MHT CET · Chemistry · Chemical Kinetics
Half-life for a first order reaction is 6.93 hour. What is the time required for \(80 \%\) completion of the reaction?
- A 12 hours
- B 18 hours
- C 6 hours
- D 16 hours
Answer & Solution
Correct Answer
(D) 16 hours
Step-by-step Solution
Detailed explanation
For first order reaction,
\(
\mathrm{k}=\frac{0.693}{\mathrm{t}_{\frac{1}{2}}}=\frac{0.693}{6.93}=0.1 \text { hour }^{-1}
\)
Here, \([\mathrm{A}]_0=100,[\mathrm{~A}]_{\mathrm{t}}=100-80=20\)
\(\text {Now, } \mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \)
\( \therefore \mathrm{t}=\frac{2.303}{0.1} \log _{10} \frac{100}{20}=23.03 \times \log _{10}\) \(5=23.03 \times 0.699 \)
\( =16.10 \text { hours.}\)
\(
\mathrm{k}=\frac{0.693}{\mathrm{t}_{\frac{1}{2}}}=\frac{0.693}{6.93}=0.1 \text { hour }^{-1}
\)
Here, \([\mathrm{A}]_0=100,[\mathrm{~A}]_{\mathrm{t}}=100-80=20\)
\(\text {Now, } \mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \)
\( \therefore \mathrm{t}=\frac{2.303}{0.1} \log _{10} \frac{100}{20}=23.03 \times \log _{10}\) \(5=23.03 \times 0.699 \)
\( =16.10 \text { hours.}\)
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