Question

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

• A. $\alpha =\frac{1+2\sqrt{K_P}}{2\sqrt{K_P}}$
• B. $\alpha =\sqrt{\frac{1+2K_P}{2}}$
• C. $\alpha =\sqrt{\frac{2K_P}{1+2K_P}}$
• D. $\alpha =\frac{2\sqrt{K_P}}{1+2\sqrt{K_P}}$

Answer 1

The correct option is D $\alpha =\frac{2\sqrt{K_P}}{1+2\sqrt{K_P}}$

Let the inital moles be $1$ so, at equilibrium

$\begin{array}{ccccc}2HI\left(g\right)& \rightleftharpoons & H_2\left(g\right)& +& I_2\left(g\right)\\ 1-\alpha & & \frac{\alpha }{2}& & \frac{\alpha }{2}\end{array}$

total moles at equilibirum $=1$
we know partial pressure is given as
= (mole fraction) $\times$ (total pressure)

hence $K_P$ is given as:
$K_P=\frac{{(\frac{\alpha }{2}\times P_T)}^2}{(1-\alpha {)}^2{P}_T^2}$
where $P_T$ represents the total pressure.

Simplifying the values,
$\frac{\alpha }{1-\alpha }=2\sqrt{K_P}$
$\alpha =\frac{2\sqrt{K_P}}{1+2\sqrt{K_P}}$



Theory:

Playing with $\alpha ,kandx$ :
Case 1 :
$A_n\left(g\right)\rightleftharpoons nA\left(g\right)$
Before dissociation, $a-$
(mole)
At $t=t_{eq}a-xnx$
(mole)
Unknowns after dissociation $a-a\alpha na\alpha$

Case 2 :
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
Before dissociation
$\text{moles}1--$

At $t=t_{eq}$
$\text{moles}1-x\frac{x}{2}\frac{x}{2}$

Moles after dissociation $1-\alpha \frac{\alpha }{2}\frac{\alpha }{2}$

Case 3 :

$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
At $t=0$ (moles)
$2--$
At $t=t_{eq}$ (moles)
$2-x\frac{x}{2}\frac{x}{2}$
Moles dissociated in terms of $\alpha$
$2-2\alpha \alpha \alpha$


Case 4:

$2A{B}_2\left(g\right)\rightleftharpoons 2AB\left(g\right)+B_2\left(g\right)$
At $t=0$ (moles)
$c--$
At $t=t_{eq}$ (moles )
$c-c\alpha c\alpha \frac{c\alpha }{2}$

Case 5:

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
At $t=0$ (moles)
$a3a-$
At $t=t_{eq}$ (moles)
$a-x3a-3x2x$

Case 6:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
At $t=0$ (moles) $15-$
At $t=t_{eq}$ (moles)
$1-x5-3x2x$


Case 7:
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$
At $t=0$ (moles)
$2-1$
At $t=t_{eq}$ (moles)
$2-x\frac{x}{2}1+\frac{x}{2}$
Moles dissociated in terms of $\alpha$
$2-2\alpha \alpha 1+\alpha$

Case 8:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
At $t=0$ (moles)
$152$
At $t=t_{eq}$ (moles)
$1-x5-3x2+2x$