a first order reaction is 15% complete in 20 min.how long will it take to b 60% complete????

for first order reaction

t = 2.303 / K (Log a / a-x)

Let a = 100

initially reaction is 15% completed, so x = 15 , t = 20 mn.

so,    20 = 2.303 / K ( log 100 / 100-15 )

          K = 2.303 / 20 ( log 100 / 85 )

          K = 2.303 / 20 (log 1.176 )

          K = 2.303 / 20 ( 0.706 )

          K = 0.11515 (0.706 )

          K = 0.00813

now reaction is 60% completed

so,     t = 2.303 / K (log a / a-x )

          t = 2.303 / 0.00813 ( log 100 / 100-60 )                                      ( putting value of K )

          t = 283.27 ( log 100 / 40)

          t = 283.27 ( log 2.25)

          t = 283.27( 0.3979 )

          t = 112.71 min.

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