Question

A schematic plot of $\ln K_{eq.}$ versus 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

Answer: (A)

Exothermic

The reaction must be Exothermic.

$lnK=-\frac{\Delta H}{RT}+C$

The slope is positive.
Slope $=-\frac{\Delta H}{R}>0$
$\frac{\Delta H}{R}<0$
$\Delta H<0$
The negative value of the enthalpy change suggests exothermic reaction.