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A first order reaction is `15%` compelte in 20 minutes. How long it take to be `60%` complete? |
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Answer» For the first order reaction, `t=(2.303)/k log (a)/(a-x)` `20 min=2.303/(k) log 100/85 = 2.303/k log a/(a-x)` 20 min=`(2.303)/(k) log (100/85) = (2.303)/k log 1.1765`...........(i) Iind case: `a=100%, (a-x)= 40%` `t=(2.303)/k log 100/40 = 2.303/k log 2.5`.............(ii) Dividing eqn (ii) by eqn. (i), `t/(20 "min")=(log 2.5)/(log 1.1765)= (0.3979)/(0.0706)` or `t=(0.3979)/(0.0706) xx (20"min") = 112.7 min` |
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