Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.87 × 10⁻³ at a certain temperature. If 5.00 g of solid PH₃BCl₃ and 0.0700 g of BCl₃ are added to a 4.500 L reaction vessel, what will the equilibrium concentration of PH₃ be? SO CONFUSED BE VERY SPECIFIC especially WITH FINDING Qc.

[PH₃(g)] = 0.0432M at equilibrium

Explanation:

Given:       PH₃BCl₃(s) => PH₃(g) + PH₃(g); Kc = 1.87 x 10⁻³

*Let solids have concentration value = 1.0M and apply common ion (compound) effect to Ci b/c of 0.070g BCl₃ added initially.

        PH₃BCl₃(s)  =>         BCl₃(g)              +            PH₃(g

Ci:        5.00g                    0.07g                            0.00g    

         *≅1.0M(s)           =0.07g/115.81g·mol⁻¹

                                   = 6 x 10⁻⁶ mol

                                   = 6 x 10⁻⁶mol/4.5L

                                   = 1.3 x 10⁻⁴M                    = 0.00M

ΔC:       -x                            +x                                 +x

Ceq:    1M - x ≅ 1M      = (1.3 x 10⁻⁴ + x)M               = x

Kc = [BCl₃(g)][PH₃(g)] = (1.3 x 10⁻⁴ + x)(x) = 1.87 x 10⁻³

=> 1.3 x 10⁻⁴·x + x² = 1.87 x 10⁻³

=> x² + 1.3 x 10⁻⁴·x  + 1.87 x 10⁻³ = 0

=> x = -b ± [Sqrt(b² - 4ac)] / 2a

=> x = -1.3 x 10⁻⁴ ± Sqrt[(1.3 x 10² - 4(1)(1.87 x 10⁻³)] / 2(1)

=> x = 0.0432M in PH₃ at equilibrium

Check:  

Kc = [BCl₃(g)][PH₃(g)] = (1.3 x 10⁻⁴ + 0.0432)(0.0432) = 1.87 x 10⁻³ (QED)                            

h let me think

20 drops hope this