Question

$K_c$ for the synthesis of $HI$ (g) from $H_2\left(g\right)$ and $I_2\left(g\right)$ is 50. The degree of dissociation of $HI$ is:
$H_2+I_2\rightleftharpoons 2HI$

• A. 0.11
• B. 0.54
• C. 0.48
• D. 0.79

Answer 1

The correct option is A 0.11

Given, $K_c=50$ for $H_2+I_2\rightleftharpoons 2HI$
equilibrium constant is reversed for the formation of HI i.e. $K_c=\frac{1}{50}$
$2HI\rightleftharpoons H_2+I_2$
$1-2\alpha$ $\alpha$ $\alpha$ -
$\frac{1}{50}=\frac{{\alpha }^2}{(1-2\alpha {)}^2}$
Taking square root both sides.
$\sqrt{\frac{1}{50}}=\frac{\alpha }{(1-2\alpha )}0.1414=\frac{\alpha }{(1-2\alpha )}0.1414-0.2828\alpha =\alpha \alpha =\frac{0.1414}{1.2828}=0.11$

Hence degree of dissociation is 0.11

Theory:

Degree of dissociation :
Degree of dissociation ( $\alpha$) is the fraction of moles of the reactant
that undergoes dissociation to form the products.

Degree of dissociation ( $\alpha$ is unitless)
For eg. Suppose 5 moles of $PC{l}_5$ are taken and if 2 moles of $PC{l}_5$ are dissociated, then:
$\alpha =\frac{2}{5}=0.4\Rightarrow 40$ % dissociated.
Maximum number of moles that can dissociate = Total number of moles taken.

Maximum value of alpha = 1 ($\alpha =1$)
Note :- Value of alpha varies from 0 to 1 and it can be fractional.

Calculation of degree of dissociation :
Degree of dissociation $\alpha =\frac{\text{Total of moles dissociated}}{\text{Initial moles of reactant}}$

Percentage dissociation$=\alpha \times 100$