Q77 / 200 · NEET 2003 · Chemistry

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 $\mathrm{B}$ ?

  • Correct answer
    A
    1 hour
  • B
    0.5 hour
  • C
    0.25 hour
  • D
    2 hour

Solution

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, \(\begin{aligned} \mathrm{t} & =\frac{2.303}{\mathrm{k}} \log \frac{0.9}{\mathrm{o} .225} \\ \mathrm{t} & =\frac{2.303}{1.3865} \log (4) \ldots . . (4) \\ \mathrm{t} & =1.661 \times 0.6020 \\ \mathrm{t} & =0.999=1 \mathrm{hr} \end{aligned}\)

Hence, the correct answer is option: 'A'.

Relevant Questions