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

Options

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

Answer

A

Explanation

(A) `In ""(k_(2))/(k_(1))= (DeltaH)/( R ) [(1)/(T_(1))-(1)/(T_(2))]`
In `(6)/(2) = (DeltaH)/(R ) [1.5xx10^(-3) - 2 xx10^(-3) ]`
`Delta` H of the reaction comes out to be - ve Therefore the reaction is exothermic .