Question

A schematic plot of in $K_{eq}$ v/s temperature for a reaction is shown below :
The reaction must be:

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

Answer 1

Answer: (B)

In $K_{eq}=-\frac{\triangle H}{R}\frac{1}{ T}+C$
For exothermic reaction
$\triangle H=-ve$
$\therefore $ plot of
In $K_{eq}$ vs $1/T$ gives a straight line with positive slope