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Question

Chemistry Question on Chemical Kinetics

What is the two third life of a first order reaction having k=5.48×10−14s−1k=5.48\times 10^{-14}s^{-1}

A

2.01×1011s2.01\times 10^{11}s

B

2.01×1013s2.01\times 10^{13}s

C

8.08×1013 s8.08\times 10^{13}\,s

D

16.04×1011 s16.04\times 10^{11}\,s

Answer

2.01×1013s2.01\times 10^{13}s

Explanation

Solution

For two-third of a reaction:
[A]0=a,[A]=a−23a=a3{[A]_{0}=a,[A]=a-\frac{2}{3} a=\frac{a}{3}}
t2/3=2.303klog⁡[A]0[A]t_{2 / 3}=\frac{2.303}{k} \log \frac{[A]_{0}}{[A]}
=2.303klog⁡aa3=2.303klog⁡3=\frac{2.303}{k} \log \frac{a}{\frac{a}{3}}=\frac{2.303}{k} \log 3
t2/3=2.303×0.47715.48×10−14t_{2 / 3}=\frac{2.303 \times 0.4771}{5.48 \times 10^{-14}}
=2.01×1013s=2.01 \times 10^{13} s