A schematic plot of In Keq versus inverse of temperature for a reaction is shown below The reaction must be

Question

A schematic plot of In Keq versus inverse of temperature for a reaction is shown below <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/190d36befaee380efa06c8ba6b4bce22-.png" /> The reaction must be (A) highly spontaneous at ordinary temperature (B) one with negligible enthalpy change (C) endothermic (D) exothermic

Tardigrade Admin · Accepted Answer

Variation of Ke q with temperature T is given by van't-Hoff equation. log K eq =-(Δ H °/2.303 RT )+(Δ S°/ R ) Slope of the given line is positive indicating that term A is positive thus Δ H° is negative. Thus. reaction is exothermic.

Q. A schematic plot of In $K_{e q}$ versus inverse of temperature for a reaction is shown below

The reaction must be

highly spontaneous at ordinary temperature

one with negligible enthalpy change

endothermic

exothermic

Variation of $K_{e q}$ with temperature $T$ is given by van't-Hoff equation.
$lo g K_{e q} = - \frac{Δ H ^{\circ}}{2.303 RT} + \frac{Δ S ^{\circ}}{R}$
Slope of the given line is positive indicating that term $A$ is positive thus $Δ H^{\circ}$ is negative.
Thus. reaction is exothermic.

Q. A schematic plot of In Keq​ versus inverse of temperature for a reaction is shown below The reaction must be