$A$ $\quad$ $B$
$\Rightarrow \quad$ To calculate $:\left[ A _{t}\right]=16 \times\left[ B _{ t }\right] \ldots .(1)$ time $=?$
$\Rightarrow \quad$ For $I$ order kinetic : $\left[ A _{ t }\right]=\frac{ A _{0}}{(2)^{ n }}$
$n \rightarrow$ no of Half lives
$\Rightarrow \quad$ Now from the relation $(1)$
$\left[ A _{ t }\right]=16 \times\left[ B _{ t }\right]$
$\Rightarrow \quad \frac{x}{(2)^{n_{1}}}=\frac{x}{(2)^{n_{2}}} \times 16 \Rightarrow(2)^{ n _{2}}=(2)^{n_{1}} \times(2)^{4}$
$\Rightarrow \quad n_{2}=n_{1}+4 \quad \Rightarrow \quad \frac{t}{\left(t_{1 / 2}\right)_{2}}=\frac{t}{\left(t_{1 / 2}\right)_{1}}+4$
$\Rightarrow \quad t\left(\frac{1}{18}-\frac{1}{54}\right)=4 \Rightarrow t=\frac{4 \times 18 \times 54}{36}$
$\Rightarrow \quad t=108\, \min$
$2 {NO}_{({g})}+2 {H}_{2({~g})} \rightarrow {N}_{2({~g})}+2 {H}_{2} {O}_{({g})}$
$[NO]$ ${mol} {L}^{-1}$ |
${H}_{2}$ ${mol} {L}^{-1}$ |
વેગ ${mol}L^{-1}$ $s^{-1}$ |
|
$(A)$ | $8 \times 10^{-5}$ | $8 \times 10^{-5}$ | $7 \times 10^{-9}$ |
$(B)$ | $24 \times 10^{-5}$ | $8 \times 10^{-5}$ | $2.1 \times 10^{-8}$ |
$(C)$ | $24 \times 10^{-5}$ | $32 \times 10^{-5}$ | $8.4 \times 10^{-8}$ |
${NO}$ના સંદર્ભમાં પ્રક્રિયાનો ક્રમ $....$ છે.
(આપેલ : $\ln 10=2.303\,\log 2=0.3010$ )