Question

Variation of $lo{g}_{10}K$ with $\frac{1}{T}$ is shown by the following graph in which straight line is at ${45}^o$, hence $\Delta H^o$ 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 Equation

$\log K=-\frac{\Delta H}{2.303RT}+A$

$y=mx+c⟶$ Equation for straight line

On comparing we have, $m=-\frac{\Delta H}{2.303R}$= slope

as, slope= $\tan \theta =\tan 45=1$

$\therefore \frac{-\Delta H}{2.303R}=\tan 45=1$

$\Rightarrow R=2$ calories

$\Rightarrow \Delta H=2\times 2.303=-4.606$ cal