Browse Questions

# $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2 (g)$ In the dissociation of $PCl_5, x$ varies with $\frac{D}{d}$ according to:

(a)

(b)

(c)

(d)

Since,
Degree of Dissociation or association = $\frac{\text{Number of moles dissociated or associated}}{\text{Total number of moles present initially}}$

Degree of Dissociation of compounds like $PCl_5$ or $N_2O_4$, $\alpha = \frac{D-d}{d(y-1)} = \frac{M_c - M_o}{M_o (y-1)}$

Where D and d are the theoretical vapour density before and after dissociation.
$M_c$ and $M_o$ are the theoretical or calculatedd molecular mass and observed molecular mass respectively.
Here y is the number of moles of products formed from one kole of the reactant.

For the reaction,

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

Here y = 2

$\therefore$ Degree of Dissociation,
$x = \frac{D-d}{d(2-1)}$
$x = \frac{D - d}{d}$
$x = \frac{D}{d} - 1$

If graph between x (along Y-axis) and $\frac{D}{d}$ (along X-axis) is a straight line with slope, m=1 and intercepts on Y-axis, c = -1.
( $y = mx + c$ type)

edited Nov 23, 2013