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Show that in case of a first order reaction, the time taken for completion of 99.9% reaction is ten times the time required for half change of the reaction. |
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Answer» SOLUTION :To show that`t_(99.9%) = 10 xx t_(1//2)` `K = (2.303)/(t) LOG ""([R]_(0))/([R])` `(t_(99.9))/(t_(1//2)) = ((2.303)/(k)log""(100)/(0.1))/((2.303)/(k)xxlog""(100)/(50))rArr (t_(99.9%))/(t_(1//2)) = (3)/(0.3)` `t_(99.9%) = 10 t_(1//2)` |
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