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Q.

The graph ln $K_{(eq .)} vs \frac{1}{T}$ relates for a reaction. The reaction must be

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a

Exothermic

b

Endothermic

c

$∆ H$ is negligible

d

High spontaneous at ordinary temperature

answer is A.

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Detailed Solution

$\ln \frac{K_{2}}{K_{1}} = \frac{∆ H}{R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

Since K is increased or decreased with temperature, thus $∆ H = - ve$

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