For a first order reaction k=t2.303loga−xa When t=60 and x=60% k=602.303log100−60100=602.303log40100=0.0153 Now, t1/2=0.01532.303log100−50100=0.01532.303×log2=0.01532.303×0.3010=45.31 min
If 60%, of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately: (log4=0.60,log5=0.69)
Held on 30 Apr 2007 · Verified 9 Jul 2026.
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