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A schematic plot of log K(eq) vs inverse...

A schematic plot of log `K_(eq)` vs inverse of temperature for a reaction is shown in the figure. The reaction must be:

A

Exothermic

B

Endothermic

C

`Delta H` is negligible

D

High spontaneous at ordinary temperature

Text Solution

Verified by Experts

The correct Answer is:
A
In `(K_(2))/(K_(1))` = `(Delta H)/(R)` `[(T_(2) - T_(1))/(T_(1) T_(2))]`
Since K is increase or decrease with temperature, thus `Delta H = -ve`
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