InterviewSolution
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InterviewSolution
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Derive order of reaction for the decomposition of `H_(2)O_(2)` from the following data. `{:("Time (in minutes)",0,10,20,30),("Volume of" KMnO_(4),25,16,10.5,7.09):}` needed for `H_(2)O_(2)` |
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Answer» The volume of `KMnO_(4)` used at anytime anytime is proportional to conc. `H_(2)O_(2)` at that time. `{:(,"At t=0",V=25,:.,a,prop,25),((i),"At t=10",V=16,:.,(a-x),prop,16),((ii),"At t=20",V=10.5,:.,(a-x),prop,10.5),((iii),"At t=30",V=7.09,:.,(a-x),prop,7.09):}` Now use, `K=2.303/t log a/((a-x))` For (i), `K=2.303/10 log 25/16` `=4.46xx10^(-2) min^(-1)` For (ii), `K=2.303/20 log 25/10.5` `=4.34xx10^(-2) min^(-1)` For (iii), `K=2.303/30 log 25/7.09` `4.20xx10^(-2) min^(-1)` The values of `K` come almost constant and thus confirming first order reaction. For velocity constant take average of all value of `K`. |
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