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The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol of Cl2 in a 1.00-L reaction vessel at 230°C, what is the concentration of PCl3 when equilibrium has been established?

Respuesta :

Answer:

Concentration of PCl₅ at equilibrium = 24.01 mole/lit

Explanation:

Given

    Equilibrium constant  Kc = 49

                      PCl₃ (g) + Cl₂(g) ⇄ PCl₅(g)

According to Law of Mass action

Given that

Concentration of PCl₃ = 0.7 mole/lit & conc. of Cl₂ = 0.7 mole/lit

∴ volume of the vessel constant = 1 lit

                           

                                [tex]K_{c}=\frac{[PCl_{5} ]}{[PCl_{3} ][Cl_{2} ]} \\\\49= \frac{x}{0.7X0.7}[/tex]

    ∴ Concentration of PCl₅ at equilibrium = 49 x 0.7 x 0.7 = 24.01

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