Question

$0.3moles$ of $H_2$ and $0.3$ moles of $l_2$ are allowed to react in $5lit$. Vessel at ${500}^{o}C$. If the equilibrium constant of the reaction $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ is $64$, the no of moles of unreacted $I_2$ at equilibrium is:

• A. $0.03$
• B. $0.06$
• C. $zero$
• D. $0.24$

Answer 1

The correct option is A $0.03$

$H_2+I_2\rightleftharpoons 2HI$

$0.3-x$ $0.3-x$ $2x$

$K_C=\frac{{\left(2x\right)}^2}{{(0.3-x)}^2}\Rightarrow 64=\frac{{4x}^2}{0.09+x^2-0.6x}\Rightarrow x=0.24$

No. of moles of unreacted $I_2$ at equilibrium $=0.3-0.24=0.06$