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Q. A schematic plot of ln $k_{eq}$ versus inverse of temperature for a reaction is shown in the figure. The reaction must beChemistry Question Image

Chemical Kinetics

Solution:

$ln \frac{k_{2}}{k_{1}}=\frac{\Delta H}{R} \left[\frac{1}{T_{1}}-\frac{1}{T_{2}}\right]$
$ln \frac{6}{2}=\frac{\Delta H}{R} \left[1.5\times10^{-3}-2\times10^{-3}\right]$
$\Delta H$ comes to be negative. Hence the reaction is exothermic.