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NVAMediumJEE 2023Integrated Rate Laws

JEE Chemistry 2023 Question with Solution

An organic compound undergoes first order decomposition. If the time taken for the 60%60\% decomposition is 540 s540 \, \text{s}, then the time required for 90%90\% decomposition will be ...... s\text{s}. (Nearest integer).

Answer

Correct answer:1350

Step-by-step solution

Standard Method

Given: The reaction is first order. Time for 60%60\% decomposition is t1=540 st_1 = 540 \, \text{s}.

Find: Time for 90%90\% decomposition, say t2t_2.

For a first order reaction,

t=1kln⁡(a0a)t = \frac{1}{k} \ln\left(\frac{a_0}{a}\right)

For 60%60\% decomposition, 40%40\% remains, so

t1=1kln⁡(a00.4a0)=1kln⁡(104)t_1 = \frac{1}{k} \ln\left(\frac{a_0}{0.4a_0}\right) = \frac{1}{k} \ln\left(\frac{10}{4}\right)

For 90%90\% decomposition, 10%10\% remains, so

t2=1kln⁡(a00.1a0)=1kln⁡(10)t_2 = \frac{1}{k} \ln\left(\frac{a_0}{0.1a_0}\right) = \frac{1}{k} \ln(10)

Using Ratio of First Order Times

Taking the ratio,

t2t1=1kln⁡(10)1kln⁡(104)\frac{t_2}{t_1} = \frac{\frac{1}{k}\ln(10)}{\frac{1}{k}\ln\left(\frac{10}{4}\right)}

From the extracted working,

t2540=ln⁡10ln⁡(10)−ln⁡(4)=11−0.6\frac{t_2}{540} = \frac{\ln 10}{\ln(10) - \ln(4)} = \frac{1}{1 - 0.6}

Therefore,

t2540=0.4\frac{t_2}{540} = 0.4

the solution concludes:

t2=5400.4=1350 st_2 = \frac{540}{0.4} = 1350 \, \text{s}

Therefore, the required time is 13501350.

Common mistakes

  • Using the percentage decomposed directly in the logarithm is incorrect. For first order kinetics, the formula uses the amount remaining, not the amount decomposed. For 60%60\% decomposition, use 40%40\% remaining; for 90%90\%, use 10%10\% remaining.

  • Applying the zero order or second order integrated rate law is wrong because the question explicitly states first order decomposition. Use

    t=1kln⁡(a0a)t = \frac{1}{k} \ln\left(\frac{a_0}{a}\right)

    instead.

  • Cancelling logarithms carelessly can lead to an inverted ratio. Set up t2t1\frac{t_2}{t_1} in the same order as the corresponding logarithmic terms, then substitute the remaining fractions carefully before solving.

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