1.00 mol of PCl5 was sealed into an evacuated 1.00 liter flask andallowed to rea

Question

1.00 mol of PCl5 was sealed into an evacuated 1.00 liter flask andallowed to reach equilibrium at some temperature, according to thereaction: PCl5 (g) <-> PCl3 (g) + Cl2 (g) It was found that 2.00% of the PCl5 has decomposed whenequilibrium was reached. what is the numerical value of Kc? A. 5.00x10-3 B. 4.08x10-4 C. 2.00x10-2 D. 4.10x10-2 E. 2.04x10-4 PCl5 (g) <-> PCl3 (g) + Cl2 (g) It was found that 2.00% of the PCl5 has decomposed whenequilibrium was reached. what is the numerical value of Kc? A. 5.00x10-3 B. 4.08x10-4 C. 2.00x10-2 D. 4.10x10-2 E. 2.04x10-4

Explanation / Answer

Given data Moles of PCl5 = 1.0 mol Volume = 1.0 L Molarity = 1.0mol/1.0 L                    = 1.0 M                                   PCl5 (g) <-> PCl3 (g) + Cl2 (g) Initial(M)              1.0               0               0 Change(M)          -x                  +x             +x Equi(M)              1.0-x              x                x    Given that    2.0 % of PCl5 has decomposed                    x = 1.0M *2/100                          = 0.02 M       [PCl5]eq= 1.0 -x M                    = 1.0 - 0.02 M                    = 0.98 M        [PCl3]eq= x = 0.02 M          [Cl2]eq= x = 0.02 M               Kc =  [PCl3]eq*  [Cl2]eq /[PCl5]eq                        = 0.02M *0.02 M / 0.98M                        = 4.08*10-4    Hence option B is correct.    Hence option B is correct.

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