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Chemistry Question on Chemical Kinetics

Calculate the half-life of a first order reaction from their rate constants given below:

  1. 200 s−1200\ s^{-1 }
  2. 2 min−12 \ min^{-1 }
  3. 4 years−14\ years^{-1}
Answer

(i)(i) Half lifeHalf \ life t12=0.693kt_{\frac 12} = \frac {0.693}{k}

= 0.693200 s−1\frac {0.693}{200\ s^{-1}}

= 3.47×10−3s (approximately)3.47 \times 10^{-3} s \ (approximately)

(ii)(ii) Half lifeHalf \ life t12=0.693kt_{\frac 12} = \frac {0.693}{k}

=0.6932 min−1\frac { 0.693}{2 \ min^{-1}}

= 0.35 min (approximately)0.35 \ min \ (approximately)

(iii)(iii) Half lifeHalf \ life t12=0.693kt_{\frac 12} = \frac {0.693}{k}

= 0.6934 year−1\frac {0.693}{4 \ year^{-1}}

= 0.173 years (approximately)0.173\ years\ (approximately)