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a first order reaction takes 16 minutes for 50 % completion. how much time will it take for 75% completion
first we have to find out the rate constant of the reaction i.e. K So, for qa 1st order reaction K=(1/16)X2.303X log[a/(a-50a/100) where a is the initial substrate concentration.i.e. K= (1/16) X 2.303 X log 2i.e. K= (1/16) X 0.693i.e. K= 0.043 Now, we have to find out the time required for 75% completion t=(1/0.043)X2.303X log[a/(a-75a/100)t=(1/0.043)X2.303X log 4t=(1/0.043) X 2 X 2.303 X log 2t=(1/0.043)X 2 X 0.693t=32.23min or 32 min ThanksSouvik MukherjeeM.Sc. Chemistry, IIT Hyderabad
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