In case of first order reaction, the ratio of the time required for 99.9% completion to 50% completion is

t99.9%=2.303klog100100−99.9 =2.303klog1000.1=2.303klog1000=2.303k×3 t50.0%=2.303klog100100−50 =2.303klog2=2.303k×0.3010 t99.9%t50.0%=2.303×3k×k2.303×0.3010=30.3=10 t99.9%=10×t50.0%