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Show that is case of first order reaction , the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction. |
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Answer» SOLUTION :Let `[A_0]=100, ` When `t=t_(99.9%),[A]=(100-99.9)=0.1` `k=(2.303)/t LOG (([A_0])/([A]))` `t_(99.9%)=(2.303)/k(100)/(0.1)impliest_(99.9%)=(2.303)/klog100 ` `t_(99.9%)=(2.303)/k(3)impliest_(99.9%)=(6.906)/k` `t_(99.9%)~=10xx(0.69)/k` `t_(99.9%)~=10t_(1//2)` |
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