The graph In K(eq.)vs (1/T) relates for a reaction. The reaction must be:

Question

The graph In K(eq.)vs (1/T) relates for a reaction. The reaction must be: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/2a587ad6a6ae54484b7652392c80c05b-.jpeg" /> (A) Exothermic (B) Endothermic (C) Δ H is negligible (D) High spontaneous at ordinary temperature

Tardigrade Admin · Accepted Answer

In (K2/K1) =(Δ H/R)[(T2-T1/T1T2)] Since K is increased or decreased with temperature, thus Δ H = -v e

Q. The graph In $K_{(e q .) v s} \frac{1}{T}$ relates for a reaction. The reaction must be:

Exothermic

Endothermic

$Δ H$ is negligible

High spontaneous at ordinary temperature

$I n \frac{K _{2}}{K _{1}} = \frac{Δ H}{R} [\frac{T _{2} - T _{1}}{T _{1} T _{2}}]$
Since $K$ is increased or decreased with temperature, thus $Δ H = - v e$

Q. The graph In K(eq.)vs​T1​ relates for a reaction. The reaction must be: