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For a first reaction , show that time required for 99% completion is twice the time required for the completion of 90% of reaction. |
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Answer» SOLUTION :For a FIRST REACTION, `t =k =(2.303)/t log.a/(a-x)` 99% completion means that `x =99% ` of a = 0.99a `:. t_(99%)=(2303)/k log. a/(a-0.99a)=(2.303)/klog 10^2=2xx(2303)/k` 90% completion means that `x=90% " of " a = 0.90a` `:. t_(99%)=(2303)/k log. a/(a-0.99a)=(2.303)/klog 10=2303/k` `(t_(99%))/(t_(90%))=((2xx2303)/k)//(2.303)/k=2 " or " t_90%=2xxt_(90%)` |
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