Molecular iodine, I2 (g), dissociates into iodine atoms at625 K with a first-ord

Question

Molecular iodine, I2 (g), dissociates into iodine atoms at625 K with a first-order rate constant of 0.271 s-1. If you start with3.8×102 M I2 at thistemperature, how much will remain after 7.68 s assuming thatthe iodine atoms do not recombine to form I2? Molecular iodine, I2 (g), dissociates into iodine atoms at625 K with a first-order rate constant of 0.271 s-1. If you start with3.8×102 M I2 at thistemperature, how much will remain after 7.68 s assuming thatthe iodine atoms do not recombine to form I2? If you start with3.8×102 M I2 at thistemperature, how much will remain after 7.68 s assuming thatthe iodine atoms do not recombine to form I2?

Explanation / Answer

I2 ---> I + I For the first order reaction   K = ( 2.303/ t ) log( a / ( a-x)) Where K = rate constant = 0.271 s^-1 t = time taken = 7.68 s a = initial concentration = 3.8 * 10^-2 M a-x = concentration of I2 after time t Plug te values we get log ( a/(a-x)) = Kt / 2.303                                                      = 0.9037                                      a / (a-x) = 10^0.9037                                                    = 8.0117                                             (a-x) = a / 8.0117                                                      = 3.8*10^-2 / 8.0117                                                       = 4.743 * 10^-3 M

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