Question

variation of ${\log }_{10}K$ with $\frac{1}{T}$ is shown by the following graph in which straight line is at ${45}^{\circ }$ hence $△H^{\circ }$ is

• A. (+4.606) cal
• B. -4.606 cal
• C. 2 cal
• D. -2 cal

Answer 1

The correct option is B -4.606 cal

According to van't Hoff isotherm

$\log K=\log A-\frac{△H}{2.303RT}$

Where log A is the intercept w

$\log K=\log A-\frac{△H}{2.303R}\times \frac{1}{T}\log K=[-\frac{△H}{2.303R}]\times \frac{1}{T}+\log A.\therefore \text{Slope of plot}=\tan \theta ={45}^{\circ }=1=\frac{-△H}{2.303R}R=2\text{calories}△H=-2.303R△H=-2.303\times 2=-4.606cal$