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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2 (g)$ In the dissociation of $PCl_5, x $ varies with $\frac{D}{d}$ according to:

(a)

(b)

(c)

(d)

 

Can you answer this question?
 
 

1 Answer

0 votes

Answer:

 
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)

answered Nov 16, 2013 by mosymeow_1
edited Nov 23, 2013 by mosymeow_1
 

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