Search any question & find its solution
Question:
Answered & Verified by Expert
After how many seconds will the concentration of the reactant in a first order reaction be halved if the rate constant is $1.155 \times 10^{-3} \mathrm{~s}^{-1}$ ?
Options:
Solution:
1288 Upvotes
Verified Answer
The correct answer is:
600
Rate constant $k=1.155 \times 10^{-3} \mathrm{~s}^{-1}$
$k=\frac{2.303}{t} \log \frac{a}{(a-x)} \quad \because a=a,(a-x)=\frac{a}{2}$
$\begin{aligned} t_{1 / 2} &=\frac{2.303}{k} \log \frac{a}{a / 2} \\ &=\frac{2.303}{1.155 \times 10^{-3}} \log 2 \\ &=\frac{2.303}{1.155 \times 10^{-3}} \times 0.3010 \end{aligned}$
$\begin{aligned} &=\frac{0.693 \times 10^{3}}{1.155} \\ \text { or } t_{1 / 2} &=\frac{0.693}{k}=\frac{0.693}{1.155 \times 10^{-3}} \\ &=600 \mathrm{~s} \end{aligned}$
$k=\frac{2.303}{t} \log \frac{a}{(a-x)} \quad \because a=a,(a-x)=\frac{a}{2}$
$\begin{aligned} t_{1 / 2} &=\frac{2.303}{k} \log \frac{a}{a / 2} \\ &=\frac{2.303}{1.155 \times 10^{-3}} \log 2 \\ &=\frac{2.303}{1.155 \times 10^{-3}} \times 0.3010 \end{aligned}$
$\begin{aligned} &=\frac{0.693 \times 10^{3}}{1.155} \\ \text { or } t_{1 / 2} &=\frac{0.693}{k}=\frac{0.693}{1.155 \times 10^{-3}} \\ &=600 \mathrm{~s} \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.