Question

$4.5$ moles each of hydrogen and iodine heated in a sealed $10$ litre vessel. At equilibrium, $3$ moles of $HI$ was found. The equilibrium constant for. $H_{2}(g)+I_{2}(g) \rightleftharpoons 2HI(g)$ is

• A. $1$
• B. $10$
• C. $5$
• D. $0.33$

Answer 1

Answer: (A)

Total mole at equilibrium $=(1-\alpha)+2\alpha =1 +\alpha$
$H_{2} +I_{2} \rightleftharpoons 2HI$
Initial cone. $\underset{x}{4.5} \underset {x}{4.5} \underset {2x}{0}$
From question $2x = 3$
$x=\frac {3}{2} =1.5$
So cone, at equilibrium $4.5 - 1.5$ of $H_{2}$
$= 4.5 - 1.5$ of $I_{2}$ and $3$ of HI
$K= \frac{\left[HI\right]^{2}}{\left[I_{2}\right]\left[H_{2}\right]}=\frac{3\times3}{3\times3}=1$