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QJEEM23Ch1157: JEE Main 2023 Chemistry - Class 12 JEEM23Ch1157 Solution: An organic compound undergoes first-order decomposition. If the …

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

Given: ln⁡10=2.3;log⁡2=0.3\ln 10=2.3 ; \log 2=0.3

Step-by-Step Verified Solution

For the first order reaction,

k=1tln⁡aa−xk=\frac{1}{t} \ln \frac{a}{a-x}

⇒\Rightarrow t=1kln⁡aa−xt=\frac{1}{k} \ln \frac{a}{a-x}

When reaction is 60%60 \% completed,

x=60100a=0.6a,t=540 seconds ;x=\frac{60}{100} a=0.6 a, t=540 \text { seconds } ;

k=1t1ln⁡aa−0.6ak= \frac{1}{t_1} \ln \frac{a}{a-0.6 a}

∴\therefore t1=1kln⁡a0.4at_1=\frac{1}{k} \ln \frac{a}{0.4 a}

When reaction is 90%90 \% completed, i.e., x=0.9ax=0.9 a

k=1t2ln⁡aa−0.9a⇒t2=1kln⁡a0.1a∴t1t2=1kln⁡a0.4a1kln⁡a0.1a⇒540t2=ln⁡104ln⁡10\begin{aligned}k= \frac{1}{t_2} \ln \frac{a}{a-0.9 a} \\\\\Rightarrow t_2=\frac{1}{k} \ln \frac{a}{0.1 a} \\\\\therefore \frac{t_1}{t_2} =\frac{\frac{1}{k} \ln \frac{a}{0.4 a}}{\frac{1}{k} \ln \frac{a}{0.1 a}} \\\\\Rightarrow \frac{540}{t_2} =\frac{\ln \frac{10}{4}}{\ln 10}\end{aligned}

⇒\Rightarrow 540t2=2.3log⁡10−2.3log⁡42.3log⁡10\frac{540}{t_2} =\frac{2.3\log 10-2.3\log 4}{2.3\log 10}

⇒\Rightarrow 540t2=2.3−2.3(0.6)2.3\frac{540}{t_2} =\frac{2.3-2.3(0.6)}{2.3}

⇒\Rightarrow 540t2=0.4\frac{540}{t_2} =0.4

⇒\Rightarrow t2=5400.4=1350t_2=\frac{540}{0.4}=1350 sec

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