Question

For a reaction: $H\left)2\right(g)+I_2(g)\rightleftharpoons 2HI(g)$ at certain temperature, the value of equilibrium constant is $50$. If the volume of the vessel is reduced to half of its original volume, the value of new equilibrium constant will be :

• A. $25$
• B. $50$
• C. $100$
• D. unpredictable

Answer 1

The correct option is B $50$

$\underset{\underset{\frac{1-\alpha }{v}}{1}}{H_2}+\underset{\underset{\frac{1-\alpha }{v}}{1}}{I_2}\rightleftharpoons \underset{\underset{\frac{2\alpha }{v}}{0}}{2HI}$
Since $\Delta n=0[2-(1+1\left)\right]$
$\therefore$ K does not depend on volume.
So on reducing volume to half K does not change.
$\therefore K=50$