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  • A and B decompose via first order kinetics with half lives <img alt="\mathrm{58.0 \mathrm{~min}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B58.0%20%5Cmathrm%7B%7Emin%7D%7D" /

A and B decompose via first order kinetics with half lives \mathrm{58.0 \mathrm{~min}} and 22.0 \mathrm{~min} respectively. starting from an equimolar non - reactive mixture of A and B, the time taken for the concentration of A to become 32 times that of B in how many min.

Option: 1

200 min


Option: 2

177 min


Option: 3

150 min


Option: 4

100 min


Answers (1)

best_answer

Given that:- 

\mathrm{\begin{aligned} & \left(t_{1 / 2}\right)_1=58 \mathrm{~min} \\ & \left(t_{1 / 2}\right)_2=22 \mathrm{~min} \end{aligned}}

\mathrm{A}                                   \mathrm{B}

\mathrm{t=0, X^{\prime} M}           \mathrm{t=0 ; X^{\prime} M}

to Calculate;\mathrm{ \left[A_t\right]=32 \times\left[B_t\right] \ldots-(i)}

for \mathrm{\text { for } I^{s t} \text { order Reaction }\left[A_t\right]=\frac{A_0}{(2)^n}}

\mathrm{\text { from Eq }(i)\left[A_t\right]=32 \times\left[B_t\right]}

\mathrm{ \frac{x}{(2)^{n_1}}=\frac{x}{(2)^{n_2}} \times 32 }

\mathrm{\begin{aligned} & (2)^{n_2}=(2)^{n_1} \times(2)^5 \\ \\& n_2=n_1+5 \\ \\& \frac{t}{\left(t_{1 / 2}\right)_2}=\frac{t}{\left(t_{1 / 2}\right)_1}+5 \\ \\& t\left(\frac{1}{22}-\frac{1}{58}\right)=5 \end{aligned}}

\mathrm{\begin{aligned} & t\left(\frac{18}{638}\right)=5 \\\\ & t=177 \mathrm{~min} \end{aligned}}

Posted by

HARSH KANKARIA

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