For the reaction, Cl 2 (g) <--->2Cl (g),K c =0.0168 at 200 degrees Celcius. 0.22

Question

For the reaction, Cl2 (g) <--->2Cl (g),Kc=0.0168 at 200 degrees Celcius. 0.222 mol of Clwere placed in a 2.50L flask at 20 degrees C; no Cl2 waspresent initially. What are the equilibrium concentrations ofCl2 and Cl? <---> = Double Arrow Thanks!! For the reaction, Cl2 (g) <--->2Cl (g),Kc=0.0168 at 200 degrees Celcius. 0.222 mol of Clwere placed in a 2.50L flask at 20 degrees C; no Cl2 waspresent initially. What are the equilibrium concentrations ofCl2 and Cl? <---> = Double Arrow Thanks!!

Explanation / Answer

Given that   The concentration of the Cl2 presentinitaially = 0.222 mole /2.50L                                                                       = 0.0888 M Therefore                    Cl2(g) <------> 2Cl(g) , Kc = 0.0168                     I(M)        0.0888               0                     C               -x                   +2x                      Eq          0.0888-x              2x                         Kc = [Cl]2 / [Cl2]                                = (2x)2 / (0.0888-x)                  0.0168   = 4x2 / (0.0888-x) By solving it we get x = 0.017326 M      Therefore at equillibrium [Cl2] = 0.0888 - 0.017326 = 0.0714 M                                            [Cl] = 2*0.017326 = 0.0346 M                   

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