\(=-\frac{1}{\text { stoichiometric coefficient of reactant }} \frac{\text {d[reactant]}}{\mathrm{d} t}\)
\(=+\frac{1}{\text { stoichiometric coefficient of product }} \frac{\text {d[product]}}{\mathrm{dt}}\)
For the reaction
\(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\)
\(-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{N}_{2}\right]}{\mathrm{dt}}\)
\(=+\frac{2 \mathrm{d}\left[O_{2}\right]}{\mathrm{dt}}\)
therefore, \(\frac{\mathrm{d}\left[ \mathrm{NO}_{2}\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}\)
\(=2 \times 6.25 \times 10^{-3}\, \mathrm{mol} \,L^{-1}\, \mathrm{s}^{-1}\)
\(=12.5 \times 10^{-3} \,\mathrm{mol} \,\mathrm{L}^{-1}\, \mathrm{s}^{-1}\)
\(=1.25 \times 10^{-2}\, \mathrm{mol} \,\mathrm{L}^{-1}\, \mathrm{s}^{-1}\)
\(\frac{\mathrm{d}\left[O_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{d} t} \times \frac{1}{2}\)
\(\frac{6.25 \times 10^{-3} \,\mathrm{mol} \,\mathrm{L}^{-1} \,\mathrm{s}^{-1}}{2}\)
\(=3.125 \times 10^{-3} \,\mathrm{mol}\, \mathrm{L}^{-1}\, \mathrm{s}^{-1}\)
પ્રયોગ |
$[A]$ ($mol\, L^{-1})$ |
$[B]$ ($mol\, L^{-1})$ |
પ્રક્રિયાની શરૂઆતનો દર $(mol\, L^{-1}$ $min^{-1})$ |
$I$ | $0.10$ | $0.20$ | $6.93 \times {10^{ - 3}}$ |
$II$ | $0.10$ | $0.25$ | $6.93 \times {10^{ - 3}}$ |
$III$ | $0.20$ | $0.30$ | $1.386 \times {10^{ - 2}}$ |
$A$ અડધો વપરાય તે માટેનો સમય મિનિટમાં કેટલો થાય
$2MnO_4^ - + 10{I^ - } + 16{H^ + } \to 2M{n^{2 + }} + 5{I_2} + 8{H_2}O$ તો $I_2$ ના ઉત્પન્ન થવાનો દર......$\times {10^{ - 2}}\,M{s^{ - 1}}$ જણાવો