A schematic plot of ln `K_(eq)` versus inverse of temperature for a reaction is shown below: The reaction must be:

Options

A. Highly spontaneous at ordinary temperature
B. One with negligible enthalpy change
C. endothermic
D. exothermic

Answer

D

Explanation

Variation of `K_(eq)` with temperature t is given by van't Hoff equation ln `K_(eq) = -(DeltaH^(@))/(R/A) .1/T + (DeltaS^(@))/(R/R)`
Slope of the given line is positive indicating that term A is positive, thus `DeltaH^(@)` is negative. Thus, reaction is exothermic.