For the second order reaction, EX2-->E+2X, the concentration ofspecies EX2 falls

Question

For the second order reaction, EX2-->E+2X, the concentration ofspecies EX2 falls from 0.040 M to 0.0050 M in 12 hours.
a) What is the rate constant for this reaction? b) What is the half-life when the initial concentration of EX2is 0.040 M? c) What concentration of EX2 would be left after 1 day(starting with 0.040 M as the initial)?
a) What is the rate constant for this reaction? b) What is the half-life when the initial concentration of EX2is 0.040 M? c) What concentration of EX2 would be left after 1 day(starting with 0.040 M as the initial)?

Explanation / Answer

Formula :                  1/ [A]   = kt + 1/[A]0 Where [A] is the concentration at time t             [A]0is the initial concentration              k is the rate constant              t is the time a ) Data :                [A]0 = 0.040 M               [A]    = 0.0050 M                  t     = 12 h Upon substitution ,          1/ 0.0050M   = k * 12h + 1 / 0.040 M 200 / M - 25 / M = k * 12h                   175 / M = k * 12 h                           k   = 175 /M /12h                                = 14.58 L / mol.h b) t1/2 = 1 / k [A]            = 1 /  14.58 L / mol.h * 0.040 M            = 1.71 h -1 c)   t     = 24 h     [A]0 = 0.040 M     [A]   = ?      k     =14.58 L / mol.h Upon substituting in the above formula,        1 / [A] =14.58 L / mol.h * 24 h + 1 / 0.040 M                   = 349.92/ M + 25 / M                    = 374.92 / M             [A]  =0.00266 M c)   t     = 24 h     [A]0 = 0.040 M     [A]   = ?      k     =14.58 L / mol.h Upon substituting in the above formula,        1 / [A] =14.58 L / mol.h * 24 h + 1 / 0.040 M                   = 349.92/ M + 25 / M                    = 374.92 / M             [A]  =0.00266 M

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