Which of the following options are correct? [closed]








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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.


 1

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.


 1

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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