Question

The two common chlorides of phosphorus, PCl3, and PCl5, both important for the production of the other phosphorus compounds, coexist in equilibrium through the reactionPCl3(g) + Cl2(g) = PCl5(g)At 250 ᵒC , an equilibrium mixture in a 25.0 L flask contains 0.105 g PCl5, 0.220 g PCl3 and 2.12 g of Cl2. What are the values of(a) Kc(b) Kp for this reaction at 250 ᵒC ?

Answer 1

(a) Kc = 264

(b) Kp = 6.15

Step-by-step explanation:

(a) To calculate Kc we need the molar concentrations of each substance.

`M=(molesofsolute)/(litresofsolution) =(massofsolute)/(molarmassofsolute * litresofsolution)`

`[PCl_(3)]=(0.220g)/(137.5g/mol * 25.0L ) =6.40 * 10^(-5) M`

`[Cl_(2)]=(2.12g)/(71.0g/mol * 25.0L ) =1.19 * 10^(-3)M`

`[PCl_(5)]=(0.105g)/(208.5g/mol * 25.0L ) =2.01 * 10^(-5) M`

Then, we replace these concentrations in the Kc expression.

`Kc=([PCl_(5)])/([PCl_(3)]* [Cl_(2)] ) =(2.01 * 10^(-5)  )/(6.40 * 10^(-5) * 1.19 * 10^(-3)  ) =264`

(b) To find out Kp we can use the following expression:

`Kp = Kc.(R.T)^(\Delta n(g) )`

where,

R is the ideal gas constant (0.08206 atm .L /mol . K)

T is the absolute temperature (250°C + 273.15 = 523.15 K)

Δn is gaseous moles of products - gaseous moles of reactants (1 - 2 = -1)

If we replace with the values we have:

`Kp = Kc.(R.T)^(\Delta n(g) )=264 * (0.08206 * 523.15)^(-1) =6.15`