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A schematic plot of ln K(eq) versus inve...

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

Text Solution

Verified by Experts

The correct Answer is:
A
At equilibrium `DeltaG = 0 rArr DeltaG^@ = - RT ln K_(eq)`
`DeltaH^@ - T Delta S^@ = - RT ln K_(eq)`
`=ln(K_(eq)) = (- DeltaH^@//R)(1//T) + (DeltaS^@//R)`
Slope `(- DeltaH^@//R) gt 0 rArr DeltaH^@ lt 0`
Reaction is exothermic.
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