Question

3 moles of pcl5 kept in 1l closed reaction vessel was allowed to attain equilibrium at 380 k . The composi

Answer 1

Answer:The composition is HCl,P(OH)2

Explanation:

Answer 2

Given:

• Temperature = 380 k
• Moles of $PCl _5 = 3 moles$

• $K_c= 1.80$

To Find:  

• Composition of the mixture at equilibrium Kc=1.80

Solution:

Reaction of $PCl _5(g)$ is

           $PCl _5(g)$   ⇄$PCl_3(g) + Cl_2_(g)$​  

Initial     3 mole            0                0

Let  x $mol/litre$ of $PCl_5$ be dissociated,

At equilibrium:

                 (3−x),           x,            x

​      $K _C =\frac{ [PCl_3][Cl_2]}{[PCl _5]}$

​            $= \frac{x^2}{3-x}$

            $1.8 = \frac{x^2}{3-x}$

            $x^2 = 1.8(3-x)$

           $x^2+1.8x-5.4=0$

             $x=1.59$

         $[PCl_5]$ $= 3-x$

                    $=3-1.59\\ =1.41M$

         $[PCl_3]=[Cl_2] = x=1.59M .$  

Thus, Composition of the mixture at equilibrium

                                $[PCl_5] = 1.41M$

                                $[PCl_3]=[Cl_2] = x=1.59M$