Question

For the reaction $2HI\left(g\right)\leftrightharpoons H_2\left(g\right)+I_2\left(g\right),$ the degree of dissociation α of $HI\left(g\right)$ is related to the equilibrium constant $\left(K_p\right)$ by the expression

• A. $\frac{1+2\sqrt{Kp}}{2}$
• B. $\frac{\sqrt{1+2Kp}}{\sqrt{2}}$
• C. $\frac{\sqrt{2Kp}}{\sqrt{2+2Kp}}$
• D. $\frac{2\sqrt{Kp}}{1+2\sqrt{Kp}}$

Answer 1

The correct option is D $\frac{2\sqrt{Kp}}{1+2\sqrt{Kp}}$


$\begin{array}{cccc}& 2HI\left(g\right)\leftrightharpoons & H_2\left(g\right)+& l_2\left(g\right)\\ Atequilibrium& 1-\alpha & \alpha /2& \alpha /2\end{array}$
Total moles at equilibrium =1
$PHI=(1-\alpha )\times P,P{H}_2=\alpha /2\times PP{l}_2=\alpha /2\times P$
$K_p=P{H}_2\times P{l}_2/(PHI{)}^2$

Substituting above values we get,

$K_p=a^2/[4\times (1-{\alpha }^2\left)\right]$
solving,$\alpha /1-\alpha =2(K_p{)}^{\frac{1}{2}}$
$\alpha =\frac{2\sqrt{Kp}}{1+2\sqrt{Kp}}$