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 $\Delta H$ is :

• A. $+4.606$ cal
• B. $-4.606$ cal
• C. $+2$ cal
• D. $-2$ cal

Answer 1

Answer: (B)

$-4.606$ cal

Slope of plot $=\frac{-\Delta H^{\circ }}{2.303R}=1$

$\Rightarrow \Delta H^{\circ }=-2.303\times 2=4.606cal$.