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Chemistry Question on Chemical Kinetics

The rate of a first order reaction is 0.693×10−2 mol L−1 min⁡−10.693\times {{10}^{-2}}\,mol\,{{L}^{-1}}\,{{\min }^{-1}} and the initial concentration of the reactants is 1 M, t1/2{{t}_{1/2}} is equal to:

A

6.92 minute

B

100 minutes

C

0.693×10−30.693\times {{10}^{-3}} minute

D

0.693×10−20.693\times {{10}^{-2}} minute

Answer

100 minutes

Explanation

Solution

For 1st order reaction, rate=K×[A]\text{rate}=K\times [A] ∴\therefore 0.693×10−2=K×10.693\times {{10}^{-2}}=K\times 1 or K=0.693×10−2min⁡−1K=0.693\times {{10}^{-2}}{{\min }^{-1}} We know that, t1/2=0.693K=0.6930.693×10−2{{t}_{1/2}}=\frac{0.693}{K}=\frac{0.693}{0.693\times {{10}^{-2}}} = 100 minute