Question

A schematic plot of ln $k_{eq}$ vs inverse of temperature for a reaction is shown below:

The reaction must be

• A. Exothermic
• B. Endothermic
• C. One with negligible enthalpy change
• D. Highly spontaneous at ordinary temperature

Answer 1

The correct option is A Exothermic

$ln\frac{k_2}{k_1}=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$
$ln\frac{6}{2}=\frac{\Delta H}{R}(1.5\times {10}^{-3}-2\times {10}^{-3})$
$\Delta H$ of the reaction comes out to be negative. The reaction is exothermic