For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction
The time required for 99 percent completion of a first order reaction is t1=2.303klog100100−99t1=2.303klog100100−99 =2.303klog100=2.303klog100 The time required for 90 percent completion of a first order reaction is t2=2.303klog100100−99t2=2.303klog100100−99 =2.303klog10=2.303klog10 =2.303k=2.303k t1=2t2t1=2t2 As a result, the time required to complete a first order reaction at 99 percent is twice the time required to complete the reaction at 90%.
