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

1.274 moles

Step-by-step explanation:

The equation for the reaction can be represented as follows:

`PCl_5(g)` ⇄
`PCl_3(g)` +
`Cl_2(g)`

K = 0.060

K =
`([PCl_3][Cl_2])/([PCl_5])`

Concentration of
`PCl_5(g)` =
`(numbers of moles)/(volume)`

Concentration of
`PCl_5(g)` =
`(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 =
`([PCl_3][Cl_2])/([PCl_5])`

K =
`([x][x])/([0.398-x])`

K =
`(x^2)/(0.398-x)`

0.060 =
`(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;

=
`(-b+/-√(b^2-4ac) )/(2a)`

=
`(-0.06+/-√((0.06)^2-4(1)(-0.02388)) )/((2*1))`

=
`(-0.060+/-√(0.0036+0.09552) )/(2)`

=
`(-0.06+/-√(0.09912) )/(2)`

=
`(-0.06+/-0.3148)/(2)`

=
`(-0.060+0.3148)/(2)` or
`(-0.060-0.3148)/(2)`

=
`(0.2548)/(2)` or
`(-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