Question

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

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

Answer 1

The correct option is B $-4.606cal$

Given,

Angle the line makes with the horizontal$={45}^{\circ }$

Therefore, slope of the line $m=\tan {45}^{\circ }=1$ $...\left(i\right)$

Now, we know that

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

$\Rightarrow -2.303RT\left(\log K\right)=△H-T\Delta S^{\circ }$

Simplifying the above expression, we get

$\Rightarrow \log K=\frac{-\Delta H}{2.303RT}+\frac{T\Delta S^{\circ }}{2.303RT}$

Simplifying the equation further and writing it in the form of equation of a line we get,

$\Rightarrow \log K=\frac{-\Delta H}{2.303R}\left(\frac{1}{T}\right)+\frac{\Delta S^{\circ }}{2.303R}$

Comparing the above equation with the general equation of a straight line $y=mx+c$ , we get

$m=\frac{-\Delta H}{2.303R}$

Also, from equation $\left(i\right)$ we get

$m=1$ , therefore we get $1=\frac{-\Delta H}{2.303R}$

Substituting the value of $R=2$ , we get $\Delta H=-2.303\left(2\right)=-4.606cal$

Therefore, the correct answer will be option B