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A second order reaction requires 140 \mathrm{~min} to change the concentration of reactants from 0.16 \mathrm{M} to 0.02 \mathrm{M} The time required to become 0.08 \mathrm{M}=2 x min. What will be the value of x.

Option: 1

4 min


Option: 2

6 min


Option: 3

8 min


Option: 4

10 min


Answers (1)

best_answer

For second order reaction

\mathrm{ { [R]_{initial} }={ }^{\prime} a^{\prime}=0.16\, \mathrm{M}}
\mathrm{[R]_{final}=0.02 \, \mathrm{M} }

\mathrm{x=R _{initial}-R_{\text {final }} }
    \mathrm{=0.16-0.02=0.14 M }

\mathrm{ (a-x)=0.16-0.14=0.02 \mathrm{~M} }
\mathrm{ K_{2}=\frac{1}{t} \times \frac{x}{a(a-x)} }

\mathrm{ k_{2}=\frac{1}{140} \times \frac{0.14}{0.16 \times 0.02} }\quad -(1)

Now, time required to become concentration

\mathrm{ =0.08 \mathrm{M} \text { ie } x=0.08 \mathrm{~M} }
\mathrm{ k_{2}=\frac{1}{t} \times \frac{0.08}{0.16 \times(0.16-0.08)}}\quad -(2)

Now put the value of  \mathrm{ k_{2}} form (1)  in (2), we get

\mathrm{ \frac{1}{140} \times \frac{0.14}{0.16 \times 0.02}=\frac{1}{t } \times \frac{0.08}{0.16 \times 0.08} }
\mathrm{ t=20 \text { min }=2 x \text { min. } }
\mathrm{ x=10 \text { min } }.
 

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