Question

Van’t Hoff equation is
A) $\left( {d/dT} \right)\ln K = - \Delta H/R{T^2}$
B) $\left( {d/dT} \right)\ln K = + \Delta H/R{T^2}$
C) $\left( {d/dT} \right)\ln K = - \Delta H/RT$
D) $K = A{e^{\Delta H/RT}}$

Answer 1

To solve this we must know that the equation which relates the change in the equilibrium constant of any chemical reaction to the change in the temperature and the standard enthalpy of the reaction is known as the van’t Hoff equation. van’t Hoff equation is commonly used to explore the changes in the state functions of thermodynamic systems.

Complete solution:
We know that the equation which relates the change in the equilibrium constant of any chemical reaction to the change in the temperature and the standard enthalpy of the reaction is known as the van’t Hoff equation.
The expression for the van’t Hoff equation is as follows:
$\left( {\dfrac{d}{{dT}}} \right)\ln K = + \dfrac{{\Delta H}}{{R{T^2}}}$ …… (1)
Where $K$ is the equilibrium constant of the reaction,
$\Delta H$ is the enthalpy of the reaction,
$R$ is the universal gas constant,
$T$ is the temperature of the reaction.
Thus, the equation $\left( {d/dT} \right)\ln K = + \Delta H/R{T^2}$ is correct. Thus, option (B) is correct.
The equation (1) can be written as follows:
$K = A{e^{\Delta H/RT}}$
Thus, the equation $K = A{e^{\Delta H/RT}}$ is correct. Thus, option (D) is correct.
Thus, the van’t Hoff equation is $\left( {d/dT} \right)\ln K = + \Delta H/R{T^2}$ and $K = A{e^{\Delta H/RT}}$.
Thus, the correct options are (B) and (D).

Note: Using the van’t Hoff equation, we can determine the equilibrium constant for reversible reactions at a variety of temperatures. For an endothermic reaction, in which heat is absorbed the enthalpy of the reaction is positive the van’t Hoff plot has a negative value. And for an endothermic reaction, in which heat is evolved the enthalpy of the reaction is negative the van’t Hoff plot has a positive value.