Question

Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined in a 0.5 L flask, what is the reaction quotient (Q) before the reaction ? g

Answer 1

The reaction quotient (Q) before the reaction is 0.32

Step-by-step explanation:

Being the reaction:

aA + bB ⇔ cC + dD

`Q=([C]^(c) *[D]^(d) )/([A]^(a)*[B]^(b) )`

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

`Concentration=(number of moles of solute)/(Volume)`

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

`Q=([PCl_(3) ] *[Cl_(2) ] )/([PCl_(5) ])`

The concentrations are:

• [PCl₃]=
  `(0.04 moles)/(0.5 L) =0.08 (moles)/(L)`

• [Cl₂]=
  `(0.08 moles)/(0.5 L) =0.16 (moles)/(L)`

• [PCl₅]=
  `(0.02 moles)/(0.5 L) =0.04 (moles)/(L)`

Replacing:

`Q=(0.08*0.16)/(0.04)`

Solving:

Q= 0.32

The reaction quotient (Q) before the reaction is 0.32