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