Question

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

• A. exothermic
• B. endothermic
• C. one with negligible enthalpy change
• D. highly spontaneous at ordinary temperature.

Answer 1

Answer: (A)

exothermic

Answer 2

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