Question

The graph relates $\ln K_{eq.}vs.\frac{1}{T}$ for a reaction. The reaction must be :

• A. exothermic
• B. endothermic
• C. $\Delta H$ is negligible
• D. spontaneous

Answer 1

The correct option is A exothermic

The graph relates $\ln K_{eq.}vs.\frac{1}{T}$ for a reaction. The reaction must be exothermic. Adding temperature would cause the reactant concentrations to increase; thus, equilibrium constant K will decrease.

$\ln \frac{K_2}{K_1}=\frac{\Delta H}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

Since, $K$ increases with decrease of temperature and thus, $\Delta H=-ve$.

An exothermic reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released into the surroundings, resulting in an overall negative quantity for the heat of reaction. A system that releases heat to the surroundings, an exothermic reaction, has a negative $\Delta H$ by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system.