1.

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

Answer»

EXOTHERMIC
endothermic
one with negligible ENTHALPY change highly spontaneous at ordinary temperature .
high spontaneous at ordinary temperature .

Solution :(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 .


Discussion

No Comment Found

Related InterviewSolutions