For a first order reaction, the rate constant is $6.909\ \text{min}^{-1}$. The time taken for 75% conversion in minutes is (a) $\left(\dfrac{3}{2}\right)\log2$ (b) $\left(\dfrac{2}{3}\right)\log2$ (c) $\left(\dfrac{3}{2}\right)\log\left(\dfrac{3}{4}\right)$ (d) $\left(\dfrac{2}{3}\right)\log\left(\dfrac{4}{3}\right)$

For a first order reaction, the rate constant is 6.909min⁻¹. The time taken for 75% conversion in minutes is (

Choose the Best Answer · Q13

6.909\ \text{min}^{-1}

(a)

\left(\dfrac{3}{2}\right)\log2

(b)

\left(\dfrac{2}{3}\right)\log2

(c)

\left(\dfrac{3}{2}\right)\log\left(\dfrac{3}{4}\right)

(d)

\left(\dfrac{2}{3}\right)\log\left(\dfrac{4}{3}\right)

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Step 1. 75% conversion means 25% of A remains, so $[A]_0/[A]=100/25=4$ .

Step 2. $t=\dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}=\dfrac{2.303}{6.909}\log4$ .

Step 3. Note $k=6.909=3\times2.303$ , so $\dfrac{2.303}{6.909}=\dfrac{1}{3}$ ; also $\log4=\log(2^2)=2\log2$ . …