Question

Calculate the number of moles of Cl2 produced at equilibrium when 3.98 mol of PCl5 is heated at 283.9 deg celsius in a vessel having a capacity of 10.0L. At 283.9, K = 0.060 for this dissociation.

Answer 1

Answer:

1.274 moles

Explanation:

The equation for the reaction can be represented as follows:

                       $PCl_5(g)$           ⇄       $PCl_3(g)$     +         $Cl_2(g)$

K = 0.060

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Concentration of   $PCl_5(g)$  = $\frac{numbers of moles}{volume}$

Concentration of   $PCl_5(g)$  = $\frac{3.98}{10.0}$

Concentration of   $PCl_5(g)$  = 0.398 moles

If we construct an ICE table for the above equation; we have:

                       $PCl_5(g)$           ⇄       $PCl_3(g)$     +         $Cl_2(g)$

Initial                0.398                          0                          0

Change            - x                               + x                        + x

Equilibrium    (0.398 - x)                      x                          x

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

K = $\frac{[x][x]}{[0.398-x]}$

K = $\frac{x^2}{0.398-x}$

0.060 = $\frac{x^2}{0.398-x}$

0.06(0.398-x) = x²

0.02388 - 0.060x = x²

x² + 0.060x - 0.02388 = 0               (quadratic equation)

a = 1;       b= 0.06;      c= -0.02388

Using quadratic formula;

=  $\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

= $\frac{-0.06+/-\sqrt{(0.06)^2-4(1)(-0.02388)} }{(2*1)}$

= $\frac{-0.060+/-\sqrt{0.0036+0.09552} }{2}$

= $\frac{-0.06+/-\sqrt{0.09912} }{2}$

= $\frac{-0.06+/-0.3148}{2}$

= $\frac{-0.060+0.3148}{2}$   or $\frac{-0.060-0.3148}{2}$

= $\frac{0.2548}{2}$  or $\frac{-0.3748}{2}$

= 0.1274 or -0.1874

We go by the positive value which says:

[x] = 0.1274 M

number of moles = 0.1274 × 10.0

= 1.274 moles

∴ the number of moles  of Cl₂ produced at equilibrium = 1.274 moles