# Which of the following options are correct? [closed]

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?

## closed as off-topic by andselisk♦, user55119, Mithoron, Todd Minehardt, Buck ThornMay 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.
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.