MHT CET · Chemistry · Chemical Kinetics
What time is required for 100 g of reactant to reduce to 25 g in a first order reaction having half life 5760 year?
- A 4760 year
- B 8640.26 year
- C 2880.15 year
- D 11526.48 year
Answer & Solution
Correct Answer
(D) 11526.48 year
Step-by-step Solution
Detailed explanation
For a first order reaction,
\({\left[\mathrm{A}_0\right]=100 \mathrm{~g} ;\left[\mathrm{A}_{\mathrm{t}}\right]=25 \mathrm{~g} ; \mathrm{t}_{1 / 2}=5760 \text { years } } \)
\( \mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{5760 \mathrm{yr}}=1.203 \times 10^{-4} \text { year }^{-1} \)
\( \mathrm{t}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{\left[\mathrm{~A}_0\right]}{\left[\mathrm{A}_{\mathrm{t}}\right]} \)
\( \therefore \mathrm{t}=\frac{2.303}{1.203 \times 10^{-4} \text { year }^{-1}} \log _{10} \frac{100}{25} \)
\( =19141.8 \times 0.6021=11526.5 \text { years}\)
\({\left[\mathrm{A}_0\right]=100 \mathrm{~g} ;\left[\mathrm{A}_{\mathrm{t}}\right]=25 \mathrm{~g} ; \mathrm{t}_{1 / 2}=5760 \text { years } } \)
\( \mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{5760 \mathrm{yr}}=1.203 \times 10^{-4} \text { year }^{-1} \)
\( \mathrm{t}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{\left[\mathrm{~A}_0\right]}{\left[\mathrm{A}_{\mathrm{t}}\right]} \)
\( \therefore \mathrm{t}=\frac{2.303}{1.203 \times 10^{-4} \text { year }^{-1}} \log _{10} \frac{100}{25} \)
\( =19141.8 \times 0.6021=11526.5 \text { years}\)
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