Question

Two moles of HI were heated in a sealed tube at ${440}^{o}C$ till the equilibrium was reached. $HI$ was found to be $22\%$ decomposed. The equilibrium constant for dissociation is:

• A. $0.282$
• B. $0.0796$
• C. $0.198$
• D. $1.99$

Answer 1

The correct option is C $0.198$

$HI$,
there are 2 equilibrium constant $Kc$ and $Kp$
The reaction, $2HI⟶H_2+I_2$
moles initially 2 0 0
moles of equilibrium $2-2x$ $x$ $x$
$Kc$$=\frac{x\times x}{{(2-2x)}^2}$
$=\frac{x\times x}{{(2-2x)}^2}=\frac{{\left(0.22\right)}^2}{4\times {\left(0.78\right)}^2}$
=Concentration(C) =$\frac{P}{RT}=$ for gas phase reaction
$RT\times X=P_{H_2}=P_{I_2}=0.19888$