Question

If 0.64 mol PCl5 is placed in a 1.0 L flask and allowed to reach equilibrium at a given temperature, what is the final concentration of Cl2 in the flask?

PCl5 (g)→ PCl3 (aq) + Cl2 (g) Kc= 0.47

Answer 1

Answer: The final concentration of $Cl_2$ at equilibrium is 0.36 M

Explanation:

Moles of  $PCl_5$ = 0.64 mole  

Volume of solution = 1.0 L

Initial concentration of $PCl_5$ = $\frac{moles}{Volume}=\frac{0.64mol}{1.0L}=0.64 M$

The given balanced equilibrium reaction is,

                     $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

Initial conc.       0.64 M         0M        0M  

At eqm. conc.     (0.64-x) M   (x) M      (x) M

The expression for equilibrium constant for this reaction will be,

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

Now put all the given values in this expression, we get

$0.47=\frac{(x)\times (x)}{(0.64-x)}$

By solving the term 'x', we get :

x = 0.36

Thus, the final concentration of $Cl_2$ at equilibrium is x = 0.36 M

Answer 2

The concentration of Cl2 is 0.36M.

Number of moles of PCl5 = 0.64 mol

Volume =  1.0 L

Concentration of PCl5 = 0.64 mol/1.0 L  = 0.64  M

The equation of the reaction is;

          PCl5 (g)→ PCl3 (aq) + Cl2 (g)

I         0.64            0                0

C       -x                 +x               +x

E     0.64 - x           x                  x

Kc = [PCl3] [Cl2]/[PCl5]

0.47 = x^2/0.64 - x

0.47(0.64 - x) =  x^2

0.3 - 0.47x =x^2

x^2 +  0.47x - 0.3 = 0

x = 0.36 M

The concentration of Cl2 is 0.36M.

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