Question

According to Van't Hoff equation, which of the following statement is correct regarding the plot of $\ln K_{eq}\text{v/s}\frac{1}{T}$ ?

• A. For an endothermic reaction, the slope is $\frac{\Delta S^{\circ }}{RT}$
• B. For an endothermic reaction, the slope is $\frac{-\Delta H^{\circ }}{R}$
• C. For an exothermic reaction, the slope is $\frac{\Delta S^{\circ }}{R}$
• D. For an exothermic reaction, the slope is $\frac{-\Delta H^{\circ }}{RT}$

Answer 1

The correct option is B For an endothermic reaction, the slope is $\frac{-\Delta H^{\circ }}{R}$

We know that,

$\Delta G^{\circ }=-RTln\left(K_{eq}\right)$

$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$

Using above two equations,
$-RTln\left(K_{eq}\right)=\Delta H^{\circ }-T\Delta S^{\circ }$

$\Rightarrow ln{K}_{eq}=\frac{-\Delta H^{\circ }}{RT}+\frac{\Delta S^{\circ }}{R}$

Comparing it with,
y= mx +c

we get,
Graph between $\ln K_{eq}\text{v/s}\frac{1}{T}$ has a slope of $\frac{-\Delta H^{\circ }}{R}$ and an intercept of $\frac{\Delta S^{\circ }}{R}$
R is the universal gas constant having a positive value.
For an endothermic reaction, $\Delta H>0$ and for an exothermic rection $\Delta H<0$
Therefore,
$\text{S}lope=-\frac{\Delta H^{\circ }}{R}<0$ for endothermic reaction.
$\text{S}lope=-\frac{\Delta H^{\circ }}{R}>0$ for exothermic reaction.
Option B is correct.


Theory:

Case 1 : plot of $lnKvs\frac{1}{T}$ for endothermic reaction


Case 2 : plot of $lnKvs\frac{1}{T}$ for exothermic reaction


We can clearly see from above plots that in case 1 plot of $lnKvs\frac{1}{T}$ for endothermic,$\Delta H^o$ is positive and reaction slope is negative with a positive intercept.

For case 2 plot of $lnKvs\frac{1}{T}$ for exothermic reaction, $\Delta H^o$ is negative and has a positive slope with a positive intercept.


If at temperature $T_1$, equilibrium constant is $K_1$ and at temperature $T_2$, it is
$K_2$ then,
$ln\left(K_1\right)=\frac{\Delta H^o}{R{T}_1}+\frac{\Delta S^o}{R}$....................(1)
$ln\left(K_2\right)=\frac{\Delta H^o}{R{T}_2}+\frac{\Delta S^o}{R}$...................(2)


Equation $\left(1\right)-\left(2\right)$
$ln\left(\frac{K_1}{K_2}\right)=\frac{\Delta H^o}{R}[\frac{1}{T_2}-\frac{1}{T_1}]$