1. Tardigrade
  2. Question
  3. Chemistry
  4. The rate of a first order reaction is 0.693× 10-2 mol L-1 min -1 and the initial concentration of the reactants is 1 M, t1/2 is equal to:

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Q. The rate of a first order reaction is $ 0.693\times {{10}^{-2}}\,mol\,{{L}^{-1}}\,{{\min }^{-1}} $ and the initial concentration of the reactants is 1 M, $ {{t}_{1/2}} $ is equal to:

JIPMERJIPMER 2001Chemical Kinetics

Solution:

For 1st order reaction, $ \text{rate}=K\times [A] $ $ \therefore $ $ 0.693\times {{10}^{-2}}=K\times 1 $ or $ K=0.693\times {{10}^{-2}}{{\min }^{-1}} $ We know that, $ {{t}_{1/2}}=\frac{0.693}{K}=\frac{0.693}{0.693\times {{10}^{-2}}} $ = 100 minute