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Q.The half life period for a radioactive decay of C-14 is 5730 years. An archaeological artifact containing wood had only 60% of the C-14 found in a living tree. Estimate the age of the sample. OR In general, it is observed that the rate of a chemical reaction becomes double with every 10° rise in temperature. If this generalisation holds for a reaction in the temperature range of 298K to 308K, what would be the value of activation energy for this reaction? (R = 8.314 J K⁻¹ mol⁻¹)

Mizoram MbseMizoram Board of School Education HSSLC 2023Subjective· 3mImportance★★★★★
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First find the first-order rate constant from the half-life, then use the first-order integrated rate law to find the time elapsed for the C-14 fraction to fall to 60%.

Given: t1/2=5730t_{1/2} = 5730 years for C-14 decay (a first-order process); the sample retains 60% of the C-14 found in a living tree.

Step 1 — Rate constant from half-life:

k=0.693t1/2=0.6935730=1.209×10−4 yr−1k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{5730} = 1.209\times10^{-4}\ yr^{-1}

Step 2 — Apply the first-order integrated rate law, with [A]0[A]_0 = C-14 in a living tree (100%) and [A][A] = C-14 remaining in the sample (60%): …

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