Question

At a certain temperature, the equilibrium constant, K c , is 0.00376 for the reaction Cl 2 ( g ) − ⇀ ↽ − 2 Cl ( g ) A. If 3.45 g Cl 2 is placed in a 3.50 L flask at this temperature, what are the equilibrium concentrations of Cl 2 and Cl ?

Answer 1

Answer:

Equilibrium concentration of Cl₂ = 0.013 molar

Equilibrium concentration of Cl = 0.0072 molar

Explanation:

                            Cl ₂ ( g ) ⇄ 2 Cl ( g )

No. of moles of Cl₂(g) = $\frac{Mass given}{Molar mass}$

                                    = $\frac{3.45}{71}$

                                    = 0.048 mole

Concentration of Cl₂ = $\frac{0.048 }{3.5}$ M = 0.013 molar

                        From the above equation we get

                     $K_{c}$ = $\frac{[Cl]^{2} }{[Cl_{2} ]}$

                      [Cl]² = Kc X [Cl₂]

                               = 0.00376 x 0.013

                               = 5.22 x 10⁻⁵ molar

                     [Cl] = 0.0072 molar