Search any question & find its solution
Question:
Answered & Verified by Expert
Reaction $\mathrm{A} \rightarrow \mathrm{B}$ follows first order kinetics. The time taken for 0.8 mole of $A$ to produce 0.6 mole of $B$ is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of $\mathrm{B}$ ?
Options:
Solution:
1773 Upvotes
Verified Answer
The correct answer is:
1 hour
For first order reaction,
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \ldots \ldots .(1)\)
Where \(k\) is rate constant
when 0.8 mol of \(A\) produces 0.6 mol of \(B\) then, \(0.8-0.6=0.2\) moles of A is left .
final moles \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.2\)
initial moles \(\mathrm{a}=0.8\)
\(\begin{aligned}
& \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{0.8}{\mathrm{o} .2} \\
& \mathrm{k}=\frac{2.303}{1} \log [4] \ldots \ldots . . (2) \\
& \mathrm{k}=2.303 \times 0.6020 \\
& \mathrm{k}=1.3865 \ldots \ldots . .(3)
\end{aligned}\)
In second case, 0.9 moles of A produces 0.675 moles of \(B\) so remaining moles of \(A\) are \(0.9-0.675=0.225\)
final moles of \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.225\)
initial mole \(=0.9\)
By using first order equation,
\(\begin{aligned}
\mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \\
\mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \\
\mathrm{t} & =1.661 \times 0.6020 \\
\mathrm{t} & =0.999=1 \mathrm{hr}
\end{aligned}\)
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \ldots \ldots .(1)\)
Where \(k\) is rate constant
when 0.8 mol of \(A\) produces 0.6 mol of \(B\) then, \(0.8-0.6=0.2\) moles of A is left .
final moles \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.2\)
initial moles \(\mathrm{a}=0.8\)
\(\begin{aligned}
& \mathrm{t}=\frac{2.303}{\mathrm{k}} \log \frac{0.8}{\mathrm{o} .2} \\
& \mathrm{k}=\frac{2.303}{1} \log [4] \ldots \ldots . . (2) \\
& \mathrm{k}=2.303 \times 0.6020 \\
& \mathrm{k}=1.3865 \ldots \ldots . .(3)
\end{aligned}\)
In second case, 0.9 moles of A produces 0.675 moles of \(B\) so remaining moles of \(A\) are \(0.9-0.675=0.225\)
final moles of \(\mathrm{A}=\mathrm{a}-\mathrm{x}=0.225\)
initial mole \(=0.9\)
By using first order equation,
\(\begin{aligned}
\mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \\
\mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \\
\mathrm{t} & =1.661 \times 0.6020 \\
\mathrm{t} & =0.999=1 \mathrm{hr}
\end{aligned}\)
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.