is related to the standard enthalpy change and the standard entropy change as follows: \mathrm{\Delta}G^o=\mathrm{\Delta}H^o-T\mathrm{\Delta}S^o…………….(2) On using equation (1) and (2) we get the following,

-RTln\ K_{eq}=\mathrm{\Delta}H^o-T\mathrm{\Delta}S^o ln\ K_{eq}=\left(-\frac{\mathrm{\Delta}H^o}{RT}\right)+\frac{\mathrm{\Delta}S^o}{R} log\ K_{eq}=\left(-\frac{\mathrm{\Delta}H^o}{2.303\ RT}\right)+\frac{\mathrm{\Delta}S^o}{2.303\ R}…………….(3)

If at temperature T_1, the equilibrium constant is K_1, we can say, ln\ K_1=\left(-\frac{\mathrm{\Delta}H^o}{RT_1}\right)+\frac{\mathrm{\Delta}S^o}{R}……(4) If at temperature T_2, the equilibrium constant is K_2, we can say, ln\ K_2=\left(-\frac{\mathrm{\Delta}H^o}{RT_2}\right)+\frac{\mathrm{\Delta}S^o}{R}……(5) Subtracting equation (4) from (5), we get as follows: ln\ K_1-ln\ K_2=\left(-\frac{\mathrm{\Delta}H^o}{RT_1}\right)+\frac{\mathrm{\Delta}S^o}{R}-\left[\left(-\frac{\mathrm{\Delta}H^o}{RT_2}\right)+\frac{\mathrm{\Delta}S^o}{R}\right] ln\ K_1-ln\ K_2=\left(-\frac{\mathrm{\Delta}H^o}{RT_1}\right)+\left[\left(\frac{\mathrm{\Delta}H^o}{RT_2}\right)\right] ln\ K_1-ln\ K_2=\frac{\mathrm{\Delta}H^o}{\ R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) ln\ \frac{K_1}{K_2}=\frac{\mathrm{\Delta}H^o}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) log\ \frac{K_1}{K_2}=\frac{\mathrm{\Delta}H^o}{2.303\ R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) This is known as Van't Hoff’s equation.

Graphical Representation

We know that the equation of a straight line is given as y = mx + c ... (6) Where m is the slope of the line and c is the y-intercept. From equation 3, log\ K_{eq}=\left(-\frac{\mathrm{\Delta}H^o}{2.303\ RT}\right)+\frac{\mathrm{\Delta}S^o}{2.303\ R} If log\ K_{eq} vs\frac{1}{T} is plotted, Comparing equations (3) and (6) y = mx + c y=log\ K_{eq} x=\frac{1}{T} m=\left(-\frac{\mathrm{\Delta}H^o}{2.303\ R}\right) c=\frac{\mathrm{\Delta}S^o}{2.303\ R} The slope can be negative or positive depending on whether the \mathrm{\Delta}H^o for the reaction is positive or negative i.e. if the reaction is endothermic or exothermic. Intercept is positive for both kinds of reactions. (a) In the case of an endothermic reaction If the reaction is endothermic, then \mathrm{\Delta}H^o will be positive, and hence, the slope will be negative with a positive intercept. Thus, the equilibrium constant will increase with increasing temperature (or decrease). (b) In the case of an exothermic reaction If the reaction is exothermic, then \mathrm{\Delta}H^o will be negative, and hence, the slope will be positive with a positive intercept. Thus, the equilibrium constant will decrease with increasing temperature (or decrease in).

Practice Problems

Q1. The value of log_{10}\ K_{eq}for a reaction A\ \rlhar\ B is: Given:\mathrm{\Delta}H^o = –54.07 kJmol^{-1}; \mathrm{\Delta}S^o = 10 JK^{-1}mol^{-1} , R = 8.314 JK^{-1}mol^{-1}(2.303 × 8.314 × 298 = 5705; T = 298 K). 5 10 95 100 Solution Given, A\ \rlhar\ B \mathrm{\Delta}H^o = –54.07 kJmol^{-1} \mathrm{\Delta}S^o = 10 JK^{-1}mol^{-1} R = 8.314 JK^{-1}mol^{-1} 2.303 × 8.314 × 298 = 5705; T = 298 K. log\ K_{eq}=\left(-\frac{\mathrm{\Delta}H^o}{2.303\ RT}\right)+\frac{\mathrm{\Delta}S^o}{2.303\ R} log\ K_{eq}=-–54.07×10002.303 × 8.314 × 298 +102.303 × 8.314 log\ K_{eq}= 9.4+0.6 = 10 Therefore, the correct answer is an option (B).

Q2. The entropy and enthalpy of a reaction are given as \mathrm{\Delta}H^o = –27 kJmol^{-1}; \mathrm{\Delta}S^o = 63.18JK^{-1}mol^{-1}. Calculate the value of Equilibrium constant \ K_{eq}for a reaction at 300 K. 100 {10}^{10} {10}^8 8 Solution \mathrm{\Delta}H^o = –27 kJmol^{-1} \mathrm{\Delta}S^o = 63.18 JK^{-1}mol^{-1} R = 8.314 JK^{-1}mol^{-1} T = 300 K log\ K_{eq}=\left(-\frac{\mathrm{\Delta}H^o}{2.303\ RT}\right)+\frac{\mathrm{\Delta}S^o}{2.303\ R} log\ K_{eq}=-–27×10002.303 × 8.314 × 300 +63.182.303 × 8.314 log\ K_{eq}= 4.7+3.3 = 8 K_{eq}={10}^8 Therefore, the correct answer is option (C).

Q3. If log\ K_{eq} vs\frac{1}{T} is plotted, Calculate the entropy of the reaction from the given graph 57.4 JK^{-1}mol^{-1} 29 JK^{-1}mol^{-1} 39.2 JK^{-1}mol^{-1} 44.98 JK^{-1}mol^{-1} Solution : From the given graph, it is understood that it is a straight line graph and the equation can be written as y = mx + c….(1) log\ K_{eq}=\left(-\frac{\mathrm{\Delta}H^o}{2.303\ RT}\right)+\frac{\mathrm{\Delta}S^o}{2.303\ R}…(2) Comparing (1) and (2), The intercept c=\frac{\mathrm{\Delta}S^o}{2.303\ R}=3 \frac{\mathrm{\Delta}S^o}{2.303\ \times8.314}=3 \mathrm{\Delta}S^o=57.4JK^{-1}mol^{-1} Hence, the correct option is (A).

Q4. For an equilibrium, K_P=0.04 at {27}^oC and \mathrm{\Delta}H^o = 4.606 kcalmol^{-1}. Calculate the K_P at {37}^oC 0.06 0.03 0.08 0.05 Solution: Given : K_{P_1}=0.04 T_1={27}^oC T_1(K)=\ 273+27\ =\ 300\ K \mathrm{\Delta}H^o = 4.606 Kcalmol^{-1} T_2={37}^oC T_2(K)=\ 273+37\ =\ 310\ K R\ =\ 2\ cal\ K^{-1}mol^{-1} K_{P_1}=x Van't Hoff’s equation : log\ \frac{K_{P_1}}{K_{P_2}}=\frac{\mathrm{\Delta}H^o}{2.303\ R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) log\ \frac{K_{P_1}}{K_{P_2}}=\frac{4.606\times1000\ }{2.303\ \times2}\left(\frac{1}{310}-\frac{1}{300}\right) log\ \frac{0.04}{K_{P_2}}=1000\left(\frac{-10}{310\times300}\right) log\ \frac{0.04}{K_{P_2}}=\frac{-10}{93}=-0.107 \frac{0.04}{K_{P_2}}=0.78 K_{P_2}=0.05 So, the correct option is (D).

Frequently Asked Questions

Q1. What is the van't Hoff equation's significance?
The Van 't Hoff equation has been widely used to investigate changes in thermodynamic state functions. The Van 't Hoff plot, which is derived from this equation, is particularly useful for assessing a chemical reaction's change in enthalpy and entropy.

Q2.What is the van't Hoff equation's limitation?
The Van't Hoff equation cannot be utilized to calculate enthalpy and entropy values as a single source. All data received in this manner should be considered errored data.

Q3.What is the assumption in Van't Hoff equation?
Van't Hoff equation assumes constance of enthalpy over the temperature range of calculation, which may be true for small temperature differences. For longer temperature differences, change in the enthalpy with temperature also need to be included in the calculation.

Q4. Who was the first person to receive the Nobel Prize in chemistry?
Van 't Hoff, Jacobus Henricus Jacobus Henricus van 't Hoff received the Nobel Prize in Chemistry in 1901 "in recognition of extraordinary services" in chemistry.

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