Question

Carbon disulfide and chlorine react according to the following equation: CS2(g) + 3Cl2(g) S2Cl2(g) + CCl4(g) When 2.94 mol of CS2 and 5.60 mol of Cl2 are placed in a 2.00-L container and allowed to come to equilibrium, the mixture is found to contain 0.580 mol of CCl4. How many moles of Cl2 are present at equilibrium?

Answer 1

Initial concentration of CS₂ = number of moles / volume in L

                                        = 2.94 mol / 2.00L

                                        = 1.47 M

Initial concentration of Cl₂ = number of moles / volume in L

                                      = 5.60 mol / 2.00L

                                      = 2.80 M

                              CS₂(g) + 3Cl₂(g) S₂Cl₂(g) + CCl₄(g)

initial conc               1.47        2.80           0              0

change                     -a            -3a          +a            +a

equb conc            1.47-a      2.80-3a         a               a

Given equilibrium moles of CCl₄ is = 0.580 mol

        Equilibrium concentration of CCl₄ = number of moles / volume in L

                                                         = 0.580 mol / 2.00L

                                                         = 0.29 M

                                                      a = 0.29 M

So Equilibrium concentration of Cl₂ = 2.80-3a

                                                    = 2.80 - (3x0.29)

                                                    = 1.93 M

So Equilibrium moles of Cl₂ , n = Molarity x volume in L

                                              = 1.93 M x 2.00 L

                                              = 3.86 moles