Q. $ 2$ moles of $ PCl_{5} $ were heated in a closed vessel of $2\,L$ capacity. At equilibrium, $40\%$ of $ PCl_{5} $ is dissociated into $ PCl_{3} $ and $ Cl_{2} $ . The value of equilibrium constant is
Haryana PMTHaryana PMT 2011
Solution:
$PCl_5$
$PCl_3$
$Cl_2$
Initial
2
0
0
Change
-0.8
0.8
0.8
Equilibrium
1.2
0.8
0.8
Molar concentration
0.6
0.4
0.4
The expression for the equillibrium constant is
$K_c = \frac{[PCl_3][Cl_2]}{[PCl_2]} $
$ = \frac{0.4 \times 0.4}{0.6} = 0.266$.
| $PCl_5$ | $PCl_3$ | $Cl_2$ | |
|---|---|---|---|
| Initial | 2 | 0 | 0 |
| Change | -0.8 | 0.8 | 0.8 |
| Equilibrium | 1.2 | 0.8 | 0.8 |
| Molar concentration | 0.6 | 0.4 | 0.4 |