At 80 ∘C, Kc=1.87×10^−3 for the reaction PH3BCl3(s)⇌PH3(g)+BCl3(g)

A.) Calculate the equilibrium concentration of PH3 if a solid sample of PH3BCl3 is placed in a closed vessel at 80 ∘C and decomposes until equilibrium is reached.

B.) Calculate the equilibrium concentration of BCl3 if a solid sample of PH3BCl3 is placed in a closed vessel at 80 ∘C and decomposes until equilibrium is reached.

C.) If the flask has a volume of 0.26 L , what is the minimum mass of PH3BCl3(s) that must be added to the flask to achieve equilibrium?

The equilibrium concentration of [tex]\rm PH_3[/tex] is 0.0432 M, and [tex]\rm BCl_3[/tex] is 0.0432 M. The minimum mass of [tex]\rm PH_3BCl_3[/tex] to achieve equilibrium is 0.59 g.

What is an equilibrium?

Equilibrium is the condition when the rate of reaction is constant. The equilibrium concentration of the reaction is given in the ICE table attached.

The equilibrium constant for the reaction is given as:

[tex]\rm K_c={[PH_3]\;[BCl_3]}[/tex]

Substituting the values for the equilibrium concentration:

[tex]1.87\;\times\;10^{-3}=[x][x]\\x^2=1.87\;\times\;10^{-3}\\x=0.0432[/tex]

(A) The concentration of [tex]\rm PH_3[/tex] at equilibrium, is equivalent to x. Thus, the equilibrium concentration of [tex]\rm PH_3[/tex] is 0.0432 M.

(B) The concentration of [tex]\rm BCl_3[/tex] at equilibrium, is equivalent to x. Thus, the equilibrium concentration of [tex]\rm BCl_3[/tex] is 0.0432 M.

(C) The concentration of [tex]\rm PH_3BCl_3[/tex] is x. The concentration is 0.0432 M.

The concentration can be given as:

[tex]\rm Concentration =\dfrac{moles}{volume}\\ Concentration=\dfrac{mass}{molar\;mass\;\times\;volume}[/tex]

The volume of solution is 0.26 L.

The molar mass of [tex]\rm PH_3BCl_3[/tex] is 50.59 g/mol.

The weight of [tex]\rm PH_3BCl_3[/tex] is given as:

[tex]\rm Weight \;of\;PH_3BCl_3=0.0432\;\times\;50.59\;\times\;0.26\;g\\Weight \;of\;PH_3BCl_3=0.56\;g[/tex]

The mass of [tex]\rm PH_3BCl_3[/tex] to achieve equilibrium is 0.59 g.

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