\(=-\frac{1}{\text { stoichiometric coefficient of reactant }} \frac{\text {d[reactant]}}{\mathrm{d} t}\)
\(=+\frac{1}{\text { stoichiometric coefficient of product }} \frac{\text {d[product]}}{\mathrm{dt}}\)
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, \(\frac{\mathrm{d}\left[ \mathrm{NO}_{2}\right]}{\mathrm{dt}}=-2 \frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{dt}}\)
\(=2 \times 6.25 \times 10^{-3}\, \mathrm{mol} \,L^{-1}\, \mathrm{s}^{-1}\)
\(=12.5 \times 10^{-3} \,\mathrm{mol} \,\mathrm{L}^{-1}\, \mathrm{s}^{-1}\)
\(=1.25 \times 10^{-2}\, \mathrm{mol} \,\mathrm{L}^{-1}\, \mathrm{s}^{-1}\)
\(\frac{\mathrm{d}\left[O_{2}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\mathrm{d} t} \times \frac{1}{2}\)
\(\frac{6.25 \times 10^{-3} \,\mathrm{mol} \,\mathrm{L}^{-1} \,\mathrm{s}^{-1}}{2}\)
\(=3.125 \times 10^{-3} \,\mathrm{mol}\, \mathrm{L}^{-1}\, \mathrm{s}^{-1}\)
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$ તો $\frac{d[NH_3]}{dt}$ અને $\frac{d[H_2]}{dt}$ વચ્ચેનો સમાનતાનો સંબંધ ............ થશે.
$2 A + B \longrightarrow C + D$
| પ્રયોગ | $[ A ] / molL ^{-1}$ | $[ B ] / molL ^{-1}$ | પ્રાથમિક $rate/molL$ $^{-1}$ $\min ^{-1}$ |
| $I$ | $0.1$ | $0.1$ | $6.00 \times 10^{-3}$ |
| $II$ | $0.1$ | $0.2$ | $2.40 \times 10^{-2}$ |
| $III$ | $0.2$ | $0.1$ | $1.20 \times 10^{-2}$ |
| $IV$ | $X$ | $0.2$ | $7.20 \times 10^{-2}$ |
| $V$ | $0.3$ | $Y$ | $2.88 \times 10^{-1}$ |
આપેલા ટેબલ માં $X$ અને $Y$ અનુક્રમે શું હશે ?
[આપેલ: $\ln 10=2.3$$R =8.3\, J \, K ^{-1}\, mol ^{-1}$]