પ્રયોગ | $\frac{[ X ]}{ mol \;L ^{-1}}$ | $\frac{[ Y ]}{ mol\; L ^{-1}}$ | $\frac{\text { Initial rate }}{ mol\; L ^{-1}\; min ^{-1}}$ |
$I$ | $0.1$ | $0.1$ | $2 \times 10^{-3}$ |
$II$ | $.2$ | $0.2$ | $4 \times 10^{-3}$ |
$III$ | $0.4$ | $0.4$ | $M \times 10^{-3}$ |
$IV$ | $0.1$ | $0.2$ | $2 \times 10^{-3}$ |
$M$ મૂલ્યનો સંખ્યાત્મક ગુણોત્તર $........$ છે. (નજીકનો પૂર્ણાંક)
Using \(I\) and \(II\)
\(\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\left(\frac{ L }{0.1}\right) \Rightarrow L =0.2\)
Using \(I\) and \(III\)
\(\frac{M \times 10^{-3}}{2 \times 10^{-3}}=\frac{0.4}{0.1} \Rightarrow M=8\)
\(\frac{ M }{ L }=\frac{8}{0.2}=40\)
Ans. \(40\)
$p ( mm Hg )$ | $50$ | $100$ | $200$ | $400$ |
સાપેક્ષ $t _{1 / 2}( s )$ | $4$ | $2$ | $1$ | $0.5$ |
પ્રક્રિયાનો ક્રમ શોધો.
$\left( {{\rm{R}} = 8.3\;{\rm{Jmo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}},\ln \left( {\frac{2}{3}} \right) = 0.4,\left. {{e^{ - 3}} = 4.0} \right)} \right.$