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The equilibrium constant $K$ for the reaction $\ce{2HI <=> H2 + I2}$ at room temperature is $2.85$ and at $\pu{698 K}$ is $0.014$. This implies that:

a) $\ce{HI}$ is exothermic compound
b) $\ce{HI}$ is very stable at room temperature
c) $\ce{HI}$ is relatively less stable than $\ce{H2}$ and $\ce{I2}$
d) $\ce{HI}$ is resonance stabilized

I was getting both a and c (only one of the options is supposed to be correct). Which is the most appropriate answer?

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closed as off-topic by andselisk, user55119, Mithoron, Todd Minehardt, Buck Thorn May 5 at 10:35

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For an exothermic reaction $\Delta H \lt 0$ and $K_p$ must decrease as the temperature increases. (It is the opposite way round for an endothermic reaction, i.e. more dissociation at higher temperature which makes sense if the reaction is endothermic: 'more energy more product'.)

Using the integrated Van Hoff equation you can calculate what happens; $\displaystyle \ln\left(\frac{K_2}{K_1} \right) = -\frac{\Delta H}{R}\left( \frac{1}{T_2}-\frac{1}{T_1} \right)$ the slope is $-\Delta H/R$. If you plot your data at $298$ and $698$ K vs $1/T$ the slope is positive so $\Delta H$ is negative and the reaction exothermic.

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You can understand this as follows : Le Chatelier's principle says that on changing some factor ( like temp.,pressure,find.,etc ) the equilibrium will shift in a direction so as to undo the effect of change . Now in this case on increasing the temperature the reaction equilibrium is shifting in backward direction (descrease in value of K ) so reverse reaction must lower the temp. i.e.,the reaction must consume heat which means endothermic process and hence 'option a' is correct.

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Option (a) is more appropriate as in case of (c) the temperature or the condition is not mentioned. As in case of 698 K, since $K_\mathrm{eq}$ is reasonably less, the equilibrium will shift in the backward direction, so HI will be stable.

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