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Chemistry Question on Chemical Kinetics

The rate constant for a first order reaction is 60 s−160 \,s^{-1}. How much time it will take to reduce the concentration of the reactant to 1/10 th1 / 10^\text{ th} of its initial value?

A

3.8×10−2 s3.8\times 10^{-2}\,s

B

1.26×1013 s1.26\times 10^{13}\,s

C

2.01×1013 s2.01\times 10^{13}\,s

D

1.097×1031.097\times 10^{3}

Answer

3.8×10−2 s3.8\times 10^{-2}\,s

Explanation

Solution

For a first order reaction, t=2.303klog⁡[A]0[A]t=\frac{2.303}{k} \log \frac{[A]_{0}}{[A]}

Here, A0=A_{0}= initial concentration A=A= final concentration

Given, k=60s−1,[A]=[A]010k=60 s^{-1},[A]=\frac{[A]_{0}}{10}

Now, t=2.30360log⁡[A]0[A]010t=\frac{2.303}{60} \log \frac{[A]_{0}}{\frac{[A]_{0}}{10}}

=2.30360log⁡10=\frac{2.303}{60} \log 10

=0.038s=0.038 s or 3.8×10−2s3.8 \times 10^{-2} s