Question

A 1 litre container contains 2 moles of pcl5 initially if at equilibrium kc is found to be 1 the degree of dissociation of pcl5 is

Answer 1

dissociation of $PCl_5$ is ....

$PCl_5\Leftrightarrow PCl_3+Cl_2$

initially, $[PCl_5]$ = 2mol/L = 2M

$[PCl_3]$ = 0

$[Cl_2]$ = 0

at equilibrium,

$[PCl_5]$ = $2(1-\alpha)$

$[PCl_3]$ = $2\alpha$

$[Cl_2]$ = $2\alpha$

now, $K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

1 = $\frac{2\alpha.2\alpha}{2(1-\alpha)}$

1 = $\frac{4\alpha^2}{2-2\alpha}$

2 - 2$\alpha$ = 4$\alpha^2$

$4\alpha^2+2\alpha-2=0$

$2\alpha^2+\alpha-1=0$

$\alpha=1, -1$

hence, degree of dissociation of PCl5 is 1 or 100% .

Answer 2

Answer:

1/2

here's the answer and hope this helps you.