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Chemical Kinetics and Nuclear Chemistry question

2023 · 13 Apr · Shift 1 · Q16
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Chemical Kinetics and Nuclear Chemistry question

2023 · 13 Apr · Shift 1 · Q16

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
t87.5\mathrm{t}_{87.5}t87.5​ is the time required for the reaction to undergo 87.5%87.5 \%87.5% completion and t50\mathrm{t}_{50}t50​ is the time required for the reaction to undergo 50%50 \%50% completion. The relation between t87.5\mathrm{t}_{87.5}t87.5​ and t50\mathrm{t}_{50}t50​ for a first order reaction is t87.5=x×t50\mathrm{t}_{87.5}=x \times \mathrm{t}_{50}t87.5​=x×t50​ The value of xxx is ‾\underline{\hspace{2cm}}​. (Nearest integer)
Numerical answer
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Correct answer: 3

  1. For a first-order reaction, the integrated rate law is t=2.303klog⁡aa−xt = \frac{2.303}{k} \log \frac{a}{a-x}t=k2.303​loga−xa​ where:

    • aaa = initial amount
    • xxx = amount reacted at time ttt

  2. Time for 50% completion: At 50%50\%50% completion, a−x=0.5aa-x = 0.5aa−x=0.5a So, t50=2.303klog⁡a0.5at_{50} = \frac{2.303}{k} \log \frac{a}{0.5a}t50​=k2.303​log0.5aa​ t50=2.303klog⁡2t_{50} = \frac{2.303}{k} \log 2t50​=k2.303​log2

    This is the half-life of a first-order reaction.

  3. Time for 87.5% completion: At 87.5%87.5\%87.5% completion, fraction left is 100−87.5=12.5%=0.125=18100 - 87.5 = 12.5\% = 0.125 = \frac{1}{8}100−87.5=12.5%=0.125=81​ Thus, a−x=a8a-x = \frac{a}{8}a−x=8a​

    Therefore, t87.5=2.303klog⁡aa/8t_{87.5} = \frac{2.303}{k} \log \frac{a}{a/8}t87.5​=k2.303​loga/8a​ t87.5=2.303klog⁡8t_{87.5} = \frac{2.303}{k} \log 8t87.5​=k2.303​log8

  4. Find the ratio: t87.5t50=log⁡8log⁡2\frac{t_{87.5}}{t_{50}} = \frac{\log 8}{\log 2}t50​t87.5​​=log2log8​ Since 8=238 = 2^38=23, log⁡8log⁡2=3\frac{\log 8}{\log 2} = 3log2log8​=3

    Hence, t87.5=3 t50t_{87.5} = 3\, t_{50}t87.5​=3t50​

  5. Therefore, x=3x = 3x=3

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