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MCQEasyJEE 2024Integrated Rate Laws

JEE Chemistry 2024 Question with Solution

The half-life of radioisotopic bromine-82 is 36 hours36 \, \text{hours}. The fraction that remains after one day is:

  • A

    0.630.63

  • B

    0.50.5

  • C

    0.750.75

  • D

    0.250.25

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The half-life of bromine-82 is T1/2=36 hoursT_{1/2} = 36 \, \text{hours} and the time elapsed is t=24 hourst = 24 \, \text{hours}.

Find: The fraction remaining after one day.

Use the radioactive decay relation:

F=(12)tT1/2F = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}

First, calculate the ratio:

tT1/2=2436=23\frac{t}{T_{1/2}} = \frac{24}{36} = \frac{2}{3}

Now substitute into the formula:

F=(12)23F = \left(\frac{1}{2}\right)^{\frac{2}{3}}

Using logarithmic form as shown in the solution:

log⁡10F=(23)log⁡10(12)\log_{10} F = \left(\frac{2}{3}\right) \log_{10}\left(\frac{1}{2}\right)

Since

log⁡10(12)=−log⁡102≈−0.3010\log_{10}\left(\frac{1}{2}\right) = -\log_{10} 2 \approx -0.3010

therefore

log⁡10F=(23)(−0.3010)≈−0.2006\log_{10} F = \left(\frac{2}{3}\right)(-0.3010) \approx -0.2006

Taking antilog:

F=antilog(−0.2006)=11.587≈0.6301F = \text{antilog}(-0.2006) = \frac{1}{1.587} \approx 0.6301

Therefore, the fraction remaining after one day is 0.630.63. The correct option is A.

First-Order Kinetics Method

Given: t1/2=36 hourst_{1/2} = 36 \, \text{hours} and t=24 hourst = 24 \, \text{hours}.

Find: The fraction of bromine-82 remaining after one day.

For first-order radioactive decay:

t1/2=0.693Kt_{1/2} = \frac{0.693}{K}

So,

K=0.69336=0.01925 hr−1K = \frac{0.693}{36} = 0.01925 \, \text{hr}^{-1}

Now use the first-order equation:

t=2.303Klog⁡aa−xt = \frac{2.303}{K} \log \frac{a}{a-x}

Hence,

log⁡aa−x=tK2.303\log \frac{a}{a-x} = \frac{tK}{2.303}

Substitute the values:

log⁡aa−x=24×0.019252.303=0.2006\log \frac{a}{a-x} = \frac{24 \times 0.01925}{2.303} = 0.2006

Taking antilog,

aa−x=1.587\frac{a}{a-x} = 1.587

If a=1a = 1, then

11−x=1.587\frac{1}{1-x} = 1.587

So,

1−x=0.63011-x = 0.6301

Thus, the fraction remaining is 0.6301≈0.630.6301 \approx 0.63. The solution also mentions 63%63\%, which corresponds to the same fraction 0.630.63. Therefore, the correct option is A.

Common mistakes

  • Using one full half-life for one day is incorrect because one day is 24 hours24 \, \text{hours}, not 36 hours36 \, \text{hours}. Use the ratio tT1/2=2436=23\frac{t}{T_{1/2}} = \frac{24}{36} = \frac{2}{3} instead.

  • Taking the remaining fraction as 12\frac{1}{2} is wrong because that value applies only after exactly one half-life. Here the elapsed time is less than one half-life, so the fraction remaining must be greater than 0.50.5.

  • Confusing fraction with percentage leads to answer-format errors. The solution shows about 63%63\%, but the question asks for the fraction remaining, so write 0.630.63, not 6363.

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