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Chemical Kinetics and Nuclear Chemistry question

2023 · 30 Jan · Shift 1 · Q15
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Chemical Kinetics and Nuclear Chemistry question

2023 · 30 Jan · Shift 1 · Q15

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
If compound A reacts with B following first order kinetics with rate constant 2.011×10−3 s−12.011 \times 10^{-3} \mathrm{~s}^{-1}2.011×10−3 s−1. The time taken by A\mathrm{A}A(in seconds) to reduce from 7 g7 \mathrm{~g}7 g to 2 g2 \mathrm{~g}2 g will be ‾\underline{\hspace{2cm}}​. (Nearest Integer) [log⁡5=0.698,log⁡7=0.845,log⁡2=0.301][\log 5=0.698, \log 7=0.845, \log 2=0.301][log5=0.698,log7=0.845,log2=0.301]
Numerical answer
View written solutionFree

Correct answer: 623

  1. Use the first-order integrated rate law

For a first-order reaction,

k=2.303tlog⁡([A]0[A]t)k = \frac{2.303}{t}\log\left(\frac{[A]_0}{[A]_t}\right)k=t2.303​log([A]t​[A]0​​)

So,

t=2.303klog⁡([A]0[A]t)t = \frac{2.303}{k}\log\left(\frac{[A]_0}{[A]_t}\right)t=k2.303​log([A]t​[A]0​​)

Here,

  • k=2.011×10−3 s−1k = 2.011 \times 10^{-3}\ \text{s}^{-1}k=2.011×10−3 s−1
  • initial amount [A]0=7 g[A]_0 = 7\ \text{g}[A]0​=7 g
  • final amount [A]t=2 g[A]_t = 2\ \text{g}[A]t​=2 g

Thus,

t=2.3032.011×10−3log⁡(72)t = \frac{2.303}{2.011 \times 10^{-3}}\log\left(\frac{7}{2}\right)t=2.011×10−32.303​log(27​)

  1. Evaluate the logarithm

log⁡(72)=log⁡7−log⁡2=0.845−0.301=0.544\log\left(\frac{7}{2}\right) = \log 7 - \log 2 = 0.845 - 0.301 = 0.544log(27​)=log7−log2=0.845−0.301=0.544

So,

t=2.303×0.5442.011×10−3t = \frac{2.303 \times 0.544}{2.011 \times 10^{-3}}t=2.011×10−32.303×0.544​

  1. Calculate the numerator

2.303×0.544=1.2528322.303 \times 0.544 = 1.2528322.303×0.544=1.252832

Hence,

t=1.2528322.011×10−3=1.2528320.002011t = \frac{1.252832}{2.011 \times 10^{-3}} = \frac{1.252832}{0.002011}t=2.011×10−31.252832​=0.0020111.252832​

  1. Final calculation

t≈622.99 st \approx 622.99\ \text{s}t≈622.99 s

Nearest integer:

623\boxed{623}623​

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