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

Equilibrium

Solution:

Slope of plot $= \frac {-\Delta H^{o}}{ 2.303R}=1$
$\Rightarrow \Delta H^{o} =-2.303 \times 2=-4.606$ cal