IIT-JEE 2012 · Paper 1 · Chemistry Q40

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ChemistryInteger+4Question 40 of 60

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\dfrac{[t_{1/8}]}{[t_{1/10}]}\times10$ ? (take $\log_{10}2=0.3$ )

Your answer

Answer and worked solution

Official answer: 9

First order. $t=\dfrac{2.303}k\log\dfrac{[A]_0}{[A]}$ .

To 1/8: $t_{1/8}=\dfrac{2.303}k\log8=\dfrac{2.303}k(0.9)$ .
To 1/10: $t_{1/10}=\dfrac{2.303}k\log10=\dfrac{2.303}k(1)$ .

Ratio. $\dfrac{t_{1/8}}{t_{1/10}}\times10=0.9\times10=9$ .

Answer: 9

Topics:Integrated rate equations

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IIT-JEE 2012 · Paper 1, Chemistry question 40