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Question

A schematic plot of $$\ln{{K}_{eq.}}$$ versus inverse of temperature for a reaction is shown below:
The reaction must be:
657063_2ea75ce555ea40c09dac4c72b451efc9.png


A
Exothermic
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B
Endothermic
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C
One with negligible enthalpy change
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D
Highly spontaneous at ordinary temperature
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Solution

The correct option is C Exothermic
The reaction must be Exothermic.
$$ \displaystyle lnK =- \dfrac {\Delta H}{RT}+C$$
The slope is positive.
Slope $$ \displaystyle   =- \dfrac {\Delta H}{R} > 0  $$
$$ \displaystyle     \dfrac {\Delta H}{R}< 0  $$
$$ \displaystyle    \Delta H< 0  $$
The negative value of the enthalpy change suggests exothermic reaction.

Chemistry

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