If the rate constant for a first order reaction is 2 . 303 × 10 - 3 s - 1 , find the time required to reduce…

If the rate constant for a first order reaction is 2.303×10-3 s-1, find the time required to reduce 4 g of the reactant to 0.2 g.
  1. 1.30 hours 
  2. 21.60 hours 
  3. 0.36 hours 
  4. 2.60 hours 

Solution

Given: - Rate constant, \(k=2.303 \times 10^{-3} \mathrm{~s}^{-1}\) for a first-order reaction. - Initial concentration of the reactant, \([A]_0=4 \mathrm{~g}\). - Final concentration of the reactant, \([A]=0.2 \mathrm{~g}\). We want to find the time required to reduce \(4 \mathrm{~g}\) of the reactant to \(0.2 \mathrm{~g}\). Using the first-order reaction kinetics formula: \(\ln \left(\frac{[A]}{[A]_0}ight)=-k t\) Substituting the given values: \(\ln \left(\frac{0.2}{4}ight)=-\left(2.303 \times 10^{-3}ight) \cdot t\) Solving for \(t\) : \(\begin{aligned} & t=\frac{-\ln (20)}{-2.303 \times 10^{-3}} \\ & t \approx \frac{2.9957}{2.303 \times 10^{-3}} \\ & t \approx 1300 \mathrm{~s} \end{aligned}\) Converting to hours: \(1300 \mathrm{~s} \times \frac{1 \text { hour }}{3600 \mathrm{~s}} \approx 0.361 \text { hours }\) ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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