For a first order reaction
\(k=\frac{2.303}{t} \log \frac{a}{(a-x)}\)
Given \(a=0.5,(a-x)=0.125,\,t=50\,min\)
\(k=\frac{2.303}{50} \log \frac{0.5}{0.125}\)
\(=2.78 \times 10^{-2} \,\mathrm{min}^{-1}\)
\(r=k\left[H_{2} O_{2}\right]=2.78 \times 10^{-2} \times 0.05\)
\(=1.386 \times 10^{-3} \,\,\mathrm{mol} \,\mathrm{min}^{-1}\)
Now
\(-\frac{d\left|H_{d} O_{2}\right|}{d t}=\frac{d\left|H_{2} O\right|}{d t}=\frac{2 d\left|O_{2}\right|}{d t}\)
\(\frac{2 d\left[O_{2}\right]}{d t}-\frac{d\left[H_{2} O_{2}\right]}{d t}\)
\(\frac{d\left[O_{2}\right]}{d t}=\frac{1}{2} \times \frac{d\left[H_{2} O_{2}\right]}{d t}\)
\(=\frac{1.386 \times 10^{-3}}{2}=6.93 \times 10^{-4}\,\, \mathrm{mol}\, \mathrm{min}^{-1}\)
$\mathop {2{N_2}{O_5}}\limits_{{\rm{(in}}\,\,{\rm{CC}}{{\rm{l}}_4}{\rm{)}}} \to \mathop {4N{O_2}}\limits_{{\rm{(in}}\,\,{\rm{CC}}{{\rm{l}}_4}{\rm{)}}} + {O_2}$
$[X]$ $0.1\,M$, $[Y]$ $0.1\,M$ દર $\rightarrow 0.002\,Ms^{-1}$
$[X]$ $0.2\,M$, $[Y]$ $0.1\,M$ દર $\rightarrow 0.002\,Ms^{-1}$
$[X]$ $0.3\,M$, $[Y]$ $0.2\,M$ દર $\rightarrow 0.008\,Ms^{-1}$
$[X]$ $0.4\,M$, $[Y]$ $0.3\,M$ દર $\rightarrow 0.018\,Ms^{-1}$
તો દર નિયમ ......
$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$
$\text { rate }=\mathrm{k}[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^{1 / 2}$
$A$ અને $B$ એમ દરેક ની સાદ્રતા $1 M$ લઇ ને પ્રક્રિયા શરૂ કરવામાં આવે છે. જો વેગ અયળાંક ($k$) એ $4.6 \times 10^{-2} \mathrm{~s}^{-1}$, હોય તો $A$ ને $0.1 \mathrm{M}$ થવા માટે જરૂરી સમય .................. sec છે. (નજીક નો પૂર્ણાંક)
$ O_3 $ $\rightleftharpoons$ $ O_2 + O$ ...... (ઝડપી) ;
$O + O_3 \rightarrow 2O_2$ ...... (ધીમી)